Business Logistics Supply Chain Management Instructors Manual

Business Logistics Supply Chain Management Planning Organizing and Controlling the Supply Chain Fifth Edition Instructors Manual Ronald H Ballou Weatherhead School of Management Case Western Reserve University ii CONTENTS Preface iii Chapter 1 Business LogisticsSupply ChainA Vital Subject 1 2 LogisticsSupply Chain Strategy and Planning 2 3 The LogisticsSupply Chain Product 4 4 LogisticsSupply Chain Customer Service 9 5 Order Processing and Information Systems 13 6 Transport Fundamentals 14 7 Transport Decisions Fowler Distributing Company Metrohealth Medical Center Orion Foods Inc R T Wholesalers 17 35 41 48 52 8 Forecasting Supply Chain Requirements World Oil Metro Hospital 65 84 88 9 Inventory Policy Decisions Complete Hardware Supply Inc American Lighting Products American Red Cross Blood Services 94 121 124 131 10 Purchasing and Supply Scheduling Decisions Industrial Distributors Inc 134 144 11 The Storage and Handling System 147 12 Storage and Handling Decisions 148 13 Facility Location Decisions Superior Medical Equipment Company Ohio Auto Drivers License Bureau Southern Brewery 162 186 190 198 14 The Logistics Planning Process Usemore Soap Company Essen USA 204 208 217 15 LogisticsSupply Chain Organization 229 16 LogisticsSupply Chain Control 230 iii PREFACE This instructors guide provides answers to the more quantitatively oriented problems at the end of the textbook chapters If the questions or problems are for discussion or they involve a substantial amount of individual judgment they have not been included Solutions to the cases and exercises in the text are also included These generally require computer assistance for solution With the text you are provided with a collection of software programs called LOGWARE that assist in the solution of the problems cases and exercises in the text The LOGWARE software along with a users manual is available for downloading from the Prentice Hall website or this book The users manual is in Microsoft Word or Acrobat pdf formats This software along with the users manual may be freely reproduced and distributed to your classes without requiring permission from the copyright holder This permission is granted as long as the use of the software is for educational purposes If you encounter difficulty with the software direct questions to Professor Ronald H Ballou Weatherhead School of Management Case Western Reserve University Cleveland Ohio 44106 Tel 216 3683808 Fax 216 3686250 Email RonaldBallouCASEedu Web site wwwprenhallcomballou 1 CHAPTER 1 BUSINESS LOGISTICSSUPPLY CHAINA VITAL SUBJECT 12 a This problem introduces the student to the evaluation of alternate channels of production and distribution To know whether domestic or foreign production is least expensive the total of production and distribution costs must be computed from the source point to the marketplace Two alternatives are suggested and they can be compared as follows Production at Houston Total cost Production cost at Houston Transportation and storage costs 8shirt100000 shirts 5cwt 1000 cwt 805000year Production at Taiwan Total cost Production cost in Taiwan Transportation and storage costs from Taiwan to Chicago Import duty Raw material transportation cost from Houston to Taiwan 4shirt100000 shirts 6cwt 1000 cwt 05shirt100000 shirts 2cwt 1000 cwt 458000year Producing in Taiwan would appear to be the least expensive b Other factors to consider before a final decision is made might be i How reliable would international transportation be compared with domestic transportation ii What is the business climate in Taiwan such that costs might change in favor of Houston as a production point iii How likely is it that the needed transportation and storage will be available iv If the market were to expand would there be adequate production capacity available to support the increased demand 2 CHAPTER 2 LOGISTICSSUPPLY CHAIN STRATEGY AND PLANNING 13 The purpose of this exercise is to allow the student in an elementary way to examine the tradeoffs between transportation and inventoryrelated costs when an incentive transportation rate is offered Whether the incentive rate should be implemented depends on the shipment size corresponding to the minimum of the sum of transportation inven tory and order processing costs These costs are determined for various shipping quantities that might be selected to cover the range of shipment sizes implied in the problem Table 21 gives a summary of the costs to Monarch for various shipment sizes From Monarchs point of view the incentive rate would be beneficial Shipment sizes should be approximately doubled so that the 40000 lb minimum is achieved It is important to note that the individual cost elements are not necessarily at a minimum at low shipment sizes whereas orderprocessing costs are low at high shipment sizes They are in cost conflict with each other Transportation costs are low at high shipment sizes but exact costs depend on the minimum volume for which the rate is quoted In preparation for a broader planning perspective to be considered later in the text the student might be asked what the place of the supplier is in this decision How does he affect the decision and how is he affected by it This will focus the students attention on the broader issues of the physical distribution channel 3 TABLE 21 Evaluation of Alternative Shipment Sizes for the Monarch Electric Company Current Proposed Type of cost 57 motors or 10000 lb 114 motors or 20000 lb 171 motors or 30000 lb 228 motors or 40000 lb 285 motors or 50000 lb Transportation RD 98750 78750 58750 43750 58750 43750 38750 26250a 38750 26250 Inventory carryingb ICQ2 025200572 1425a 0252001142 2850 0252001712 4275 0252002282 5700 0252002852 7125 Order processingc DSQ 50001557 1316 500015114 658 500015171 439 500015228 329 500015285 263a Handling HD 0308750 2625 0308750 2625 0308750 2625 0308750 2625 0308750 2625 Total 84116 49883 51089 34904a 36263 a Minimum values b Students should be informed that average inventory can be approximated by one half the shipment size c Demand D has been converted to units per year LEGEND R transportation rate cwt D annual demand cwt I inventory carrying cost year C cost of a motor motor Q shipment size in motors where Q2 represents the average number of motors maintained in inventory S order processing costs order H handling costs cwt 4 CHAPTER 3 THE LOGISTICSSUPPLY CHAIN PRODUCT 3 The 8020 principle applies to sales and items where 80 percent of the dollar volume is generated from 20 percent of the product items While this ratio rarely holds exactly in practice the concept does We can apply it to these data by ranking the products by sales and the percentage that the cumulative sales represent of the total The following table shows the calculations The 8020 rule cannot be applied exactly since the cumulative percent of items does not break at precisely 20 percent However we might decide that only products 08776 and 12121 should be ordered directly from vendors The important principle derived from the 8020 rule is that not every item is of equal importance to the firm and that dif ferent channels of distribution can be used to handle them The 8020 rule gives some rational basis for deciding which products should be shipped directly from vendors and which are more economically handled through a system of warehouses 6 a Reading the ground transport rates for the appropriate zone as determined by zip code and the weight of 27 lb rounding upward of 265 lb gives the following total cost table for the four shipments Product code Dollar sales Cumulative sales Cumulative sales as of total Cumulative items as of total 08776 71000 71000 182 83 12121 63000 134000 343 167 10732 56000 190000 486 250 11693 51000 241000 616 333 10614 46000 287000 734 417 12077 27000 314000 803 500 07071 22000 336000 859 583 10542 18000 336000 905 667 06692 14000 354000 941 750 09721 10000 368000 967 833 14217 9000 378000 989 917 11007 4000 391000 1000 1000 Total 391000 5 b The transport rate structure is reasonably fair since ground rates generally follow distance and size of shipment These are the factors most directly affecting transport costs They are not fair in the sense that customers within a zone are all charged the same rate regardless of their distance from the shipment origin point However all customers may benefit from lower overall rates due to this simplified zonerate structure 10 a This is a delivered pricing scheme where the seller includes the transport charges in the product price The seller makes the transport arrangements b The seller prices the product at the origin but prepays any freight charges however the buyer owns the goods in transit c This is a delivered pricing scheme where the freight charges are included in the product price however the freight charges are then deducted from the invoice and the seller owns the goods in transit d The seller initially pays the freight charges but they are then collected from the buyer by adding them to the invoice The buyer owns the goods in transit since the pricing is fob origin e The price is fob origin The buyer pays the freight charges and owns the goods in transit Regardless of the price policy the customer will ultimately pay all costs If a firm does not consider outbound freight charges the design of the distribution system will be different than if it does Since pricing policy is an arbitrary decision it can be argued that transport charges should be considered in decision making whether the supplying firm directly incurs them or not 11 This shows how Paretos law 8020 principle is useful in estimating inventory levels when a portion of the product line is to be held in inventory An empirical function that approximates the 8020 curve is used to estimate the level of sales for each product to be held in inventory According to Equation 32 the constant A is determined as follows To zip code Catalog price UPS zone Transport costa Total cost 11107 9995 2 737 10732 42117 9995 5 1046 11041 74001 9995 6 1317 11312 59615 9995 8 1829 11824 a Use 27 lb 6 A X Y Y X 1 0 25 1 75 0 75 0 25 0125 The 8020 type curve according to Equation 31 is Y A A X X X 1 1 0125 0125 This formula can be used to estimate the cumulative sales from the cumulative item proportion For example item 1 is 005 of the total number of items 20 so that Y 1 0125 0 05 0125 0 05 0321 Of the 2600000 in total annual warehouse sales item 1 should account for 03212600000 835714 By applying this formula to all items the following inventory investment table can be developed which shows sales by item The average inventory investment by item is found by dividing the turnover ratio into the item sales The sum of the average inventory value for each item gives a total projected inventory of 380000 Inventory Investment Table Pro duct Cumulative item pro portion X Cumulative sales Y Projected item sales Turnover ratio Average inventory value 1 005 835714 835714 8 104464 2 010 1300000 464286 8 58036 3 015 1595454 295454 8 36932 4 020 1800000 204546 8 25568 5 025 1950000 150000 6 25000 6 030 2064705 114706 6 19118 7 B 035 2155263 90558 6 15093 8 040 2228571 73308 6 12218 9 045 2289130 60559 6 10093 10 050 2340000 50870 6 8478 11 055 2383333 43333 4 10833 12 060 2420689 37356 4 9339 13 065 2453226 32537 4 8134 14 070 2481818 28592 4 7148 15 C 075 2507142 25324 4 6331 16 080 2529719 22587 4 5647 17 085 2550000 20271 4 5068 18 090 2568293 18293 4 4473 19 095 2584884 16591 4 4148 20 100 2600000 15116 4 3779 Total 380000 A 7 12 This problem involves the application of Equations 31 and 32 We can develop an 80 20 curve based on 30 percent of the items accounting for 70 percent of sales That is A X Y Y X 1 030 1 0 70 0 70 030 0 225 Therefore the sales estimating equation is Y X X 1 0 225 0 225 By applying this estimating curve we can find the sales of A and B items For example 20 percent of the items or 0220 4 items will be A items with a cumulative proportion of sales of YA 1 0 225 0 20 0 225 0 20 05765 and 300000005765 1729412 The AB item proportion will be YA B 1 0 225 050 0 225 050 08448 and 300000008448 2534400 The product group B sales will AB sales less A sales or 2534400 1729412 804988 The product group C will be the remaining sales but these are not of particular interest in this problem The average inventories for A and B products are found by dividing the estimated sales by the turnover ratio That is A 17294129 192157 B 8049885 160988 Total inventory 353155 cases The total cubic footage required for this inventory would be 35315515 529732 cu ft The total square footage for products A and B is divided by the stacking height That is 52973116 33108 sq ft 8 13 This problem is an application of Equations 31 and 32 We first determine the constant A That is A X Y Y X 1 0 20 1 0 65 0 65 0 20 0156 and 0 75 1 0156 0156 X X Solving algebraically for X we have X AxY A Y x 1 0156 0 75 1 0156 0 75 0 288 That is about 29 of the items 02885000 1440 items produce 75 of the sales 14 The price would be the sum of all costs plus an increment for profit to place the automotive component in the hands of the customer This would be 2510585transportation cost or 53T Based on the varying transportation cost the following price schedule can be developed Quantity Price per unit Discount 1 to 1000 units 53558 0 1001 to 2000 units 5340057 17a 2000 units 5330056 35 a 58 575810017 9 CHAPTER 4 LOGISTICSSUPPLY CHAIN CUSTOMER SERVICE 6 a This company is fortunate to be able to estimate the sales level that can be achieved at various levels of distribution service Because of this the company should seek to maximize the difference between sales and costs These differences are summarized as follows Percent of orders delivered within 1 day Contribution to 50 60 70 80 90 95 100 profit 18 20 35 40 34 28 20 The company should strive to make deliveries within 1 day 80 percent of the time for a maximum contribution to profit b If a competing company sets its delivery time so that more than 80 percent of the orders are delivered in 1 day and all other factors that attract customers are the same the company will lose customers to its competitor as the sales curve will have shifted downward Cleanco should adjust its service level once again to the point where the profit contribution is maximized Of course there is no guarantee that the previous level of profits can be achieved unless the costs of supplying the service can correspondingly be reduced 7 a This problem solution requires some understanding of experimental design and statistical inference which are not specifically discussed in the text Alert the students to this The first task is to determine the increase in sales that can be attributed to the change in the service policy To determine if there is a significant change in the control group we set up the following hypothesis test z X X s N s N 2 1 2 2 2 1 2 1 2 2 224 185 61 102 79 102 39 36 48 6118 394 Now referring to a normal distribution table in Appendix A of the text there is a significant difference at the 001 level in the sales associated with the control group That is some factors other than the service policy alone are causing sales to increase Next we analyze the test group in the same manner 10 z 2 295 1342 576 56 335 56 953 5 924 2 004 10 7 2 2 This change is also significant at the 001 level The average increase in sales for the control group is 224185 121 or 21 The average sales increase in the test group is 22951342 171 or 71 If we believe that 21 of the 71 increase in the test group is due to factors other than service policy then 71 21 50 was the true service effect Therefore for each sales unit an incremental increase in profit of 04095050 19 can be realized Since the cost of the service improvement is 2 the benefit exceeds the cost The service improvement should be continued Note If the students are not well versed in statistical methodology you may wish to instruct them to consider the before and after differences in the mean values of both groups as significant The solution will be the same b The use of the beforeafterwithcontrolgroup experimental design is a methodology that has been used for some time especially in marketing research studies The outstanding feature of the design is that the use of the control group helps to isolate the effect of the single service variable On the other hand there are a number of potential problems with the methodology The sales distributions may not be normal The time that it takes for diffusing the information that a service change has taken place may distort the results The products in the control group may not be mutually exclusive from those in the test group The method only shows the effect of a single step change in service and does not develop a salesservice relationship It may not always be practical to introduce service changes into ongoing operations to test the effect 8 a The optimum service level is set at that point where the change in gross profit equals the change in cost The change in gross profit P Trading margin Sales response rate Annual sales 10000015100000 150 per year per 1 change in the service level The change in cost C Annual carrying cost Standard product cost z 11 Demand standard deviation for order cycle 0301000400z Now set P C and solve for z 150 1200z z 0125 From the tabulated changes in service level with those changes in z the service level should be set between 9697 b The weakest link in this analysis is estimating the effect that a change in service will have on revenue This implies that a salesservice relationship is known 9 The methodology is essentially the same as that in question 7 except that we are asked to find X instead of Y That is P 0750001580000 90 and C 0251000500z 1250z Then P C 90 1250z z 0072 From the normal distribution see Appendix A the z for an area under the curve of 93 is 148 and for 92 z is 141 Since the difference of 148 141 007 we can conclude that the instock probability should be set at 9293 Of course the change in z is found by taking the difference in z values for 1 differences in the area values under the normal distribution curve for a wide range of area percentages 10 Apply Taguchis concept of the loss function First estimate the loss per item if the target level of service is not met We know the profit per item as follows 12 Sales price 595 Cost of item 425 Other costs 030 Profit per item 140 Since onehalf of the sales are lost the opportunity loss per item would be 0 70item 880 140 12880 Opportunity loss Next find k in the loss function L k y m k k k 2 2 0 70 10 5 0 70 25 0 03 Finally the point where the marginal supply cost equals the marginal sales loss is 1 67 2 0 03 0 10 2 5 k B y 6 67 5 1 67 y The retailer should not allow the outofstock percentage to deviate more than 167 and should not allow the outofstock level to fall below 167 5 667 Profit per item Sales lost Current sales Target outofstock at point where ½ sales are lost 13 CHAPTER 5 ORDER PROCESSING AND INFORMATION SYSTEMS All questions in this chapter require individual judgment and response No answers are offered 14 CHAPTER 6 TRANSPORT FUNDAMENTALS 14 The maximum that the power company can pay for coal at its power plant location in Missouri is dictated by competition Therefore the landed cost at the power plant of coal production costs plus transportation costs cannot exceed 20 per ton Since western coal costs 17 per ton at the mine the maximum worth of transportation is 20 17 3 per ton However if the grade of coal is equal to the coal from the western mines eastern coal can be landed in Missouri for 18 per ton In light of this competitive source transportation from the western mines is worth only 18 17 1 per ton 15 Prior to transport deregulation it was illegal for a carrier to charge shippers less for the longer haul than for the shorter haul under similar conditions when the shorter haul was contained within the longer one To be fair the practice probably should be continued If competitive conditions do not permit an increase in the rate to Z then all rates that exceed 1 per cwt on a line between X and Z should not exceed 1 per cwt Therefore the rate to Z is blanketed back to Y so that the rate to Y is 1 per cwt By blanketing the rate to Z on intervening points no intervening point is discriminated against in terms of rates 16 a From text Table 64 the item number for place mats is 474500 For 2500 lb the classification is 100 since 2500 lb is less than the minimum weight of 20000 lb for a truckload shipment From text Table 65 the rate for a shipment 2000 lb is 8727cwt The shipping charges are 8727 25 cwt 218175 b This is an LTL shipment with a classification of 100 item number 498000 in text Table 64 From Table 65 the minimum charge is 9351 and the rate for a 500 lb shipment is 5401cwt Check the charges using the 500 lb rate and compare it to the minimum charge That is 5401 15 cwt 8102 Since this is less than the minimum charge of 9351 pay the minimum charge c From Table 64 the item number is 205500 with a classification of 55 for LTL and 375 for TL at a minimum weight of 36000 lb There are three possibilities that need to be examined 1 Ship LTL at class 55 and 27000 lb shipment 2 Ship at class 55 and 30000 lb rate 3 Ship at class 375 and 36000 lb rate 15 Try 1 Rate is 565cwt 565 270 152550 Try 2 Rate is 387cwt 387 300 116100 Lowest cost Try 3 Rate is 370cwt 370 360 133200 d The shipment is a truckload classification 207000 of 65 The rate at 30000 lb is 421cwt The charges are 421 300 126300 e Classification of this product is 55 486000 for a truckload of 24000 lb Check the break weight according to Equation 61 Break weight 387 30000 565 lb 20 549 Since current shipping weight of 24000 lb exceeds the break weight ship as if 30000 lb Hence 387 300 116100 Now discount the charges by 40 percent That is 1161 1 040 69660 21 The question involves evaluating two alternatives The first is to compute the transport charges as if there are three separate shipments The next is to see if a stopoff privilege offers any cost reduction The comparison is shown below With stopoff Ship direct to B and split deliver thereafter Rate Stopoff Loadingunloading Route cwt charge Charges 25000 A to B 120 30000 Direct shipment 40000 B to D 220 88000 Stopoff C 2500 2500 Stopoff D 2500 2500 Total charges 123000 Separate shipments Rate Stopoff Loadingunloading Route cwt charge Charges 22000 A to D 320 70400 3000 A to C 250 7500 15000 B to C 150 22500 Total charges 100400 16 Split deliver at all stops Rate Stopoff Loadingunloading Route cwt charge Charges 40000 A to D 320 128000 Stopoff B 2500 2500 Stopoff C 2500 2500 Stopoff D 2500 2500 Total charges 133500 Other combinations may be tried In this case there appears to be no advantage to using the stopoff privilege 17 CHAPTER 7 TRANSPORT DECISIONS 1 Selecting a mode of transportation requires balancing the direct cost of transportation with the indirect costs of both vendor and buyer inventories plus the intransit inventory costs The differences in transport mode performance affect these inventory levels and therefore the costs for maintaining them as well as affect the time that the goods are in transit We wish to compare these four cost factors for each mode choice as shown in Table 71 of the manual The symbols used are R transportation rate unit D annual demand units C item value at buyers inventory C item value at vendors inventory T time in transit days Q Shipping quantity units Rail has the lowest total cost 2 As in question 1 this problem is one of balancing transport costs with the indirect costs associated with inventories However in this case we must account for the variability in transit time as it affects the warehouse inventories We can develop the following decision table Service type TABLE 71 An Evaluation of the Transport Alternatives for the Wagner Company Cost type Method Rail Piggyback Truck Transport RD 2550000 1250000 4450000 2200000 8850000 4400000 Intransit inventorya ICDt365 02547550000 16365 260274 02545650000 10365 156164 02541250000 4365 56438 Wagers inventorya ICQ2 025475100002 593750 02545670002 399000 02541250002 257500 Electronics inventory ICQ2 02550010002 625000 02550070002 437500 02550050002 312500 Total 2729024 3192664 5026438 a C refers to price less transport cost per unit 18 Cost type Method A B Transport RD 129600 115200 11809600 114048 Intransit inventory ICDt365 020509600 4365 1052 020509600 5365 1315 Plant inventory ICQ2 0305032182 2684 0305035782 2684 Warehouse inventory ICQ2 ICr 0306232132 03062505 3927 030618032182 0306180606 4107 Total 122863 122154 Recall that Q DS IC 2 2 9 600 100 03 50 3578 cwt for the plant assuming the order cost is the same at plant and warehouse However for the warehouse we must account for safety stock r and for the transportation cost in the value of the product Therefore For A Q DS IC 2 2 9 600 100 03 62 3213 cwt and for z 128 for an area under the normal distribution of 090 the safety stock is 50 5 cwt 9 600 365 51 1 28 d zs r LT For B Q 2 9 600 100 03 6180 3218 cwt and 60 6 cwt 9 600 365 81 1 28 r Service B appears to be slightly less expensive 3 The shortest route method can be applied to this problem The computational table is shown in Table 72 The shortest route is defined by tracing the links from the destination node They are shown in Table 72 as A D F G for a total distance of 980 miles TABLE 72 Tabulation of Computational Steps for the Shortest Route Method Applied to Transcontinental Trucking Company Problem 19 Step Solved nodes directly connected to unsolved nodes Its closest connected unsolved node Total time involved nth nearest node Its minimum time Its last connectiona 1 A B 186 mi B 186 mi AB A D 276 2 A D 276 D 276 AD B C 186110 296 3 B C 186110 296 C 296 BC D C 276 58 334 D F 276300 576 4 C E 296241 537 E 537 CE C F 296350 646 D F 276300 576 5 C F 296350 646 E G 5374791016 D F 276300 576 F 576 DF 6 E G 5374791016 F G 576404 980 G 980 FG a Asterisk indicates the shortest route 4 In this actual problem the US Army used the transportation method of linear programming to solve its allocation problem The problem can be set up in matrix form as follows Origin Destination Cleve land South Charleston San Jose Demand Letterkenny 150 150 100 150 800 300 Fort Hood 325 50 350 300 50 100 Fort Riley 275 100 325 350 100 Fort Carson 375 400 275 100 100 Fort Benning 300 100 250 0 450 100 Supply 400 150 150 The cell values shown in bold represent the number of personnel carriers to be moved between origin and destination points for minimum transportation costs of 153750 An alternative solution at the same cost would be 20 5 This problem can be used effectively as an inclass exercise Although the problem might be solved using a combination of the shortest route method to find the optimum path between stops and then a traveling salesman method to sequence the stops it is intended that students will use their cognitive skills to find a good solution The class should be divided into teams and given a limited amount of time to find a solution They should be provided with a transparency of the map and asked to draw their solution on it The instructor can then show the class each solution with the total distance achieved From the leastdistance solutions the instructor may ask the teams to explain the logic of their solution process Finally the instructor may explore with the class how this and similar problems might be treated with the aid of a computer Although the question asks the student to use cognitive skills to find a good route a route can be found with the aid of the ROUTER software in LOGWARE The general approach is to first find the route in ROUTER without regard to the rectilinear distances of the road network Because this may produce an infeasible solution specific travel distances are added to the database to represent actual distances traveled or to block infeasible paths from occurring A reasonable routing plan is shown in Figure 71 and the ROUTER database that generates it is given in Figure 72 The total distance for the route is 905 miles and at a speed of 20 miles per hour the route time is approximately 30 minutes Origin Destination Number of carriers Cleveland Letterkenny 150 S Charleston Letterkenny 150 Cleveland Fort Hood 50 San Jose Fort Hood 50 Cleveland Fort Riley 100 San Jose Fort Carson 100 Cleveland Fort Benning 100 05 10 15 20 0 5 0 05 10 15 20 0 5 0 17 19 20 21 0 05 10 15 20 0 5 0 05 10 15 20 0 5 0 05 10 15 20 0 5 0 05 10 15 20 0 5 0 17 19 20 21 0 21 FIGURE 72 Input Data for ROUTER for School Bus Routing Problem PARAMETERS AND LABELS Problem label School Bus Routing Exercise Grid corner with 00 coordinates NW SW SE or NE NW DEPOT DATA Depot description Atlanta Located in zone 0 Horizontal coordinate 014 Vertical coordinate 045 Earliest starting time min 0 Latest return time min 9999 Default vehicle speed miles per hour 20 After how many clock hours will overtime begin 9999 GENERAL DATA Percent of vehicle in use before allowing pickups 0 Horizontal scaling factor 1 Vertical scaling factor 1 Maximum TIME allowed on a route hours 9999 Maximum DISTANCE allowed on a route miles 9999 LOADUNLOAD TIME FORMULA Fixed time per stop 0 Variable time per stop by weight 0 By cube 0 BREAK TIMES Duration of 1st break minutes 0 To begin after 9999 Duration of 2nd break minutes 0 To begin after 9999 Duration of 3rd break minutes 0 To begin after 9999 Duration of 4th break minutes 0 To begin after 9999 22 STOP DATA NO STOP DESCRIPTION TY LOAD WGHT VOL CUBE HCRD VCRD ZN LOAD TIME BEG1 END1 BEG2 END2 1 Stop 1 D 1 0 014 080 0 0 0 9999 9999 9999 2 Stop 2 D 1 0 014 114 0 0 0 9999 9999 9999 3 Stop 3 D 1 0 014 131 0 0 0 9999 9999 9999 4 Stop 4 D 1 0 035 131 0 0 0 9999 9999 9999 5 Stop 522 D 1 0 052 061 0 0 0 9999 9999 9999 6 Stop 6 D 1 0 058 131 0 0 0 9999 9999 9999 7 Stop 7 D 1 0 080 131 0 0 0 9999 9999 9999 8 Stop 8 D 1 0 103 061 0 0 0 9999 9999 9999 9 Stop 9 D 1 0 103 096 0 0 0 9999 9999 9999 10 Stop 10 D 1 0 103 131 0 0 0 9999 9999 9999 11 Stop 11 D 1 0 136 131 0 0 0 9999 9999 9999 12 Stop 12 D 1 0 148 131 0 0 0 9999 9999 9999 13 Stop 13 D 1 0 180 131 0 0 0 9999 9999 9999 14 Stop 14 D 1 0 187 131 0 0 0 9999 9999 9999 15 Stop 15 D 1 0 184 061 0 0 0 9999 9999 9999 16 Stop 16 D 1 0 195 061 0 0 0 9999 9999 9999 17 Stop 17 D 1 0 129 010 0 0 0 9999 9999 9999 18 Stop 18 D 1 0 126 061 0 0 0 9999 9999 9999 19 Stop 19 D 1 0 115 010 0 0 0 9999 9999 9999 20 Stop 20 D 1 0 069 023 0 0 0 9999 9999 9999 21 Stop 21 D 1 0 014 026 0 0 0 9999 9999 9999 VEHICLE DATA CAPACITY VEHICLE DRIVER NO VEHICLE DESCRIPTION TP NO WGHT CUBE FIXED COST PER MI COST FIXED COST PER HR COST OVER TIME COST 1 Bus 1 1 9999 9999 0 0 0 0 0 SPECIFIED STOPTOSTOP DISTANCES NO STOP NO STOP DESCRIPTION STOP NO STOP DESCRIPTION DISTANCE IN MILES 1 14 Stop 14 16 Stop 16 078 2 14 Stop 14 15 Stop 15 090 3 15 Stop 15 17 Stop 17 106 4 16 Stop 16 17 Stop 17 118 5 18 Stop 18 9 Stop 9 058 6 19 Stop 19 8 Stop 8 076 7 19 Stop 19 20 Stop 20 059 8 19 Stop 19 522 Stops522 114 9 19 Stop 19 18 Stop 18 053 10 9 Stop 9 20 Stop 20 108 11 9 Stop 9 19 Stop 19 111 12 9 Stop 9 21 Stop 21 169 13 522 Stops 522 1 Stop 1 056 14 522 Stops 522 21 Stop 21 105 15 522 Stops 522 20 Stop 20 114 16 522 Stops 522 9 Stop 9 097 17 20 Stop 20 21 Stop 21 084 18 20 Stop 20 0 School 103 19 20 Stop 20 522 Stops 522 055 23 20 17 Stop 17 0 School 243 21 0 School 522 Stops 522 137 22 2 Stop 2 522 Stops 522 103 6 Strategy 1 is to stay at motel M2 and serve the two routes on separate days Using the ROUTESEQ module in LOGWARE gives us the sequence of stops and the coordinate distance The routes originating at M2 would be Route Stop sequence Distancea 1 861423579 9555 mi 2 101314171816121511 8645 18200 mi a Includes map scaling factor The total cost of this strategy would be Motel 3 nights 4900 14700 Travel 182 miles 30mi 5460 Total 20160 Strategy 2 is a mixed strategy involving staying at motels closest to the center of the stops clusters The route sequences from different motels are Route Stop sequence Distance 1 423579862 9850 mi 2 181713141011151216 8030 17880 mi The total cost of this strategy is Strategy 2 appears to be most economical 7 a Since distances are asymmetrical we cannot use the geographically based traveling salesman method in LOGWARE Rather we use a similar module in STORM that allows such asymmetrical matrices or the problem is small enough to be solved by inspection For this problem the minimal cost stop sequence would be Motel M1 1st night 4000 M1 2nd night 4000 M1 3rd night 4500 Travela 21480 mi 030mi 6444 Total 18944 a17880 36 21480 24 BakeryStop 5Stop 3Stop 4Stop 2Stop 1Bakery with a tour time of 130 minutes b Loadingunloading times may be added to the travel times to a stop The problem may then be solved as in part a c The travel times between stop 3 and all other nodes are increased by 50 The remaining times are left unchanged Optimizing on this matrix shows no change in the stop sequence However the tour time increases to 14750 minutes 8 This may be solved by using the ROUTER module in LOGWARE The screen set up for this is as follows 25 FIGURE 73 Input Data for ROUTER for Sima Donuts Making a run with ROUTER will give the route design PARAMETERS AND LABELS Problem label Sima Donuts Grid corner with 00 coordinates NW SW SE or NE NE DEPOT DATA Depot description Atlanta Located in zone 0 Horizontal coordinate 2084 Vertical coordinate 7260 Earliest starting time min 180 Latest return time min 9999 Default vehicle speed miles per hour 45 After how many clock hours will overtime begin 168 GENERAL DATA Percent of vehicle in use before allowing pickups 0 Horizontal scaling factor 0363 Vertical scaling factor 0363 Maximum TIME allowed on a route hours 40 Maximum DISTANCE allowed on a route miles 1400 LOADUNLOAD TIME FORMULA Fixed time per stop 0 Variable time per stop by weight 0 By cube 0 BREAK TIMES Duration of 1st break minutes 60 To begin after 720 Duration of 2nd break minutes 60 To begin after 1200 Duration of 3rd break minutes 60 To begin after 2160 Duration of 4th break minutes 60 To begin after 2640 STOP DATA NO STOP DESCRIPTION TY LOAD WGHT VOL CUBE HCRD VCRD ZN LOAD TIME BEG1 END1 BEG2 END2 1 Tampa FL D 20 0 1147 8197 0 15 360 1440 1800 2880 2 Clearwater FL P 14 0 1206 8203 0 45 360 1440 1800 2880 3 Daytona Beach F D 18 0 1052 7791 0 45 360 1440 1800 2880 4 Ft Lauderdale FL D 3 0 557 8282 0 45 180 1440 1800 2880 5 N Miami FL D 5 0 527 8341 0 45 360 1440 1800 2880 6 Oakland Park FL P 4 0 565 8273 0 45 180 1440 1800 2880 7 Orlando FL D 3 0 1031 7954 0 45 180 1440 1800 2880 8 St Petersburg FL P 3 0 1159 8224 0 45 180 1440 1800 2880 9 Tallahassee FL D 3 0 1716 7877 0 15 600 1440 1800 2880 10 W Palm Beach F D 3 0 607 8166 0 45 360 1440 1800 2880 11 Puerto Rico D 4 0 527 8351 0 45 360 1440 1800 2880 VEHICLE DATA CAPACITY VEHICLE DRIVER NO VEHICLE DESCRIPTION TP NO WGHT CUBE FIXED COST PER MI COST FIXED COST PER HR COST OVER TIME COST 1 Truck 120 1 3 20 9999 0 130 0 0 0 2 Truck 225 2 1 25 9999 0 130 0 0 0 3 Truck 330 3 1 30 9999 0 130 0 0 0 26 FIGURE 74 Graphical Display of Route Design for Sima Donuts The route design involves 3 routes for a total distance of 3830 miles a cost of 497871 and a total time of 1004 hours The route details are as follows Route 1 with 20pallet truck Depot Start time 300AM of day 1 Daytona Beach Deliver 18 pallets Clearwater Pickup 14 pallets Depot Return time 548AM of day 2 Route 2 with 20pallet truck Depot Start time 300AM of day 1 Orlando Deliver 3 pallets W Palm Beach Deliver 3 pallets Ft Lauderdale Deliver 3 pallets N Miami Deliver 5 pallets MiamiPuerto R Deliver 4 pallets Depot Return time 443PM of day 2 Pickup Pickup 27 Route 3 with 30pallet truck Depot Start time 413AM of day 1 Tallahassee Deliver 3 pallets Tampa Deliver 20 pallets St Petersburg Pickup 3 pallets Oakland Park Pickup 4 pallets Depot Return time 403PM of day 2 9 Given sailing times and dates when deliveries are to be made loadings need to be accomplished no later than the following dates To A B C D From 1 16 40 1 2 69 25 5 The problem can be expressed as a transportation problem of linear programming There will be 6 initial states 11 25 116 225 140 and 269 and 6 terminal states D10 C15 A36 B39 C52 and A86 The linear program is structured as shown in Figure 74 Using a transportation solution method we determine one of the optimum solutions There are several The solution is read by starting with the slack on initial loading state 1 This tells us to next select the cell of terminal state 1 In turn this defines initial state 3 and hence terminal state 3 And so it goes until we reach the terminal state slack column This procedure is repeated until all initial state slacks are exhausted Our solution shows two routings The first is 11D10 116A36269A86 The second is 25C15225B39140C52 Two ships are needed 28 FIGURE 75 Transportation Matrix Setup and Solution for the Queens Lines Tanker Scheduling Problem 10 This is a problem of freight consolidation brought about by holding orders so they can be shipped with orders from subsequent periods The penalty associated with holding the orders is a lost sales cost i Orders shipped as received Weight Rate Cost Haysa 10000 00519 51900 Manhattan 14000 00519 72600 Salina 13000 00408 53000 Great Bend 10000 00498 49800 Transportation 227400 Lost sales 00 Total 227400 a Ship 8000 lb as if 10000 lb Average period cost is 227400 Loading points and dates Load date Discharge date 1 1 2 5 1 16 2 25 1 40 2 69 Slack Rim re striction D 10 100 XXa 100 XX 1 1 1 1 1 10 1 C 15 100 XX 100 XX 1 1 1 1 1 10 1 A 36 100 XX 100 XX 100 XX 100 XX 1 1 1 10 1 B 39 100 XX 100 XX 100 XX 100 XX 1 1 1 10 1 C 52 100 XX 100 XX 100 XX 100 XX 100 XX 1 10 1 1 A 86 100 XX 100 XX 100 XX 100 XX 100 XX 100 XX 10 1 1 Slack 10 1 10 1 10 10 10 10 10 4 6 Rim re striction 1 1 1 1 1 1 6 a XX inadmissible cells given a high cost 29 ii Consolidate first period orders with second period orders Weight Rate Cost Hays 16000 00519 83040 Manhattana 40000 00222 88800 Salina 26000 00342 88920 Great Bend 10000 00498 49800 Transportation 310560 Lost sales 105000 Total 415560 a Ship 28000 lb as if 40000 lb The lost sales cost is 1000 cases 105 105000 to hold one group of orders for 2 weeks Average cost per period is 4155602 207780 iii Hold all orders until the third period Weight Rate Cost Hays 24000 00426 102240 Manhattan 42000 00222 93240 Salinaa 40000 00246 98400 Great Bend 15000 00498 74700 Transportation 368580 Lost sales 315000 Total 683580 a Ship as if 40000 lb Lost sales Hold 1st period orders for 2 periods 10001052 2100 Hold 2nd period sales for 1 period 1000105 1050 3150 Average period cost is 6835803 227860 Summary Ship immediately 227400 Hold orders 1 period 207780 Optimum Hold orders 2 periods 227860 11 Routes are built by placing the trips endtoend throughout the day from 4AM until 11PM respecting the times that a warehouse can receive a shipment This is a 19hour block of time per day or there are 95 hours per week per truck in which a truck may 30 operate If there were no delivery time restrictions on warehouses and trips could be placed endtoend for a truck without any slack at the end of the day the absolute minimum number of trucks can be found multiplying the number of trips by the route time and then dividing the total by the 95 hours allowed per week That is Warehouse location 1 Number of trips 2 Total time per trip hr 312 Total time hr Flint 43 125 5375 Alpena 5 1050 5250 Saginaw 8 225 1800 Lansing 21 375 7875 Mt Pleasant 12 550 6600 W Branch 5 600 3000 Pontiac 43 275 11825 Traverse City 6 1050 6300 Petoskey 5 1175 5875 Total 53900 For 539 trip hours 53995 567 rounded to six trucks needed per week Now it is necessary to adjust for the problem constraints A good schedule can be found by following a few simple rules that can be developed by examining the data First begin the day with a trip where the driving time to a warehouse is just long enough for the truck to arrive at the warehouse just after it opens Onehalf the driving time should exceed 630 400 230 or 2½ hr Trips to Alpena Traverse City and Petoskey qualify Second use the short trips at the end of the day to avoid slack time Third allocate the trips to the days using the longest ones first Make sure that the total trip time for a day does not exceed 19 hours For a minimum of six trucks the following feasible schedule can be developed by inspection Day 1 Day 2 Day 3 Day 4 Day 5 Truck 1 Petoskey 1175 W Branch 600 Flint 125 Total 1900 hr Petoskey 1175 W Branch 600 Flint 125 Total 1900 hr Petoskey 1175 W Branch 600 Flint 125 Total 1900 Petoskey 1175 5 Flint 625 Total 1800 hr Petoskey 1175 5 Flint 625 Total 1800 hr Truck 2 T City 1050 2 Lansing 750 Total 1800 hr T City 1050 2 Lansing 750 Total 1800 hr T City 1050 2 Lansing 750 Total 1800 T City 1050 2 Lansing 750 Total 1800 hr T City 1050 2 Lansing 750 Total 1800 hr Truck 3 T City 1050 2 Lansing 750 Total 1800 hr Alpena 1050 2 Lansing 750 Total 1800 hr Alpena 1050 2 Lansing 750 Total 1800 hr Alpena 1050 2 Lansing 750 Total 1800 hr Alpena 1050 2 Lansing 750 Total 1800 hr Truck 4 Alpena 1050 Lansing 375 3 Flint 375 Total 1800 hr M Pleasant 550 4 Pontiac 1100 2 Flint 250 Total 1900 hr M Pleasant 550 4 Pontiac 1100 2 Flint 250 Total 1900 hr M Pleasant 550 4 Pontiac 1100 2 Flint 250 Total 1900 hr M Pleasant 550 4 Pontiac 1100 2 Flint 250 Total 1900 Truck 5 M Pleasant 550 4 Pontiac 1100 2 Flint 250 Total 1900 hr M Pleasant 550 4 Pontiac 1100 2 Flint 250 Total 1900 hr M Pleasant 550 4 Pontiac 1100 Flint 125 Total 1775 M Pleasant 550 4 Pontiac 1100 Total 1650 hr M Pleasant 550 4 Pontiac 1100 Total 1650 Truck 6 M Pleasant 550 4 Pontiac 1100 2 Flint 250 Total 1900 hr M Pleasant 550 3 Pontiac 825 2 Flint 250 Total 1625 hr M Pleasant 550 6 Saginaw 1350 Total 1900 hr W Branch 600 2 Saginaw 450 Total 1050 hr W Branch 600 10 Flint 1250 Total 1850 hr 31 Although this schedule meets the requirements of the problem it might be improved by better balancing the workload across the trucks and the days 12 a A sweep method solution is shown on the following figure Five trucks are needed with a total route distance of 3029394419510 1615 miles b The sweep method is a fast and relatively simple method for finding a solution to rather complex vehicle routing problems Solutions can be found graphically without the aid of a computer However there are some limitations Namely The method is heuristic and has an average error of about 10 to 15 percent This error is likely to be low if the problem contains many points and the weight of each point is small relative to the capacity of the vehicle The method does not handle timing issues well such as time windows Too many trucks may be used in the route design 13 This problem may be solved with the aid of ROUTER in LOGWARE The model input data may be formatted as shown in Figure 76 5 3 0 6 4 5 Warehouse 5 3 3 4 3 3 4 2 5 4 3 4 3 2 3 1 4 0 2 4 6 8 10 12 14 16 18 20 22 24 26 Miles x 10 2 4 Miles x 10 20 18 16 14 12 10 8 2 6 2 Route 1 Load 19 Route 2 Load 20 Route 3 Load 17 Route 4 Load 8 Route 5 Load 9 32 a The solution from ROUTER shows that four routes are needed with a minimum total distance of 492 miles The route design is shown graphically in Figure 77 A summary for these routes is given in following partial output report Route Run Stop Brk Stem Route time time time time time Start Return No of Route Route no hr hr hr hr hr time time stops distMi cost 1 12 10 3 0 4 0859AM 1012AM 3 29 00 2 89 66 13 10 11 0832AM 0525PM 19 199 00 3 62 37 14 10 9 0842AM 0254PM 14 112 00 4 75 51 15 10 14 0830AM 0402PM 12 152 00 Total 238 164 44 30 38 48 492 00 b Note that route 1 is short and that a driver and a station wagon would be used for a route that takes 12 hours to complete By attaching route 1 to route 3 the same driver and station wagon may be used and the constraints of the problems are still met The refilled station wagon can leave the depot by 330345PM and still meet the customers time windows and return to the depot by 6PM Thus only three drivers and station wagons are actually needed for this problem FIGURE 76 Input Data for ROUTER for Medic Drugs PARAMETERS AND LABELS Problem label Medic Drugs Grid corner with 00 coordinates NW SW SE or NE SW DEPOT DATA Depot description Pharmacy Located in zone 0 Horizontal coordinate 137 Vertical coordinate 212 Earliest starting time min 480 Latest return time min 9999 Default vehicle speed miles per hour 30 After how many clock hours will overtime begin 168 GENERAL DATA Percent of vehicle in use before allowing pickups 0 Horizontal scaling factor 46 Vertical scaling factor 46 Maximum TIME allowed on a route hours 168 Maximum DISTANCE allowed on a route miles 9999 LOADUNLOAD TIME FORMULA Fixed time per stop 0 Variable time per stop by weight 0 By cube 0 BREAK TIMES Duration of 1st break minutes 60 To begin after 720 Duration of 2nd break minutes 0 To begin after 9999 Duration of 3rd break minutes 0 To begin after 9999 Duration of 4th break minutes 0 To begin after 9999 STOP DATA NO STOP DESCRIPTION TY LOAD WGHT VOL CUBE HCRD VCRD ZN LOAD TIME BEG1 END1 BEG2 END2 1 Covington House D 1 0 2340 1290 0 2 540 1020 9999 9999 2 Cuyahoga Falls D 9 0 1340 1340 0 18 540 1020 9999 9999 3 Elyria D 1 0 630 1680 0 5 540 1020 9999 9999 33 46 Westbay D 6 0 840 1800 0 10 630 690 9999 9999 47 Westhaven D 2 0 850 1810 0 5 540 1020 9999 9999 48 Broadfiels Mnr D 6 0 1820 2290 0 2 540 1020 9999 9999 VEHICLE DATA CAPACITY VEHICLE DRIVER NO VEHICLE DESCRIPTION TP NO WGHT CUBE FIXED COST PER MI COST FIXED COST PER HR COST OVER TIME COST 1 Station wagon 1 50 63 9999 0 0 0 0 0 FIGURE 77 Graphical Solution for Medic Drugs 14 There is no exact answer to this problem nor is one intended Several approaches might be taken to this problem We could apply the savings method or the sweep method to solve the routing problem for each day of the week given the current demand patterns However we can see that there is much overlap in the locations of the customers by delivery day of the week We might encourage orders to be placed so that deliveries form tight clusters by working with the sales department and the customers Perhaps some incentives could be provided to help discipline the order patterns The orders should form a general pattern as shown below Currently the volume for Thursday exceeds the available truck capacity of 45 caskets Maybe the farthest stops could be handled by a forhire service rather than acquiring another truck for such little usage 34 Monday Tuesday Wednesday Thursday Friday Depot It appears that the truck capacity is about right given that some slack capacity is likely to be needed Once the pattern orders are established either as currently given or as may be revised apply principles numbers 1 3 4 5 and 7 35 FOWLER DISTRIBUTING COMPANY Teaching Note Fowler Distributing Company involves a problem of routing and scheduling a fleet of trucks to serve customers for beer and wine products on a daily basis Determining an efficient design for making daily deliveries when stop sequence number of trucks and their sizes warehouse location and time window restrictions are variables is the objective of this exercise For the most part students should consider that they are to prepare a consulting report to management about this problem The several questions provided at the end of the case study should help direct the analysis The ROUTER model is used to cost out and to optimize the various design scenarios Run results are tabulated in Table 1 Q1 The current design may not be as efficient as it might be Therefore our first task is establish a profile of the current design and then to plan the routes so as to minimize the total miles driven and the number of trucks needed to serve the customers subject to the restrictions of truck capacity time windows total time on the route etc Running the current route design in ROUTER establishes a benchmark as shown in Figure 1 The design requires 334 miles of travel with 5 trucks The daily routing cost is 76462 Next optimizing this problem given no change in data or restrictions gives the revised benchmark design as shown in Figure 2 Now costs are reduced to 73131 per day for a total routing distance of 311 miles Six trucks are needed There appears to be a substantial change in the design of the routes that can achieve a 334 311334100 7 reduction in the total miles traveled to make the deliveries The saving in cost is 76462 73131250 832750 per year This is not a large savings but it can be achieved without changing the restriction on deliveries or incurring additional investment Q2 Time window restrictions often force route design into stop sequences that are not very economical that is routes cross themselves This is the case for the revised benchmark design as shown in Figure 2 To determine the impact of the time windows we can make an optimizing ROUTER run where there are no time window restrictions The daily time windows are all set 800AM to 500PM The optimized routes are shown in Figure 3 The route mileage is 270 miles per day for five trucks The total cost is 67364 per day From the revised benchmark this is an additional cost reduction of 73131 67364 250 1441750 per year 36 FIGURE 1 Current Route Design The questions for management are How restrictive are the time windows Can deliveries be made outside of the accounts open hours such as by giving the driver the key to a safe storage area Can management offer a small incentive to widen the time window when it otherwise would not be convenient for the account Relaxing such time windows is often one of the important sources for cost savings in routing problems Q3 When there are no truck capacity restrictions on a routing problem the most efficient route design would be to use one large truck to serve all accounts Therefore we would expect that trucks of larger capacity would reduce the total distance traveled A ROUTER optimizing run was made with 600 case capacity trucks to find out All other conditions were set at the current design 37 FIGURE 2 Optimized Current Route Design Compared with the optimized current design the potential savings is 73131 72298 250 208250 per year Two 600case trucks would be needed along with 3 of the smaller trucks Although there is a positive savings associated with using larger trucks the savings seems quite small and perhaps not worth switching to some larger trucks at this time We should note that these savings are a result of comparing full costs which include truck depreciation Although a present value analysis would be appropriate here we would need some additional information which might include 1 the proportion of truck costs allocated to equipment depreciation 2 the life of the trucks 3 the estimated salvage value of the trucks and 4 and the required rate of return on investments of this type 38 FIGURE 3 Route Design with Relaxed Time Window Constraints Q4 From the incremental costs in the detailed report of the optimized current design we can see that account 14 costs 3905 and account 19 costs 3504 to make the deliveries Since the volume of account 19 is 90 cases and exceeds the 50 case capacity of an outside transport service account 19 cannot be considered for alternate delivery If it only costs 3500 for an outside delivery service to handle account 14 then a cost savings can be realized Dropping this account and redesigning the routes shows that only 5 trucks are needed for a route cost of 69017 and route miles of 286 The total cost for handling all accounts would be 69017 3500 72517 Compared with the optimized current design this is an annual savings of 73131 72517 250 153500 Economically Roy should use the outside transportation However losing direct control over the deliveries and possible adverse reactions from route salesmen may give him second thoughts about it Q5 Two ROUTER runs were made here in order to determine the effect of a shorter workday before overtime begins If no overtime is allowed then six trucks are required for a total daily routing cost of 75477 If some overtime is permitted then the route cost can be reduced to 73328 with 6 trucks required Compared with the optimized current design Roy Fowler would seem to be putting himself at a disadvantage by not wanting to pay overtime Q6 Moving the warehouse to a more central location would have the appeal of being closer to all stops and would result in shorter routes Testing this shows that total route distance can be reduced to 305 miles with a daily cost of 71691 However in order to meet the time window and other constraints an extra truck is needed There is a potential annual savings of 73131 71691 250 360000 Since it costs 15000 to make 39 the move a simple return on investment would be 360015000 100 24 At this ROI the move should be seriously considered Q7 Routing and scheduling beer trucks from a central depot is quite similar to delivery problems found in manufacturing retail and service industries Several examples are listed below Serving branch banks by making pickups and deliveries of canceled checks supplies and other paper work from a central domicile point for the trucks Making Federal Express pickups and deliveries around an airport Dispatching school buses around a school for student pick up and drop off Dispatching service personnel for various service companies such as electric water gas and telephone Making retail deliveries from a warehouse Making deliveries from a plant location to warehouses and then picking up supplies from vendors on the return trip Making deliveries and pickups of medical records to doctors offices and other locations from a central recordkeeping location such as a hospital Summary Roy Fowler should consider optimizing his route design This can be done at no investment to him and he can gain about a 7 reduction in operating costs on the average A substantial benefit can be realized by widening the time windows He should especially see if those time windows that cause routes to overlap themselves can be widened The potential for doing this is up to 14418 per year He should explore the possibility and reasonableness of serving accounts 14 and 19 by an outside transport service Finally he should consider relaxing his rigid policy of not wanting to pay overtime especially if the union is successful in negotiating a 7½hour workday There seems to be little opportunity for reducing cost by increasing truck capacity or relocating the warehouse Although there are operating cost reductions available they do not seem sufficient to justify a change in truck size unless Roy can make good use of the trucks in other ways 40 TABLE 1 Results of Various ROUTER Runs Run type Miles Cost Trucks Comments Current design 334 76462 5 Cost out current design Optimized current 311 73131 6 Improves on current route design No time windows restrictions 270 67364 5 Time windows opened Trucks at 600case capacity 299 72298 5 Two routes require larger trucks 7½ hr work day no overtime 332 75477 6 Route time allowed to be no longer than 80 hours 7½ hr work day with overtime 311 73328 6 Route time allowed to exceed 80 hours New warehouse location 305 71691 7 Warehouse moved to more central location 41 METROHEALTH MEDICAL CENTER Teaching Note Strategy MetroHealth Medical Centers mission is to provide transportation to and from its facility for those patients that cannot afford the expense or are unable to use other modes of transportation MMC currently has a fleet of vans and drivers to serve this patient population In addition some transportation can be provided as part of the hospitals contract ambulance service The primary service provided by Transportation Services is to pick patients up and bring them to MMC for their scheduled appointments and then to return them to their origin points Careful management of the fleet is needed through vehicle assignment to patients routing the vans and assigning some of the patients to the ambulance service for transportation It is the purpose of this case to show how a basic routing and scheduling model can be used to assist in answering the typical questions surrounding fleet utilization in a service setting The ROUTER module in LOGWARE is used as the analytical tool This case does require some insight on the part of the student to deal with the complexities of this problem Considering the amount of data provided and the richness of the issues involved this case is probably best assigned as a project rather than as a homework exercise Answer to Questions 1 Which vans from the current fleet should be used To what extent should subcontracting be used This problem is complex considering that it is quite dynamic Although the patient pick up list appears to be fixed for a particular day changes take place in the form of cancellations and occasional additions Patient appointment times are met a high percentage of the time but not always Patients to be returned to their origins may find alternate means of transportation so that 100 percent of the patients picked up may not have to be returned For simplicity it will be assumed that the patient pick up list is fixed for a particular day and appointment times are rigid In addition the returns are not directly considered in designing the routes Rather they are expected to be seeded into the pick up routes This need not be a serious limitation as long as van capacity number of pickup patients to the number of available seats is not highly utilized on a pickup route and van utilization for pick ups drops as appointments are concentrated in the morning hours and return times are shifted toward the afternoon hours as shown in Figure 1 42 Appointment times Return times FIGURE 1 Appointment Time and Return Time Distributions A typical days patient list can be submitted to ROUTER Coordinate points for zip code areas are scaled from the map in Figure 2 of the case study Appointment times are represented as time windows with a 2hour gap for pick up A large number of 15 passenger and 6passenger vans are assumed available to start the ROUTER model Since ROUTER is a basic vehicle routing and scheduling model it simply assigns patients to vans and gives a sequence to pick them up It assumes that a single van leaves and returns to MMC only once in a day Analysis must account for this since vans make several out and back trips per day It is not possible to accurately recreate the current routing patterns for the vans from the data given However optimized routing and scheduling can be found through the application of ROUTER First the typical daily appointment list is solved with ROUTER using a 2hour time window where the ending window time is the appointment time An average speed of 275 miles per hour variable costs of 011 per mile1 and a large number of 15passenger and 6passenger vans are used A complete database is shown in the appendix of this note Using the appointment list for the typical day as representative with all 56 patients ROUTER shows that 6 vans are needed where one is a 15passenger type route 1 This is routing where no patient is transported by the contract transport service The routing represents a total of 329 miles driven at a variable cost of 329 mi 011mi 3619 The routes are placed endtoend to cover the full day and then ranked first by the number of patients transported throughout the day on a van and second by the total miles driven as shown in Table 1 A reasonable procedure for allocating routes to vans is to assign routes with the largest number of patients first to the largest vans This is because 1Per mile variable cost is determined as repair cost30000 mi per yr 100030000 003 per mi plus fuel cost at 100 per gallonaverage miles per gallon 113 008 per mi for a total of 008 003 011 per mile 43 the ambulance service charges on a per patient basis and these routes would be most costly to subcontract If the number of patients is tied for a van rank first the van with the shortest total route distance TABLE 1 Route Design for Representative Patient Appointments with No Transport Subcontracting Van no Start from MMC Return to MMC Patients on route Route distance Route time 1 607AM 628AM 1 7 mi 03 hr 1 652AM 828AM 81 22 16 1 855AM 1030AM 4 33 16 1 1104AM 1131AM 1 9 04 1 1141AM 127PM 5 35 18 2 724AM 905AM 5 32 17 2 1059AM 1157AM 5 13 10 3 741AM 916AM 5 29 16 3 1045AM 1227PM 5 33 17 4 737AM 912AM 5 30 16 4 1152AM 134PM 5 33 17 5 724AM 844AM 4 25 13 6 743AM 903AM 3 28 16 56 329 mi 176 hr 1A 15passenger van needed The cost of this initial route configuration can be summarized as follows Annual cost of 1 15passenger van 5750 1 Annual cost of 5 6passenger vans 28500 2 Annual repair cost of 6 vans 6000 Annual cost for 6 drivers 141000 Variable cost 8686 3 Total cost 189936 1230004 yr 5750 2190004 yr 6 28500 3329 mi x 011mi 240 daysyr 8686 The question now is whether substituting contract carriage for some of the routes will significantly lower vehicle and driver costs while only slightly increasing variable costs Since the routes are ranked in Table 1 a reasonable rule for dropping routes would be to start from the bottom of the route list in Table 1 and work upward This will determine the number of vans needed and the patients to be transported by each mode These results are shown in Table 2 An unlimited number of subcontracting trips is assumed TABLE 2 Optimal Number of Vans 44 Vans Driver Vehicle Variable gas repair Subcon tracting Total 6 141000 34250 14686 0 189936 5 117500 29500 12946 6235 1 166181 4 94000 20000 11286 14549 139835 3 70500 15250 8623 35333 129706 2 47000 10500 5986 56117 119603 1 23500 5750 3798 76901 109949 0 0 0 0 116390 116390 1Annual subcontracting cost is 866 per round trip 240 days per year 3 patients 623520 2 How many subcontracted trips should MMC negotiate and at what price One 15passenger van appears to be the optimal number but this requires 37 patients 20 days per month 740 subcontracting trips per month This is more than the 500 allowed If the 500trip limit 50020 days per mo 25 patients per day is to be respected approximately one additional van will be needed On the other hand the cost of the additional van is the annual cost of a van the annual drivers salary the gas and repair cost 190004 23500 11882 29438yr to transport 10 patients per day At this rate MMC could afford to pay 2943824010 1227 per round trip find 10 patients for van 2 in Table 1 Possibly the subcontractor would be willing to offer this additional transport for a price between the 866 and 1227 per trip It would be in Macs interest to do this Renegotiating all trips at the 750trip level is another possibility MMC can afford to pay a subcontracting cost of up to 189936 310483 158888yr This would require transporting 37 patients per day at a maximal average trip cost of 15888824037 1789 Since 1789 is the optimal worth of subcontracting MMC probably can negotiate a rate much less than this while still encouraging the subcontractor to provide service at the higher level 3 MMC is considering using all 6passenger vans Would this be a good decision Developing daily routes and assigning them to 6passenger vans is the same as for Table 1 This initial allocation without subcontracting is shown in Table 3 Considering the subcontracting trip limit of 500 patients per month 25 patients per day two vans are needed in the MMC fleet This number of vans is determined by counting the patients from the bottom of Table 3 until approximately 25 patients are found considering increments of full vans and noting the number of vans remaining at the top of the list Assume 27 an extra 2 patients can be accommodated by the subcontractor at no additional cost The variable cost for two vans is shown below Variable cost type Two 6passenger One 15 one 6passenger van 245 miles x 011mile 240 daysyr 1188 3Driver vehicle variable costs for one van or 190004 23500 2798 31048 45 vans Annual cost of 2 drivers 47000 47000 Vehicle mileage repair cost 59601 5986 Subcontracting cost 561172 56117 Total 109077 109103 1150 mi 011mi 240 daysyr 2000 3960 227 patients 866trip 240 daysyr 56117 In contrast the 15passenger and 6passenger van configuration compared with the two 6passenger van configuration saves 109103 109077 26 per year The initial outlay is less for the smaller van by 4000 From purely an economic standpoint using just 6passenger vans would appear to be the best choice however there may be customer service reasons for having the larger van available to meet peak demand needs TABLE 3 Route Design for 6Passenger Vans Only Van no Start from MMC Return to MMC Patients on route Route distance Route time 1 607AM 628AM 1 7 mi 03 hr 1 652AM 737AM 3 12 08 1 737AM 912AM 5 30 16 1 1104AM 1131AM 1 9 04 1 1152AM 134PM 5 33 17 2 722AM 819AM 5 13 10 2 855AM 1030AM 4 33 16 2 1059AM 1157AM 5 13 10 3 741AM 916AM 5 29 16 3 1045AM 1227PM 5 33 17 4 724AM 905AM 5 32 17 4 1141AM 127PM 5 35 18 5 724AM 844AM 4 25 13 6 743AM 903AM 3 28 13 56 332 mi 178 hr 46 APPENDIX A Database for ROUTER PARAMETERS AND LABELS Problem label METROHEALTH MEDICAL HOSPITAL Grid corner with 00 coordinates NW SW SE or NE SW DEPOT DATA Depot description Medical Center Located in zone 0 Horizontal coordinate 740 Vertical coordinate 613 Earliest starting time min 360 Latest return time min 960 Default vehicle speed miles per hour 28 After how many clock hours will overtime begin 680 GENERAL DATA Percent of vehicle in use before allowing pickups 100 Horizontal scaling factor 3102 Vertical scaling factor 2820 Maximum TIME allowed on a route hours 20 Maximum DISTANCE allowed on a route miles 99990 LOADUNLOAD TIME FORMULA Fixed time per stop 600 Variable time per stop by weight 000 By cube 000 BREAK TIMES Duration of 1st break minutes 0 To begin after 9999 Duration of 2nd break minutes 0 To begin after 9999 Duration of 3rd break minutes 0 To begin after 9999 Duration of 4th break minutes 0 To begin after 9999 STOP DATA STOP LOAD VOLUME COORDINATES LOAD TIME WINDOWS NO DESCRIPTION TY WGHT CUBE HCRD VCRD ZN TIME BEG1 END1 BEG2 END2 1 Baker P 1 0 650 890 0 0 390 510 9999 9999 2 Boyd P 1 0 620 900 0 0 390 510 9999 9999 3 Carver P 1 0 1100 500 0 0 420 540 9999 9999 4 Ivey P 1 0 1000 600 0 0 420 540 9999 9999 5 Rashed P 1 0 1000 1000 0 0 420 540 9999 9999 6 Walsh P 1 0 350 500 0 0 420 540 9999 9999 7 Johnson P 1 0 800 550 0 0 450 570 9999 9999 8 Burgess P 1 0 900 750 0 0 465 585 9999 9999 9 Delgado P 1 0 820 720 0 0 465 585 9999 9999 10 Fairrow P 1 0 860 780 0 0 480 600 9999 9999 11 Middlebrooks P 1 0 800 540 0 0 480 600 9999 9999 12 Suech P 1 0 500 650 0 0 480 600 9999 9999 13 Lawson P 1 0 630 880 0 0 510 630 9999 9999 14 Reed P 1 0 905 730 0 0 510 630 9999 9999 15 Bongiovanni P 1 0 900 580 0 0 525 645 9999 9999 16 Miller P 1 0 780 540 0 0 540 660 9999 9999 17 Talley P 1 0 650 1020 0 0 570 690 9999 9999 18 Williams P 1 0 750 700 0 0 585 705 9999 9999 19 Dumas P 1 0 720 590 0 0 630 750 9999 9999 20 Taylor P 1 0 980 700 0 0 645 765 9999 9999 21 Barker P 1 0 720 600 0 0 660 780 9999 9999 22 Lhota P 1 0 800 620 0 0 660 780 9999 9999 23 Manco P 1 0 1010 900 0 0 660 780 9999 9999 24 Webb P 1 0 970 700 0 0 660 780 9999 9999 25 Wilson P 1 0 900 580 0 0 660 780 9999 9999 26 Arrington P 1 0 1020 590 0 0 720 840 9999 9999 27 Staunton P 1 0 640 880 0 0 720 840 9999 9999 28 Wall P 1 0 950 510 0 0 720 840 9999 9999 29 Williams P 1 0 820 720 0 0 720 840 9999 9999 30 Caruso P 1 0 700 500 0 0 375 495 9999 9999 31 West P 1 0 560 600 0 0 390 510 9999 9999 32 Amaro P 1 0 600 660 0 0 420 540 9999 9999 33 Brown P 1 0 720 420 0 0 420 540 9999 9999 34 Ciesicki P 1 0 650 300 0 0 420 540 9999 9999 35 Pinkevich P 1 0 660 510 0 0 420 540 9999 9999 36 Staufer P 1 0 580 600 0 0 420 540 9999 9999 37 Winterich P 1 0 730 480 0 0 420 540 9999 9999 38 Brown P 1 0 400 500 0 0 435 555 9999 9999 39 Ball P 1 0 620 630 0 0 450 610 9999 9999 40 Lanza P 1 0 620 590 0 0 450 610 9999 9999 41 Mayernik P 1 0 710 700 0 0 450 610 9999 9999 42 Suech P 1 0 400 650 0 0 465 625 9999 9999 43 Heffner P 1 0 520 450 0 0 480 640 9999 9999 44 Jarrell P 1 0 420 680 0 0 480 640 9999 9999 47 45 Piatak P 1 0 700 350 0 0 480 640 9999 9999 46 Swaysland P 1 0 610 670 0 0 480 640 9999 9999 47 Baer P 1 0 380 490 0 0 540 700 9999 9999 48 Wills P 1 0 470 610 0 0 540 700 9999 9999 49 Fauber P 1 0 520 680 0 0 660 780 9999 9999 50 Mullins P 1 0 720 630 0 0 660 780 9999 9999 51 Pack P 1 0 620 500 0 0 660 780 9999 9999 52 Westerfield P 1 0 590 630 0 0 675 795 9999 9999 53 Lisiewski P 1 0 500 550 0 0 690 810 9999 9999 54 McPherson P 1 0 420 620 0 0 720 840 9999 9999 55 Mykytuk P 1 0 600 650 0 0 750 870 9999 9999 56 Gutschmidt P 1 0 550 590 0 0 780 900 9999 9999 VEHICLE DATA VEHICLE DRIVER OVER VEHICLE CAPACITY FIXED PER MI FIXED PER HR TIME NO DESCRIPTION TP NO WEIGHT CUBE COST COST COST COST COST 1 15pasngr vehcl 1 20 15 99 000 011 000 000 000 2 6pasngr vehcl 2 20 6 99 000 011 000 000 000 48 ORION FOODS INC Teaching Note Strategy The purpose of this case study is to allow students to analyze a distribution problem where determining the optimal paths through a network is central to the problem solution The shortest route methodology applies and the ROUTE module in the LOGWARE software can effectively be used This module permits students to quickly find the shortest distance paths from any starting point to all other points in the network A prepared data file of the network is available for use under the ROUTE module The case has several dimensions that allow it to be used as a homework assignment a short case study project or as a basis for classroom discussion It is a simple problem in network design highlighting transportation routing issues however additional factors such as inventory consolidation and investment costs allow for an enriched analysis Note A database for this problem is available in the ROUTE module of the LOGWARE software Answers to Questions 1 Can Anita improve upon the current distribution operations It should be obvious to Anita that the current distribution system may not be performing at optimum Allowing carriers to decide the routes to use when they are being paid on a mileage basis is like asking a fox to watch the chicken coop She really cannot expect that carriers will be motivated to seek out optimal routing patterns Therefore she should determine the best routes between regional and field warehouses and insist that carriers invoice according to the mileages along these specified routes Using the map provided in Figure 1 of the case study and the current assignments of field warehouses to regional warehouses she can develop a database for the ROUTE module in LOGWARE The database is shown in the Supplement to this note Running ROUTE will give the optimal routes from which she can develop transportation costs as shown in Table 1 below Compared with the current cost level a savings of 652274 630140 22134 per year can be realized This savings does not require any investment however it will be magnified by the growth in demand in the next five years 49 2 Is there any benefit to expanding the warehouse at Burns OR Finding the optimal distances when serving all field warehouses from Burns or Fresno gives a slightly different allocation of field warehouses to regional warehouses than is currently the case That is Field warehouse If served from Burns If served from Fresno Los Angeles 806 mi 219 mi Phoenix 973 588 Salt Lake City 536 815 San Francisco 555 183 Portland 293 757 Butte 676 1120 Seattle 467 925 Indicates optimal assignment This shows that Salt Lake City would be better served out of Burns rather than Fresno Burns currently is near its capacity limit so to assign Salt Lake Citys volume to it would require expansion In the short term 350008 4375 cwt of inventory capacity is needed However 43000 5000 560008 13000 cwt of the 15000 cwt of available capacity is currently being used At minimum an additional increment of capacity is required at a cost of 300000 Reassignment of Salt Lake City to Burns would save 815 536 279 miles per trip On 1167 trips per year the annual savings would be 130 279 1167 42327 The simple return on investment ROI would be ROI 42 300 327 000 100 141 TABLE 1 Optimal Transportation Costs Under Current Distribution System Design Regional warehouse Field warehouse Optimal route miles Average number of tripsa Transport cost Fresno Los Angeles 219 3667 104399b Fresno Phoenix 588 2000 152880 Fresno Salt Lake City 815 1167 123644 Fresno San Francisco 183 2800 66612 Burns Portland 293 1433 54583 Burns Butte 676 167 14676 Burns Seattle 467 1867 113346 Totals 3241 630140 a Determined from the warehouse throughput divided by the average shipment size eg 110000 cwt300 cwt 3667 trips b 130 per mi 219 mi 3667 trips 104399 50 If the anticipated growth in demand is realized the number trips to Salt Lake City would increase to 56000300 1867 The projected savings in the fifth year would be 130 279 1867 67716 The average annual savings would be 42327 677162 55022 The average annual ROI is ROI 55 300 022 000 100 183 Anita must now compare this return to other worthy investments in the firm to see if this opportunity is worth the risk 3 Is there any merit to consolidating the regional warehousing operation at Reno NV If Reno were to replace the Burns and Fresno warehouses an initial cost of 2000000 would be incurred to establish the new location and shut down the existing warehouses Against this cost would be a savings of 40 percent of the inventory in the two regional warehouses That is the total inventory is 3930008 49125 cwt now and 5280008 66000 cwt in five years The inventory cost savings now would be 040 035 60 49125 412650 and in five years 040 035 60 66000 554400 However transportation costs will increase compared with the twowarehouse distribution system If Reno is used the transportation costs would be Field warehouse served from Reno Optimal route miles Current no of trips Current transport cost 5thyear no of trips 5thyear transport costs Los Angeles 472 3667 225007a 4400 269984 Phoenix 732 2000 190320 2800 266448 Salt Lake City 520 1167 78889 1867 126209 San Francisco 228 2800 82992 3500 103740 Portland 542 1433 100969 1900 133874 Butte 823 167 17867 500 53495 Seattle 716 1867 173780 2633 245080 Totals 4033 869824 1198830 a 472 3367 130 225007 We cannot make a fair comparison with the current system design since there is inadequate capacity at Burns to handle the growth in volume Two additional units of capacity will be needed for a total of 600000 Then the net investment attributable to Reno is 2000000 600000 1400000 The revised transportation cost for the fifth year is as follows 51 Now Current year 5th year Transport costs 587813 828817 BurnsFresno Transport costs 869824 1198830 Reno Net increase 282011 370013 Less inventory savings 412650 554400 Net cost savings 130639 184387 The average annual savings for Reno now is 130639 1843872 157513 The relevant return on investment is ROI 157 1 513 400 000 100 1125 This is probably not a sufficient return to justify the warehouse at Reno Rather if Orion wishes to serve the increasing demandthere is no requirement to do sothen the better strategy would be to expand Burns by 20000 cwt and serve Salt Lake City from this location An interesting question to pose to students is What does it mean to only serve demand up to the limits of capacity Orion could serve only the more profitable demand and avoid the risks of expansion Regional warehouse Field warehouse Optimal route miles 5thyear number of trips Transport cost Fresno Los Angeles 219 4400 125268 Fresno Phoenix 588 2800 214032 Burns Salt Lake City 536 1867 130092 Fresno San Francisco 183 3500 83265 Burns Portland 293 1900 72371 Burns Butte 676 500 43940 Burns Seattle 467 2633 159849 Totals 2962 828817 52 R T WHOLESALERS Teaching Note4 The objective of this assignment is to minimize the total monthly delivery costs for RT Wholesalers a company distributing general products throughout India The focus is on one warehouse acting as a truck depot that delivers merchandise to retailers located in surrounding towns Delivery expenses are minimized through the optimal utilization of the trucks crews and related expenses The constraints and other considerations listed below were used when designing the solution methodology and identifying the optimal solution Constraints Operating Schedule Trucks make deliveries every day of the week except Saturday or Sunday Normal operation is for trucks to be loaded overnight and leave from the warehouse in the morning Trucks make deliveries within towns from 9 am to 6 pm Earliest start time for trucks is 12 am in the morning of delivery Trucks returned to the depot require 2 hours for reloading and subsequent same day delivery Visits Per Month Every town has a known number of visits per month Truck Capacity T407 trucks have a capacity of Rs500000 T310 trucks have a capacity of Rs350000 Truck Operating Costs Trucks operate at an average speed of 40 kmhr T407 trucks have an operating cost of Rs13500 T310 trucks have an operating cost of Rs7000 T407 trucks have a running cost of Rs5 per kilometer T310 trucks have a running cost of Rs3 per kilometer Each truck has a crew of two a driver and a helper The driver is paid Rs2200 per month The helper is paid Rs1400 per month Each crewmember receives Rs60 per day for meals and other expenses while on the road Working Schedule Flexibly planned breaks for crewmembers are at approximately 6 am 12 pm and 6 pm Breakfast and lunchtime breaks are 30 minutes each and dinner is 60 minutes 4 The solution to this exercise is provided by Nutthapol Dussadeenoad Inderjot Gandhi Earle Keith Lisa Kuta Jan Shahan and Piyanuch Vichitakul who were students in the MBA program of the Weatherhead School of Management 53 Crewmembers are allowed at least an 8hour overnight break before starting a route on the following day No overtime is paid and the company policy is to return the crews to the warehouse each day rather than plan for overnight layovers Other Considerations Assumptions Delivery demand is fixed by quantity and location and city demand cannot be subdivided Opportunities Subcontracting is available at an estimated rate of Rs15 per km one way to the town Methodology The methodological options leading to a good solution are outlined in Figure 1 START 1 Route 43 cities altogether 2 Route all 4visit cities and outsource all 2visit cities 3 Route 2 sets of data all 4visit cities in one seat and 2visit cities in another set 4 Route 2 sets of data mix between 4visit cities and 2visit cities 5 Outsource all cities whose demand is Rs350000 and route 2 sets of data mix between 4visit cities and 2visit cities Satisfy STOP N Y Cost Rs116678 Cost Rs135254 Cost Rs109092 Cost Rs78328 Cost Rs65780 FIGURE 1 Computational Options to Find a Good Solution 54 Detail on each methodological step is offered as follows 1 Route all 43 cities altogether Weeks 13 visit all cities 14 routesweek Weeks 24 visit only 4visit cities 7 routesweek Results No of truck used T401 2 T301 2 Cost Rs116678 2 Route all 4visit cities outsource all 2visit cities All week 7 routesweek Weeks 1 3 visit half of 2visit cities Weeks 2 4 visit another half of 2visit cities Results No of truck used T401 1 T301 1 Cost Rs135254 3 Route 2 set of data separate between 4visit cities and 2visit cities 4visit cities all weeks 9 routesweek 2visit cities Week 13 visit half of 2visit cities 4 routes week Week 24 visit another half of 2visit cities 3 routesweek Result No of truck used T4011 T301 2 Cost Rs109092 4 Route 2 sets of data mix between 4visit cities and 2visit cities 4visit cities half of 2visit cities visit in weeks 13 10 routesweek 4visit cities another half of 2visit cities visit in week 24 9 routesweek Note Arrange 2visit cities into 2 sets by following the below steps Separate those 2visit cities into 2 sets by area eastwest Sort cities in each set by sales per visit Set all cities in odd row as set 1 and all cities in even row as set 2 The logic behind this method is that each set should be balanced in terms of demand and area Result No of truck used T401 1 T301 1 Cost Rs78328 5 Route 2 sets of data mix 4visit cities and 2visit cities and outsource all cities whose demand greater than Rs350000 4visit cities half of 2visit cities visit in weeks 13 10 routesweek 4visit cities another half of 2visit cities visit in weeks 24 10 routesweek Note 55 1 Arrange 2visit cities into 2 sets by following the below steps Separate those 2visit cities into 2 sets by sales per visits low demand and high demand Sort cities in each set by distance from the central warehouse Set all cities in odd row as set 1 and all cities in even row as set 2 The logic behind this method is each set should be balanced in terms of demand and distance from the central warehouse Using distance should reflect the better route than roughly separating cities by eastwest area like in method 4 2 Outsource cities whose demand is greater than 350000 so that only a T310 truck is required Guntur and Rajahmundry are the only 2 cities whose demand is greater than 350000 Both of them are weekly visit cities therefore outsource to these two cities every week Result No of truck used T301 2 Outsource 2 cities for all weeks Guntur and Rajahmundry Cost Rs65780 which is the best solution Best Solution The best solution per method 5 completes all deliveries for a total monthly cost of Rs65780 using only 2 of T301 trucks together with 2 sets of crew members The cost calculation is shown in Figure 2 The daily schedule shown in Figure 3 represents the number of trucks needed the truck routes with stop sequence the schedule of truck usage through the month and the schedule for using the crews The Gantt chart of truck deliveries throughout the month is shown in Figure 4 See the router solution reports generated from LOGWARE ROUTER module for individual route detail The reports divide the month into weeks 1 3 Figure 5 and weeks 2 4 Figure 6 56 FIGURE 2 Cost Calculation Sheet for Best Solution Total monthly cost fixed cost 2 of truck 2 T310 2 x 7000 14000 2 of driver 2 x 2200 4400 2 of helper 2 x 1400 2800 allowance Week 13 4 people x 2 weeks x 5 days x Rs 60 2400 Week 24 2 people x 2 weeks x 5 days x Rs 60 1200 2 people x 2 weeks x 4 days x Rs 60 960 Running cost Week 1 3 3146 km x 3 Rskm x 2 weeks 18876 Week 2 4 2414 km x 3 Rskm x 2 weeks 14484 Common Carrier To Guntur 35 km x 15 Rskm x 4 weeks 2100 To Rajahmundry 76 km x 15 Rskm x 4 weeks 4560 Total 65780 57 FIGURE 3 Daily Schedule WEEK 1 3 Truck Route Start Time End Time Crew DAY 1 I 1 Depot 427 AM Driver I Helper I Podili Kondulur Tanguturu Depot 704 PM 405 II 25 Depot 546 AM 339 Driver II Helper II Jangareddygudem Kakinada Depot 545 PM DAY 2 I 5 Depot 639 AM Driver I Helper I Chilakalurupet Narasaraopet Macheria Depot 818 PM 306 II 7 Depot 759 AM Driver II Helper II Tadikonda Tenali Chirala Vuyyuru Depot 847 PM 232 DAY 3 I 13 Depot 647 AM Driver I Helper I Kaikalur Bhimavaram Tadepallegudem Depot 707 PM 253 II 8 Depot 701 AM Driver II Helper II Sattenapalle Ongole Depot 812 PM 307 DAY 4 I 19 Depot 519 AM Driver I Helper I Narasapur Mandapeta Depot 703 PM 349 II 11 Depot 719 AM Driver II Helper II Pamarru Machilipatnam Palakolu Depot 519 PM 240 Daily Schedule Summary Route Miles 58 DAY 5 I 24 Depot 656 AM Driver I Helper I Eluru Chintalapuidi Depot 311 PM 190 II 2 Depot 344 AM Driver II Helper II Kani Giri Bestavaipetta Giddalur Markapur Depot 1121 PM 525 Outsource Guntur Rajahmundry WEEK 2 4 Truck Route Start Time End Time Crew DAY 1 I 3 Depot 501 AM Driver I Helper I Ongole Depot 328 PM 278 II 18 Depot 634 AM Driver II Helper II Jaggayyapeta Eluru Depot 500 PM 276 DAY 2 I 13 Depot 741 AM Driver I Helper I Vuyyuru Machilipatnam Gudivada Depot 330 PM 153 II 19 Depot 826 AM Driver II Helper II Hunuman Junction Chirala Bapatia Repalie Depot 749 PM 175 Route Miles 59 DAY 3 I 14 Depot 735 AM Driver I Helper I Tenali Palakolu Depot 440 PM 243 II 24 Depot 512 AM Driver II Helper II Tanuku Nidadvole Kovvur Depot 504 PM 315 DAY 4 I 16 Depot 727 AM Driver I Helper I Nuzvid Tadepallegudem Bhimavaram 551 PM 236 Depot II 29 Depot 623 AM Driver II Helper II Piduguralia Addanki Vinukonda Narasaraopet Depot 749 PM 318 DAY 5 I 23 Depot 423 AM Driver I Helper I Amaiapuram Kakinada Depot 752 PM 420 II Open Outsource Guntur Rajahmundry Note Truck represents the number of type T301 truck 60 FIGURE 4 Gantt Chart of Truck Deliveries Throughout the Month 61 FIGURE 5 ROUTER Solution Report Weeks 1 and 3 Label Untitled Date 11232002 Time 24717 AM SUMMARY REPORT TIMEDISTANCECOST INFORMATION Route Run Stop Brk Stem Route time time time time time Start Return No of Route Route no hr hr hr hr hr time time stops distMi cost 1 146 101 25 20 78 0427AM 0704PM 3 405 121500 2 196 131 45 20 94 0344AM 1121PM 4 525 157500 5 136 76 40 20 54 0639AM 0818PM 3 306 91800 7 128 58 50 20 14 0759AM 0847PM 4 232 69600 8 132 77 35 20 50 0701AM 0812PM 2 307 92100 11 100 60 30 10 33 0719AM 0519PM 3 240 72000 13 123 63 40 20 45 0647AM 0707PM 3 253 75900 19 137 87 30 20 72 0519AM 0703PM 2 349 104700 24 82 48 25 10 36 0656AM 0311PM 2 190 57000 25 120 85 25 10 77 0546AM 0545PM 2 339 101700 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle no typ capcty weight weight util capcty cube cube util description 1 1 350 180 0 514 9999 0 0 0 T310 2 1 350 134 0 383 9999 0 0 0 T310 5 1 350 342 0 977 9999 0 0 0 T310 7 1 350 337 0 963 9999 0 0 0 T310 8 1 350 350 0 1000 9999 0 0 0 T310 11 1 350 350 0 1000 9999 0 0 0 T310 13 1 350 326 0 931 9999 0 0 0 T310 19 1 350 330 0 943 9999 0 0 0 T310 24 1 350 266 0 760 9999 0 0 0 T310 25 1 350 296 0 846 9999 0 0 0 T310 DETAIL REPORT ON ROUTE NUMBER 1 A T310 leaves at 427AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 62 2 Podili 0900AM 1 0930AM 1 30 2430 162 YES 6 Kondulur 1128AM 1 1228PM 1 60 1185 79 YES Break 30 minutes 1 Tanguturu 0119PM 1 0219PM 1 60 210 14 YES Break 60 minutes Depot 0704PM 1 2250 150 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 514 0 2 Podili 24 0 18300 76 446 0 6 Kondulur 90 0 5100 6 189 0 1 Tanguturu 66 0 4800 7 0 0 Totals Weight Del 180 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 101 hr To 1st stop 162 mi Loadunload 25 From last stop 150 Break 20 On route 93 Total 146 hr Total 405 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 121500 Fixed 00 Total 121500 DETAIL REPORT ON ROUTE NUMBER 2 A T310 leaves at 344AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 5 Kani Giri 0900AM 1 1130AM 1 150 2865 191 YES Break 30 minutes 9 Bestavaipetta 0121PM 1 0151PM 1 30 810 54 YES 7 Giddalur 0241PM 1 0341PM 1 60 495 33 YES 4 Markapur 0512PM 1 0542PM 1 30 915 61 YES Break 60 minutes Depot 1121PM 1 2790 186 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 383 0 5 Kani Giri 24 0 8700 36 314 0 9 Bestavaipetta 25 0 1500 6 243 0 7 Giddalur 25 0 21000 84 171 0 4 Markapur 60 0 2100 4 0 0 Totals Weight Del 134 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 131 hr To 1st stop 191 mi Loadunload 45 From last stop 186 Break 20 On route 148 Total 196 hr Total 525 mi Max allowed 240 hr Max allowed 9999 mi 63 Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 157500 Fixed 00 Total 157500 DETAIL REPORT ON ROUTE NUMBER 5 A T310 leaves at 639AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 11 Chilakalurupet 0900AM 1 1000AM 1 60 1110 74 YES 12 Narasarapet 1031AM 1 1131AM 1 60 315 21 YES Break 30 minutes 42 Macheria 0146PM 1 0346PM 1 120 1050 70 YES Break 60 minutes Depot 0818PM 1 2115 141 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 977 0 11 Chilakalurupet 92 0 5700 6 714 0 12 Narasarapet 100 0 00 0 429 0 42 Macheria 150 0 40500 27 0 0 Totals Weight Del 342 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 76 hr To 1st stop 74 mi Loadunload 40 From last stop 141 Break 20 On route 91 Total 136 hr Total 306 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 91800 Fixed 00 Total 91800 DETAIL REPORT ON ROUTE NUMBER 7 A T310 leaves at 759AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 14 Tadikonda 0900AM 1 1000AM 1 60 315 21 YES 19 Tenali 1058AM 1 1158AM 1 60 585 39 YES Break 30 minutes 8 Chirala 0200PM 1 0400PM 1 120 915 61 YES 18 Vuyyuru 0557PM 1 0657PM 1 60 1170 78 YES Break 60 minutes Depot 0847PM 1 495 33 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 64 963 0 14 Tadikonda 60 0 7200 12 791 0 19 Tenali 140 0 3300 2 391 0 8 Chirala 98 0 21900 22 111 0 18 Vuyyuru 39 0 6600 17 0 0 Totals Weight Del 337 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 58 hr To 1st stop 21 mi Loadunload 50 From last stop 33 Break 20 On route 178 Total 128 hr Total 232 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 69600 Fixed 00 Total 69600 DETAIL REPORT ON ROUTE NUMBER 8 A T310 leaves at 701AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 15 Sattenapalle 0900AM 1 1000AM 1 60 885 59 YES Break 30 minutes 3 Ongole 0113PM 1 0343PM 1 150 1635 109 YES Break 60 minutes Depot 0812PM 1 2085 139 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 1000 0 15 Sattenapalle 45 0 8700 19 871 0 3 Ongole 305 0 56700 19 0 0 Totals Weight Del 350 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 77 hr To 1st stop 59 mi Loadunload 35 From last stop 139 Break 20 On route 109 Total 132 hr Total 307 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 92100 Fixed 00 Total 92100 DETAIL REPORT ON ROUTE NUMBER 11 A T310 leaves at 719AM on day 1 from the depot at Vijayawada Stop Drive Distance Time 65 Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 20 Pamarru 0900AM 1 1000AM 1 60 705 47 YES 22 Machilipatnam 1036AM 1 1136AM 1 60 360 24 YES Break 30 minutes 38 Palakolu 0212PM 1 0312PM 1 60 1260 84 YES Depot 0519PM 1 1275 85 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 1000 0 20 Pamarru 62 0 900 1 823 0 22 Machilipatnam 108 0 12600 12 514 0 38 Palakolu 180 0 28500 16 0 0 Totals Weight Del 350 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 60 hr To 1st stop 47 mi Loadunload 30 From last stop 85 Break 10 On route 108 Total 100 hr Total 240 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 72000 Fixed 00 Total 72000 DETAIL REPORT ON ROUTE NUMBER 13 A T310 leaves at 647AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 23 Kaikalur 0900AM 1 1000AM 1 60 1020 68 YES 39 Bhimavaram 1054AM 1 1224PM 1 90 540 36 YES Break 30 minutes 36 Tadepallegudem 0148PM 1 0318PM 1 90 540 36 YES Break 60 minutes Depot 0707PM 1 1695 113 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 931 0 23 Kaikalur 48 0 1200 2 794 0 39 Bhimavaram 148 0 3900 3 371 0 36 Tadepallegudem 130 0 12300 9 0 0 Totals Weight Del 326 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 63 hr To 1st stop 68 mi Loadunload 40 From last stop 113 Break 20 On route 72 Total 123 hr Total 253 mi Max allowed 210 hr Max allowed 9999 mi 66 Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 75900 Fixed 00 Total 75900 DETAIL REPORT ON ROUTE NUMBER 19 A T310 leaves at 519AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 30 Narasapur 0900AM 1 1000AM 1 60 1905 127 YES 29 Mandapeta 1130AM 1 0130PM 1 120 900 60 YES Break 30 minutes Break 60 minutes Depot 0703PM 1 2430 162 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 943 0 30 Narasapur 160 0 7500 5 486 0 29 Mandapeta 170 0 28500 17 0 0 Totals Weight Del 330 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 87 hr To 1st stop 127 mi Loadunload 30 From last stop 162 Break 20 On route 60 Total 137 hr Total 349 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 104700 Fixed 00 Total 104700 DETAIL REPORT ON ROUTE NUMBER 24 A T310 leaves at 656AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 37 Eluru 0900AM 1 1100AM 1 120 945 63 YES Break 30 minutes 41 Chintalapuidi 1241PM 1 0111PM 1 30 705 47 YES Depot 0311PM 1 1200 80 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 760 0 37 Eluru 198 0 9000 5 194 0 41 Chintalapuidi 68 0 19200 28 0 0 Totals Weight Del 266 Pickups 0 Cube Del 0 Pickups 0 67 Route time Distance Driving 48 hr To 1st stop 63 mi Loadunload 25 From last stop 80 Break 10 On route 47 Total 82 hr Total 190 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 57000 Fixed 00 Total 57000 DETAIL REPORT ON ROUTE NUMBER 25 A T310 leaves at 546AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 40 Jangareddygudem 0900AM 1 0930AM 1 30 1635 109 YES 32 Kakinada 1015AM 1 1215PM 1 120 450 30 YES Break 30 minutes Depot 0545PM 1 3000 200 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 846 0 40 Jangareddygudem 68 0 18300 27 651 0 32 Kakinada 228 0 36300 16 0 0 Totals Weight Del 296 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 85 hr To 1st stop 109 mi Loadunload 25 From last stop 200 Break 10 On route 30 Total 120 hr Total 339 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 101700 Fixed 00 Total 101700 68 FIGURE 6 ROUTER Solution Report Weeks 2 and 4 Label Untitled Date 11232002 Time 42132 AM SUMMARY REPORT TIMEDISTANCECOST INFORMATION Route Run Stop Brk Stem Route time time time time time Start Return No of Route Route no hr hr hr hr hr time time stops distMi cost 3 104 70 25 10 70 0501AM 0328PM 1 278 83400 13 78 38 30 10 20 0741AM 0330PM 3 153 45900 14 91 61 20 10 30 0735AM 0440PM 2 243 72900 16 104 59 35 10 38 0727AM 0551PM 3 236 70800 18 48 38 5 5 38 0634AM 500PM 2 276 82800 19 114 44 50 20 17 0826AM 0749PM 4 175 52500 23 155 105 30 20 91 0423AM 0752PM 2 420 126000 24 119 79 30 10 70 0512AM 0504PM 3 315 94500 29 134 80 35 20 40 0623AM 0749PM 4 318 95400 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle no typ capcty weight weight util capcty cube cube util description 3 1 350 305 0 871 9999 0 0 0 T310 13 1 350 327 0 934 9999 0 0 0 T310 14 1 350 320 0 914 9999 0 0 0 T310 16 1 350 315 0 900 9999 0 0 0 T310 18 1 350 235 0 671 9999 0 0 0 T310 19 1 350 280 0 800 9999 0 0 0 T310 23 1 350 318 0 909 9999 0 0 0 T310 24 1 350 229 0 654 9999 0 0 0 T310 29 1 350 305 0 871 9999 0 0 0 T310 DETAIL REPORT ON ROUTE NUMBER 3 A T310 leaves at 501AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 3 Ongole 0900AM 1 1130AM 1 150 2085 139 YES Break 30 minutes Depot 0328PM 1 2085 139 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 871 0 69 3 Ongole 305 0 83400 27 0 0 Totals Weight Del 305 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 70 hr To 1st stop 139 mi Loadunload 25 From last stop 139 Break 10 On route 0 Total 104 hr Total 278 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 83400 Fixed 00 Total 83400 DETAIL REPORT ON ROUTE NUMBER 13 A T310 leaves at 741AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 18 Vuyyuru 0900AM 1 1000AM 1 60 495 33 YES 22 Machilipatnam 1058AM 1 1158AM 1 60 585 39 YES Break 30 minutes 26 Gudivada 0120PM 1 0220PM 1 60 510 34 YES Depot 0330PM 1 705 47 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 934 0 18 Vuyyuru 39 0 600 2 823 0 22 Machilipatnam 108 0 15300 14 514 0 26 Gudivada 180 0 2100 1 0 0 Totals Weight Del 327 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 38 hr To 1st stop 33 mi Loadunload 30 From last stop 47 Break 10 On route 73 Total 78 hr Total 153 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 45900 Fixed 00 Total 45900 DETAIL REPORT ON ROUTE NUMBER 14 A T310 leaves at 735AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 70 19 Tenali 0900AM 1 1000AM 1 60 540 36 YES Break 30 minutes 38 Palakolu 0133PM 1 0233PM 1 60 1830 122 YES Depot 0440PM 1 1275 85 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 914 0 19 Tenali 140 0 21900 16 514 0 38 Palakolu 180 0 51300 28 0 0 Totals Weight Del 320 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 61 hr To 1st stop 36 mi Loadunload 20 From last stop 85 Break 10 On route 122 Total 91 hr Total 243 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 72900 Fixed 00 Total 72900 DETAIL REPORT ON ROUTE NUMBER 16 A T310 leaves at 727AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 21 Nuzvid 0900AM 1 0930AM 1 30 630 42 YES 36 Tadepallegudem 1045AM 1 1215PM 1 90 750 50 YES Break 30 minutes 39 Bhimavaram 0139PM 1 0309PM 1 90 540 36 YES Depot 0551PM 1 1620 108 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 900 0 21 Nuzvid 37 0 6300 17 794 0 36 Tadepallegudem 130 0 6600 5 423 0 39 Bhimavaram 148 0 9300 6 0 0 Totals Weight Del 315 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 59 hr To 1st stop 42 mi Loadunload 35 From last stop 108 Break 10 On route 86 Total 104 hr Total 236 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 70800 Fixed 00 Total 70800 71 DETAIL REPORT ON ROUTE NUMBER 18 A T310 leaves at 634AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 24 Jaggayyapeta 0900AM 1 0930AM 1 30 1155 77 YES Break 30 minutes 37 Eluru 1324AM 1 324PM 1 120 945 63 YES Depot 500PM 1 1155 77 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 106 0 24 Jaggayyapeta 37 0 46200 125 0 0 Totals Weight Del 37 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 38 hr To 1st stop 77 mi Loadunload 5 From last stop 63 Break 5 On route 136 Total 48 hr Total 276 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 82800 Fixed 00 Total 82800 DETAIL REPORT ON ROUTE NUMBER 19 A T310 leaves at 826AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 25 Hanuman Junctio 0900AM 1 1000AM 1 60 30 2 YES 8 Chirala 1111AM 1 0111PM 1 120 720 48 YES Break 30 minutes 27 Bapatia 0201PM 1 0301PM 1 60 195 13 YES 16 Repalie 0408PM 1 0508PM 1 60 675 45 YES Break 60 minutes Depot 0749PM 1 1005 67 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 800 0 25 Hanuman Junctio 50 0 11700 23 657 0 8 Chirala 98 0 5700 6 377 0 27 Bapatia 82 0 00 0 143 0 16 Repalie 50 0 10800 22 0 0 Totals Weight Del 280 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 44 hr To 1st stop 2 mi Loadunload 50 From last stop 67 72 Break 20 On route 106 Total 114 hr Total 175 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 52500 Fixed 00 Total 52500 DETAIL REPORT ON ROUTE NUMBER 23 A T310 leaves at 423AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 31 Amaiapuram 0900AM 1 1000AM 1 60 2475 165 YES 32 Kakinada 1122AM 1 0122PM 1 120 825 55 YES Break 30 minutes Break 60 minutes Depot 0752PM 1 3000 200 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 909 0 31 Amaiapuram 90 0 6000 7 651 0 32 Kakinada 228 0 27000 12 0 0 Totals Weight Del 318 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 105 hr To 1st stop 165 mi Loadunload 30 From last stop 200 Break 20 On route 55 Total 155 hr Total 420 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 126000 Fixed 00 Total 126000 DETAIL REPORT ON ROUTE NUMBER 24 A T310 leaves at 512AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 34 Tanuku 0900AM 1 1000AM 1 60 1980 132 YES 35 Nidadvole 1033AM 1 1133AM 1 60 330 22 YES 33 Kovvur 1155AM 1 1255PM 1 60 225 15 YES Break 30 minutes Depot 0504PM 1 2190 146 73 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 654 0 34 Tanuku 134 0 6000 4 271 0 35 Nidadvole 50 0 900 2 129 0 33 Kovvur 45 0 8100 18 0 0 Totals Weight Del 229 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 79 hr To 1st stop 132 mi Loadunload 30 From last stop 146 Break 10 On route 37 Total 119 hr Total 315 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 94500 Fixed 00 Total 94500 DETAIL REPORT ON ROUTE NUMBER 29 A T310 leaves at 623AM on day 1 from the depot at Vijayawada Stop Drive Distance Time Stop Arrive Depart time to stop to stop wind No description time Day time Day Min Min Miles met Break 30 minutes 43 Piduguralia 0900AM 1 1000AM 1 60 1275 85 YES 10 Addanki 1157AM 1 1227PM 1 30 1170 78 YES Break 30 minutes 13 Vinukonda 0152PM 1 0252PM 1 60 555 37 YES 12 Narasarapet 0356PM 1 0456PM 1 60 630 42 YES Break 60 minutes Depot 0749PM 1 1140 76 Stop Stop volume Inc cost to serve stop Capacity in use No description Weight Cube In In unit Weight Cube 871 0 43 Piduguralia 80 0 14700 18 643 0 10 Addanki 60 0 15000 25 471 0 13 Vinukonda 65 0 5400 8 286 0 12 Narasarapet 100 0 00 0 0 0 Totals Weight Del 305 Pickups 0 Cube Del 0 Pickups 0 Route time Distance Driving 80 hr To 1st stop 85 mi Loadunload 35 From last stop 76 Break 20 On route 157 Total 134 hr Total 318 mi Max allowed 210 hr Max allowed 9999 mi Route costs Driver reg time 00 Driver over time 00 Vehicle mileage 95400 Fixed 00 Total 95400 74 CHAPTER 8 FORECASTING SUPPLY CHAIN REQUIREMENTS 4 a The answer to this question is aided by using the FORECAST module in LOGWARE A sample calculation is shown as carried out by FORECAST The results are then summarized from FORECAST output An example calculation for an 01 is shown Other values would be used ranging 001 to 10 We first calculate a starting forecast by averaging the first four weekly requirements That is 2056 2349 1895 15144 195350 Now we backcast this value and start the forecast at time 0 Thus the forecasts and the associated errors would be The standard error of the forecast is S N F Total squared error 1 677 71376 6 528 79 Note FORECAST does not use N1 in the denominator Repeating this type of analysis the following table can be developed The results from FORECAST are shown Forecast Error Squared error F1 195350 F2 12056 9195350 196375 F3 12349 9196375 200228 F4 11895 9200220 199148 F5 11514 9199148 194373 74973 56209507 F6 11194 9194373 186876 39924 15939258 F7 12268 9186876 190800 74500 55502500 F8 12653 9190800 198250 5650 319225 F9 12039 9198250 198815 41085 16879772 F10 12399 9198815 202924 47876 22921114 F11 12508 9202924 207712 Total squared error 167771376 75 SF 01 52872 05 52842 1 52846 2 52889 5 53555 10 56607 The that minimizes SF is 005 b The forecast errors are computed in part a c If we assume that the errors are normally distributed around the forecast we can then construct a 95 confidence band on the forecast That is if Y is the actual volume in period 11 then the range of the forecast F11 201781 for 005 will be Y F11 z SF 201781 19652842 Then 98211 Y 305351 All values are in thousands 5 a b The solution to this problem was aided by the use of the exponential smoothing module in FORECAST Using the first four weeks data to initialize the leveltrend version of the exponential smoothing model and setting and equal to 02 the forecast for next week is F11 202447 with a standard error of the forecast of SF 17128 c Assuming that the forecast errors are normally distributed around F11 a 95 statistical confidence band can be constructed The confidence band is Y F11 z SF 202447 19617128 where z 196 for 25 of the area under the two tails of a normal distribution The range of the actual weekly volume is expected to be 168876 Y 236018 6 76 a The data may be restated as shown below Sales S t St t2 Trend valuea St Seasonal indexb 27000 1 27000 1 41087 066 70000 2 140000 4 41192 170 41000 3 123000 9 41298 099 13000 4 52000 16 41403 032 30000 5 150000 25 41508 072 73000 6 438000 36 41613 175 48000 7 336000 49 41719 115 15000 8 120000 64 41824 036 34000 9 306000 81 41929 081 82000 10 820000 100 42035 195 51000 11 561000 121 42140 121 16000 12 192000 144 42245 038 500000 78 3265000 650 a Computed from the linear trend line For example for period 1 S1 409816 10531 41087 b The ratio of the actual sales S to the trend line value St For example for period 1 the seasonal index is 2700041087 066 Given the values from the above table and that t 7812 65 N 12 and S 50000012 41666 the coefficients in the regression trend line would be 105 3 650 12 6 5 265 000 12 41666 6 5 3 2 2 2 t N t N S t S t b and a S b t 41 666 1053 65 40 9816 Therefore the trend value St for any period t would be St 409816 1053t b The seasonal factors are determined by the ratio of the actual sales in a period to the trend value for that period For example the seasonal factor for period 12 4th quarter of last year would be 1600042245 038 This and the seasonal factors for all past quarters are shown in the previous table c The forecasts using the seasonal factors from the last 4 quarters are as follows Seasonal 77 t St factors Forecast 13 42351 081 34304 14 42456 195 82789 15 42561 121 51499 16 42666 038 16213 7 An exponential smoothing model is used to generate a forecast for period 13 January of next year The sales for January through April are used to initialize the model and an 02 is used as the smoothing constant The FORECAST module is used to generate the forecast The results are summarized as follows Region 1 Region 2 Region 3 Combined Forecast F13 21973 40704 30330 93826 Forecast error SE 2689 2550 1754 6141 Note that the sum of the forecasts by region nearly equals the forecast of the combined usage However whether a byregion forecast is better than an overall forecast that is disaggregated by region depends on the forecast error The standard error of the forecast is the best indicator A comparison of a bottomsup forecast developed from regional forecasts to that of a forecast from combined data can be based on the law of variances That is if the usage rates within the regions are independent of each other the estimate of the total error can be built from the individual regions and compared to that of the combined usage data The total forecast error variance from the individual regions 2 T S might be estimated as the weighted average of the variances as follows 2 3 2 2 2 1 2 3 2 1 E C E C E C T F S F F S F F S F S where Fi forecasts of each region FC forecast based on combined data SEi 2 variance of the forecast in each region ST 2 total variance of the forecast based on regional data Therefore ST 2 2 2 2 219 73 930 07 2689 407 04 930 07 2550 30330 930 071754 0 236 72307 0 438 650 25 0326 307 65 55574 Then 78 ST 55574 2357 Since ST SC it appears that a bottomup or regional forecast will have a lower error than a topdown forecast 9 a See the plot in Figure 81 It shows that there is a seasonal component with a very slight trend to the data as well as some random or unexplained variation 0 50 100 150 200 250 300 Jan Apr Jly Oct Jan Apr Jly Oct Jan Apr Jly Oct Jan Apr Jly Oct Jan Apr Jly Oct Average monthly unit prices Time months FIGURE 81 Plot of time series data for Problem 9 b A time series model typically will involve only two components trend and seasonality Using 2 years of data should be sufficient to establish an accurate trend line and the seasonal indices We can develop the following table for computing a regression line and seasonal indices 79 We also have N 24 t 30024 125 and P 557524 23229 Now b P t N P t t N t 2 2 2 69 678 24 232 29 125 4 900 24 125 0 008 and a P b t t 232 29 0 008 125 23239 Therefore the trend equation is Prices Pt Time t Pt t2 Trenda Tt Seasonal indexb St 211 1 211 1 2324 091 210 2 420 4 2324 090 214 3 642 9 2324 092 208 4 832 16 2324 090 276 5 1380 25 2322 119 269 6 1614 36 2323 116 265 7 1855 49 2323 114 253 8 2024 64 2323 109 244 9 2196 81 2323 105 202 10 2020 100 2323 087 221 11 2431 121 2323 095 210 12 2520 144 2322 090 215 13 2795 169 2323 093 092 225 14 3150 196 2322 097 093 230 15 3450 225 2323 099 096 214 16 3424 256 2323 092 091 276 17 4692 289 2322 119 119 261 18 4698 324 2322 112 114 250 19 4750 361 2322 108 111 248 20 4960 400 2322 107 108 229 21 4809 441 2322 099 102 221 22 4862 484 2322 095 091 209 23 4807 529 2322 090 092 214 24 5136 576 2322 092 091 5575 300 69678 4900 a Computed from the trend regression line For example the period 1 trend is T1 23239 00081 2324 b The seasonal index is the ratio of the actual price to the trend for the same period For example the period 1 seasonal index is 211232 091 80 T t t 232 29 0 008 Note that the trend is negative for the last two years of data even though the 5year trend would appear to be positive Now computing the trend value Tt for each value of t gives the results as shown in the previous table The seasonal index is a result of dividing Pt by Tt for each period t The indices are averaged for corresponding periods that are one year apart Forecasting into the 5th year shows the potential error in the method That is for January of the 5th year the forecast is Ft TtSt12 or F25 23239 000825092 2136 Repeating for each month we have t Actual price Forecast price Forecast error Squared error Revised seasonala 25 210 2136 36 130 091 26 223 2156 71 504 27 204 2229 189 3572 28 244 2113 327 10693 29 274 2763 23 53 30 246 2646 186 3459 31 237 2577 207 4285 32 267 2507 163 2657 33 212 2368 248 6150 34 211 2112 02 00 35 188 2135 255 510 36 188 2112 232 5382 Total squared error 37395 a The seasonal index for period 25 is 90 The average of the seasonal index for period 25 12 13 and this period is 092 0902 091 The standard error of the forecast is S F 3 7395 12 2 1934 Now the forecast for period 37 would be F37 23239 0 008 37 091 21121 c Using the exponential smoothing module in the FORECAST software the forecast for the coming period is F 20126 with SF 1727 The smoothing constants given in the problem are the best that FORECAST could find d Each model should be combined according to its ability to forecast accurately We can give each a weight in proportion to its forecast error or standard error of the forecast SF Hence the following table can be developed 81 Model type 1 Forecast error 2 13661 Proportion of total error 312 Inverse of error proportion 434013 Model weights Regression 1934 0528 1894 0472 Exp smooth 1727 0472 2119 0528 Total 3661 1000 4013 1000 Therefore each of the model results is weighted according to the model weights The weighted forecast for the upcoming January would be Model type 1 Forecast 2 Model weight 312 Weighted proportion Regression 21121 0472 9969 Exp smooth 20126 0528 10627 Weighted forecast 20596 In a similar fashion we can weight the forecast error variances to come up with a weighted forecast error standard deviation SFw That is SFw 0 472 1934 0528 17 27 18 28 2 2 A 95 confidence band using the combined results might be constructed as Y 20596 z1828 where z is 196 for 95 of the area under the normal distribution Y 20596 1961828 Hence we can be 95 sure that the actual price Y will be within the following range 17013 Y 24179 10 The plot of the sales data is shown in Figure 82 The plot reveals a high degree of seasonality with a noticeable downward trend A leveltrendseasonal model seems reasonable b Using the search capability within the FORECAST software a LevelTrendSeasonal form of the exponential smoothing model was found to give the lowest forecast error A 14period initialization and 6 periods to compute error statistics were used The respective smoothing constants were 001 008 and 060 This produced 82 a forecast for the upcoming period January 2004 of F 632760 and a standard error of the forecast of SF 112081 c Assuming that the forecast errors are normally distributed around the forecast a 95 confidence band on the forecast is given by Y F zSF Y 632760 196112081 where z 196 for 95 of the area under the normal distribution curve Therefore we can be 95 sure that the actual sales Y should fall within the following limits 41308 Y 85244 11 a For A569 the BIAS 165698 and the RMSE 126567 when using the 3month moving average However if a level only exponential smoothing model with an 010 the BIAS drops to 9556 and the RMSE is 118689 The model fits the data better and there is a slight improvement in the forecasting accuracy For A366 the BIAS 18231 and the RMSE 144973 when using the 3month moving average A leveltrendseasonal model offers the best fit but it is suspect since the data show a high degree of random variability rather than seasonality Overall a simple levelonly model is probably better in practice The model has an 008 a BIAS 3227 and a RMSE 136256 This is an improvement over the 3month moving average 0 5000 10000 15000 20000 25000 30000 Jan Apr Jly Oct Jan Apr Jly Oct Jan Apr Jly Oct Jan Apr Jly Oct Jan Apr Jly Oct Aggregate sales in 000s Time months FIGURE 82 Plot of Time Series Data for Hudson Paper Company 83 b Using the levelonly models the forecast for October for A569 193230 and for A366 603671 c The 3sigma 997 confidence band on the forecasts would be For A569 Y 193230 3118689 or 0 Y 549297 For A366 Y 603671 3136256 or 194903 Y 1012439 The actual October usage falls within the 3sigma confidence bands for each of these products The difference of the actual from the forecast for each product is attributable to the substantial variability in the data which is characteristic of purchasing in the steel processing industry 84 WORLD OIL Teaching Note Strategy The purpose of this case study is to allow students to develop an appropriate forecasting model for some time series data Discussion may begin with the nature of this productone with which most students should be very familiar Based on the many available forecasting approaches students should be encouraged to select several for consideration In this note both exponential smoothing and time series decomposition are evaluated Both are appropriate here because 1 they can project from historical time series data 2 they can handle seasonality which appears to be present in the data 3 there is enough data to construct and test the models and 4 the forecast is for a short period into the future Assistance with the computational aspects of this problem is available with the use of the FORECAST module in the LOGWARE software Answers to Questions 1 Develop a forecasting procedure for this service station Why did you select your method Both exponential smoothing and time series decomposition forecasting methods are tested using the FORECAST module in LOGWARE For exponential smoothing an initialization period of one seasonal cycle 52 weeks plus two weeks are used for a total of 54 weeks a minimum requirement in FORECAST The last 30 weeks of data is used for computing the error statistics This number of periods is arbitrary but seems reasonably large so as to give stable statistical values We wish to minimize the forecast error over time and FORECAST computes both MAD and RMSE statistics that can be used to make comparisons among model types Testing the various exponential smoothing model types and the time series gives the following statistics Model type Smoothing constants MAD BIAS RMSE Forecast week 6 of this year Level only 4 3782 527 6761 81735 Leveltrend 2 5 4585 713 6780 86026 Levelseasonal 3 10 3897 1130 4571 64875 Leveltrend seasonal 01 2 4 3027 605 4417 77074 TS decomp 5946 3718 7185 73133 The MAD and RMSE statistics show how well the forecast has been able to track historical fuel usage rates They are an indication of the accuracy of the forecasting process in the future on the average We favor forecasting methods that can minimize these statistics In this case the LevelTrendSeasonal version of the exponential 85 smoothing model seems to do this best Both MAD and RMSE are the lowest for this model type among the alternatives Further evidence of the performance of a forecasting method is obtained from a plot of the forecast against the actual usage rates This is shown in Figure 1 Note that the LevelTrendSeasonal model tracks the usage rates quite well especially in the more recent weeks The modeling process has likely stabilized in the last 30 weeks of the data and is now tracking quite well FIGURE 1 Fit of LevelTrendSeasonal Exponential Smoothing Model to Fuel Usage Data on Mondays of the Week 2 How should the periods of promotions holidays or other periods where usage rates deviate from normal patterns be handled in the forecast If the deviations occur at the same time within the seasonal cycle and with the same relative intensity no special procedures are required The adaptive characteristic of the exponential smoothing process will automatically incorporate these deviations into the forecast However when the deviations are not regular as promotions may be timed irregularly they may best be handled as outliers in the time series and eliminated from the time series The model may be fit without the outliers and then the effect of them treated as modifications to the forecast These modifications can be handled manually 86 3 Forecast next Mondays fuel usage and indicate the probable accuracy of the forecast From the LevelTrendSeasonal exponential smoothing model developed in question 1 where the smoothing constants are 001 02 and 04 the forecast for Monday of week 6 would be 771 gallons However this forecast only represents the average fuel usage Determining the accuracy of the forecast requires that the forecast track the mean of the actual usage ie a bias of 0 and that the forecast errors be normally distributed While the BIAS sum of the forecast errors over the last 30 weeks is not exactly 0 and will not likely ever be so it is low 605 such that we will assume good tracking by the forecast model A histogram of the forecast errors can reveal whether they follow the familiar bellshaped pattern Such a histogram is given below We can conclude that while the errors are not precisely normally distributed we cannot reject the idea that they did not come from a normally distributed population A goodnessoffit test could be used to check this assumption Although this test is not performed here it is quite forgiving such that the normal distribution of errors assumption is not likely to be rejected where the data show a reasonably normal distribution pattern The distribution here qualifies We can now proceed with developing a 95 confidence band around the forecast The forecast of the actual fuel usage rate Y will be F z Y F z F F where F is the standard error of the forecast F is the forecast and z is the number of standard deviations for 95 of the area under a normal distribution FORECAST computes the root mean squared error RMSE as RMSE A F N t t t N 2 1 87 Since RMSE is uncorrected for degrees of freedom lost we apply a correction factor CF as a multiplier to RMSE to get the unbiased estimate of the standard error of the forecast F ˆ CF N N n where n is the number of degrees of freedom lost in the model building process We estimate n to be the number of smoothing constants in the model or three in this case Hence F RMSE CF 4417 30 30 3 4417 1054 4656 Now with z95 196 from a normal distribution table we can be 95 confident that the true 87octane fuel usage Y on Monday of week 6 will be 771 1964656 Y 771 1964656 680 Y 862 gallons HISTOGRAM FOR FORECAST ERROR OF LAST 30 WEEKS Class Width 200000 Number of Classes 10 0 50 100 MID CLASS 800000 700000 500000 300000 100000 100000 300000 500000 700000 900000 1100000 1200000 88 METRO HOSPITAL You are the materials manager at Metro Hospital Approximately one year ago the hospital began stocking a new drug Ziloene that helps the healing process for wounds and sutures It is your responsibility to forecast and order the monthly supply of Ziloene The goal is to minimize the combined cost of overstocking and understocking the drug Orders are placed and received at the beginning of the month and demand occurs throughout the month The following demand and cost data have been compiled Costs If more is ordered than is demanded a monthly holding cost of 100 per case is incurred If less is ordered than is demanded a 200 per case lost sales cost is incurred The drug has a short shelf life and any overstocked product at the end of the month is worthless and no longer available to meet demand Demand The demand for the twelve months of last year was Last years demand Month 1 2 3 4 5 6 7 8 9 10 11 12 Cases 43 36 24 69 34 75 90 67 59 51 77 50 You believe this demand to be representative of Metros normal usage pattern FIGURE 1 Plot of last years monthly demand in cases 89 Decision Worksheet Month Cases ordered Actual demand Over 1case Short 2case Cost 1 13 2 14 3 15 4 16 5 17 6 18 7 19 8 20 9 21 10 22 11 23 12 24 Total Month Cases ordered Actual demand Over 1case Short 2case Cost 1 25 2 26 3 27 4 28 5 29 6 30 7 31 8 32 9 33 10 34 11 35 12 36 Total 90 METRO HOSPITAL Exercise Note Purpose Metro Hospital is an inclass exercise designed to illustrate the relationship between good forecasting and the control of inventory related costs It shows that accurate forecasting is a primary factor in minimizing inventory costs Participants in this exercise use a variety of methods often intuition to forecast demand and to come up with a purchase quantity Their performance is measured as over or understock costs Using a simple exponential smoothing forecasting model and an understanding of the standard deviation of the forecast an effective purchase plan can be constructed This process results in costs that are significantly lower than the majority of the participants are able to achieve using intuitive methods Administration The descriptive material and the decision worksheet are to be distributed to the class at the time that the exercise is conducted To hand out the material ahead of time may take away much of the drama from the exercise About one half hour should be scheduled for running the exercise The instructor asks the class to make a decision regarding the size of the order to be placed in the upcoming period and to record it on the worksheet The participants are then informed of the demand for that period from Table 1 after the simulated time of one month has passed Given that they now know the actual demand for the period the participants are asked to record their costs and then to place an order for the next period The pattern is repeated for at least twelve months a full seasonal cycle The participants are asked to sum their costs and to report them to the exercise leader They are displayed in a public place such as a chalkboard for all to see Then the exercise leader announces his or her cost level that was achieved using a disciplined approach using a simple forecasting procedure and some basic statistics TABLE 1 Actual Demand for Period 13 Through 36 Period 13 14 15 16 17 18 19 20 21 22 23 24 Demand 47 70 55 38 90 24 65 65 23 55 85 66 Period 25 26 27 28 29 30 31 32 33 34 35 36 Demand 53 64 61 63 65 38 80 88 45 70 50 56 Quantitative Analysis The demand series was generated using a normal distribution with a reasonably high variance and a very slight upward trend To illustrate the use of a quantitative approach to forecasting an exponential smoothing model was selected although other methods such as time series decomposition would also be appropriate The twelve historical data points were submitted to the FORECAST module in LOGWARE A 3month initialization period and a 3month time period for computing error statistics were chosen The smoothing constants for the level leveltrend and leveltrendseasonal models were examined Based on the root mean squared error RMSE the best model 91 was the leveltrendseasonal RMSE 1359 but the level model with 019 and RMSE 1389 performed very well and is used here The model is t t t F A F 081 019 1 where t F t A t F t t t for current period forecast demand for current period actual 1 forecast for next period 1 LOGWARE gives a forecast value of 581 and this is used as the forecast value for period 13 Applying this simple level only model to the second year demand as it is revealed in each period gives the following forecast values TABLE 2 Simple Exponential Smoothing Forecast Values for the Next Year Period Actual demand Forecast 13 47 581 14 70 560 15 55 587 16 38 580 17 90 542 18 24 610 19 65 540 20 65 561 21 23 578 22 55 512 23 85 519 24 66 546 Now we must determine the order quantity It can be calculated from zRMSE Ft Q 1 Recall the RMSE was 1389 for this model To be precise we calculate z by trial and error The following order quantity and cost computations can be made for a z value of 08 Table 3 92 TABLE 3 Purchase Order Quantity and Associated Inventory Costs Period Actual demand Forecast Order quantity Units over Units short Cost 13 47 581 69 22 22 14 70 560 67 3 6 15 55 587 70 15 15 16 38 580 69 31 31 17 90 542 65 25 50 18 24 610 72 48 48 19 65 540 65 0 20 65 561 67 2 2 21 23 578 69 46 46 22 55 512 62 7 7 23 85 519 63 22 44 24 66 546 66 0 271 Q 581 081389 6921 or 69 We see from the following graph Figure 1 that z 08 is optimal 265 270 275 280 285 290 295 300 305 310 0 02 04 06 08 1 12 14 Z Cost FIGURE 1 Plot of Total Annual Costs Against the Factor z Figure 2 graphically shows the good purchase pattern of Table 3 93 0 10 20 30 40 50 60 70 80 90 100 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Time period Cases Order quantity Demand Forecast FIGURE 2 Plot of Forecast and Purchase Order Quantity on Product Demand Summary The exercise leader should discuss that one of the problems with intuitively forecasting demand is overreacting to randomness in the demand pattern This has the effect of causing extreme over and short costs in inventories A model for short term forecasting that is integrated into the purchasing and inventory control process can help to avoid these extremes and give lower costs Several forecasting models may perform well such as exponential smoothing a simple moving average a regression model or a times series decomposition model One of the most practical for inventory control purposes is the exponential smoothing model The results from a simple level only model were illustrated above using the same information that was available to the participants Recognizing that it is less costly to order too much than to order too little the purchase quantity should exceed the forecast by some margin The astute participant will likely approximate the standard deviation of demand from the range of the demand values that is Max Min6 Then one or two might be used to add a margin of safety to the forecast and size of the purchase order This simple approximation procedure can lead to reasonable results 94 CHAPTER 9 INVENTORY POLICY DECISIONS 1 The probability of finding all items in stock is the product of the individual probabilities That is 095093 087 085 094 090 055 2 a The order fill rate is the weighted average of filling the item mix on an order We can setup the following table Order 1 Item mix probabilities 2 Frequency of order 312 Marginal probability 1 9595959090 69 020 0139 2 959595 86 015 0129 3 95959090 73 005 0037 4 95959595959090 62 015 0094 5 959590909090 59 030 0178 6 9595959595 77 015 0116 Order fill rate 0693 Since 693 92 the target order fill rate is not met b The item service levels that will give an order fill rate of 92 must be found by trial and error Although there are many combinations of item service levels that can achieve the desired service level a service level of 99 for items A B C D E and F and 97 to 98 for the remaining items would be about right The order fill rates can be found as follows Order 1 Item mix probabilities 2 Frequency of order 31 2 Marginal probability 1 9939752 922 020 0184 2 993 970 015 0146 3 9929752 932 005 0047 4 9959752 904 015 0136 5 9929754 886 030 0266 6 995 951 015 0143 Order fill rate 0922 3 95 This is a problem of push inventory control The question is one of finding how many of 120000 sets to allocate to each warehouse We begin by estimating the total requirements for each warehouse That is Total requirements Forecast zForecast error From Appendix A we can find the values for z corresponding to the service level at each warehouse Therefore we have Ware house 1 Demand forecast sets 2 Forecast error sets 3 Values for z 4123 Total require ments sets 1 10000 1000 128 11280 2 15000 1200 104 16248 3 35000 2000 118 37360 4 25000 3000 141 29230 Total 85000 94118 We can find the net requirements for each warehouse as the difference between the total requirements and the quantity on hand The following table can be constructed There is 120000 89118 30882 sets to be prorated This is done by assuming that the demand rate is best expressed by the forecast and proportioning the excess in relation to each warehouses forecast to the total forecast quantity That is for warehouse 1 the proration is 100008500030882 3633 sets Prorations to the other warehouses are carried out in a similar manner The allocation to each warehouse is the sum of its net requirements plus a proration of the excess as shown in the above table 4 a The reorder point system is defined by the order quantity and the reorder point quantity Since the demand is known for sure the optimum order quantity is Q DS IC 2 2 3 200 35 015 55 164 78 or 165 cases The reorder point quantity is Ware house 1 Total require ments 2 On hand quantity 312 Net require ments 4 Proration of excess 534 Allocation 1 11280 700 10580 3633 14213 2 16248 0 16248 5450 21698 3 37360 2500 34860 12716 47576 4 29230 1800 27430 9083 36513 94118 89118 30882 120000 96 ROP d LT 3 200 52 15 92 units b The total annual relevant cost of this design is TC D S Q I C Q 1 2 3 200 35 164 78 015 55 164 78 2 679 69 67997 359 66 c The revised reorder point quantity would be ROP 3 200 52 3 185 units The ROP is greater than Q It is possible under these circumstances the reorder quantity may not bring the stock level above the ROP quantity In deciding whether the ROP has been reached we add any quantities on order or in transit to the quantity on hand as the effective quantity in inventory Of course we start with an adequate instock quantity that is at least equal to the ROP quantity 5 a The economic order quantity formula can be used here That is Q DS IC 2 2 300 8 500 010 8 500 775 or 78 students b The number of times that the course should be offered is N D Q 300 775 39 or about 4 times per year 6 This is a singleperiod inventory control problem We have Revenue 350unit Profit 350 250 100unit Loss 02250 50unit Therefore CPn 100 100 50 0 667 Developing a table of cumulative frequencies we have 97 Quantity Frequency Cumulative frequency 50 010 010 55 020 030 60 020 050 65 030 080 Q 70 015 095 75 005 100 100 CPn lies between quantities of 60 and 65 We round up and select 65 as the optimal purchase order size 7 This question can be treated as a singleorder problem We have Revenue 1 001 101 CostLoss 0102365 000055 which is the interest expense for 2 days Profit 101 100055 000945 and CPn 0 00945 0 00945 0 00055 0945 For an area under the normal curve of 0945 see Appendix A z 160 The planned number of withdrawals is Q D z D 120 16020 15200 The amount of money to stock in the teller machine over 2 days would be Money Q75 1520075 11400 8 This is a singleperiod inventory control problem a We have Profit 400 320 Loss 320 300 Then 98 CPn 400 320 400 320 320 300 080 We now need to find the sales that correspond to a cumulative frequency of 080 In the following table Q lies between 1000 and 1200 in the cumulative frequency table We choose to roundup to Q 1250 units b Carrying the excess inventory to next year CPn 80 80 0 2 320 0556 where the loss is the cost of holding a unit until the next year The Q now lies between 750 and 1000 units We choose 1000 units Holding the excess units means a potential loss of 02320 64unit whereas discounting the excess units represents a loss of only 320 300 20unit Therefore Cabot will need fewer units if they are held over in inventory 9 a The optimum order quantity is 556 cases 2 1 2505240 30 56 2 DS IC Q and the reorder point quantity is ROP d LT z sd where s s LT d d 475 25 751 and zP0 80 084 Sales Frequency Cumulative frequency 500 02 02 750 02 04 1000 03 07 1250 02 09 Q 1500 01 10 10 99 Now ROP 1 250 25 084 751 3 756 cases Policy When the amount of inventory on hand plus any quantities on order or in transit falls below ROP reorder an amount Q b For the periodic review system we first estimate the order review time T Q d 556 1 250 0 44 weeks The max level is M d T LT z sd where sd now is s s T LT d d 475 0 44 25 814 cases Hence M 1 250 0 44 25 084 814 4 359 cases Policy Find the amount of stock on hand every 044 weeks and place a reorder for the amount equal to the difference between the quantity on hand plus on order and the max level M of 4359 cases c The total annual relevant cost for these policies is Q kDs E ICzs ICQ DS Q TC z d d 2 For the reorder point system TCQ 12505240556 3565562 35684751 101250527511120556 467626 467040 1059811 9833237 11827714 For the periodic review system TCP 12505240556 3565562 100 35684814 101250528141120556 467626 467040 1148717 10658129 12741512 d The actual service level achieved is given by SL s E Q d z 1 For the reorder point system SLQ 1 751 01120 556 1 015 or demand is met 85 of the time For the periodic review system SLP 1 814 01120 556 1 016 or demand is met 84 of the time e This requires an iterative approach as follows Compute Q DS IC 2 Compute P QIC Dk 1 then z then Ez Compute Q D S ks E IC d z 2 Go back and stop when there is no change in either P or Q After the initial value of Q 5563 the process can be summarized in tabular form 101 Step Q P z Ez 1 7784 09856 219 00050 2 8600 09799 206 00072 3 8899 09778 201 00083 4 8996 09777 200 00085 5 9028 09767 199 00087 6 9028 09767 199 00087 Now for P 09767 z 199 ROP 125025 199751 4620 cases and the total relevant cost is 40 275 902 8 10 65000 751 0 0087 0 3 56 1 99 751 2 0 3 56 902 8 902 8 000 40 65 kDs E Q ICzs ICQ2 DSQ TC z d d Q This is considerably less than the 11827714 for the preset P at 080 If you solve this problem using INPOL you will get a slightly different answer That is Q 858 This simply is because z is carried to two significant digits rather than the 4 significant digits used in the above calculations 10 Refer to the solution of problem 109 for the general approach a Q 5563 cases and ROP d LT z LT s d s d LT 2 2 2 2 2 2 1 250 25 084 25 475 1 250 05 3125 084 977 08 3 946 cases b An approximation for T Qd or T 5561250 044 weeks and approximating sd as 102 s T LT s d s d d LT 2 2 2 2 2 2 0 44 25 475 1 250 05 1 027 cases So Max d T LT z sd 1 250 0 44 25 084 1 027 4 537 cases c According to INPOL TCQ 4686 4686 128195 13862 151429 TCP 4686 4686 134751 14571 158694 d According to INPOL SLQ 8028 SLP 7927 e According to INPOL Q 930 cases ROP 5128 cases TCQ 49532 SLQ 9922 T 076 weeks MAX 6257 cases TCP 52894 SLP 9918 11 a The production run quantity is 1 000 units 100 300 300 0 2575 2100250250 2 d p p IC DS Qp b The production run cycle is 3 33 days 1 000 300 p Qp 103 c The number of production runs is D Qp 100 250 1 000 25 runs per year 12 a The order quantity is Q DS IC 2 2 2 000 250 100 030 35 309 valves and the reorder point quantity is ROP d LT z s LT d but sd 0 Therefore 250 valves 2 000 8 1 LT d ROP b Boxes are set up that contain 309 valvesthe optimum order quantity When an order arrives from a supplier 250 valves are set aside in a separate box and are treated as the backup stock The residual 309 250 59 valves are used on the production line When the 59 valves at the production line are used up the backup box containing 250 valves is brought to the production line and the empty box is sent to the supplier for refilling When the order arrives one hour later there will be 0 valves remaining at the production line Then 250 valves are set aside and 59 are sent to the production line The cycle is then repeated This problem is similar to the KANBAN system Lead times are very short so that lead times are virtually certain Demand is certain since it is fixed by the production schedule Boxes or cards are used to assure movement of the most economic quantity KANBAN is essentially classic economic reorder point inventory control under certainty 13 a The economical quantity of cars to be called for at a time is found by the economic order quantity formula Q DS IC 2 2 40 52 500 0 25 90 000 30 2 000 785 or 79 cars b This is the reorder point quantity ROP d LT z s LT d where z 128 from Appendix A for an area under the curve equal to 090 Therefore 104 ROP 40 1 128 333 1 443 cars or 44390000 2000 1994 tons of soda ash 14 a This is a reorder point design under conditions of uncertainty for both demand and leadtime We assume that the probability of an out of stock is given Therefore the order quantity is Q DS IC 2 2 50 365 50 030 45 367 7 units and ROP d LT z sd where z 104 see Appendix A for the area under the curve equal to 085 and s s LT d s d d LT 2 2 2 2 2 2 15 7 50 2 107 6 units Therefore ROP 50 7 104 107 6 4619 units b This is the periodic review system design under uncertainty The complexity requires us to make some approximations here The time interval for review of the stock level is T Q d 367 7 50 735 days The MAX level is MAX d T LT z sd where z 104 and sd is approximated as s T LT s d s d d LT 2 2 2 2 2 2 735 7 15 50 2 1150 units 105 Therefore MAX 50735 7 1041150 8371 units c Since the service level is specified the probability is not set at the optimum level Knowing the outofstock cost allows us to find the most appropriate service level Since this is an iterative process we use INPOL to carry out the calculations The optimized service level yields a reorder point design of Q 410 units and ROP 571 units and the total relevant cost drops from 12642 in part a to 8489 The demand in stock in part a was 9774 and it now increases to 9981 15 a Find the common review time 52 weeks 500 1 90 2 000 52 2 25 0 30 100 2 2 i i I C D I s O T Then M d T LT z s T LT A A A d A where zA 128 for P 090 8 256 units 51 52 1 28100 51 2 000 52 M A and M B 500 25 15 0842 70 25 15 2118 units where zB 0842 for P 080 The control system works as follows the stock levels of both items are reviewed every 25 weeks The reorder size for A is the difference between the amount on hand includes onorder and 8256 units The reorder size for B is the difference between the amount on hand includes onorder and 2118 units b The average amount in inventory is expected to be 106 AIL d T z s T LT d 2 For A AILA 2 000 25 2 128 100 25 15 2 756 units For B AILB 500 25 2 0842 70 25 15 743 units c The service level is given by SL s E d T d z 1 For A SLA 1 100 25 15 0 0475 2 000 25 0998 For B SLB 1 70 25 15 01120 500 25 0987 d We set T 4 and cycle through the previous calculations Thus we have M M A B 11301 2 888 units units AILA 4301 AILB 1138 SLA 0999 SLB 0 991 16 This problem is one of comparing the combined cost of transportation and intransit inventory In tabular form we have the following annual costs 107 Cost type Formula Rail Truck Transportation RD 640000125 300000 1140000125 550000 Intransit inventory ICDT365 0 25 250 40 000 21 365 143836 365 0 25250400007 47945 Total 443836 597945 Select rail 17 The two transport options from the consolidation point are diagrammed in Figure 91 Whether to choose one mode other the other depends more than transportation costs alone Because the transport modes differ in the time in transit the cost of the money tied up in the goods while in transit must be considered in the choice decision This in transit inventory cost is estimated from 365 ICDt The following design matrix can be developed Cost type Method Air Ocean Transportation RD 180800 98800 Intransit inventory ICDt365 3447 34467 Total 184247 133267 ICDt365 017185200002365 3447 Ocean appears to be the lowest cost option even when a substantial intransit inventory cost is included The ocean option assumes that the trucking cost to move the goods from the consolidation point to the Port of Baltimore is included in the ocean carrier rate FIGURE 91 The Consolidation Operation for a Hydraulic Equipment Manufacturer Multiple sourcing points Baltimore Sao Paolo 2 days 20 days Consolidation point 108 18 The demand pattern is definitely lumpy since sd 327 d 169 To develop the min max system of inventory control we first find Q That is Q DS IC 2 2 169 12 10 0 20 096 0 048 4485 units The ROP is ROP d LT z s ED d where z 104 from Appendix A ED 8 unitsthe average daily demand rate and s s LT d s d d LT 2 2 2 2 2 2 327 4 169 08 6678 units So ROP 1694 1046678 8 13785 units The max level is M ROP Q ED 13785 4485 8 1819 units 19 a The basic relationship is I I n T i 109 We know that IT 5000000 If there are 10 warehouses the amount of inventory in a single one would be I IT 1 10 5 000 000 3162 1581139 The inventory in all 10 warehouses would be 158113910 15811390 b The inventory in a single warehouse would be IT 1 000 000 9 3 000 000 In each of 3 warehouses we would have I 3 000 000 3 732 051 1 and in all 3 warehouses we would have 17320513 5196152 20 a The turnover ratio is the annual demand throughput divided by the average inventory level These ratios for each warehouse and for the total system are shown in the table below Ware house Annual warehouse thruput Average inventory level Turnover ratio 21 2586217 504355 513 24 4230491 796669 531 Avg 559 20 6403349 1009402 634 13 6812207 1241921 549 2 16174988 2196364 736 11 16483970 1991016 828 4 17102486 2085246 820 1 21136032 2217790 953 23 22617380 3001390 754 9 24745328 2641138 937 18 25832337 3599421 718 12 26368290 2719330 970 15 28356369 4166288 681 14 28368270 3473799 817 6 40884400 5293539 772 7 43105917 6542079 659 22 44503623 2580183 1725 8 47136632 5722640 824 17 47412142 5412573 876 16 48697015 5449058 894 110 10 57789509 6403076 903 19 75266622 7523846 1000 3 78559012 9510027 826 Avg 866 5 88226672 11443489 771 818799258 97524639 840 The overall turnover ratio is 840 Ranking the warehouses by throughput and averaging turnover ratios for the top 3 and the bottom 3 warehouses shows that the lowest volume warehouses have a lower turnover ratio 559 than the highest volume warehouses 866 There are several reasons why this may be so The larger warehouses contain the highervolume items such as the A items in the line These may carry less safety stock compared with the sales volume Conversely the lowvolume warehouses may have more dead stock in them There may be startup fixed stock in the warehouses needed to open them that becomes less dominant with greater throughput b A plot of the inventorythroughput data is shown in Figure 101 A linear regression line is also shown fitted to the data The equation for this line is Inventory 200168 01132Throughput FIGURE 101 Plot of Inventory and Warehouse Thruput for California Fruit Growers Association 0 2 4 6 8 10 12 0 20 40 60 80 100 Annual warehouse thruput Millions Average inventory level Millions Estimating line 111 c The total throughput for the three warehouses is Using this total volume and reading the inventory level from Fig 101 or using the regression equation we have Inventory 200168 0113270121702 8137945 d Warehouse 5 has a throughput of 88226672 Splitting this throughput by 30 and 70 we have 03088226672 26468002 07088226672 61758670 88226672 Estimating the inventory for each of the new warehouses using the regression equation we have Inventory 200168 0113226468002 3196346 and Inventory 200168 0 113261758670 7191249 for at total inventory in the two warehouses of 10387595 21 The order quantity for each item when there is no restriction on inventory investment is Q DS IC 2 We first find the unrestricted order quantities Q Q Q A B C 2 51 000 10 0 25 17 1527 2 25 000 10 0 25 325 784 2 9 000 10 0 25 250 537 units units units Warehouse Throughput 1 21136032 12 26368290 23 22617380 Total 70121702 112 The total inventory investment for these items is IV C Q C Q C Q A A B B C C 3 2 2 2 175 1527 2 325 784 2 250 537 2 28138 Since the total investment limit is exceeded we need to revise the order quantities For each product Q DS C I 2 For product A QA 2 51 000 10 175 0 25 For product B QB 2 25 000 10 325 0 25 For product C QC 2 9 000 10 250 0 25 Now the investment limit must be respected so that 3 000 2 2 2 C Q C Q C Q A A B B C C Expanding we have 3 000 175 2 51 000 10 175 0 25 325 2 25 000 10 325 0 25 250 2 9 000 10 250 0 25 We now need to find an value by trial and error that will satisfy this equation We can set up a table of trial values 113 Investment in A B C Total inventory value 003 126244 120453 63387 310084 004 124048 118358 62284 304690 0045 122992 117351 61754 302097 0049 122167 116563 61340 300070 005 121963 116369 61237 299569 010 112916 107736 56695 277347 When the term I is the same for all products as in this case may be found directly from Equation 1030 We can substitute the value for 0049 into the equation for Q and solve Hence we have Q Q Q A B C 2 51000 10 175 0 25 0 049 1396 2 25 000 10 325 0 25 0 049 717 2 9 000 10 250 0 25 0 049 491 units units units Checking 17513962 3257172 2504912 3000 22 We first check to see whether truck capacity will be exceeded Since three items are to be placed on the truck at the same time the items are jointly ordered The interval for ordering follows Equation 1022 or T O S I C D i i i 2 2 60 0 0 25 50 100 52 30 300 52 25 200 52 120 0 25 988 000 0 022 years or 1144 weeks Now from D T w i i i Truck capacity 10070 30060 200251144 34320 lb 114 The truck capacity of 30000 lb has been exceeded and the order quantity or the order interval must be reduced Given the revised Equation 1030 the increment to add to I can be found That is 2 2 60 30 000 100 52 70 300 52 60 200 52 10 50 10 52 30 30 52 25 20 52 0 25 120 30 000 2 340 000 988 000 0 25 0 73895 0 25 0 48895 2 2 2 O D w C D I i i i i Truck capacity Revise T the order interval by T O S I C D i i i 2 2 60 0 0 25 0 48895 50 100 52 30 300 52 25 200 52 120 0 73895 988 000 0 01282 years or 06667 weeks Once again we check that the truck capacity has not been exceeded 10070 30060 20025066667 30000 lb Therefore place an order every 47 or approximately 5 days 23 The average inventory for each item is given by 2 ds z Q AIL where s s LT d d and Q is found by Q DS IC 2 z 95 165 from the normal distribution in Appendix A The results of these computations can be tabulated 115 Summing the AIL for each product gives a total inventory of 1022 cases 24 The peak quantity of an item to appear on a shelf can be approximated as the order quantity plus safety stock or Q z sd 250 boxes where z93 148 from Appendix A and s s LT d d 19 1 19 boxes The economic order quantity is Q DS IC 2 2 123 52 125 019 129 25542 boxes Checking to see if the shelf space limit will be exceeded by this order quantity 25542 14819 28354 boxes The quantity is greater than the 250 allowed Subtracting the safety stock from the limit gives 250 28 222 boxes The order quantity should be limited to this amount 25 The plot of average inventory to period facility throughput shipments gives an overall indication of how the company is managing collectively its inventory for all stocked items We can see that the relationship is linear with a zero intercept This suggests that the company is establishing its inventory levels directly to the level of demand throughput An inventory policy such as stocking to a number of weeks of demand may be in effect Overall the inventory policy seems to be well executed in that the regression line fits the point for each warehouse quite well The terminal with an inventory level of 6000 seems to be an outlier and it should be investigated If its low turnover ratio were brought in line with the other terminals an inventory reduction from 6000 to 4000 on the average could be achieved The stocktodemand inventory policy should be challenged An appropriate inventory policy should show some economies of scale ie the inventory turnover ratio should increase as terminal throughput increases Whereas the current policy is of the form D I 0 012 a better policy would be I kD 70 where D represents terminal throughput and I is the average inventory level The coefficient 0012 for the current policy is found as the ratio of 6000500000 0012 for the last data point in the plot A B C D E sd 775 1549 1936 1162 2711 Q 18838 23828 42123 36198 56514 AIL 10698 14470 24256 20016 32730 116 The k value for the improved policy needs to be estimated From the cluster of the lowest throughput facilities the average inventory level is approximately 2000 with an average throughput of about 180000 Therefore from 0 419 4 771894 2 000 4 771894 2 000 180000 2 000 70 70 k k k k kD I Reading values from the plot the following table can be developed showing the inventory reduction that might be expected from revised inventory policy Note If the inventorythroughput values cannot be adequately read from the plot the values in the following table may be provided to the students Terminal Actual Inventory Shipments Estimated inventory D I 0 012 Revised inventory 70 0 419 D I 1 2000 150000 1800 1760 2 1950 195000 2340 2115 3 2000 200000 2400 2152 4 2050 200000 2400 2152 5 3900 320000 3840 2991 6 6000 330000 3960 3056 7 4500 390000 4680 3435 8 4300 410000 4920 3558 9 5500 500000 6000 4088 Totals 32200 2695000 32340 25307 Revising the inventory control policy has the potential of reducing inventory from the linear policy by 21 7 100 32340 25307 32340 x 26 We can use the decision curves of Figure 923 in the text answer this question since it applies to a fill rate of 95 and an 07 First determine K for an inventory throughput curve for the item which is 1 466 6 117 12 30 1 x TO D K Next 117 0 90 0 20400 1 466 12117 12 30 1 70 x ICK tD X and with z 196 from Appendix A 0 18 1 466117 12 9615 2 1 70 x KD zs LT Y a The demand ratio r is 42177 036 The intersection of r and X lies below the curve Y use curve Y 025 so do not cross fill 27 Regular stock For two warehouses estimate the regular stock for the three products Product A 457 units 2 0 0215 5 00025 2 354 units 2 0 0215 2 3 00025 2 2 2 2 1 A A RS RS IC dS Q RS Product B 445 units 2 0 0230 9 50025 2 408 units 2 0 0230 8 00025 2 2 1 B B RS RS Product C 612 units 2 0 0225 1500025 2 559 units 2 0 0225 1250025 2 2 1 C C RS RS Regular system inventory for two warehouses is RS2W 354 457 408 445 559 612 2835 118 Regular stock for a central warehouse 829 units 2 0 0225 2750025 2 604 units 2 0 0230 1750025 2 577 units 2 0 0215 8 00025 2 C B A RS RS RS Total central warehouse regular stock is RS1W 577 604 828 2009 units Safety Stock Product A units SS units SS LT zs SS A A d 1 000 0 75 1 65700 714 0 75 65500 1 2 1 where z095 165 from Appendix A Product B units SS units SS B B 479 0 75 1 65335 357 0 75 65250 1 2 1 Product C units SS units SS LT zs SS C C d 3 572 0 75 1 65 2 500 5 001 0 75 65 3 500 1 2 1 System safety stock is SS2W 714 1000 357 479 5001 3572 11123 units For each product the estimated standard deviation of demand on the central warehouse is 4 301 units 2 500 3 500 418 units 335 250 860 units 700 500 2 2 2 2 2 2 2 2 2 1 B B A s s s s s The safety stock is 119 6146 units 1 65 4 301 75 597 units 1 65418 75 1 229 units 65860 75 1 C B A SS SS SS zs LT SS Total safety stock in the central warehouse SS1W 1229 597 6146 7972 units Total inventory with two warehouses RS2W SS2W 2835 11123 13958 units and for a central warehouse RS1W SS1W 2009 7972 9981 units Centralizing inventories reduces them by 13958 9981 3977 units 28 The solution to this multiechelon inventory control problem is approached by using the basestock control system method The idea is that inventory at any echelon is to plan its inventory position plus the inventory from all downstream echelons First compute the average inventory levels for each customer This requires finding Q and the safety stock Q is found from the EOQ formula For customer 1 units x Q 270 20 35 425 1250 2 1 211 units 50 1 6565 2 270 2 1 1 1 1 LT zs Q AIL d where z095 165 from Appendix A For customer 2 units x Q 239 20 35 2333 1250 2 180 units 50 1 6552 2 239 2 2 2 2 2 LT zs Q AIL d For customer 3 units x Q 218 20 35 2276 1250 3 159 units 50 1 6543 2 218 2 3 3 3 3 LT zs Q AIL d 120 Total customer echelon inventory is AILC 211 180 159 550 units For the distributors echelon units as given QD 2 000 1120 units 01 1 2894 2 2 000 2 D LT zs Q AIL dD D D where z090 128 from Appendix A The expected inventory that the distributor will hold is the distributor echelon inventory less the combined inventory for the customers or 1120 550 570 units 121 COMPLETE HARDWARE SUPPLY INC Teaching Note Strategy Complete Hardware Supply is an exercise involving the control of inventoried items collectively Data for a random sample of 30 items from the companys total of 500 items held in inventory are given The objective is to manage the total dollar value allowed to be held as inventory Several alternatives can be considered for changing inventory levels some of which require an investment other than in inventory The number of items that must be analyzed and the multiple scenarios that are to be examined can be computationally time consuming It is strongly suggested that students use the INPOL module within LOGWARE to aid analysis The current database has been prepared and is available in the LOGWARE software The Base Case We begin with the current data optimized as a reorder point design The optimum order quantities and associated inventory levels are found The base case costs are shown as follows Fixed order quantity policy Purchase cost 556912 Transport costa 0 Carrying cost 4425 Order processing cost 4425 Outofstock cost 0 Safety stock cost 2529 Total cost 568291 Total investment 27801 aIncluded in the purchase cost We note that optimizing the current design shows that investment of 27801 exceeds the allowed investment level of 18000 Ways need to be explored to reduce this Transmit Orders More Rapidly Instead of mailing orders to vendors Tim OHare can buy a facsimile machine and transmit orders electronically This scenario can be tested by reducing the lead times in the base case by 2 days or 25 040 weeks and increasing order processing costs by 2 and then optimizing again INPOL shows that there will be a slight increase in operating costs from 568291 to 568640 an incremental increase of 349 Projecting this to all 500 items we have 34950030 5817 Since both operating cost and inventory investment level increase there is no economic incentive to implement this change Faster Transportation Suggesting that vendors who are located some distance 600 miles from the warehouse use premium transportation is a possible way of reducing lead times and therefore safety 122 stock levels Of course the increase in transportation cost for those affected vendors is likely to lead to a price increase to cover these costs This scenario is tested by reducing the leadtime in weeks to 22 for those vendors over 600 miles from the warehouse For these same vendors a 5 price increase is made Compared with the base case there is little change in the inventory investment 27801 vs 27746 however operating costs increase The total costs now are 585490 compared with the base case of 568291 an increase of 27199 The major portion 17159 of this comes from the increase in price We conclude that this is not a good option for Tim Reduce Forecast Error Reducing the forecast error involves reducing the standard deviation of the forecast error Testing this option requires taking 70 of the basecase forecast error standard deviations and optimizing the design once again These changes have a positive impact on operating costs and inventory investment Operating cost now is 567529 and inventory investment is 24739 This is a saving in operating costs of 762 per year For all 500 we can project the savings to be 76250030 12700 Based on a simple return on investment we have ROI 12 700 50 000 0 25 or 25 year This would appear to be attractive since carrying costs are 25 per year and the companys return on investment probably makes up about 80 of this value Reduce Customer Service At this point we have only accepted the idea of reducing the forecast error However inventory investment remains too high We can now try to reduce it by reducing the service levels This is tested by dropping the service index from its current 098 level to a level where inventory investment approximates 18000 This is done assuming the forecast software will be purchased and the forecast error reduced by 30 By trial and error the service index is found to be 054 which gives an investment level of 18028 The revised service level compared with the base case is summarized below for the 30 items 123 Notice how little the service level changes even with a substantial reduction in the service index Conclusions Tim can make a good economic argument for purchasing software that will reduce the forecast error The only questions here are whether the software can truly produce at least the error reduction noted and whether a 25 return on investment is adequate for the risks involved Arguing to accept a service reduction in order to lower the investment level is a little less obvious since we do not know the effect that service levels have on sales However Tim may point out that the service levels need to be changed so little that it is unlikely that customers will detect the change He might also raise the question as to whether customer service levels were too high initially and suggest that customers be surveyed as to the service levels that they do need Item Base case Revised Item Base case Revised 1 9988 9626 16 9998 9956 2 9992 9802 17 9990 9757 3 9996 9854 18 9995 9781 4 9998 9915 19 9989 9596 5 9998 9945 20 9997 9815 6 9996 9860 21 9969 8953 7 9997 9884 22 9997 9896 8 9996 9861 23 9997 9896 9 9992 9729 24 9996 9758 10 9998 9926 25 9992 9933 11 9999 9970 26 9997 9668 12 9999 9943 27 9993 9745 13 9992 9730 28 9989 9878 14 9998 9914 29 9997 9692 15 9996 9884 30 9991 9678 124 AMERICAN LIGHTING PRODUCTS Teaching Note Strategy American Lighting Products is a manufacturer of fluorescent lamps in various sizes for industrial and consumer use As frequently happens in business top management has requested that inventories be reduced across the board but it does not want to sacrifice customer service Sue Smith and Bryan White have been asked to eliminate 20 percent of the finished goods inventory Their plan is to reduce the number of stocking locations and thereby eliminate the amount of inventory needed Of course they must recognize that with fewer stocking points transportation costs are likely to increase and customer delivery times may increase as well On the other hand facility fixed cost may be reduced The purpose of this case is to allow students to examine inventory policy and planning through aggregate inventory management procedures They also can see the connection between location and inventory levels Answer to Questions 1 Evaluate the companys current inventory management procedures The companys procedures for controlling inventory levels are at the heart of whether inventory reductions are likely to be achieved through inventory consolidation The company appears to be using some form of reorder point control for the entire system inventory but it is modified by the need to produce in production lot sizes It is not clear how the reorder point is established If it is based on economic order quantity principles then the effect of the principles becomes distorted by the need to produce to a lot size that is different from the economic order quantity Therefore average inventory levels in a warehouse will not be related to the square root of the warehouses throughput demand ie throughput raised to the 05 power5 Rather the throughput will be raised to a higher exponent between 05 and 10 The above ideas can be verified by plotting the data given in Table 1 of the case and then fitting a curve of the form I TP Note The curve can be found from standard linear regression techniques when the equation is converted to a linear form through a logarithmic transformation ie lnI ln lnTP The results are shown in Figure 1 The inventory curve is I 299TP 0 816 with r 086 where I and TP are in lamps The projected inventory reduction can be calculated by using this formula From the plot of the inventory data we can see that there is substantial variation about the fitted inventory curve There is not a consistent turnover ratio between the warehouses This probably results from the centralized control policy On the other hand improved control may be achieved by using a pull procedure at each MDC The data available in the case do not let us explore this issue 5Based on the economic order quantity formula the average inventory level AIL for an item held in inventory can be estimated as AIL Q DS IC 2 2 2 Collecting all constants into K we have AILKD05 where D is demand or throughput 125 FIGURE 1 Plot of MDC average inventory vs annual throughput 2 Should establishing the LOC be pursued One of the ideas proposed in the case is to consolidate all Consumer product line items into one large order center LOC Evaluating the impact of the LOC on inventory reduction requires that an assumption be made as to how much demand and associated inventory of the total belongs to Consumer products Table 2 of the case gives the order and back order breakdown by sales channel Using these data total consumer demand is 312211 line items or 334 of the total line items The assumption is that the same percentage applies to total demand Hence Consumer demand is 334169023000 56453682 lamps From the inventorythroughput curve we can estimate the amount of inventory needed at the single LOC That is I 2997564536820816 6339684 lamps If Consumer products account for 334 of total inventory then there are 33423093500 7713229 lamps in Consumer inventory The reduction that can be projected is 7713229 6339684 1373545 lamps for a reduction of 178 100 7713229 1373545 Reduction in Consumer inventory levels but only a 6 reduction in overall inventory levels The 20 reduction goal is not achieved Other alternatives need to be explored 3 Does reducing the number of stocking locations have the potential for reducing system inventories by 20 Is there enough information available to make a good inventory reduction decision The second alternative proposed in the case is to reduce the number of MDCs from eight to a smaller number In order to evaluate this proposal it needs to be determined which MDCs will be consolidated and the associated total demand flowing through the consolidated facilities The inventorythroughput relationship can then be used to estimate the resulting inventory levels For example if the Seattle and Los Angeles MDCs are combined the consolidated demand would be 4922000 21470000 26392000 lamps The combined inventory is projected to be I 2997263920000816 126 3408852 lamps compared with the inventory for the two locations of 4626333 as shown in Table 1 This yields a 263 reduction from current levels Table 1 shows other possible MDC consolidations and the resulting inventory reductions that can be projected TABLE 1 Inventory Reduction for Selected MDC Combinations in Lamps MDC combination Combined demand Combined inventory Inventory reduction SeattleLos Angeles 26392000 3408852 1217481 Kansas CityDallas 29194000 3701403 50181 ChicagoRavenna 49174000 5664257 557590 AtlantaDallas 39314000 4718862 1224721 Kansas CityChicago 39271000 4714650 933900 RavennaHagerstown 64046000 7027231 1715607 K CityDallasChicago 52515000 5976377 36377 RavennaHtownChicago 87367000 7508054 3423196 AtlantaDallasK City 55264000 5242351 2293566 From the MDC combinations in Table 1 proximity to each other is a primary consideration in order to not increase transportation costs or jeopardize delivery service any more than necessary Several options can be identified that yield a 20 inventory reduction These are Option MDC combinations Inventory reduction lamps Total inventory reduction 1 LASeattle 1217481 RavennaHtownChicago 3423196 Total reduction 4640677 201 2 LASeattle 1217481 Kansas CityHagerstown 1224721 RavennaHagerstown 1715602 Total reduction 4157804 180 3 LASeattle 1217481 RavennaHagerstown 1715602 AtlantaDallasK City 2293566 Total reduction 5226649 226 Options 1 and 3 achieve the 20 reduction goal although other MDC combinations not evaluated may also do so The maximum reduction would be achieved with one MDC The total inventory would be I 29971690230000816 15512812 lamps for a system reduction of 328 However we must recognize that as the number of warehouses is decreased outbound transportation costs will increase Inbound transportation costs to the combined MDC will remain about the same since replenishment shipments are 127 already in truckload quantities Some difference in cost will result from differences in the length of the hauls to the warehouses On the other hand outbound costs may substantially increase since the combined MDC locations are likely to be more removed from customers then they are at present Outbound transportation rates will be higher as they are likely to be for shipments of lessthantruckload quantities If the sum of the inbound and outbound transportation cost increases is greater than the inventory carrying cost reduction then the decision to reduce inventories must be questioned Calculating all transportation cost changes is not possible since the case study does not provide sufficient data on outbound transportation rates However they should be determined before and after consolidation to assess the tradeoff between inventory reduction and transportation costs increases On the other hand inbound transportation costs can be found as shown below for option 1 where the consolidation points are Los Angeles and Hagerstown Location TL rate TL Annual demand lamps Transport cost Combined annual demand lamps Transport cost Seattle 1800 4922000 253131a Los Angeles 1800 21470000 1104171 26392000 1357302 Ravenna 250 25853000 184664 Hagerstown 475 38193000 518334 87367000 1185695 Chicago 350 23321000 233210 Total 113759000 2293510 113759000 2542997 a4922000350001800 253131 There will be a net increase in inbound transportation costs of 2542997 2293510 249487 for option 1 In addition the annual fixed costs for the MDCs will be less since the total space needed in the consolidated facilities should be less than that for the existing facilities Again the case study does not estimate the fixed costs for existing or potential locations We do know that taking them into account would favor consolidation In summary the costs associated with option 1 that just meets the 20 inventory reduction goal would be Although Sue and Bryan could report a substantial savings in inventory related costs they should be encouraged to include fixed costs and transportation costs so as to report the true benefits of the inventory reduction plan 4 How might customer service be affected by the proposed inventory reduction Cost type Cost savings Inventory carrying cost reduction 02008824640677 818615 Warehouse cost 0104640677 464068 Warehouse fixed cost Unknown but may be included in warehouse cost Outbound transportation cost Unknowndata not given Inbound transportation cost 249487 128 The general effect of inventory consolidation is to reduce the number of stocking points and make them more remote from customers That is the delivery distance will be increased if inventory consolidation is implemented Therefore delivery customer service may be jeopardized and must be considered before deciding to consolidate inventories From Table 3 of the case it can be seen that customer lead times remain constant for a variety of locations with the exception of Kansas City Since consolidation points will be selected among the existing locations outbound lead times will remain unaffected Customer service due to location should be constant at least for a moderate degree of consolidation Customer service due to stock availability will be affected if safety stock levels are reduced after consolidation Although the inventorythroughput relationship projects adequate safety stock to maintain the current firsttime delivery levels it does not account for any increase in lead times that may occur between the current system of MDCs and the consolidated ones By comparing the weighted inbound lead times for the existing distribution system and option 1 as shown in Table 2 the average inbound lead time is slightly reduced through consolidation Leadtime variability is usually related to average leadtime This should have a favorable affect on inventory levels since uncertainty is reduced Firsttime deliveries should not be adversely affected by consolidation according to option 1 TABLE 2 A Comparison of Inbound Lead Times for the Existing Distribution System and a Consolidated Distribution System Option 1 a Current Distribution System Master Distribution Center Shipments Inbound lead time days Weighted lead time days Atlanta 26070000 2 0308 Chicago 23321000 1 0138 Dallas 13244000 3 0235 Hagerstown 38193000 1 0226 Kansas City 15950000 2 0094 Los Angeles 21470000 5 0635 Ravenna 25853000 1 0153 Seattle 4922000 6 0175 Total 169023000 1964 129 b Consolidation Option 1 Master Distribution Centera Shipments Inbound lead time days Weighted lead time days Atlanta 26070000 2 0308 Dallas 13244000 3 0235 HtownRavennaChicago 87367000 1 0517 Kansas City 15950000 2 0094 Los AngelesSeattle 26392000 5 0781 Total 169023000 1935 aConsolidation is assumed to take place at the MDC with the largest number of current shipments 130 AMERICAN RED CROSS BLOOD SERVICES Teaching Note Strategy The American Red Cross Blood Services has a mission to provide the highest quality blood components at the lowest possible cost High quality blood products are provided to regional hospitals but managing the inventory to meet demand as it occurs is a difficult problem Blood is considered a precious product especially by those who give it voluntarily So managing this perishable product carefully is a foremost concern Blood is a vital product to those in need of it for emergencies and a precious product to those requiring it for elective surgery and other treatments The goal is to always have what is needed but never so much that this perishable product has to outdated Managing the blood inventory is quite difficult because 1 forecasting demand is not particularly accurate 2 the planning horizon for collections can be up to a year long with uncertain yields 3 the life of blood products ranges from 42 days to as short as 5 days 4 once scheduled blood donors are never turned away except for medical reasons and 5 there is a limited opportunity to sell blood outside of the local region if too much is on hand Overall this situation has many characteristics of a supply driven inventory management problem which requires inventory management techniques different from those for typical consumer products The intended purpose of this case study is for students to examine an inventory situation where there is limited control over the amount of the product flowing into inventory This supplydriven inventory situation is likely to be quite different from that discussed on the introductory level Students are encouraged to consider the various elements that affect inventory levels of individual products and how they interact These elements are 1 demand forecasting 2 collections 3 decision rules for creating blood derivatives 4 product prices and 5 inventory policy It is expected that students will be able to make general suggestions for improvement Questions 1 Describe the inventory management problem facing blood services at the American Red Cross One of the major problems facing the American Red Cross ARC is that the availability of blood is supplydriven meaning that quantities of blood received for processing to meet demand in the short term are unknown yet they must be placed in inventory if demand is less than the collected quantities Blood availability is a function of number of factors that cannot be wellcontrolled by the regional blood center in the short run causing wide variability in supply The usage of blood at hospital blood banks which creates the demand on ARCs blood inventories is also uncertain and varies from day to day and between hospital facilities The yield of blood at the point of collection is random and does not necessarily give the product mix needed to meet demand Different blood types can only be known by a probability distribution as to the percentage of the blood types that exist in the general population In the short term the demand for blood types may differ from the collected 131 quantities resulting in a potential for under and overstocking since blood is drawn from all qualified donors as they arrive at collection sites Forecasting demand for blood products will likely be reasonably accurate for a base load Surgery loads on hospitals are scheduled in advance so that blood needs will be known with a fair degree of certainty although each operation will not typically use the full amount of blood allocated to it However emergency blood needs are not well predicted and they can cause spikes in demand and unplanned draws on inventory A problem is establishing how much accuracy is needed for good inventory management Inventory policy for managing inventory levels is a mixed strategy of product pricing derivative product selection for processing at the time of collection conversion to other products later in the product life cycle product sell off emergency supply call for blood discount pricing and stocking rules for hospitals Although there are many avenues to controlling inventory levels shortages and outdating cannot always be avoided It is not clear that these procedures lead to an optimal control of inventory levels Competition from local independent blood banks that sell selected blood products at low prices makes it difficult for ARC to cover costs ARC provides a wider range of products but it has difficultydifferentiating price among derivative products so that it might compete effectively Given pressures for hospitals to increase efficiency they will shop around for the lowestpriced blood products ARC is having difficulty maintaining its position as the dominant supplier of blood products in the region which results in the greater uncertainty in managing inventory levels In summary blood is a precious product given by volunteers for the benefit of others Donors have the right to expect that their contribution will be handled responsibly To ARC this means managing the blood supply so that recipients receive a highquality product at the lowest possible price To achieve this goal ARC manages the blood supply through four interconnected elements 1 estimating the blood product needs over time 2 planning the collection of whole blood 3 deciding which derivative products and their amounts should be created from whole blood and 4 controlling the inventory levels to avoid outdating The volunteer nature of the blood giving and donor attitudes surrounding it long planning lead times and the associated uncertainties rising competition among some products from local blood banks and the uncertainties of blood needs all make blood supply management a unique inventory management problem 2 Evaluate the current inventory management practices in light of ARCs mission Performance of blood management can be evaluated on two levels customer service and cost Tables 8 and 9 of the case show that in March standards were not quite met overall Within specific product types there was up to an 8 percent deficit Both order fill rate and item fill rate were less than 100 percent for most products There would seem to be some room for improvement especially in managing the variation among product types From a cost standpoint it is not known how efficiently the blood supply is managed since no costs are reported In addition the revenue that the blood products generate is not known We would like to know how prices of the various products are set so that revenues might be maximized considering competition among some of the product line We do expect that demand is price elastic since hospitals do shop around for blood 132 products that are available from local commercial and community blood banks On the other hand ARC is the sole regional supplier of certain products such as platelets Setting product fillrate standards at various levels can influence costs We do not know this effect Setting inventory levels by a number of days of inventory rule of thumb is simple but not as effective as planning inventory levels based on the uncertainties that occur in demand forecasts and supply lead times The numberofdaysofinventory rule does tend to lead to too much inventory or to too many outofstock situations The plan for evaluation if enough data were available would be to establish a base case of cost and service This then would provide a basis for evaluating the effect of change in the supply procedures 3 Can you suggest any changes in ARCs inventory planning and control practices that might lead to cost reduction or service improvement Suggestions for improvement in blood supply management stem from a basic understanding of the nature of the demandsupply relationship When supply is uncertain and all supply must be taken that is available there is the possibility that significant excess inventory will occur The goal is to manage the demand in the short run to reduce inventory levels when overstocking occurs rather than focusing on managing supply Several approaches for doing this are Aggressively price selected products that are in excess supply and are nearing their expiration dates eg run a sale or offer price discounts Sell off excess supply to secondary demand sources or other regions of the ARC Temporarily adjust return rules for hospitals Bring demand more in line with supply by converting products into derivative ones that have excess demand eg reprocess whole blood into plasma Encourage hospitals to buy certain products in excess supply for a more favorable status in buying other products that are in short supply such as phersis platelets and rare whole blood types Try to create excess demand for all products especially those items that are available from local blood banks through promotion of ARCs distinct advantages such as quality high service levels and a wide range of blood derivative products Offer twoforone sales such that if a hospital buys one blood product it may receive another at a favorable price Pool the risk of uncertain demand by maintaining a central inventory for all hospitals or managing the inventories at all hospitals as well at ARC collectively Provide quick deliveries or transfers among inventory locations ARC should attempt to be the premier provider of blood products and leverage the advantage This will allow it to maintain a degree of control over the demand for blood Effectively controlling demand in turn allows it to control its costs and avoid product outdating 133 4 Is pricing policy an appropriate mechanism to control inventory levels If so how should price be determined From the previous discussion it can be seen that price plays a role in controlling demand Since there appears a relationship between demand and price for some products especially among those products offered by local blood banks that compete with ARC blood products price may be an effective weapon to meet competition Rather than setting price based on the cost of production ARC might consider raising the price on products for which it is the sole provider such as platelets and then meeting the price of competitors on whole blood Although ARC strives to be a nonprofit organization the increased volume that an effective pricing strategy promotes would allow more of the fixed costs to be covered This may lead to lower overall average prices for ARCs products Blood could also be priced as a function of its freshness at two or more levels Although blood that has been donated within 42 days legally can be utilized the quality of blood does not remain the same for the entire 42day period A chemical compound found in blood called 23DPG decreases with the age of the stored blood and is believed to be important in oxygen delivery For this reason certain procedures such as heart transplants and neonatal procedures require that blood be fresh usually donated within 10 days or less Thus a simple pricing policy could be to charge a higher price for blood that is less than 10 days old and a lower price for blood that is between 10 and 42 days old Price differences here are based on product quality 134 CHAPTER 10 PURCHASING AND SUPPLY SCHEDULING DECISIONS 1 a The following requirements schedules will lead to the proper timing and quantities for the purchase orders Desk style A Week 1 2 3 4 5 6 7 8 Sales forecast 150 150 200 200 150 200 200 150 Receipts 200 300 300 300 300 Qty on hand 0 50 200 0 100 250 50 150 0 Releases to prod 300 300 300 300 Desk style B Week 1 2 3 4 5 6 7 8 Sales forecast 60 60 60 80 80 100 80 60 Receipts 100 100 100 100 100 Qty on hand 80 20 60 0 20 40 40 60 0 Releases to prod 100 100 100 100 100 Desk style C Week 1 2 3 4 5 6 7 8 Sales forecast 100 120 100 80 80 60 60 80 Receipts 100 100 100 100 100 Qty on hand 200 100 80 80 0 20 60 0 60 Releases to prod 100 100 100 100 100 Summing the releases for these three desk release schedules gives a production requirements schedule for desks in general and sheets of plywood in particular That is Week 1 2 3 4 5 6 7 8 Desk requirements 500 100 400 500 200 400 100 0 Plywood sheetsa 1500 300 1200 1500 600 1200 300 0 a Desk requirements times 3 Now find the purchase order releases for the plywood sheets Week 1 2 3 4 5 6 7 8 Sales forecast 1500 300 1200 1500 600 1200 300 0 Receipts 600 1000 1000 1000 1000 Qty on hand 2400 900 1200 1000 500 900 700 400 400 Releases to prod 1000 1000 1000 1000 Therefore purchase orders should be placed in weeks 1 2 3 and 4 for 1000 sheets each 135 b Using Equation 102 in the text the probability of not having the plywood sheets at the time needed would be P P C P r c c c 1 1 5 01 5 0 02 From Appendix A z102 205 Therefore the leadtime should be T LT z sLT 14 2 05 2 181 days Another ½ week should be added to the current leadtime of 2 weeks 2 a Using Equation 102 the probability of not having the item when needed for production is 0 9999 150 35365 20 150 c c c r P C P P The time to place an order ahead of need is 28 days 63 4 14 sLT z LT T where z09999 36 from Appendix A b Use part period cost balancing The unit carrying cost is 025235 0134 Then Q250 Week 4 0134500 2002 469 Q1350 Weeks 4 5 01341350 10502 1050 2002 2446 The carrying cost closest to the order cost of 50 is Q 250 Order this amount 3 Using the requirements planning procedure we can develop a schedule of material flows through the network over the next 10 weeks Whse 1 1 2 3 4 5 6 7 8 9 10 Requirements 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 Schd receipts 7500 7500 Onhand qty 1700 500 6800 5600 4400 3200 2000 800 7100 5900 4700 Releases 7500 7500 136 Whse 2 1 2 3 4 5 6 7 8 9 10 Requirements 2300 2300 2300 2300 2300 2300 2300 2300 2300 2300 Schd receipts 7500 7500 7500 Onhand qty 3300 1000 6200 3900 1600 6800 4500 2200 7400 5100 2800 Releases 7500 7500 7500 Whse 3 1 2 3 4 5 6 7 8 9 10 Requirements 2700 2700 2700 2700 2700 2700 2700 2700 2700 2700 Schd receipts 7500 7500 7500 7500 Onhand qty 3400 700 5500 2800 100 4900 2200 7000 4300 1600 6400 Releases 7500 7500 7500 7500 Regnl whse A 1 2 3 4 5 6 7 8 9 10 Requirements 22500 0 0 15000 0 7500 15000 0 7500 0 Schd receipts 15000 15000 Onhand qty 52300 29800 29800 29800 14800 14800 7300 7300 7300 1300 1300 Releases to plant 15000 15000 Whse 4 1 2 3 4 5 6 7 8 9 10 Requirements 4100 4100 4100 4100 4100 4100 4100 4100 4100 4100 Schd receipts 7500 7500 7500 7500 7500 Onhand qty 5700 1600 5000 900 4300 200 3600 7000 2900 6300 2200 Releases 7500 7500 7500 7500 7500 Whse 5 1 2 3 4 5 6 7 8 9 10 Requirements 1700 1700 1700 1700 1700 1700 1700 1700 1700 1700 Schd receipts 7500 7500 Onhand qty 2300 600 6400 4700 3000 1300 7100 5400 3700 2000 300 Releases 7500 7500 Whse 6 1 2 3 4 5 6 7 8 9 10 Requirements 900 900 900 900 900 900 900 900 900 900 Schd receipts 7500 7500 Onhand qty 1200 300 6900 6000 5100 4200 3300 2400 1500 600 7200 Releases 7500 7500 Regnl whse B 1 2 3 4 5 6 7 8 9 10 Requirements 22500 0 7500 0 15000 7500 0 7500 7500 0 Schd receipts 15000 15000 15000 Onhand qty 31700 9200 24200 16700 16700 1700 9200 1700 9200 9200 9200 Releases to plant 15000 15000 Plant 1 2 3 4 5 6 7 8 9 10 Requirements 0 0 0 15000 15000 0 30000 0 0 0 Schd receipts 20000 20000 20000 Onhand qty 0 0 0 0 5000 10000 10000 0 0 0 0 Releasesmatls 20000 20000 20000 Summing the releases to the plant shows that the plant should place into production 15000 cases in weeks 4 and 5 and 30000 cases in week 7 Orders for materials should be placed in weeks 1 2 and 4 in an order size to make 20000 units Because demand is shown to be constant the average inventory must be onehalf the order quantity For the six field warehouses and a shipping quantity of 7500 the average long run inventory would be 750026 22500 cases For the regional warehouses 137 the average inventory would be 1500022 15000 cases For the plant the average inventory would be 200002 10000 cases The total system average inventory would be 22500 15000 10000 47500 cases 4 a The leverage principle shows the relative change that must be made in cost price or sales volume to affect a given change in the profit level Usually it is used in reference to the cost of goods sold to show the impact that small changes in the cost of goods will have on profits and the important role that purchasing plays in the profitability of the firm The following simple profit and loss statements will show how much change is needed in various activities to increase profits to 5000000 Sales Price LS OH COG Current 4 1 3 6 2 Sales 550 572a 555 550 550 550 Cost of goods 275 286 275 275 275 270 Labor salaries 150 156 150 145 150 150 Overhead 80 80 80 80 75 80 Profit 45 50 50 50 50 50 aSales 7727xSales 8 5 where LS is 02727 of Sales and COG is 05 of Sales So Sales 5 81 07727 572 Due to the magnitude of cost of goods sold it requires less than a 2 percent change in COG to increase profits to 5000000 b The current ROA as Profit margin 4555100 82 Investment turnover 5520 275 ROA 27582 226 Reducing cost of goods by 7 will increase profits to 55 275093 15 8 643 and the profit margin now is 64310055 117 Inventory at 20 of total assets is 4 million If the cost of goods is reduced by 7 inventory value will decline to 4093 372 Total assets will be 372 16 1972 million The investment turnover is 551972 2789 The ROA now will be 1172789 3263 5 a A mixed purchasing strategy will generally be beneficial when prices show a definite seasonality they are predictable and inventory costs associated with forward buying are not excessive In the problem we should consider forward buying in the first half of the year and handtomouth buying in the last half To test the various strategies compare 1 handtomouth buying 2 forward buying every 2 months 3 forward buying every 3 months and 4 forward buying for the first 6 months The results are summarized in Table 101 Or ROA ProfitAssets 138 The inventory for the handtomouth buying strategy can be approximated as 500002 25000 The carrying cost would be 03049825000 37350 per year The carrying cost for the 2mouth forward buying strategy is 030488051000002 05500002 54900 For the 3month forward buying strategy 03456053000002 05500002 119700 From the total costs in Table 101 the best strategy is to forward buy the first six months requirements in January and handtomouth buy for the last six months b Some possible disadvantages are Prices may fall rather than rise in the first six months There may not be adequate storage space to accommodate such a large purchase The materials may be perishable and not easily stored Uncertainties in the requirements and carrying costs may void the strategy For 2nd half of year 139 TABLE 101 A Comparison of Various Forward Buying Strategies with HandtoMouth Buying Handtomouth buy 2month forward buy 3month forward buy 6month forward buy Price unit Quantity units Total Price unit Quantity units Total Price unit Quantity units Total Price unit Quantity units Total Jan 400 50000 200000 400 100000 400000 400 150000 600000 400 300000 1200000 Feb 430 50000 215000 Mar 470 50000 235000 470 100000 470000 Apr 500 50000 250000 500 150000 750000 May 525 50000 262000 525 100000 525000 Jun 575 50000 287500 Jly 600 50000 300000 600 50000 300000 600 50000 300000 600 50000 300000 Aug 560 50000 280000 560 50000 280000 560 50000 280000 560 50000 280000 Sep 540 50000 270000 540 50000 270000 540 50000 270000 540 50000 270000 Oct 500 50000 250000 500 50000 250000 500 50000 250000 500 50000 250000 Nov 450 50000 225000 450 50000 225000 450 50000 225000 450 50000 225000 Dec 425 50000 212000 425 50000 212000 425 50000 212500 425 50000 212500 Subtotals 2987500 2932500 2887500 2737500 Inventory costs 37350 54900 72150 119700 Totals 3024850 2987400 2959650 2857200 Average priceunit 498 488 481 456 140 6 a On the average a total expenditure of 11025000 27500 should be made for copper each month b For the next 4 months the dollar averaging purchases would be The average perlb cost would be 110000100970 1089 The inventory carrying cost over 4 months would be 0201089412 12622 916 If handtomouth were used we would have 1 2 312 422 Price No of Total Average Month lb lb cost inventory lb 1 132 25000 33000 12500 2 105 25000 26250 12500 3 110 25000 27500 12500 4 095 25000 23750 12500 100000 110500 12500a a 500004 12500 The average perlb cost would be 110500100000 1105 The inventory carrying cost over 4 months would be 0201105412 12500 921 If 100000 lb of copper is purchased the two strategies can be compared as follows Purchase Inventory Total Strategy cost cost cost Dollar averaging 108900 916 109816 Handtomouth 110500 921 111421 Dollar averaging buying would be preferred 7 For an inclusive quantity discount price incentive plan we first compute the economic order quantities for each range of price Using Q DS IC 2 we compute 1 2 312 422 Price No of Total Average Month lb lb cost inventory lb 1 132 20833 27500 10417 2 105 26190 27500 13095 3 110 25000 27500 12500 4 095 28947 27500 14474 100970 110000 12622a a504864 12622 141 Q1 2 500 15 0 20 4995 38 75 cases Q2 2 500 15 0 20 4495 4085 cases Since Q2 is outside of the second price bracket Q1 is the only relevant quantity Now we check the total cost at Q1 and at the minimum quantities within the price break We solve TC PD DS Q IC Q i i i i i 2 At Q 3875 TC 4995500 500153875 02499538752 25362 At Q 50 TC 4495500 5001550 024495502 22850 At Q 80 TC 3995500 5001580 023995802 20388 Floor polish should be purchased in quantities of 80 cases 8 This noninclusive price discount problem requires solving the following relevant total cost equation for various order quantities until the minimum cost is found TC PD DS Q IC Q i i i i i 2 The computations can be shown in the table below given that D 1400 S 75 and I 025 142 Q Price PD DSQ ICQ2 Total cost 20 795 111300000 525000 198750 112023750 50 795 111300000 210000 496875 112006875 100 795 111300000 105000 993750 112398750 200 795 111300000 52500 1987500 113340000 300 200795100750 109200000 35000 2925000 112160000 300 400 200795200750 108150000 26250 3862500 112038750 400 500 200795200750 106820000 21000 4768750 111609750 100725 500 550 200795200750 106336364 19091 5221875 111577327 150725 550 600 200795200750 105933333 17500 5675000 111625833 200725 600 The optimal purchase quantity is 550 motors 9 a This problem is a good application of the transportation method of linear programming We begin by determining the costs for the current sourcing arrangement Source Destination Price Transport Volume Cost Dayton Cincinnati 340 005 5000 17250 Dayton Baltimore 340 015 1000 3550 Kansas City Dallas 345 008 2500 8825 Minneapolis Los Angeles 325 024 1200 4188 Total 33813 To optimize we establish the following transportation cost matrix and solve it using any appropriate method such as the TRANLP module in LOGWARE Cincin nati Dallas Los Angeles Baltimore Capacity Minneapolis 340 344 349 1200 346 1200 Kansas City 355 353 365 363 4800 Dayton 345 5000 352 2500 367 355 1000 9999 Requirements 5000 2500 1200 1000 The total cost for this solution is 33788 or a savings of 25 over the current sourcing 143 b Because Minneapolis is at capacity this supplier should be examined further If unlimited capacity were available at Minneapolis all requirements would be met by this supplier for a total cost of 33248 or a savings of 565 for this material c The above analysis does indicate that too many suppliers are being used Only two are needed if Minneapolis continues to supply at the current level If Minneapolis can be expanded it becomes the only supplier Of course whether the company would risk a single supplier for this material must be left unanswered 10 a The dealbuying equation Equation 105 can be applied to this problem First find the optimal order quantity before the discount Q DS IC 2 2 120 000 40 030 100 566 units Next find the adjusted order quantity after the discount has been applied 21648 units 5 100 100566 5030 100 5120000 d p pQ dI p dD Q ˆ A large order size of 21648 units should placed b The time that an order of this size will be held before it is depleted is given by 0 18 years or 94 weeks 120 000 21648 ˆ D Q 144 INDUSTRIAL DISTRIBUTORS INC Teaching Note Strategy The purpose of the Industrial Distributors case study is to illustrate the computation of purchase quantities under inclusive and noninclusive price discounts and transport rate weight breaks The INPOL module of LOGWARE is helpful in conducting the analysis As a teaching strategy it may be worthwhile to begin any class discussion with the cost tradeoffs that are present in such a problem as this This will help to establish the nature of the total cost equation that needs to be solved in this problem Answers to Questions 1 What size of replenishment orders to the nearest 50 units should Walter place given the manufacturers noninclusive price policy When price discounts are offered purchase quantities are not simply determined by a single formula Due to discontinuities in the total cost curve as a function of order quantity the optimal order quantity is found by computing total costs for different quantity values In this case of both price and transport rate breaks plus warehousing costs that can be affected by the order size the following annual total cost formula is to be solved TC PD RD SD Q ICQ W Q 2 300 where TC total cost for quantity Q PD purchase cost for price P RD transport costs at rate R SDQ ordering cost at quantity Q ICQ2 carrying cost at quantity Q WQ300 public warehousing cost if Q is greater than 300 units W public warehousing rate per unit per year D annual demand units P price for orders of size Q per unit R transport per unit for shipments of size Q per unit S order processing cost per order I annual carrying cost C product value per unit Q size of purchase order units Under noninclusive price discounts price is an average determined by the number of units in each break For example if 250 units are to be ordered the average price per unit would be computed as 145 P250 100 100 50 250 00 700 680 670 686 A table of annual costs can now be developed as shown in Table 1 To the nearest 50 units the optimal purchase quantity should be 250 units 2 If the manufacturers pricing policy were one where the prices in each quantity break included all units purchased should Walter change his replenishment order size The average price per unit is more easily determined in this case than the previous one Since all units are included in the price break back to the first unit the average price is simply the price associated with a given purchase quantity Finding the optimal purchase quantity is simply a matter of determining the total cost for the quantities found by the economic order quantity formula assuming these quantities are feasible and for the quantities at the transport rateweight break The comparison is made among the total costs of these alternatives These costs are shown in Table 2 The order quantities as determined by the economic order quantity formula for the base price of 700 would be or 18 units 3 18 700 45 30 2150025 2 IC DS Q where C is the 700 price per unit at Baltimore plus the 45 transport cost from Baltimore as determined by an LTL shipment 18 units 250 lb 4500 lb at 18 25 cwt 45 per unit The Q values for the other prices in the schedule lie outside the feasible range of the price used to compute Q The optimal strategy is to purchase 201 units per order which is one unit into the last price break Yes Walter should alter his buying strategy TABLE 1 Annual Costs by Quantity Purchased for Noninclusive Price Discounts Quantity Average price Purchase cost Transport cost Ordering cost Carrying cost Warehouse cost Total cost 18a 70000 1050000 67500 2083 2013 0 1121596 50 70000 1050000 67500 750 5592 0 1123842 100 70000 1050000 67500 375 11184 0 1129059 150 69333 1039995 67500 250 16619 0 1124364 160 69250 1038750 45000 234 17355 0 1101339 200 69000 1035000 45000 188 21619 0 1101807 250 68600 1029000 45000 150 26873 0 1101023Opt 300 68333 1024995 45000 125 32128 0 1102248 400 68000 1020000 45000 94 42638 1000 1108732 a EOQ at a price of 700 45 0625 per unit b First price break c Transport rate break d Second price break 146 TABLE 2 Annual Costs by Quantity Purchased for Inclusive Price Discounts Quantity Average price Purchase cost Transport cost Ordering cost Carrying cost Warehouse cost Total cost 18 70000 1050000 67500 2083 2013 0 1121596 19 68000 Infeasible 19 67000 Infeasible 101 68000 1020000 67500 371 10993 0 1098864 160 68000 1020000 45000 234 17175 0 1082409 201 67000 1005000 45000 187 21124 0 1071311Opt a Feasible EOQ at a price of 700 45 0625 per unit b Infeasible EOQ at a price of 680 45 0625 per unit c Infeasible EOQ at a price of 670 30 0625 per unit d First price break e Transport rate break f Second price break 147 CHAPTER 11 THE STORAGE AND HANDLING SYSTEM All questions in this chapter require individual judgment and response No answers are offered 148 CHAPTER 12 STORAGE AND HANDLING DECISIONS 2 Various alternatives are evaluated in Tables 121 to 124 The annual costs of each alternative are plotted in Figure 121 The best economic choice is to use all public warehousing or a Pure Public Warehouse Strategy Privatelyoperated Rented Space equire ments sq ftb Private allo cation Monthly fixed cost Monthly variable cost Rented allo cation Monthly storage cost Monthly handling cost Monthly total cost 62500 0 0 0 100 30000c 50000d 80000 50000 0 0 0 100 24000 40000 64000 37500 0 0 0 100 18000 30000 48000 25000 0 0 0 100 12000 20000 32000 12500 0 0 0 100 6000 10000 16000 3125 0 0 0 100 1500 2500 4000 15625 0 0 0 100 7500 12500 20000 28125 0 0 0 100 13500 22500 36000 37500 0 0 0 100 18000 30000 48000 43750 0 0 0 100 21000 35000 56000 50000 0 0 0 100 24000 40000 64000 56250 0 0 0 100 27000 45000 72000 21875 0 0 202500 337500 540000 5lb ft Thruput lb 1 2 turns 1040 storage space ratio 01 cu ft 110 ft 5 lb 2 and 100 of the demand through the rented warehouse then 1000000 1002 006 30000 50000 150 TABLE 122 Costs for a Mixed Warehouse Strategy Using a 10000 Square Foot Privately Operated Warehouse Privatelyoperated Rented Month Ware house thruput lba Space require ments sq ftb Private allo cation Monthly fixed cost Monthly variable cost Rented allo cation Monthly storage cost Monthly handling cost Monthly total cost Jan 1000000 62500 16 9792d 3200e 84 25200f 42000g 80192 Feb 800000 50000 20 9792 3200 80 19200 32000 64192 Mar 600000 37500 27 9792 3200 73 13140 21900 48032 Apr 400000 25000 40 9792 3200 60 7200 12000 32192 May 200000 12500 80 9792 3200 20 1200 2000 16192 June 50000 3125 100 9792 1000 0 0 0 10792 July 250000 15625 64 9792 3200 36 2700 4500 20192 Aug 450000 28125 36 9792 3200 64 8640 14400 36032 Sept 600000 37500 27 9792 3200 73 13140 21900 48032 Oct 700000 43750 23 9792 3200 77 16170 26950 56112 Nov 800000 50000 20 9792 3200 80 19200 32000 64192 Dec 900000 56250 18 9792 3200 82 22140 36900 72032 Totals 6750000 421875 117504 36200 147930 246550 548184 a Thruput lb Sales 5lb b Space requirements sq ft Thruput lb ½ 1040 0110 5 Thruput lb 00625 c 1000062500 016 d 351000020 101000012 9792 per month e 1000000016002 3200 f Given a turnover ratio of 2 and 84 of the demand through the rented warehouse then 1000000 0842 006 25200 d 1000000 084 005 42000 151 TABLE 123 Costs for a Mixed Warehouse Size Strategy Using a 30000 Square Foot Privately Operated Warehouse Privatelyoperated Rented Month Ware house thruput lba Space require ments sq ftb Private allo cation Monthly fixed cost Monthly variable cost Rented allo cation Monthly storage cost Monthly handling cost Monthly total cost Jan 1000000 62500 48c 29375d 9600e 52 15600f 26900g 80575 Feb 800000 50000 60 29375 9600 40 9600 16000 64575 Mar 600000 37500 80 29375 9600 20 3600 6000 48575 Apr 400000 25000 100 29375 8000 0 0 0 37375 May 200000 12500 100 29375 4000 0 0 0 33375 June 50000 3125 100 29375 1000 0 0 0 30375 July 250000 15625 100 29375 5000 0 0 0 34375 Aug 450000 28125 100 29375 9000 0 0 0 38375 Sept 600000 37500 80 29375 9600 20 3600 6000 48575 Oct 700000 43750 69 29375 9600 31 6510 13020 58505 Nov 800000 50000 60 29375 9600 40 9600 16000 64575 Dec 900000 56250 53 29375 9600 47 12690 21150 72815 Totals 6750000 421875 352500 94200 61200 104170 612070 a Thruput lb Sales 5lb b Space requirements sq ft Thruput lb ½ 1040 0110 5 Thruput lb 00625 c 3000062500 048 d 353000020 103000012 29375 per month e 1000000048002 9600 f Given a turnover ratio of 2 and 52 of the demand through the rented warehouse then 1000000 0522 006 15600 d 1000000 052 005 26000 152 TABLE 123 Costs for a Mixed Warehouse Size Strategy Using a 40000 Square Foot Privately Operated Warehouse Privatelyoperated Rented Month Ware house thruput lba Space require ments sq ftb Private allo cation Monthly fixed cost Monthly variable cost Rented allo cation Monthly storage cost Monthly handling cost Monthly total cost Jan 1000000 62500 64c 39167 12800e 36 10800f 18000g 80767 Feb 800000 50000 80 39167 12800 20 4800 8000 64767 Mar 600000 37500 100 39167 12800 0 0 0 51167 Apr 400000 25000 100 39167 8000 0 0 0 47167 May 200000 12500 100 39167 4000 0 0 0 43167 June 50000 3125 100 39167 1000 0 0 0 40167 July 250000 15625 100 39167 5000 0 0 0 44167 Aug 450000 28125 100 39167 9000 0 0 0 48167 Sept 600000 37500 100 39167 12000 0 0 0 51167 Oct 700000 43750 91 39167 12800 0 1890 3150 57007 Nov 800000 50000 80 39167 12800 20 4800 8000 64767 Dec 900000 56250 71 39167 12800 29 7830 13050 72847 Totals 6750000 421875 470004 115000 30120 50200 665324 a Thruput lb Sales 5lb b Space requirements sq ft Thruput lb ½ 1040 0110 5 Thruput lb 00625 c 4000062500 064 d 354000020 104000012 39167 per month e 1000000064002 12800 f Given a turnover ratio of 2 and 52 of the demand through the rented warehouse then 1000000 0362 006 10800 d 1000000 036 005 18000 153 FIGURE 121 Total Annual Costs for a Combined Warehouse Size Using Private and Public Warehouse Space 3 The annual cost of public warehousing is Handling 600000 Storage 300000 Total 900000 The costs of private warehousing are Annual operating 250000 Annual lease payment 3150000 450000 Other fixed one time 400000 The savings in operating costs of lease vs public warehousing is Savings 900000 250000 650000yr 0 10000 30000 40000 530 550 570 590 610 630 650 670 0 10000 30000 40000 Private warehouse space sq ft Total cost 000s 154 TABLE 125 TenYear Cash Flow Stream for Public vs Leased Warehouse Comparison Year Savings Lease vs public Pretax net cash flow Depre ciation sche dule Savings less depre ciation Taxes 35 Savings less depre ciation tax Savings less tax Aftertax net cash flow Discount factor 11ij Dis counted cash flow 0 0 3050a 0 0 0 0 0 3050 3050 1 650 650 57b 593 208 385 442c 442 09009 398 2 650 650 57 593 208 385 442 442 08116 359 3 650 650 57 593 208 385 442 442 07312 323 4 650 650 57 593 208 385 442 442 06587 291 5 650 650 57 593 208 385 442 442 05935 262 6 650 650 57 593 208 385 442 442 05346 236 7 650 650 58 592 207 385 443 443 04817 213 8 650 650 0 650 228 422 442 442 04339 183 9 650 650 0 650 228 422 442 442 03909 165 10 650 650 0 650 228 422 442 442 03522 149 6500 3450 400 6100 2139 3961 4361 1311 NPV 471 a Capitalization lease plus initial cash outlay ie 2650154 400000 3050154 b Depreciation charge for each of seven years is 17 01429 such that 40000001429 57143 c Add back depreciation ie 385 57 442 155 Capitalizing the lease over ten years we have PV 450 000 1 011 1 011 1 011 650154 10 10 2 The initial investment in 000s then is Initial investment 2650 400 3050 The tenyear cash flow stream is shown in Table 125 Since the savings are expressed to favor leasing and the net present value is negative choose public warehousing 4 Given k 210sq ft S 100000 sq ft C 001ft10000 100ft The width is W C k C k S 8 2 8 100 8 210 2 100 8 210 100 000 308 ft The length is L S W 100 000 308 325 ft 5 Space layout according to text Fig 124a can be determined by the application of equations 128 and 129 These equations specify the best number of shelf spaces and the best number of double racks respectively Equations 1210 and 1211 give the length and width of the building The optimal number of shelf spaces would be m L dC aC C dC C K w a L h h s p h p 1 1 2 2 2 2 1 4 400 000 0 001 2 10 050 2 300 2 400 000 0 001 300 50 000 8 10 4 2 4 120 48 or 121 156 The optimal number of double racks would be n w a dC C dC aC C K w a L h h p h s p 1 1 2 2 2 2 1 8 10 2 400 000 0 001 300 400 000 0 001 2 10 050 2 300 50 000 8 10 4 2 4 52 The warehouse length would be u n w a 1 1 52 8 10 936 ft and the width would be v a m L 1 1 2 2 10 121 4 504 ft 6 According to Equation 1217 the number of truck doors can be estimated by N DH CS Therefore N 7512000 3312000 8 937 or 10 doors 7 Summarizing the given information as follows Three Five Seven type 1 type 2 type 3 units units units Initial investment 60000 50000 35000 Useful life 10 yr 10 yr 10 yr Salvage value 15 of initial cost 9000 7500 5250 Annual operating expenses 6000 12500 21000 Return on investment before tax 20 20 20 An initial solution to this problem can be found through a discounted cash flow analysis Three alternatives are to be evaluated 157 PV1 10 10 10 60 000 6 000 1 0 2 1 0 2 1 0 2 9 000 1 1 0 2 60 000 6 000 4 2 9 000 016 760 83 PV2 50 000 12 500 4 2 7 500 016 50 000 52 500 1 200 300 101 PV3 35 000 21 000 4 2 5 250 016 35 000 88 200 840 360 122 The low present value of the Type 1 truck indicates that from among these three alternatives this would be the best buy 8 Given Initial cost of equipment 4000 Operating costs 500 40t 12 30t 1 Salvage value Sn I1 t7 Rate of return on investment 20 Replacement is expected to be with equipment of like kind The best replacement year can be found by comparing the equivalent annual cost of a sequence of similar equipment replaced every n years The equivalent annual cost is 1 1 1 1 1 n n n j 1 n j i i i i S i C I AC n j n Solving this equation for different years is facilitated if the equation is set up in tabular form as shown in Table 126 The equipment should be replaced at the end of the third year of service although a 5 year replacement cycle is also attractive 158 9 a1 Layout by popularity involves locating the more frequently ordered items closest to the outbound dock Based on the average number of daily orders on which the item appears the items closest to the outbound dock would be ranked as follows BIEAFHJCGD The storage space might then be used as follows Inbound D J C G H F A H J A E E B B I E Outbound a2 Layout by cube places the smallest items nearest the outbound dock Using the individual item size the ranking would be as follows AEICJHGBF The layout of items in the storage bays would be TABLE 126 Equivalent Annual Cost Computations for Problem 8 Year n 1 Initial investment I 2 Operating costs C i j j 1 3 Salvage value S i n n 1 4 Factor i i i n n 1 1 1 51234 Equivalent annual cost ACn 1 4000 416 2857 120 1871 2 4000 770a 1984 065 1821 3 4000 1117b 1323 047 1783 4 4000 1488 827 039 1818 5 4000 1898 459 033 1795 6 4000 2350 191 030 1848 7 4000 2841 0 028 1915 a 416 500 40212 302110202 770 b 770 500 40312 30 3110203 1117 159 Inbound F D B H D H G B D J C E I C E A E Outbound a3 The cubeperorder index is created by ratioing the average required cubic footage of a product to the average number of daily orders on which the item is requested Hence this index is found as follows 1 2 312 Space required Daily CPO Product cu ft orders index A 5000a 56 89 B 30000 103 291 C 15000 27 556 D 17000 15 1133 E 55000 84 655 F 11000 55 200 G 7000 26 269 H 28000 45 622 I 13000 94 138 J 9000 35 257 a500 sq ft stacked 10 ft high Locating the products with the lowest index values nearest to the outbound dock results in the following ranking and layout AIFJGBCHED 160 Inbound D E E F E H C B F I A F J G B Outbound b All of the above methods assume 1 that the product is moved to the storage locations in large unit loads but retrieved from the storage locations in relatively small quantities and 2 that only one product is retrieved during an outandback trip Therefore these methods do not truly apply to the situation of multiple picks on the same trip However they may be used with some degree of approximation if the products can be aggregated as one and grouped together or zoned in the same section of the warehouse 10 This is an extra challenging problem that requires some knowledge of linear programming It may be formulated as follows Let Xij represent the amount per 1000 units of product j stored in location i Let Cij be the handling time associated with storage bay i and product j Gj is the capacity of a bay for product j and Rj is the number of units of product j required to be stored The linear programming statement is Objective function Zmin 90X11 75X12 90X13 80X21 65X22 95X23 60X31 70X32 65X33 70X41 55X42 45X43 50X51 50X52 45X53 40X61 45X62 35X63 Subject to Capacity restrictions on bays 161 20X11 333X12 167X13 100 20X21 333X22 167X23 100 20X31 333X32 167X33 100 20X41 333X42 167X43 100 20X51 333X52 167X53 100 20X61 333X62 167X63 100 and storage requirements restrictions on products X11 X21 X31 X41 X51 X61 11 X12 X22 X32 X42 X52 X62 4 X13 X23 X33 X43 X53 X63 12 Solving the linear programming problem by means of any standard transportation code of linear programming such as LNPROG in LOGWARE yields X12 1610 The total minimum X21 1020 handling time is X22 2390 13868 hours X31 5000 X43 5988 X51 4980 X53 0024 X63 5988 where Xs are in thousands of units That is product 1 should be stored in bays 3 4 and 5 in quantities of 1020 5000 and 4980 respectively Product 2 should be stored in bays 1 and 2 in quantities of 1610 and 2390 respectively Product 3 should be stored in bays 4 5 and 6 in quantities of 5988 24 and 5988 respectively Graphically this is Bay Product 1 2 3 4 5 6 Require ments 1 1020 5000 4980 11000 2 1610 2390 4000 3 5988 24 5988 12000 of bay capacity 537 1000 1000 1000 1000 1000 162 CHAPTER 13 FACILITY LOCATION DECISIONS 1 a The centerofgravity method involves finding the XY coordinates according to the formulas X V R X V R i i i i i i i and Y V R Y V R i i i i i i i These formulas can be solved for in tabular form as follows Point X Y Vi Ri ViRi ViRiXi ViRiYi P1 3 8 5000 040 2000 6000 16000 P2 8 2 7000 040 2800 22400 5600 M1 2 5 3500 095 3325 6650 16625 M2 6 4 3000 095 2850 17100 11400 M3 8 8 5500 095 5225 41800 41800 Totals 16200 93950 91425 Now X 9 3950 1 620 0 58 and Y 91425 1 620 0 564 This solution has a total cost for transportation of 5361491 This problem may also be solved using the COG module in LOGWARE b Solving for the exact centerofgravity method requires numerous computations We now use the COG module of LOGWARE to assist us A table of partial results is shown below 163 Iteration Total number X coord Y coord cost 0 5799383 5643518 5361491 COG 1 5901199 5518863 5351085 2 5933341 5446919 5348397 3 5941554 5402429 5347460 50 5939314 5317043 5346771 After 50 iterations there is no further change in total cost The revised coordinates are X 594 and Y 532 for a total cost of 5346771 c The centerofgravity solution can be one that is close to optimum when there are many points in the problem and no one point has a dominant volume that is has a larger volume relative to the others Otherwise the best single location can be at a dominant location The exact centerofgravity approach has the capability to find the minimal cost location Although the COG model only considers transportation costs that are constant per mile the transportation cost can be the major consideration in single facility location However other costs such as labor real estate and taxes can also be important in selecting one location over another These are not directly considered by the model Although the COG model may seem of limited capability it is a useful tool for locating facilities where transportation costs are dominant Location of oil wells in the Gulf truck terminals and single warehouses are examples of application It also can be quite useful to provide a starting solution to more complex location models d Finding multiple locations by means of the centerofgravity approach requires assigned supply and demand volumes to specific facilities and then solving for the center of gravity for each In this problem there are 3 market combinations that need to be considered This creates 3 scenarios that need to be evaluated They can be summarized as follows Point volumes appear in the body of the table Scenario Whse P1 P2 M1 M2 M3 1 1458a 2042 3500 I 2 3542 4958 3000 5500 1 2708 3792 3500 3000 II 2 2292 3208 5500 1 3750 5250 3500 5500 III 2 1250 1750 3000 a Allocated as a proportion of the volume to be served through the warehouse That is 500035003500 3000 5500 1458 The volumes associated with other supply points are computed similarly 164 The COG module in LOGWARE was used to find the exact centers of gravity for each warehouse in each scenario The computational results are Warehouse 1 Warehouse 2 Total Scenario X Y X Y cost I 200 500 788 780 39050 II 584 404 800 800 35699 III 706 728 600 400 46568 Scenario II appears to be the best 2 a The centerofgravity formulas Eqs 135 and 136 can be solved using the COG module of LOGWARE or they can be solved in tabular form as shown below Point X Y Vi Ri ViRi ViRiXi ViRiYi A 50 0 9000 75 6750 337500 0 B 10 10 1600 75 1200 12000 12000 C 30 15 3000 75 2250 67500 33750 D 40 20 700 75 525 21000 10500 E 10 25 2000 75 1500 15000 37500 F 40 30 400 75 300 12000 9000 G 0 35 500 75 375 0 13125 H 5 45 8000 75 6000 30000 270000 I 40 45 1500 75 1125 45000 50625 J 20 50 4000 75 3000 60000 150000 Totals 23025 600000 586500 Now X 600 000 23 025 261 and Y 586 500 23 025 255 The total cost of this location is 609765 The exact centerofgravity coordinates are X Y 2351 2698 with a total cost of 608478 b The number of points even in this small problem requires us to apply some heuristics to find which patient clusters should be assigned to which warehouses We will use a clustering technique whereby patient clusters are grouped by proximity until two clusters are found The procedure works as follows 165 There are as many clusters as there are points which is 10 in this case The closest points are found and replaced with a single point with the combined volume located at the center of gravity point There is now one less cluster The next closest two pointsclusters are found and they are further combined and located at their center of gravity The process continues until only two clusters remain The centers of gravity for these two clusters will be the desired clinic locations Applying the clustering technique we start by combining points D and F into cluster DF X 40 700 0 75 40 400 0 75 700 0 75 400 0 75 40 00 and Y 20 700 0 75 30 400 0 75 700 0 75 400 0 75 2364 Continuing this process we would form two clusters containing A C D and F and B E G H I and J The centers of gravity would be Cluster 1 ACDF X X 50 00 0 00 Cluster 2 BEGHIJ X Y 500 4500 for a total cost of 241828 These are the same results obtained from the MULTICOG module in LOGWARE c The second clinic can save 608278 241828 366458 in direct costs annually This savings does not exceed the annual fixed costs of 500000 required to maintain a second clinic On economic grounds it should not be built 3 a The centerofgravity location can be determined by forming the following table or by using the COG module in LOGWARE The coordinates for each location must be approximated 166 Point X Y Vi Ri ViRi ViRiXi ViRiYi A 15 66 10000 10 1000 1500 6600 B 47 73 5000 10 500 2350 3650 C 80 71 70000 10 7000 56000 49700 D 15 40 30000 10 3000 4500 12000 E 50 49 40000 10 4000 20000 19600 F 85 51 12000 10 1200 10200 6120 G 15 13 90000 10 9000 13500 11700 H 44 18 7000 10 700 3080 1260 I 78 18 10000 10 1000 7800 1800 Totals 27400 118930 112430 The centerofgravity coordinates are X 118 930 27 400 434 and Y 112 430 27 400 410 with a total cost of 195966 b After 100 iterations in the COG module the exact center of gravity was found to be X Y 4 75 4 62 with a total cost of 195367 In this case using the exact center of gravity coordinates as compared with the approximate ones reduced costs by only 195 966 195 367 195 966 100 03 c Additional costs can be included in the analysis although not necessarily in the model The COG model can evaluate the variable costs of location Other costs are compared with these 4 We begin by developing a 3dimensional transportation problem The cost matrix is developed in the same manner as that in text Fig 1311 The initial throughput of W1 and W2 is found by assuming that an equal amount of the customer demand flows through each warehouse The cell cost for W1C1 would be 1002000002072000002 2 4 0 32 2 4 92 167 In fact the cell costs are identical to those in text Fig 1311 except that there is no fixed cost element Using the transportation method of linear programming eg the TRANLP module in LOGWARE the cell cost and solution matrix for iteration 1 is shown in Fig 131 The solution shows that only W2 remains and the solution process can be terminated A summary of the costs is shown in Table 131 The total cost is 2213714 and the product is produced in plant P2 and stocked in warehouse W2 No further iterations are needed since only one warehouse is used and no further dropping of warehouses is possible TABLE 131 Summary Information for Solution to Problem 4 Whse 1 Whse 2 Whse throughput 0 200000 Costs Transportation Inbound 0 400000 Outbound 0 300000 Inventory 0 513714 Warehousing 0 200000 Fixed 0 0 Production 0 800000 Total 2213714 Iteration 2 A repeat of the iteration 1 solution Stop iterating Warehouses Customers W1 W2 C1 C2 C3 Capacity Plants P1 4 0 9 0 99a 99 99 60000 P2 8 0 6 200000 99 99 99 999999 Whses W1 0 60000 99a 92 0 82 0 102 0 60000 W2 99 0 799999 62 50000 52 100000 62 50000 999999 Capacity Reqmts 60000 999999 50000 100000 50000 a High rate of 99unit for an inadmissible cell FIGURE 131 Cell Cost and Solution Matrix for Iteration 1 of Problem 4 5 We begin by forming the cell cost matrix of a 3dimensional transportation problem as shown in Figure 132 It is similar to the text Figure 131 except that the capacity for warehouse 1 is set at 75000 Solving the problem by means of the transportation method shows the solution given in Figure 132 168 Warehouses Customers W1 W2 C1 C2 C3 Capacity Plants P1 4 60000 9 100 100 100 60000 P2 8 40000 6 100000 100 100 100 999999 Whses W1 0 899999 100 97 87 100000 107 999999 W2 100 0 82 50000 72 82 50000 100000 Capacity Reqmts 999999 100000 50000 100000 50000 FIGURE 132 Cell Cost and Solution Matrix for Iteration 1 of Problem 5 Given the solution from iteration 1 the perunit inventory and fixed costs are revised Inventory W1 100 3 100 000 100000 un 16 0 7 its unit W2 100 3 100 000 100 000 16 0 7 units unit Fixed W1 100000100000 100unit W2 400000100000 400unit Adding outbound transportation and warehouse handling to perunit inventory and fixed costs gives the following cell costs C1 C2 C3 W1 102 92 112 W2 102 92 102 Revising the warehousecustomer cell costs and solving gives the same warehouse throughputs so cell costs will no longer change A stopping points is reached The solution is the same as that in Figure 132 A summary of the costs is 169 Compared with the costs from the text example the cost difference is 3092456 2613714 478742 This is the penalty for restricting a warehouse with economic benefit to the network 6 Prepare a matrix for a 3dimensional transportation problem like that in text Figure 13 11 except that the perunit cell costs for warehouse 2 to customer are reduced by 1unit to reflect the reduction in that warehouses fixed costs That is 200000200000 1unit instead of 400000200000 2unit The matrix setup and first iteration solution are shown in Figure 133 FIGURE 132 Cell Cost and Solution Matrix for Iteration 1 of Problem 6 Cost type Warehouse 1 100000 cwt Warehouse 2 100000 cwt Production 600004 240000 400004 160000 1000004 400000 Inbound transportation 600000 0 400004 160000 1000002 200000 Outbound transportation 1000003 300000 500002 100000 500002 100000 Fixed 100000 400000 Inventory carrying 10010000007 316228 10010000007 316228 Handling 1000002 200000 1000001 100000 Subtotal 1476228 1616228 Total 3092456 Warehouses Customers W1 W2 C1 C2 C3 Plant warehouse capacities P1 4a 9 99b 99 99 60000 P2 8 6 200000 99 99 99 999999c W1 0 60000 99 97d 87 107 60000 W2 99b 0 799999 72e 50000 62 100000 72 50000 999999c Warehouse capacity customer demand 60000 999999c 50000 100000 50000 a Production plus inbound transport rates that is 4 0 4 b Used to represent an infinitely high cost c Used to represent unlimited capacity d Inventory carrying warehousing outbound transportation and fixed rates that is 32 2 4 05 97 e 32 1 2 1 72 Plants Warehouses 170 The results show that one warehouse is to be used Further computations are not needed as further warehouse consolidation is not possible The total network costs are the same as those in the text example minus the 200000 reduction in fixed costs for a total coat of 2413714 8 This problem requires us to rework the dynamic programming solution to the example problem given in the text The only change is that the cost of moving from one location to another is now 300000 instead of 100000 We begin with the last year and determine the best action based on the highest net profits The action will be to move M or to stay S For example given the discounted moving cost of 3000001 0204 144676 we evaluate each course of action assuming that we are in location alternative A at the end of year 4 From the location profits of text Table 136 we generate the following table for location A Alter Location Moving Net native x profit cost profit A 1336000 0 1336000 B 1398200 144676 1253524 P5A max C 1457600 144676 1312924 D 1486600 144676 1341924 E 1526000 144676 1381324 The best action in the beginning of the 5th year if we are already in location A is to move to location E This is an entry in Table 132 Once each of the five alternatives is evaluated for the 5th year then the 4th year alternatives are evaluated The moving cost is 3000001 023 173611 We now include the profits for the subsequent years in our calculations After Table 132 is completed we search the first column for the highest cumulative profit This is initially to locate in location D and remain there throughout the subsequent years 9 a Using PMED software in LOGWARE and the PMED02DAT database solve for the number of locations from 1 to 9 The best locations for each number of sites are given in the table below 171 TABLE 132 LocationRelocation Strategies Over a FiveYear Planning Horizon with Cumulative Profits Shown from Year j to Year 5 for Problem 8 Warehouse location alter natives x P1x Stra tegya P2x Stra tegya P3x Stra tegya P4x Stra tegya P5x Stra tegya A 3557767 SA 3363767 SA 3007667 MD 2268289 MD 1381324 ME B 3556167 SB 3379667 SB 3007667 MD 2268289 MD 1398000 SB C 3673200 SC 3500900 SC 3156200 SC 2319800 SC 1457600 SC D 3720300 SD 3553600 SD 3216000 SD 2459900 SD 1486600 SD E 3534100 SE 3374700 SE 3071300 SE 2355800 SE 1526000 SE a Strategy symbol refers to staying S in the designated location or moving M to a new location as indicated b Arrows indicate maximum profit location plan when warehouse is initially located at D 1st 2nd 3rd 4th 5th Year from present date j b 172 Number of sites Total cost 1 11694821 2 7634242 3 7596604 4 8381775 5 9455339 6 11536669 7 13909997 8 16581348 9 20000338 The optimal number of sites is 3 and they are to be located at Cincinnati Phoenix and Denver b The optimal cost for four sites is 8381775 as found in part a The company operates the same sites as found in the optimal solution for four sites Therefore the cost savings comes from a reassignment of customers to the sites The savings is 35000000 8381775 26618225 without any major investment All of this may not be recovered since there may be other operating costs included that are not directly associated with location The savings between the optimal 4 sites and the optimal 3 sites is 8381775 7596604 785171 There is also the reclamation of the salvage value of two incinerators close Chicago and Atlanta and the construction of a new one Cincinnati If the real estate recovery cost exceeds the new construction cost that would add to the savings otherwise a ROI estimation is needed to see there is an adequate return from the savings on the net investment Reallocation of customers among the existing incinerators is certainly attractive and the reduction of the number of incinerators from 4 to 3 is also attractive as long as there is no net investment required c Increase the annual volume for Los Angeles and Seattle markets by a factor of 10 and resolve the problem as in part a Selected results are as follows Number of sites Total cost 2 10506286 3 9831846 4 9810216 5 10595387 The optimal number of sites is four An additional site at Seattle is needed compared with the three locations found in part a 173 10 a We can apply Huffs model of retail gravitation to this problem The solution table Table 133 can be developed Summarizing branch A can be expected to attract 1173511735 11765 499 of the customers and branch B should attract the remaining 501 TABLE 133 Estimate of the Number of Customers Attracted to Each Branch Bank Time to ja Tij 2 S T j ij 2 P S T S T ij j ij j ij j 2 2 E P C ij ij i Customer i A B A B A B A B A B 1 028 072 008 052 125 14 090 010 900 100 2 010 050 001 025 1000 28 097 003 1940 60 3 028 045 008 020 125 35 078 022 3120 880 4 040 020 016 004 63 175 026 074 1820 5180 5 020 040 004 016 250 44 085 015 850 150 6 050 050 025 025 40 28 059 041 885 625 7 045 028 020 008 50 88 036 064 1440 2560 8 057 020 032 004 31 175 015 085 300 1700 9 067 054 045 029 22 24 048 052 480 520 11735 11765 a Time X X Y Y i i 2 2 50 b The economic analysis of site A would be Revenue 100No of customers 1173500 Operating expenses 300000 Profit 873000 Return on investment 873000750000 1164 The ROI seems sufficiently high so that the branch should be constructed c The size of a branch and its proximity to customers may be too simple to explain the market share of each The nature of the services offered the accessibility of the site and the reputation of the bank may be just as important in estimating patronage What will the countermoves be of the competing branch Adding another branch and locating near branch A could substantially reduce its market share How likely is this to happen Are the customer numbers stable Will a third or fourth bank be locating branches in the region Can we expect that customers will drive such long distances to seek banking services 174 11 This problem can be solved as an integer linear programming problem similar to the Ohio Trust Company example in the text First we create a table showing the counties that are adjacent to each county That is Counties under Adjacent counties consideration by number 1 Williams 256 2 Fulton 136 3 Lucas 2467 4 Ottawa 378 5 Defiance 16910 6 Henry 123571011 7 Wood 3468101112 8 Sandusky 4712 9 Paulding 51013 10 Putnam 5679111314 11 Hancock 671012141516 12 Seneca 781116 13 Van Wert 910141718 14 Allen 1011131518 15 Hardin 111416181921 16 Wyandot 11121519 17 Mercer 1318 18 Auglaize 131415172021 19 Marion 1516 20 Shelby 1821 21 Logan 151820 Next according to the problem formulation given in the Ohio Trust Company example we can build the matrix as given in the prepared database called ILP03DAT The problem formulation is shown in Table 134 This matrix can be solved by the integer programming module MIPROG in LOGWARE for solution Note that all coefficients are 1s or 0s A coefficient of 1 is given to each county and its adjacent counties The sum of all constraints must be 1 or greater The Xs take on the values of 0 or 1 An X of 1 means that the branch is located in the county Solving this problem using the integerprogramming module in LOGWARE shows that a minimum of 5 principal places of business are needed They should be located in Henry Wood Putnam Hardin and Auglaize counties 175 TABLE 134 Coefficient Matrix Setup for Problem 11 X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 RHS Obj Fun 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Constraints Williams 1 1 1 1 1 Fulton 1 1 1 1 1 Lucas 1 1 1 1 1 1 Ottawa 1 1 1 1 1 Defiance 1 1 1 1 1 1 Henry 1 1 1 1 1 1 1 1 1 Wood 1 1 1 1 1 1 1 1 1 Sandusky 1 1 1 1 1 Paulding 1 1 1 1 1 Putnam 1 1 1 1 1 1 1 1 1 Hancock 1 1 1 1 1 1 1 1 1 Seneca 1 1 1 1 1 1 Van Wert 1 1 1 1 1 1 1 Allen 1 1 1 1 1 1 1 Hardin 1 1 1 1 1 1 1 1 Wyandot 1 1 1 1 1 1 Mercer 1 1 1 1 Auglaize 1 1 1 1 1 1 1 1 Marion 1 1 1 1 Shelby 1 1 1 1 Logan 1 1 1 1 1 176 12 This problem can be solved with the aid of the PMED program in LOGWARE A database has been prepared for it called PMED04DAT The database shows the fixed costs for a single site When other numbers are to be evaluated the FOC must be recalculated and entered into the database The scaling factor is set at 1 for this problem The fixed operating cost must be calculated for each possible number of locations Using PMED in LOGWARE gives the following results based on recalculated FOC values an estimation of vendor to laboratories transportation cost and an enumerative search No of loca tions Sites Volume lb Outbound cost Inbound distance mi Inbound cost FOC Total cost 1 Chicago 680000 28350000 0 0 5000000 33350000 2 Cleveland 515000 310 31930001 35355342 L Angeles 165000 1750 5775000 3535534 Total 680000 15083500 8968000 7071068 31122568 3 New York 235000 713 3351100 2886751 Chicago 280000 0 0 2886751 L Angeles 165000 1750 5775000 2886751 Total 680000 10641999 9126100 8660253 28428352 4 New York 200000 713 2852000 2500000 Atlanta 145000 585 1696500 2500000 Chicago 170000 0 0 2500000 L Angeles 165000 1750 5775000 2500000 Total 680000 7515250 10313500 10000000 27828750 5 New York 200000 713 2852000 2236067 Atlanta 100000 585 1170000 2236067 Chicago 170000 0 0 2236067 Dallas 60000 790 948000 2236067 L Angeles 150000 1750 5250000 2236067 Total 680000 6079249 10220000 11180335 27479584 6 New York 200000 713 2852000 2041241 Atlanta 65000 585 760500 2041241 Miami 35000 1180 826000 2041241 Chicago 170000 0 0 2041241 Dallas 60000 790 948000 2041241 L Angeles 150000 1750 5250000 2041241 Total 680000 5029250 10636500 12247446 27913196 1 515000x310x0023193000 2 3 535534 5 000000 2 2 The PMED program is used to find the best combination of sites for a particular number of sites to be found The fixed cost must be adjusted for the number of sites being evaluated It should be recognized that the model handles only the outbound leg of the network sites to serve laboratories The vendor to site transportation cost is included externally as shown in the previous table Calculating the distances between vendor and the selected sites easily can be done by using the MILES module in LOGWARE with a scaling factor of 1 Then inbound transport costs are a product of site volume distance and the inbound rate 177 Searching from 1 to N sites shows that outbound transportation costs decrease while inbound and fixed costs increase with increasing numbers of sites Initially total cost declines until 5 sites are reached after which total cost increases We select 5 sites as economically the best number Their customer assignments are Location number Assignments Volume Customers 1 New York 200000 1 2 3 and 12 2 Atlanta 100000 4 and 5 3 Chicago 170000 6 7 8 9 10 and 11 4 Dallas 60000 13 14 and 16 5 Los Angeles 150000 15 17 18 19 and 20 A map of the solution is as follows The total cost for 5 sites is 27479584 13 After changing the fixed costs in the problem setup matrix the following solution is found 178 OPTIMAL SOLUTION Variable Value Rate Cost Variable label X1 0000 80000 0000 P1S1W1C1 X2 100000000 70000 700000000 P1S1W1C2 X3 500000000 90000 4500000000 P1S1W1C3 X4 0000 110000 0000 P1S1W2C1 X5 0000 100000 0000 P1S1W2C2 X6 0000 110000 0000 P1S1W2C3 X7 500000000 120000 6000000000 P1S2W1C1 X8 900000000 110000 9900000000 P1S2W1C2 X9 0000 130000 0000 P1S2W1C3 X10 0000 80000 0000 P1S2W2C1 X11 0000 70000 0000 P1S2W2C2 X12 0000 80000 0000 P1S2W2C3 X13 0000 60000 0000 P2S1W1C1 X14 300000000 50000 1500000000 P2S1W1C2 X15 200000000 70000 1400000000 P2S1W1C3 X16 0000 110000 0000 P2S1W2C1 X17 0000 100000 0000 P2S1W2C2 X18 0000 110000 0000 P2S1W2C3 X19 200000000 90000 1800000000 P2S2W1C1 X20 0000 80000 0000 P2S2W1C2 X21 400000000 100000 4000000000 P2S2W1C3 X22 0000 70000 0000 P2S2W2C1 X23 0000 60000 0000 P2S2W2C2 X24 0000 70000 0000 P2S2W2C3 X25 10000 1000000000 1000000000 zW1 X26 0000 5000000000 0000 zW2 X27 10000 1400000000 1400000000 yW1C1 X28 10000 2600000000 2600000000 yW1C2 X29 10000 2200000000 2200000000 yW1C3 X30 0000 700000000 0000 yW2C1 X31 0000 1300000000 0000 yW2C2 X32 0000 1100000000 0000 yW2C3 Objective function value 370000000 Note that warehouse 2 is no longer used in favor of all products flowing through warehouse 1 14 a The demand of customer 1 for product 1 increases to 100000 cwt In the problem matrix of ILP02DAT the following cell values are changed Cell From To Dem P1W1C1 yW1C1 50000 100000 Dem P1W2C1 yW2C1 50000 100000 CapW1 yW1C1 70000 120000 CapW2 yW2C1 70000 120000 Obj coef yW2C1 140000 240000 Obj coef yW2C1 70000 120000 The result shows that warehouse 2 is still the only warehouse used and the products are sourced from plant 2 However the costs have increased to 3500000 179 b Using the ILP02DAT file in MIPROG of LOGWARE the following changes are made to the following cells Obj Coef P2S2W1C1 9 12 Obj Coef P2S2W1C2 8 11 Obj Coef P2S2W1C3 10 13 Obj Coef P2S2W2C1 7 10 Obj Coef P2S2W2C2 6 9 Obj Coef P2S2W2C1 7 10 The result shows that warehouse 2 is still the only warehouse used and the products are sourced from plant 2 However the costs have increased to 3380000 c Using the ILP02DAT file in MIPROG of LOGWARE the changes are made to the following cells Obj Coef yW2C1 70000a 280000 Obj Coef yW2C2 130000 520000 Obj Coef yW2C3 110000 440000 a The sum of demand for the same customer for all products multiplied by the handling rate ie 50000 20000 4cwt 280000 The solution for product 1 shows that 50000 cwt flows from plant 1 through warehouse 1 and on to customer 1 The remainder flows from plant 2 through warehouse 2 and on to customers 2 and 3 For product 2 plant 1 supplies warehouse 1 and customer 1 with 20000 cwt The remaining 90000 cwt flows from plant 2 through warehouse 2 to customers 2 and 3 The total cost is 3920000 d Making some slight revisions in file ILP02DAT can adjust the capacities on plant 1 The cell changes to make are Cell From To CapP1S1 RHS 60000 150000 CapP1S2 RHS 999999 90000 Both warehouses are now used for both products The solution can be summarized as Cell From To Cell From To 180 Product Plant Warehouse Customer 1 1 50000 cwt 1 50000 1 50000 cwt 1 1 60000 2 60000 2 60000 1 2 40000 2 40000 2 40000 1 2 50000 2 50000 3 50000 2 1 20000 1 20000 1 20000 2 2 30000 2 30000 2 30000 2 2 60000 2 60000 3 60000 The total cost is 3270000 e Again revising the ILP02DAT file by changing cells Obj coef P2S1W2C3 and Obj Coef P2S2W2C3 to have a very high cost 999 these cells are locked out of consideration The solution is the same as the text example except that the customers both products are serve from warehouse 1 The total cost is 3340000 15 This problem follows the form of the Ohio Trust Company example in the text First identify the zones that are within 30 minutes of any particular zone That is Zone no Zones within 30 minutes 1 12478910 2 12347910 3 23458910 4 12345610 5 345678 6 45678 7 12567910 8 13568910 9 12378910 10 123478910 Using the MIPROG module in LOGWARE the following matrix can be defined 181 The solution from MIPROG is OPTIMAL SOLUTION Variable Value Rate Cost Variable label X1 10000 10000 10000 1 X2 0000 10000 0000 2 X3 0000 10000 0000 3 X4 10000 10000 10000 4 X5 0000 10000 0000 5 X6 0000 10000 0000 6 X7 0000 10000 0000 7 X8 0000 10000 0000 8 X9 0000 10000 0000 9 X10 0000 10000 0000 10 Objective function value 200 The optimal solution is to place claims adjuster stations in zones 1 and 4 16 This is a location problem where the dominant location factor is transportation cost is and this cost is determined from optimizing the multistop routes originating at the material yard The ROUTER module in LOGWARE can be used to generate these routes for each yard location A database file for this problem RTR13DAT has been prepared Solving this problem requires balancing the cost of transporting the merchandise to the customers with the operating cost of the material yards at various locations Optimizing the routing from the current material yard gives the route design shown in Figure 133 A minimum of 9 trucks are required to meet all constraints on the problem The total daily cost for this location is the route cost vehicle costs yard operating cost or P450087 9 x P200 P350 P665087 FIGURE 133 Optimized Routing from Current Material Yard Location SUMMARY REPORT 182 TIMEDISTANCECOST INFORMATION Route Run Stop Brk Stem Route time time time time time Start Return No of Route Route no hr hr hr hr hr time time stops distMi cost 1 99 64 25 10 50 0800AM 0556PM 4 205 60245 2 34 16 18 0 7 0800AM 1125AM 3 52 22025 3 100 65 25 10 31 0800AM 0558PM 4 207 60859 4 74 41 23 10 22 0800AM 0324PM 3 130 41605 5 96 64 23 10 53 0800AM 0537PM 3 204 59964 6 94 58 26 10 43 0800AM 0522PM 4 185 55296 7 96 67 19 10 62 0800AM 0534PM 2 214 62579 8 96 59 27 10 28 0800AM 0535PM 4 189 56348 9 59 28 22 10 15 0800AM 0156PM 3 89 31165 Total 748 461 207 80 311 30 1476 450087 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle no typ capcty weight weight util capcty cube cube util description 1 1 1000 925 0 925 9999 0 0 0 Truck 1 2 1 1000 625 0 625 9999 0 0 0 Truck 1 3 1 1000 900 0 900 9999 0 0 0 Truck 1 4 1 1000 950 0 950 9999 0 0 0 Truck 1 5 1 1000 900 0 900 9999 0 0 0 Truck 1 6 1 1000 950 0 950 9999 0 0 0 Truck 1 7 1 1000 825 0 825 9999 0 0 0 Truck 1 8 1 1000 1000 0 1000 9999 0 0 0 Truck 1 9 1 1000 850 0 850 9999 0 0 0 Truck 1 Total 9000 7925 0 881 89991 0 0 0 Substituting yard location A for the current yard location and solving for the route design in ROUTER yields Figure 134 No routes can be put endtoend so that one truck can be used instead of two so the minimum number of trucks remains at nine The total daily cost for this location is 387202 9 x 200 480 P6152092 Yard Location A FIGURE 134 Route Design for Yard Location A 183 SUMMARY REPORT TIMEDISTANCECOST INFORMATION Route Run Stop Brk Stem Route time time time time time Start Return No of Route Route no hr hr hr hr hr time time stops distMi cost 1 98 70 18 10 54 0800AM 0547PM 4 224 64988 2 84 51 23 10 38 0800AM 0426PM 3 163 49840 3 84 50 24 10 39 0800AM 0426PM 3 161 49157 4 86 48 29 10 22 0800AM 0437PM 5 152 47004 5 72 40 22 10 26 0800AM 0312PM 3 128 41015 6 63 29 24 10 16 0800AM 0218PM 3 93 32127 7 53 21 22 10 16 0800AM 0115PM 3 66 25422 8 89 57 21 10 37 0800AM 0451PM 3 183 54868 9 51 17 23 10 6 0800AM 0103PM 3 55 22782 Total 680 383 207 90 255 30 1225 387202 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle no typ capcty weight weight util capcty cube cube util description 1 1 1000 475 0 475 9999 0 0 0 Truck 1 2 1 1000 950 0 950 9999 0 0 0 Truck 1 3 1 1000 1000 0 1000 9999 0 0 0 Truck 1 4 1 1000 975 0 975 9999 0 0 0 Truck 1 5 1 1000 875 0 875 9999 0 0 0 Truck 1 6 1 1000 1000 0 1000 9999 0 0 0 Truck 1 7 1 1000 875 0 875 9999 0 0 0 Truck 1 8 1 1000 825 0 825 9999 0 0 0 Truck 1 9 1 1000 950 0 950 9999 0 0 0 Truck 1 Total 9000 7925 0 881 89991 0 0 0 Continuing with location B nine trucks are required and the total daily cost for the route design in Figure 3 is 337042 9 x 200 450 P562042 Yard Location B FIGURE 3 Design for Yard Location B SUMMARY REPORT 184 TIMEDISTANCECOST INFORMATION Route Run Stop Brk Stem Route time time time time time Start Return No of Route Route no hr hr hr hr hr time time stops distMi cost 1 71 37 24 10 23 0800AM 0306PM 4 120 38930 2 65 31 23 10 18 0800AM 0227PM 3 100 34053 3 79 46 24 10 30 0800AM 0356PM 4 146 45497 4 59 28 22 10 14 0800AM 0156PM 3 89 31184 5 59 26 23 10 11 0800AM 0155PM 3 84 30041 6 67 36 21 10 23 0800AM 0241PM 3 114 37539 7 58 25 23 10 21 0800AM 0150PM 3 80 29022 8 61 33 19 10 28 0800AM 0207PM 2 104 35046 9 97 58 28 10 30 0800AM 0540PM 5 187 55731 Total 617 320 207 90 198 30 1024 337042 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle no typ capcty weight weight util capcty cube cube util description 1 1 1000 825 0 825 9999 0 0 0 Truck 1 2 1 1000 950 0 950 9999 0 0 0 Truck 1 3 1 1000 825 0 825 9999 0 0 0 Truck 1 4 1 1000 850 0 850 9999 0 0 0 Truck 1 5 1 1000 925 0 925 9999 0 0 0 Truck 1 6 1 1000 825 0 825 9999 0 0 0 Truck 1 7 1 1000 950 0 950 9999 0 0 0 Truck 1 8 1 1000 825 0 825 9999 0 0 0 Truck 1 9 1 1000 950 0 950 9999 0 0 0 Truck 1 Total 9000 7925 0 881 89991 0 0 0 Finally the route design from location C is shown in Figure 4 Although 10 routes are in the design two of these can be dovetailed so that only nine trucks are needed The total daily cost is 447999 9 x 200 420 P669999 Yard Location C FIGURE 4 Design for Yard Location C SUMMARY REPORT 185 TIMEDISTANCECOST INFORMATION Route Run Stop Brk Stem Route time time time time time Start Return No of Route Route no hr hr hr hr hr time time stops distMi cost 1 15 6 9 0 6 0800AM 0930AM 1 20 14091 2 67 37 20 10 19 0800AM 0241PM 2 120 38921 3 80 41 29 10 15 0800AM 0400PM 5 132 42020 4 94 56 28 10 34 0800AM 0525PM 5 180 54048 5 75 42 23 10 27 0800AM 0329PM 3 134 42569 6 77 42 24 10 19 0800AM 0339PM 3 136 42910 7 70 37 23 10 28 0800AM 0259PM 3 117 38304 8 73 40 23 10 35 0800AM 0318PM 3 127 40853 9 96 72 14 10 56 0800AM 0535PM 2 229 66340 10 98 74 14 10 63 0800AM 0547PM 3 236 67942 Total 745 447 207 90 303 30 1432 447999 VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup Cube Vehicle no typ capcty weight weight util capcty cube cube util description 1 1 1000 375 0 375 9999 0 0 0 Truck 1 2 1 1000 875 0 875 9999 0 0 0 Truck 1 3 1 1000 975 0 975 9999 0 0 0 Truck 1 4 1 1000 925 0 925 9999 0 0 0 Truck 1 5 1 1000 925 0 925 9999 0 0 0 Truck 1 6 1 1000 1000 0 1000 9999 0 0 0 Truck 1 7 1 1000 950 0 950 9999 0 0 0 Truck 1 8 1 1000 950 0 950 9999 0 0 0 Truck 1 9 1 1000 550 0 550 9999 0 0 0 Truck 1 10 1 1000 400 0 400 9999 0 0 0 Truck 1 Total 10000 7925 0 793 99990 0 0 0 From an economic analysis it appears that yard location B is the best choice 186 SUPERIOR MEDICAL EQUIPMENT COMPANY Teaching Note Strategy The purpose of the Superior Medical Equipment Company case study is to encourage students to apply the centerofgravity methodology to a problem involving a single warehouse location Although the methodology is somewhat elementary it can be useful in providing some first approximations to good warehouse locations It allows students to evaluate the financial implications of alternative network designs and it is intended to be solved with the aid of the COG module provided in the LOGWARE software Answers to Questions 1 Based on information for the current year is Kansas City the best location for a warehouse If not what are the coordinates for a better location What cost improvement can be expected from the new location When a single warehouse is to be located the primary location costs are transportation both inbound to the warehouse and outbound from it and the warehouse lease which varies with the location The current location serves as a benchmark against which the costs for other locations can be compared That is based on information given in the case the total relevant cost for the Kansas City location is Inbound transportation 2162535 Outbound transportation 4819569 Lease 275sq ft 200000 sq ft 550000 Total relevant cost 7532104 Using the COG module with inbound transport rates from Phoenix set at 16731163 0014cwtmile and from Monterrey set at 9401188 0008cwtmile and the outbound transport rate from the unknown warehouse location set at 00235cwtmile the coordinates for the best location are X 761 and Y 451 or approximately Oklahoma City Total transportation cost for this location would be 6754082 The total relevant cost would be Transportation 6754082 Lease 325sq ft 200000 sq ft 650000 Total 7404082 The annual cost savings can be projected as 7532104 7404082 128022 2 In five years management expects that Seattle Los Angeles and Denver markets to grow by 5 but the remaining markets to decline by 10 Transport costs are expected to be unchanged Phoenix output will increase by 5 and Monterreys output will decrease by 10 Would your decision about the location of the warehouse change If so how 187 A new benchmark for the 5th year can be computed from the data given in Tables 1 and 2 After adjusting the plant and market volumes according to the changes indicated the 5thyear benchmark costs can be computed as follows Volume Rate Transport Point cwt cwt cost 1 64575 1673 1080340 2 108540 940 1020276 3 17850 3369 601367 4 33600 3043 1022448 5 13125 2575 337969 6 8550 1832 156636 7 26550 2524 670122 8 18900 1966 371574 9 37170 2652 985748 10 7740 2617 202556 11 9630 2798 269447 Totals 346230 6718483 The total 5thyear benchmark relevant cost would be Transportation 6718483 Lease 275sq ft 200000 sq ft 550000 Total 7268483 Optimizing the location with the 5thyear data gives a location at X 705 and Y 452 The relevant costs for this location are Transportation 6464206 Lease 325sq ft 200000 sq ft 650000 Total 7114206 The annualized cost savings would be 7268483 7114206 154277 The average annual cost savings are 128022 1542772 141150 The simple annual return on investment moving cost can be computed as ROI 141 300 150 000 100 471 Management must now judge whether 471 annual return is worth the risk of changing warehouse locations 3 If by year 5 increases are expected of 25 in warehouse outbound transport rates and 15 in warehouse inbound rates would your decision change about the warehouse location 188 It is assumed here that the 5thyear demand level applies A revised 5thyear benchmark can be recomputed by applying the cost growth factors to the overall 5thyear transport costs That is Outbound 1254617867 5772334 Lease 550000 Total 8738042 Running COG shows that the minimum transport cost location would be at coordinates X 720 and Y 462 which is near the previous location in question 2 The shift in location is minimal The cost for this location is Transportation 7939545 Lease 325sq ft 200000 sq ft 650000 Total 8589545 The annualized cost savings would be 8738042 8589545 148497 It can be concluded that 1 Location is similar to the optimized 5thyear location 2 The increase in possible cost savings further encourages relocation from Kansas City and toward a site near Oklahoma City OK 4 If the centerofgravity method is used to analyze the data what are its benefits and limitations for locating a warehouse The centerofgravity method locates a facility based on transportation costs alone This is reasonable when only one facility is being located and the general location for it is being sought Such costs as inventory carrying production and warehouse fixed are not included but they are not particularly relevant to the problem However costs such as warehouse storage and handling and other costs that vary by the particular site are not included but may be relevant in a given situation Transportation costs are assumed linear with distance This may not be strictly true although distance may be nonlinear The obvious benefits of the method are 1 it is a fast solution methodology 2 it considers all possible locations continuous 3 it is simple to use 4 its data are readily available and 5 it gives precise locations through a coordinate system Some potential limitations are 1 coordinates need to be linear 2 transportation rates on a permile basis are constant 3 volumes are known and constant for given demand and source points and 4 locations may be suggested that are not feasible such as in lakes central cities or restricted lands Concluding Comments Inbound 115 2100616 2415708 Subtotal 8188042 189 The analysis in the case seems to suggest a move from Kansas City to a region around Oklahoma City would be advantageous A return on investment of 47 percent or higher is possible however management must now seek a particular site in the area whose choice may add or detract from this savings potential In any case the COG method has assisted in the selection of good potential locations and testing their sensitivity to changes in costs and volumes APPENDIX 1 LOGWARE COG Module Input Data for the Current Year Title SUPERIOR MEDICAL EQUIPMENT COMPANY Power factor 5 Scale factor 230 Point Xcoordinate Ycoordinate Volume Rate 1 360 390 61500 00140 2 690 100 120600 00080 3 090 910 17000 00235 4 195 420 32000 00235 5 560 610 12500 00235 6 780 360 9500 00235 7 1020 690 29500 00235 8 1130 395 21000 00235 9 1400 655 41300 00235 10 1270 780 8600 00235 11 1430 825 10700 00235 190 OHIO AUTO DRIVERS LICENSE BUREAUS Teaching Note Strategy The purpose of this case study is to introduce students to a logistics problem in a service area It also provides an application for the multiple center of gravity location methodology The MULTICOG module in the LOGWARE software can effectively be applied The module allows students to quickly evaluate alternatives as to the number of bureaus to use the bureau locations and the size of the territory that each bureau should serve The case may be assigned as a homework problem a short case study project or as a case for class discussion The later would be appropriate especially if adequate attention is given to the issues of how Dan should go about solving a problem such as this what data he needs and where to obtain it and what concerns he should have about changing the existing network design This would encourage students to think beyond the computational aspects of the problem Answers to Questions 1 Do you think there is any benefit to changing the network of license bureaus in the Cleveland area If so how should the network be configured The nature of the costs and the number of possible alternative network designs make it impractical to seek an optimal solution Therefore a possible approach to the analysis is outlined as follows First establish a benchmark against which changes to the network can be compared Much of the data for this is given in the case write up The costs can simply be applied to the size of each bureau and its associated staff The cost for residents traveling to the bureaus is not known because the bureau territories are not known However an estimate can be made of travel costs by solving the problem in MULTICOG for eight bureaus Since MULTICOG attempts to optimize bureau location this travel cost is probably understated The benchmark costs are summarized in Table 1 The location costs for the current operation are estimated to be 1355706 Second what improvements can be made on the existing eight locations Besides moving the locations of the bureaus which results in resizing the facilities and adjusting the staff numbers there are no obvious improvements to be made Therefore the benchmark remains the base for comparison Third it is now necessary to estimate the approximate number of bureaus that are needed to serve the area Since the costs for a particular network design depend on the size of each bureau which cannot be known until the problem is solved an initial assumption must be made It will be assumed that all bureaus are of the same size Hence for 5 bureaus the average number of residents in each bureaus territory would be the total number of residents divided by the number of bureaus or 6917005 138340 Rent staff salaries and utility expenses can be derived from this estimate Table 2 is developed to show the bureau size and the number of staff for 1 to 10 bureaus Table 3 extends the average costs from these estimates A reduced number of bureaus in the range of two is about right 191 Bureau Size sq ft Staff Rent Salaries Utilities Customer Travel 1 1700 4 37400 84000 6800 2 1200 4 26400 84000 4800 3 2000 5 44000 105000 8000 4 1800 4 39600 84000 7200 5 1500 4 33000 84000 6000 6 2200 5 48400 105000 8800 7 2700 5 59400 105000 10800 8 1500 5 33000 105000 6000 Totals 14600 36 321200 756000 58400 220106 Approximated by running MULTICOG at 8 bureaus TABLE 2 Average Size and Staff for Various Numbers of Bureaus 1 No of bureaus 6917001 Avg no of residents per bureau Avg bureau size sq ft Staff per bureau 1 691700 4500 10 2 345850 3000 7 3 230567 2500 6 4 172925 2000 5 5 138340 2000 5 6 115283 2000 5 7 98814 1500 4 8 86463 1500 4 9 78855 1500 4 10 69170 1500 4 TABLE 1 Benchmark Costs for the Current Network of License Bureaus 192 TABLE 3 Average Costs by Number of Bureaus No of bureaus Rent Staff salaries Utilities Resident travel Annual total cost 1 99000 210000 18000 662319 989319 2 132000 294000 24000 430922 880922 3 165000 378000 30000 354239 927239 4 176000 420000 32000 298000 926965 5 220000 525000 40000 278181 1063181 6 264000 630000 48000 249287 1191287 7 231000 588000 42000 237635 1098636 8 264000 672000 48000 220106 1204106 9 297000 756000 54000 206496 1313496 10 330000 840000 60000 198600 1428600 The cost estimates can now be refined around two bureaus A comparison with the benchmark costs and a return of the initial investment costs related to changing the network design are sought A sample analysis for two bureaus based on a design provided by MULTICOG is shown below Bureau Resi dents Size sq ft Staff Rent Staff salaries Utilities Resident travel cost 1 290200 2500 6 55000 126000 10000 166332 2 401500 3500 8 77000 168000 14000 264590 Totals 691700 6000 14 132000 294000 24000 430922 The total annual variable cost is 132000 294000 24000 430922 880922 There are onetime costs due to staff separation and equipment moves Compared with the benchmark 36 14 22 staff members will be separated for a cost of 22 8000 176000 Equipment movement costs to two bureaus would be 2 10000 20000 Total movement costs would be 176000 20000 196000 Annual variable cost savings compared with the benchmark would be 1355706 880922 474784 A simple return on investment would be ROI 474 196 784 000 100 242 Similar calculations are carried out for various numbers of bureaus These results are tabulated in Table 4 193 TABLE 4 Cost Savings and Return of Investment for Alternate Network Designs as Produced by MULTICOG No of bureaus Total size sq ft Total staff Annual variable cost Moving cost Savings Return on invest ment 1 4500 10 989319 218000 366387 168 2 6000 14 880922 196000 474784 242 3 7500 18 927239 176000 428467 243 4 8500 21 960965 160000 394741 246 5 9500 24 1029181 146000 326525 223 6 11500 29 1157287 106000 198419 187 7 12000 31 1200635 110000 155071 141 8 13000 34 1272106 96000 83600 87 Benchmark 14600 36 1355706 The maximal annual savings occurs with a network containing two bureaus However the maximal return on investment occurs with four bureaus ROI is selected as the appropriate measure on which to base this economic decision The details for a design with four bureaus are given in Table 5 The design is shown pictorially in Figure 1 of this note TABLE 5 Design Details for a Network with Four Bureaus Bureau Column grid coordi nate Row grid coordi nate Residents Grid box number assignment 1 336 274 218200 12345678910111213 14151617181920212223 24252627282930313233 2 700 300 168700 34353637383940414344 45464750515253 3 974 566 195000 42484954555660616263 67686970747576778182 8384 4 1000 200 109800 57585964656671727378 7980 2 Do you think Dan Rogers study approach is sound Overall Dan can be praised for the simplicity of the methodology that he has chosen The centerofgravity approach is appropriate in this problem since there are no capacity limitations on the facilities locations across a continuous space are desired and transportation cost except for some fixed costs is the primary location variable The data requirements are relatively straightforward although they are not always easy to fulfill This would likely be a problem with any other solution approach as well The results of this methodology can only be used as a first approximation at best Several criticisms of this particular approach can be offered as follows 194 The effect of bureau location on the residents perception of service is not as well known as portrayed in the case In addition service may need to be represented by more than location Travel to the bureaus is assumed straight line However location in the area is likely to be influenced by a road network Time may be more important than distance to residents The fixed costs associated with location are not handled directly by the centerof gravity approach Residents are assumed to travel to the locations within their assigned territories They may not strictly do this Good facilities may not be available at the indicated location coordinates The analysis is particularly weak around the estimate of the resident travel cost While an exact cost is not likely to be known Dan should conduct a sensitivity analysis around this cost He may find that the design does not change a great deal over a wide range of assumed values If this is the case he can feel comfortable that his recommendation is fundamentally sound If not he should seek to find a more precise value 3 What concerns besides economic ones should Dan have before suggesting that any changes be made to the network A quantitative approach to location will rarely give the precise locations to be implemented Rather it provides a starting point for further analysis There are a number of other factors to be considered before the revised network design can be implemented First there are site selection factors to be taken into account such as the availability of adequate space near the location coordinates proximity to good highway linkages and reasonable neighborhood reactions to this type of operation Second there are political concerns Reducing the number of locations will result in a releasing some of the staff Dan may experience some political resistance to this Since staff is a large expense in the operation currently about 23 of the costs retention of a larger number of bureaus may be required Of course transferring staff to other governmental operations may be a way of dealing with this issue This assumes their willingness to be relocated although this is not likely to be a strong issue if relocation were to occur in the same area Third there may be difficulty in demonstrating the economics of network redesign Although others may appreciate the costs of rent salaries and utilities the cost of resident travel is subject to much interpretation Those favoring many bureaus may argue the high cost while those wanting to reduce the number of bureaus may perceive it as not very significant 195 SUPPLEMENT Sample Input Data File for MULTICOG in LOGWARE for the Ohio Auto Drivers License Bureau Case Study Title LICENSE BUREAU Number of sources 4 Number of demand points 84 Scaling factor 25 POINT XCOORDINATE YCOORDINATE VOLUME RATE 1 1 1 4100 12 2 1 2 6200 12 3 1 3 7200 12 4 1 4 10300 12 5 1 5 200 12 6 1 6 0 12 7 1 7 0 12 8 2 1 7800 12 9 2 2 8700 12 10 2 3 9400 12 11 2 4 11800 12 12 2 5 100 12 13 2 6 0 12 14 2 7 0 12 15 3 1 8100 12 16 3 2 10500 12 17 3 3 15600 12 18 3 4 10500 12 19 3 5 200 12 20 3 6 0 12 Current bureau locations Revised bureau locations Grid column number Grid row number Current bureau locations Revised bureau locations Grid column number Grid row number FIGURE 1 Four Bureau Locations and Their Territories 196 21 3 7 0 12 SUPPLEMENT Continued POINT XCOORDINATE YCOORDINATE VOLUME RATE 22 4 1 10700 12 23 4 2 12800 12 24 4 3 13800 12 25 4 4 15600 12 26 4 5 400 12 27 4 6 0 12 28 4 7 0 12 29 5 1 11500 12 30 5 2 13900 12 31 5 3 14500 12 32 5 4 13700 12 33 5 5 600 12 34 5 6 0 12 35 5 7 0 12 36 6 1 9300 12 37 6 2 14900 12 38 6 3 13700 12 39 6 4 10200 12 40 6 5 1200 12 41 6 6 0 12 42 6 7 0 12 43 7 1 10100 12 44 7 2 12600 12 45 7 3 16700 12 46 7 4 15800 12 47 7 5 12400 12 48 7 6 2600 12 49 7 7 0 12 50 8 1 8800 12 51 8 2 13700 12 52 8 3 15200 12 53 8 4 14100 12 54 8 5 10800 12 55 8 6 17200 12 56 8 7 500 12 57 9 1 5300 12 58 9 2 16700 12 59 9 3 13800 12 60 9 4 11900 12 61 9 5 13500 12 62 9 6 18600 12 63 9 7 12000 12 64 10 1 5100 12 65 10 2 17400 12 66 10 3 10300 12 67 10 4 9800 12 68 10 5 10300 12 69 10 6 15500 12 70 10 7 11700 12 71 11 1 7700 12 72 11 2 9200 12 73 11 3 7500 12 74 11 4 8500 12 75 11 5 7800 12 76 11 6 9900 12 77 11 7 8700 12 78 12 1 4300 12 197 79 12 2 6700 12 80 12 3 5800 12 SUPPLEMENT Continued POINT XCOORDINATE YCOORDINATE VOLUME RATE 81 12 4 6800 12 82 12 5 5400 12 83 12 6 7100 12 84 12 7 6400 12 198 SOUTHERN BREWERY Teaching Note Strategy The purpose of this case study is to provide students with the opportunity to design a distribution network where plant location is at issue They first should identify the major costs and alternatives that are important to such a design problem Second they should be encouraged to apply the transportation method of linear programming to assist in the analysis of alternatives using the TRANLP module in LOGWARE Finally they should consider factors other than those in the analysis that might alter the course of their recommendation and be sensitive to the limitations and benefits of linear programming as a solution methodology Answers to Questions 1 If you were Carolyn Carter would you agree with the proposal to build the new brewery If you do what plan for distribution would you suggest If growth is uniform over the next five years Southern can expect that demand for its products will exceed the currently available plant capacity That is demand is increasing at the rate of 0 595000 403000 5 38400 barrels per year Thus the current annual capacity of 500000 barrels will be used up in 500000 403000 38400 25 years A major concern is whether it would be profitable to construct the new plant Rough estimates of its profitability can be made projecting profits with and without the new plant We know that 595000 500000 95000 barrels of beer would not be sold annually in the 5th year if additional capacity is not constructed This represents a potential average lost revenue of 280barrel 020 0 95000 25 2128000 The figure of 25 years assumes that the new brewery can be brought on stream at approximately the time when capacity will be used up in the existing plants From the benchmark1 costs for the current system as shown in Table 1 Southern is currently producing and distributing 403000 barrels at a total cost of 60015000 This is an average cost per barrel of 60015000 403000 14892 The overhead and sales expense is 27 percent or 280 027 7560 per barrel Total costs per barrel are 14892 7560 22452 which is about 80 percent of the sales dollar The 20 percent profit margin seems valid Therefore the benefit of serving the potentially lost demand with a new brewery can be estimated using on a simple return on investment 021 or 21 10000000 2128000 ROI If management feels that this is an adequate return for such a project Carolyn should proceed with her analysis Lets assume that she has this approval Next she may wish to explore the opportunities available by improving upon the existing distribution system without the presence of the new plant This is an improved 1 A benchmark refers to the costs of producing and distributing demand as currently allocated throughout the network 199 benchmark2 and it can be found by solving a transportationtype linear programming problem of the type shown in Table 2 The results given in the TRANLP module of LOGWARE can be summarized in Table 3 This shows that with some slight reallocation of demand among the plants production and distribution costs can be reduced by 60015000 59804000 211000 annually A comparison of the benchmark results in Table 1 and the improved benchmark results in Table 3 shows that Knoxvilles demand should be shifted from Columbia to Montgomery and 28000 barrels of Columbias market should be shifted from the Columbia plant to the Montgomery plant The Columbia plant is a highcost producer and cost savings are achieved by trading production costs for transportation costs That is production costs will be reduced by 400000 while transportation costs will be increased by 189000 Note that this 211000 savings can be realized without any capital investment TABLE 1 Benchmark of Production and Transportation Costs 000s for Current Demand Market area Brewery of origin Demand in 000s barrels Produc tion costs Trans port costs Total costs 1 Richmond Richmond 56 7840 475 8315 2 Raleigh Richmond 31 4340 332 4672 3 Knoxville Columbia 22 3190 282 3472 4 Columbia Columbia 44 6380 306 6686 5 Atlanta Montgomery 94 12878 959 13837 6 Savannah Montgomery 13 1781 179 1960 7 Montgomery Montgomery 79 10823 550 11373 8 Tallahassee Montgomery 26 3562 355 3917 9 Jacksonville Montgomery 38 5206 577 5783 Total 403 56000 4015 60015 TABLE 2 TRANLP Problem Setup FromTo RICH MOND RA LEIGH KNOX VILLE COL UMBIA AT LANTA SAVAN NAH MONT GOMRY TALL AHAS JACK SONVL Supply RICHMD 14849 15070 15638 15254 15548 15464 15998 16430 15884 100 COLMBA 15754 15478 15781 15196 15685 15454 15793 16018 15727 100 MONTGM 15698 15335 15080 14993 14720 15080 14369 15065 15218 300 JACKVL 15213 14925 15048 14616 14880 14454 14880 14472 14268 0 Demand 56 31 22 44 94 13 79 26 38 2 An improved benchmark refers to a reallocation of current demand in an optimal way respecting plant capacity restrictions 200 TABLE 3 Improved Benchmark of Production and Transportation Costs 000s for Current Demand Market area Brewery of origin Demand in 000s barrels Produc tion costs Trans port costs Total costs 1 Richmond Richmond 56 7840 475 8315 2 Raleigh Richmond 31 4340 332 4672 3 Knoxville Montgomery 22 3014 304 3318 4 Columbia Columbia 16 2320 111 2431 4 Columbia Montgomery 28 3836 362 4198 5 Atlanta Montgomery 94 12878 959 13837 6 Savannah Montgomery 13 1781 179 1960 7 Montgomery Montgomery 79 10823 550 11373 8 Tallahassee Montgomery 26 3562 355 3917 9 Jacksonville Montgomery 38 5206 577 5783 Total 403 55600 4204 59804 Adding a plant at Jacksonville with a capacity of 100000 barrels per year provides enough capacity to satisfy demand out to the 5th year If the new plant were constructed and producing immediately total costs could be reduced from the improved benchmark by 59804000 59090000 714000 per year Compare the total costs in Tables 3 and 4 The Columbia plant would not be needed if the lowercost Jacksonville plant were on line We do not know the savings for the 5thyear demand level since not all demand can be served without the presence of the new plant Therefore a futureyear benchmark cannot be determined However we do know how the new plant should be utilized within the system see Table 5 and how demand allocation should be adjusted to accommodate it Also note that the Columbia plant is needed once again although its capacity is not required until the last onehalf year of the 5year planning horizon This suggests that the Columbia plant should not be sold but perhaps some alternate use could be made of the facility in the interim such as subcontracting beer production to a non competing company The new plant is not likely to be brought on stream immediately nor is it needed for 25 years so Carolyn might suggest a distribution plan similar to that in Table 5 A careful inspection of this plan shows that only 1 barrel of demand in the Knoxville region is assigned to Richmond Splitting demand to this extent is probably not practical and can be assigned to Columbia where there is excess capacity Costs will rise only slightly An interesting question is whether the Columbia plant through modernization could be made as efficient as the new brewery and what the implications for distribution might be We know that this could potentially save 145 135 10 per barrel in production costs At a 100000barrel capacity this is 1000000 in cost savings If the modernization were to cost no more than 5000000 this option might be attractive Of course we would need to resolve the linear programming problem with Columbias per barrel costs at 135 plus transportation costs This would tell us how and to what extent demand would be allocated to Columbia and give a more accurate basis for determining 201 the cost savings Similarly it would be interesting to explore what it means to expand the capacity of an existing brewery at a lower investment cost per barrel than the construction of a new facility TABLE 4 Production and Transportation Costs 000s for Current Demand with the Jacksonville Plant Market area Brewery of origin Demand in 000s barrels Produc tion costs Trans port costs Total costs 1 Richmond Richmond 56 7840 475 8315 2 Raleigh Richmond 31 4340 332 4672 3 Knoxville Montgomery 22 3014 304 3318 4 Columbia Montgomery 21 2877 272 3149 4 Columbia Jacksonville 23 3105 257 3362 5 Atlanta Montgomery 94 12878 959 13837 6 Savannah Jacksonville 13 1755 124 1879 7 Montgomery Montgomery 79 10823 550 11373 8 Tallahassee Jacksonville 26 3510 253 3763 9 Jacksonville Jacksonville 38 5130 292 5422 Total 403 55272 3818 59090 2 If the new brewery is not to be constructed what distribution plan would you propose to top management Table 6 shows a linear programming solution where the new plant is not brought on stream and the demand in the markets is set at the 5year level An interesting solution occurs when demand exceeds capacity The most costly demand region to serve is not assigned to any plant As can be seen in Table 6 portions of the demand in Tallahassee and Jacksonville should not be served and essentially the entire Knoxville market should not be served at all Top management may wish to adjust this plan for reasons other than economic ones 202 TABLE 5 Production and Transportation Costs 000s for Projected 5thYear Demand With the Jacksonville Plant Market area Brewery of origin Demand in 000s barrels Produc tion costs Trans port costs Total costs 1 Richmond Richmond 64 8960 543 9503 2 Raleigh Richmond 35 4900 375 5275 3 Knoxville Richmond 1 140 16 156 3 Knoxville Columbia 20 2900 256 3156 3 Knoxville Montgomery 12 1644 166 1810 4 Columbia Columbia 55 7975 383 8358 5 Atlanta Montgomery 141 19317 1438 20755 6 Savannah Columbia 20 2900 191 3091 7 Montgomery Montgomery 119 16303 828 17131 8 Tallahassee Montgomery 28 3836 382 4218 8 Tallahassee Jacksonville 24 3240 233 3473 9 Jacksonville Jacksonville 76 10260 584 10844 Total 595 82375 5395 87770 TABLE 6 Production and Transportation Costs 000s for Projected 5th Year Demand Without the Jacksonville Plant Market area Brewery of origin Expected demand in 000s barrels Served demand in 000s barrels Total costs 1 Richmond Richmond 64 64 9503 2 Raleigh Richmond 35 35 5275 3 Knoxville Richmond 33 1 156 4 Columbia Columbia 55 55 8358 5 Atlanta Montgomery 141 141 20755 6 Savannah Columbia 20 20 3091 7 Montgomery Montgomery 119 119 17131 8 Tallahassee Montgomery 52 40 6026 9 Jacksonville Columbia 76 25 3931 Total 595 500 74226 Indicates market demand is not fully served due to inadequate plant capacity 3 What additional considerations should be taken into account before reaching a final decision A number of assumptions have been implied in the analysis shown above For example Demand has been assumed to grow at a constant rate in the markets Production is assumed limited to exactly the values given without the possibility for expansion through overtime additional shifts or subcontracting 203 Perunit production and transportation costs are assumed to remain unchanged with the reallocation of demand throughout the network Customer service effects are not considered in reallocation of demand There is no change in perunit costs throughout the 5year planning horizon This case might end with a discussion of the appropriateness of using linear programming as a vehicle for analysis in a problem such as this Mentioning that linear programming does not consider such factors as fixed costs return on investment or the many subjective factors top managements intuition about location vested interests etc that are typically a part of such problems means that linear programming at best is a facilitating vehicle for analysis It does not provide the final answer 204 CHAPTER 14 THE LOGISTICS PLANNING PROCESS 3 The MILES module within the LOGWARE software is used to solve this problem It computes distance based on the great circle distance formula using longitude and latitude a The estimated road distance is 1380 miles b The estimated road distance is 830 miles c Since both latitudes are in the same hemisphere no adjustments need to be made The estimated distance is 244 miles or 244161 393 km d In this case one point is east and the other west of the Greenwich line Therefore we need to set a sign convention Lets set west longitudes as and east longitudes as Thus 220o E longitude is entered into MILES as 220 o The estimated distance is 250 miles or 250161 4025 km 4 Suppose that a certain linear grid coordinate system has been overlaid on a map of the United States The grid numbers are calibrated in miles and there is a road circuity factor of 121 Find the expected road distances between the following pairs of points Equation 141 in the text is used to approximate distances from linear coordinates The K factor in the equation is set at 121 a Lansing MI to Lubbock TX D 121 9243 1 488 6 1 6752 2 579 4 1 290 2 2 miles b El Paso TX to Atlanta GA D 121 1 6963 6249 2 7693 2 318 7 1 406 2 2 miles c Boston MA to Los Angeles CA Location X Coordinate Y Coordinate a From Lansing MI 9243 16752 To Lubbock TX 14886 25794 b From El Paso TX 16963 27693 To Atlanta GA 6249 23187 c From Boston MA 3747 13266 To Los Angeles CA 23654 27639 d From Seattle WA 26688 19008 To Portland OR 26742 20397 205 D 121 374 7 2 3654 1326 6 2 7639 2 971 2 2 miles d Seattle WA to Portland OR D 121 2 6688 2 674 2 19008 2 039 7 168 2 2 miles 5 The plot of the truck class rates is shown in Figure 141 The rates show a high degree of linearity A linear regression was found with aid of the MULREG module in LOGWARE The rate equation was determined to be R 51745 00041D The standard error of the estimate SE is 09766 The coefficient of determination r2 is 0928 The best single estimate of the rate at 500 miles is R 51745 00041500 723cwt Assuming the error around the regression line is normally distributed a 95 confidence band would give a range for the actual rate That is Y R 196SE 723 1914 where 196 is the normal deviate for the normal distribution representing 95 of the area in a twotailed distribution The range of the estimate is 532cwt Y 914cwt The r2 value of 0928 indicates that a linear rate equation explains about 93 of the variation in the data with distance Such a simple relationship seems to represent the rates quite well 206 0 2 4 6 8 10 12 14 16 18 20 0 500 1000 1500 2000 2500 3000 3500 Distance miles Class rate cwt Estimating line FIGURE 141 Plot of Truck Class Rates 6 A plot of the average inventory level versus warehouse throughput is shown in Figure 14 2 The multiple regression software in LOGWARE was used to test two equation forms The first was of the form I aTPb and the other was of the form I a bTP Both forms showed high r2 values with the exponential form being slightly higher at 09406 It was selected as the equation form to use This equation was I TP 0 704 0 83 where TP and I are both expressed in thousands of dollars We can now estimate that for an annual warehouse throughput of 50000000 the average inventory would be I 0 704 50 000 5 593939 0 83 or 5593939 Warehouse 22 has a much higher inventory turnover ratio than the average of the other warehouses This would suggest that the inventory control procedures might be different from the others One reason might be that person in control of the inventory in this warehouse attempts to keep inventories at low level demand may be high such that the inventory level has been restored to a normal level or lead times have been extended to the point where replenishment has been delayed The reason should be investigated 207 This type of inventorythroughput relationship is very useful in network planning especially warehouse location to estimate how inventory levels will change when sales are reallocated to a varying number of warehouses FIGURE 142 Plot of Inventory Levels and Warehouse Throughput for California Fruit Growers Association 0 2 4 6 8 10 12 0 20 40 60 80 100 Annual warehouse thruput Millions Average inventory level Millions Estimating line 208 USEMORE SOAP COMPANY Teaching Note The purpose of this case study is to provide students with the opportunity to evaluate and design a largescale productiondistribution network using real data and cost relationships To assist in the substantial amount of computational effort in this problem an interactive computer program WARELOCA is available in the LOGWARE collection of software modules Major Issues The text of the case suggests a number of questions that are critical to production distribution network design These reduce to three major issues namely 1 Should plant capacity be added and if so when and where 2 How many warehouses are optimal and where should they be located 3 Should the current customer service level be retained Although no change can be made in the network without potentially affecting other variables the attempt here will be to treat these questions sequentially to converge on a good network design Numerous computer runs were made to provide the basic information needed in the analysis The more meaningful runs are summarized in Appendix A to this note Tables 1 and 2 compare selected runs for both the currentyear and the futureyear time periods This information is used throughout the analysis of the major issues The Plant Expansion Issue An attempt to meet 5year growth goals using current plant capacity will cause the system having a total capacity of 1630000 cwt to be out of capacity in 17 years That is 5thyear demand 1908606 cwt Current demand 1477026 Net increase 431580 cwt Therefore the average annual growth rate is 4315805 86316 cwt So in 1630000 147702686316 17 years all available capacity will be depleted If no expansion of plant capacity occurs then 1908606 1630000 278606 cwt will potentially be lost by the 5th year Sales are 100 million on 1477 million cwt in volume for a product value of 677cwt With a profit margin of 20 the profit per cwt would be 20677cwt or 201477 13 Thus 27860613 316 million in lost sales The weighted profit loss over the fiveyear period would be 25 0 35 0 362 108myr 209 TABLE 1 CurrentYear Comparison of Network Alternatives 000s Improved Optimum Optimum Maximum Bench bench number number Relaxed Relaxed oppor Cost type mark mark of whses of whses service 1 service 2 tunity Production 30762 30678 30673 30675 30678 30673 30386 Warehouse operations 1578 1468 1608 1572 1296 1420 1529 Order processing 369 354 370 358 349 354 358 Inventory carrying 457 431 508 490 390 445 500 Transportation Inbound 2050 1802 1976 1860 1249 1178 1178 Outbound 6896 6991 6310 6365 7238 6698 6458 Total costs 42112 41725 41447 41321 41201 41043 40409 Customer service 300 mi 93 93 98 92 75 88 81 600 mi 98 98 100 100 98 100 94 No of stocking points 22 21 31 30 19 26 40 No of plants 4 4 4 4 4 4 6 Savings vs benchmark 0 387 665 791 911 1069 1703 Savings vs improved benchmark 0 0 278 404 524 682 1316 Comments Service 600 mi con 600 mi con Unlimited to match straint on straint on service bench current opt no of whses and mark warehouses warehouses plant cap 210 TABLE 2 FutureYear Comparison of Alternatives 000s Add plant Memphis Memphis No plant Add plant Memphis and opt no and opt no Cost type expansion Memphis Chicago of whses of whses Production 33965 39517 39548 39524 39522 Warehouse operations 1496 1842 1847 2028 1976 Order processing 393 462 454 470 460 Inventory carrying 431 505 497 591 573 Transportation Inbound 1647 2350 2000 2614 2426 Outbound 7230 9030 9036 8117 8222 Total costs 45164 53705 53382 53342 53179 Customer service 300 mi 98 94 95 98 92 600 mi 99 98 98 100 100 No of stocking points 20 21 20 31 30 No of plants 4 5 6 5 5 Comments Not all Service demand at bench met mark 211 Based on a simple rate of return on investment capturing this profit potential would yield 1084 27 annually on a 4000000 investment for expansion The return would increase to 90 per year with the full loss in the 5th year The potential seems great enough to justify one unit of expansion 1000000 cwt Two units of expansion probably cannot be justified since adequate capacity would be available from the first capacity unit to meet demand requirements The only benefit would be from the network design improvement The savings would be about 323000 per year in the 5th year see Table 2 comparing one additional plant with two additional plants and keeping the current number of warehouses The simple return on investment using 5thyear savings would only amount to about 8 3230001004000000 81 The next question is Where should the expansion take place at an existing plant or at one of the two proposed locations From a test of expanding any of the four existing plants or the two proposed plant locations runs 10 through 16 in Appendix A of this note it would appear that Memphis would be the lowest cost site in the 5th year with Chicago next at only an additional cost of 76000 per year compare runs 14 and 15 in Appendix A Adding a plant at a new location rather than expanding an existing plant site saves a minimum of 281000 annually compare runs 11 and 14 in Appendix A of this note which results from placing plant capacity closer to warehouses Selecting Warehouses A simple test on the number of warehouses in the network shows that transportation costs are dropping more rapidly than inventory related costs are increasing see Figure 1 This means that 40 active warehouses will have the lowest total cost However some of these warehouses will have low throughput In order to maintain a minimum replenishment frequency and shipment size a minimum throughput needs to be met Approximately a truckload every two weeks or 10400 cwt of throughput per year is the minimum activity needed to open a warehouse Therefore any warehouse showing less than this throughput will be eliminated from consideration Under various assumptions about plants and their capacities demand growth and service levels 30 to 31 warehouses seem most economical with no deterioration on service over the benchmark network The following table shows selected results 212 Percent Type of Plant of demand Total No of run Year capacities 300 mi cost whses Benchmark Current Current 93 42112 22 Improved benchmark Current Current 93 41725 21 Improved Current benchmark 5th yr Memphis 94 53705 31 Current yr whses Current 92 41321 30 5th yr 5th Current whses year Memphis 92 53179 30 Note that this conclusion about the number of warehouses depends on the previous conclusion that a Memphis plant should be added by the 5th year The number of warehouses should be increased from the present 22 in both the current year and the 5th year FIGURE 1 Cost and Customer Service Profiles for Alternative Network Designs More detailed economic analysis shows that if the plants are held at current throughput levels a savings realized from 30 warehouses would be 41725000 41321000 404000 see previous table If current plant capacities are used and the Memphis plant is onstream in year 5 the savings of the added warehouse would be 53705000 53179000 526000 41 412 414 416 418 42 22 26 30 31 36 40 Number of warehouses Total cost 000000s 90 91 92 93 94 95 96 97 98 99 100 of demand 300 mi Cost left scale Service right scale Practical design 213 On the average there can be savings of approximately 404000 5260002 465000 per year by increasing the number of warehouses to 30 from the current 22 Since these are public warehouses little or no investment would be required to implement the change Although the number of warehouses remains relatively unchanged from the current year to the 5th year there is some shifting among the particular warehouses in the mix The 30 warehouses in the current year should be numbers 12345781113141516171819202125283132333435363738404445 providing that the loading on the current plants is allowed up to the limits of their current capacity When the Memphis plant is brought onstream by the end of the second year the warehouse mix should begin to evolve to numbers 12345781113141517181920212528293132343536373840444547 As the Memphis plant is bought on stream the Memphis public warehouse is closed and the volume is shifted to the Memphis plant as a warehouse In addition the Richmond VA warehouse is closed and the Las Vegas NV warehouse is opened The number of warehouses remains at 30 Both in the base year and in the future year the throughputs in the plants serving as warehouses are within acceptable limits as the following summary shows Plant Current Future as a Thruput year year warehouse limits solution solution Covington 450000 cwt 254471 cwt 306478 cwt New York 380000 302043 380523 Arlington 140000 66592 66161 Long Beach 180000 95943 117288 Customer Service Currently a high proportion of demand 93 is located within 300 miles of a stocking point Since the service distance may be up to 600 miles and still meet the companys service policy should the service level be reduced somewhat to effect a cost saving For example using the improved benchmark as the base case run 2 93 of the demand is within 300 miles and 975 is within 600 miles If a 600mile constraint is applied to the current network configuration run 23 75 of the demand is within 300 miles and 98 is still within 600 miles The total costs are reduced from 41725000 to 41201000 or a savings of 524000 per year In addition if the number of warehouses in the network is optimized the costs can be reduced by another 158000 per year run 23 vs run 22 However 278000 of the total 524000 158000 628000 can be realized without a service change This leaves approximately 404000 that can be saved by a relaxed service restriction The question now becomes one of whether the higher costs associated with the more restrictive service level are justified Since there is no salesservice relationship for this 214 problem we can only estimate the worth of the service That is can enough sales be generated to cover the higher service level If physical distribution costs for the com pany are 15 percent of sales which is probably a conservative estimate then 1015 670 in sales must be generated for each dollar that is added to distribution costs Therefore to cover 404000 in cost would require 404 6 0 000 70 71 100 38124 lb lbcwt cwt increase in sales In terms of overall demand this would be 381241001477026 25 But not all customers would experience a higher service level Comparing the demand centers for 299818 cwt of demand shows a reduction in warehouse to customer miles Thus moving from a minimum cost network to one with a high service level where the percent of demand less than 300 miles increases from 75 percent to 93 percent requires that the 38124 cwt increase in demand occur in the 299818 cwt of demand affected by the change This would be a 13 percent increase The products are not highly differentiated from others in the marketplace so that service plays an important role in selling these products Whether a 93 75 18 percentage points increase in service can result in a 25 percent increase in overall sales cannot be judged by the distribution department alone The sales department must play an important part in indicating whether the additional sales are possible If they are not likely to be realized there is no incentive for a network other than the minimum cost one If this information is not available from sales the conclusion is likely to be to maintain the status quo as represented by the benchmark That is oneday service is most likely to guide the design Overall Analysis and Summary The recommended design would involve an immediate increase in the number of warehouses from 22 to 30 In addition there should be an immediate reallocation of demand among the existing plants No reduction in the customer service level seems justified at this time Therefore a total cost reduction of 42112000 41321000 791000 per year seems immediately achievable run 1 vs run 18 By the end of the 2nd year the Memphis plant should be brought on stream and the network should begin to evolve from the current design run 24 to that for the 5th year run 25 The addition of a plant is justified from the high rate of return realized from the profit potential of being able to continue meeting the growth in demand For the current year a breakdown of the service and the cost changes show the following 215 Current Bench year Change from Cost type mark design benchmark Production 30762 30675 87 03 Whse operations 1578 1572 6 04 Order processing 369 358 11 30 Inventory carrying 457 490 33 72 Transportation Inbound 2050 1860 190 93 Outbound 6896 6365 531 77 Total costs 000s 42112 41320 792 19 By the 5th year total distribution costs should be 53179000 or 531790001908606 2786 compared with the currentyear cost of 421124631477026 2851 per cwt If current year costs are projected to the 5th year demand level the 5thyear productiondistribution costs might be 28511908606 54414357 or a savings of 54414357 53179000 1235357 per year Of course these savings can only be realized through the addition of capacity at Memphis for 4000000 If this capacity is useful for at least 15 years the amortization of 400000015 267000 per year would yield a net savings of 532000 per year Overall the design change appears to be justified 216 APPENDIX A Listing of Selected Computer Runs Service Percent of Run Run No of Plant Demand con No of Total demand within no description plants capacity level straint whses costs 300 600 Comments 1 Benchmark 4 Current mi 22 42112 93 98 Current network design 2 Improved benchmark 4 Current Current 300 21 41725 93 98 No investment required 3 No serv constraint 4 Current Current 9000 18 40896 71 89 4 Max opportunity 6 Current Current 9000 40 40409 81 94 Added plants at 1m cwt 5 Future yrimp bmk 4 Crnt1m 5th yr 300 21 53777 93 98 Plant cap 1m cwt 6 Test 27 whses 4 Current Current 300 26 41744 95 100 7 Test 32 whses 4 Current Current 300 31 41615 98 100 8 Test 37 whses 4 Current Current 300 36 41501 99 100 9 Test 42 whses 4 Current Current 300 40 41486 99 100 10 Exp Covington 4 Current 5th yr 300 21 54145 94 98 Covington cap 1m cwt 11 Exp New York 4 Current 5th yr 300 21 53986 93 98 New York 1m cwt 12 Exp Arlington 4 Current 5th yr 300 21 54709 94 98 Arlington cap 1m cwt 13 Exp Long Beach 4 Current 5th yr 300 21 55251 94 98 Long Beach cap 1m cwt 14 Add Memphis 5 Current 5th yr 300 21 53705 94 98 Add Memphis at 1m cwt 15 Add Chicago 5 Current 5th yr 300 20 53781 94 98 Add Chicago at 1m cwt 16 Add Mem Chi 6 Current 5th yr 300 20 53382 95 98 Add Chi Mem at 1m cwt ea 17 No plant expansion 4 Current 5th yr 300 20 45164 98 100 Only 854 of demnd served 18 Optimum whses 4 Current Current 300 31 41447 98 100 Plants at current capacity 19 Optimum whses 4 See cmt Current 300 30 41563 97 100 Plants at current thruput 20 Optimum whses 5 Current 5th yr 300 31 53342 98 100 Memphis at 1m cwt 21 Test cust service 4 Current Current 600 31 40996 80 100 Whses at opt no 31 22 Test cust service 4 Current Current 600 26 41043 88 100 Whses from opt no 31 23 Test cust service 4 Current Current 600 19 41201 75 98 Whses from current 22 24 Optimum whses 4 Current Current 375 30 41321 92 100 Service level at bmk 25 Optimum whses 5 See cmt 5th yr 375 30 53179 92 100 Serv at bmkMem 1m cwt 217 ESSEN USA Teaching Note Strategy Essen USA is concerned with entire supply channel performance The supply channel consists of four echelons ranging from factory to customers The purpose of this case study is for the student to manipulate the supply channel variables through the use of a channel simulator in order to improve individual member and systemwide performance The channel variables include forecasting methods inventory policies transportation services production lot sizes order processing costs and stock availability levels Students should seek to optimize channel performance although it is not expected that the optimum actually can be found or verified However improving performance over existing levels is achievable The SCSIM module of LOGWARE is used to simulate the demand and product flows throughout the multiechelon supply chain SCSIM is an ordinary Monte Carlo dayto day type of simulator Using a simulator for performance improvement requires thinking of it in terms of as an experimental methodology That is a single run of the simulator is a particular event sequence generated from random numbers Changing the seed number in the simulator causes a different set of random numbers to be generated and possibly another outcome from the same input data A simulation run with a specified seed number should be viewed as a single statistical observation and multiple outcomes from various seed numbers should be treated as a statistical sample and analyzed accordingly ie comparing means and standard deviations Each simulation is run for a period of 11 years with results taken from years 2 through 11 The first year is not used since it can show unstable results due to startup conditions The results appear to reach steady state by the second year and the results for the 10 years thereafter are averaged to give a reasonable representation of channel performance for a given run The database used to represent the current performance of the channel as derived from the case study is summarized in the Appendix A of this note and a typical run report is shown in Appendix B This case provides students with the opportunity to observe the operation of a multi echelon supply channel and to assess the impact of changing key operating variables on individual members as well as on channelwide performance The effect on cost and customer service as well as sales inventory and back order levels of demand patterns demand forecasting methods inventory control methods transportation performance production lot sizing order processing procedures and item fill rates can be observed in both graphical and report forms Most importantly students can see the effects of supply chain decisions rather than project the results statistically Questions 1 What can you say about the logistics performance throughout the supply channel for Essen and its customers 218 General observations It is recognized that Essen must deal with demand that has significant seasonal peaks at gift giving times of the year as shown in Figure 1 Compared with a smooth demand pattern this can cause increasing demand variability upstream from the customers as illustrated in Figure 2 This bull whip effect is partly a result of the demand for an upstream member being derived from the order size and pattern of its immediate downstream channel member Forecast accuracy leadtime uncertainty and inventory control method also affect demand variability and the resulting cost of that variability Figure 1 Typical Demand Pattern for Essen Over the Period of One Year Retailer Distri butor Ware house Factory Retailer Factory Retailer warehouse Essen warehouse Retailer Distri butor Ware house Factory Retailer Factory Retailer warehouse Essen warehouse Figure 2 Increasing Demand Variability of Upstream Channel Members for a FourYear Period 219 Benchmark Running the simulator SCSIM with a seed number of 123456 and simulated period of 11 years with results taken from the last 10 years the channel generates average annual sales of 1095 million for a net average annual system profit contribution of 244 million as shown in Table 1 The question arises as to whether channel performance can be improved and profits increased At least two observations can be made that suggest there is room for improvement First the inventory levels for both the retailers warehouse and Essens warehouse are quite high compared with the Retailer see Figure 3 It is possible that Retailer inventories are too low However the inventory turnover ratio is about 7 for the Essens warehouse see Table 1 This is not particularly high for a food product that might have a turnover at least in the range of 10 to 12 The turnover for the retailers warehouse appears more in line with industry norms of about 13 see Table 1 Second the backorders at the Retailer level do not seem to recover well from the seasonal spike in demand Correspondingly the Retailer inventory turnover is 81 see Table 1 which is quite high The low percentage of demand filled on request 50 suggests that inadequate inventory is being maintained to meet reasonable fill rates Third customer service levels are also low for the retailers warehouse and Essens warehouse Backorder occurrences are high for both channel members Although inventory levels are adequate most of the time seasonal demand rippling through the supply chain causes a significant number of back orders before inventory can be replenished The observation is that there is an opportunity to improve channel performance especially in terms of customer service A major concern is how to mange the seasonal demand pattern that is causing the cyclical behavior throughout the echelons of the channel Current performance of the channel members is summarized in Table 1 for 4 simulation runs using different seed numbers Retailer Retailer Retailer warehouse Essen warehouse Retailer Retailer Retailer warehouse Essen warehouse Figure 3 Inventory Levels for Four Years Using Benchmark Data 220 Channel member Run 1 Run 2 Run 3 Run 4 Average Essens factory Total cost 73105904 72967088 72616192 72477376 72791640 Units produced 37918 37846 37664 37592 37755 Cost per unit 1928 1928 1928 1928 1928 Essens warehouse TO ratio 652 659 661 658 658 Fill rate 50 50 50 50 50 Cost 5578291 5549202 5521236 5540447 5547294 Cost per unit 14702 14650 14632 14665 14662 Retailers warehouse TO ratio 1295 1301 1296 1289 1295 Fill rate 50 50 50 50 50 Cost 3873236 3895406 3853890 3912699 3883808 Cost per unit 10194 10215 10214 10307 10233 Retailer TO ratio 8102 8038 8045 8060 8061 Fill rate 50 5302 5409 50 50 Units sold 38017 37983 37774 37804 37895 Cost 2884527 2768697 2939464 2981607 2893574 Cost per unit 7619 7289 7782 7887 7644 System Profit 24426593 24591053 24235498 24340563 24398426 Profit as of sales 2223 2240 2220 2228 2228 Seed Number 123456 444444 555555 666666 2 What steps would you suggest taking to improve logistics performance throughout the channel Do any of the changes involve Essen If so does the company directly realize any cost andor operating performance improvements There are a number of actions that generally can be taken to lower costs and improve customer service Improving the forecast shortening the lead times changing the inventory control policy and changing production lot sizes are all variables that can be altered for possible performance improvement The interactions among these variables and the large number of variable combinations preclude finding the optimal set However they can be explored in a systematic way to find improvement The primary focus of this analysis will be to increase the fill rates at the risk of increasing costs Ultimately revenues through improved customer service may be preserved or increased to more than compensate for reduced profits Retailer Level Start with the retailer because of the proximity to the customer Fill rates need to be improved probably in the 9599 range as specified in the database Inventory turns can Table 1 Average Annual Performance of Channel Members and the System at Benchmark 221 be guided by the industry average of 12 turns per year Where the two cannot be jointly met service will prevail Clearly putting additional inventory at the retail point will improve customer service Using the companys current inventory policy of stocking to demand the target level can be raised without changing the review time Exploring different target levels shows 14 days to offer about 35 turns and a 99 fill rate Because of the high cost of a back order total costs at the retail level drop significantly Altering the forecasting method and the settings associated with the method yield little opportunity for improvement Using an exponential smoothing model with a high smoothing constant to better follow the seasonal changes in demand results in increased costs Lowering the smoothing constant to 01 did not offer improvement either Altering the number of periods in the moving average model did not improve costs and only degraded performance Shortening the review time in the stocktodemand reorder policy did have a positive effect on fill rate but resulted in high costs and lower inventory turns The tradeoff did not seem beneficial given the fill rate and turnover targets Retail Warehouse Level Determining an improved policy at this level is difficult because a 95 fill rate and 10 to 12 inventory turns is an illusive goal Using service as the primary target a stockto demand control policy is used with a review period of 7 days and a target of 25 days of inventory The forecasting method is moving average with a period of 7 days The performance achieved at this channel level is about 9 inventory turns per year and a 97 fill rate Essens Warehouse Level The performance at Essens warehouse level seems to mimic that at the retail warehouse level except that there is more demand variability Again an inventory turnover ratio in the target range cannot be achieved while maintaining a high fill rate level Trying to achieve high service levels with high levels of inventory is difficult probably due to the extensive demand variability that filters back to this member of the channel Multiple simulation runs show that a high fill rate cannot consistently be achieved even when on the average inventory levels are high However average performance shows an 82 fill rate and 15 inventory turns per year based on a 7day moving average forecasting model and a stocktodemand inventory control policy with a review time of 7 days and an inventory target of 25 days Essens Factory The concern with the factory level in the channel is whether product should be manufactured in a larger lot size but with slightly higher production time variability The reduced costs seem to out weigh the negative effects of increased variability Producing in the larger lot size is favored Overall Using the objective of improving customer service it is not surprising that supply channel costs increase as shown from the reduced profit in Table 2 compared with Table 1 The average fill rate has increased for all members of the channel but the cost effects 222 are spread disproportionately among the members Even with a higher fill rate the retailer benefits from a substantial reduction in the cost per unit sold On the other hand the cost for handling a unit of the product at Essens warehouse is substantially increased Essen should take advantage of the cost reduction from producing in the larger batch size but this does not offset the higher cost at the companys warehouse As an upstream member of the supply channel Essen undoubtedly suffers from the variability in demand which cannot entirely be controlled The retailer benefits from the action to increase fill rates across the channel However Essen is put at a disadvantage and may take a counter action to improve its cost position Essen may simply lower its inventory level by reducing the reorder target quantity from 25 to 10 days This reduces Essens perunit warehouse cost but it also increases the costs for the retailer The reduced inventory level at the Essen warehouse causes lower fill rates for the downstream retailer Unless the retailer can find an incentive to reward Essen for its good service it will be difficult for Essen to provide the level of service that the retailer would like and that is economically beneficial to Essen Channel member Run 1 Run 2 Run 3 Run 4 Average Essens factory Total cost 68638738 74761258 78418188 75400666 74304713 Units produced 35950 39286 41268 39622 39032 Cost per unit 1909 1903 1900 1903 1904 Essens warehouse TO ratio 147 142 151 146 147 Fill rate 9065 100 6316 7292 8168 Cost 12060444 12449170 11895743 12178581 12145984 Cost per unit 31751 32628 31187 31961 31882 Retailers warehouse TO ratio 924 902 914 928 917 Fill rate 9664 9629 9760 9809 9716 Cost 4018143 4080160 4043973 3992791 4033767 Cost per unit 10570 10709 10655 10551 10621 Retailer TO ratio 3544 3536 3499 3514 3523 Fill rate 9952 9953 9970 9940 9954 Units sold 38017 37936 37979 37730 37916 Cost 727378 717363 709772 748198 725677 Cost per unit 1913 1891 1869 1975 1912 System Profit 24423849 17627666 14690765 17184464 18481686 Profit as of sales 2223 1608 1338 1569 1685 Seed Number 123456 111111 222222 333333 Table 2 Average Annual Performance of Channel Members and the System as Revised 223 3 Would shipping by airfreight from Germany be a benefit to channel performance To Essen No Selling candies to end customers at 2890 per thousand lb using airfreight shipping results in obvious loss to Essen The cost of shipping by air is 1833 per thousand lb plus 1000 material cost and 850 production cost results in a total cost much higher than selling price There is no point to using airfreight Running the simulation with the higher freight rate but lower variability confirms that channel profits would be negative 4 Is there a benefit to producing in the larger 20000pound batch size Yes This was tested in question 2 From Tables 1 and 2 it can be seen that production costs drop from 1928 per unit to 1904 per unit The overall channel cost reduction overshadows the negative effects of greater length and variability in production time Appendix A Simulation Database for Essen USA Under Current Conditions Title ESSEN USA Initialization 123456 Seed value 11 Length of simulation years 2890 Annual price unit Customer demand pattern Generate daily demand 100 Average daily demand units 15 Standard deviation of daily demand units 1 Annual demand growth increment Monthly seasonal indices Month Index Month Index Month Index Month Index 1 025 4 075 7 075 10 075 2 125 5 075 8 075 11 150 3 125 6 075 9 075 12 250 RetailerLevel 1 Product item data 2220 Item value in inventory unit 1 Customer order filling cost unit 35 Purchase order processing cost order 25 Inventory carrying cost year 1 Average customer order fill time days 0 Customer order fill time standard deviation days 98 Instock probability 670 Back order cost unit 224 Forecasting method Moving average 7 Number of periods Reorder policy Stocktodemand control method 10 Target days of inventory 7 Review time in days DistributorLevel 2 Product item data 2220 Item value in inventory unit 20 Retailer order filling cost unit 75 Purchase order processing cost order 25 Inventory carrying cost year 2 Average retailer order fill time days 02 Retailer order fill time standard deviation days 95 Instock probability 100 Back order cost unit Forecasting method Moving average 30 Number of periods Reorder policy Stocktodemand control method 45 Target days of inventory 30 Review time in days WarehouseLevel 3 Product item data 1710 Item value in inventory unit 15 Distributor order filling cost unit 75 Purchase order processing cost order 20 Inventory carrying cost year 3 Average distributor order filling time days 03 Distributor order fill time days 95 Instock probability 25 Back order cost unit Forecasting method Moving average 360 Number of periods Reorder policy Stocktodemand control method 90 Target days of inventory 30 Review time in days 225 FactorySource Product item data 850 Production cost unit 10 Minimum production lot size units 10 Warehouse order filling cost unit 8 Average production time days 2 Production time standard deviation days 1000 Purchase cost unit Transportation Transport between Distributor and Retailer 25 Transport cost unit 1 Average time intransit days 0 Transit time standard deviation days Transport between Warehouse and Distributor 70 Transport cost unit 5 Average time intransit days 1 Transit time standard deviation days Transport between Factory and Warehouse 78 Transport cost unit 9 Average time intransit days 3 Transit time standard deviation days Appendix B Benchmark Simulation Results with Seed Number 123456 and Simulation Length of 11 Years SUPPLY CHANNEL REPORT FOR SIMULATED YEARS 2 TO 11 Yearly Simulated average period FINANCIAL PERFORMANCE 109868552 1098685520 Revenue 37918000 379180000 Cost of purchased goods 71950552 719505520 Gross margin 32230300 322303000 Production cost Transportation costs 949025 9490250 Distributor to retailer 2656010 26560100 Warehouse to distributor 2957604 29576040 Factory to warehouse Sales order handling cost for 38017 380168 Customer orders 759890 7598900 Retailer orders 569145 5691450 Distributor orders Order processing cost for 1638 16380 Orders to distributor 683 6825 Orders to warehouses 675 6750 Orders to factory Inventory costs 226 260414 2604145 Retailer 1627909 16279092 Distributor 1988984 19889837 Warehouse Back order costs 2584458 25844580 Retailer 535730 5357300 Distributor 363478 3634775 Warehouse 24426593 244265928 Net profit contribution Appendix C Simulation Database for Essen USA as Revised for Serivce Improvement Title ESSEN USA Initialization 123456 Seed value 11 Length of simulation years 2890 Annual price unit Customer demand pattern Generate daily demand 100 Average daily demand units 15 Standard deviation of daily demand units 1 Annual demand growth increment Monthly seasonal indices Month Index Month Index Month Index Month Index 1 025 4 075 7 075 10 075 2 125 5 075 8 075 11 150 3 125 6 075 9 075 12 250 RetailerLevel 1 Product item data 2220 Item value in inventory unit 1 Customer order filling cost unit 35 Purchase order processing cost order 25 Inventory carrying cost year 1 Average customer order fill time days 0 Customer order fill time standard deviation days 98 Instock probability 670 Back order cost unit Forecasting method Moving average 7 Number of periods Reorder policy Stocktodemand control method 14 Target days of inventory 227 7 Review time in days DistributorLevel 2 Product item data 2220 Item value in inventory unit 20 Retailer order filling cost unit 75 Purchase order processing cost order 25 Inventory carrying cost year 2 Average retailer order fill time days 02 Retailer order fill time standard deviation days 95 Instock probability 100 Back order cost unit Forecasting method Moving average 7 Number of periods Reorder policy Stocktodemand control method 35 Target days of inventory 7 Review time in days WarehouseLevel 3 Product item data 1710 Item value in inventory unit 15 Distributor order filling cost unit 75 Purchase order processing cost order 20 Inventory carrying cost year 3 Average distributor order filling time days 03 Distributor order fill time standard deviation days 95 Instock probability 25 Back order cost unit Forecasting method Moving average 7 Number of periods Reorder policy Stocktodemand control method 25 Target days of inventory 7 Review time in days FactorySource Product item data 825 Production cost unit 20 Minimum production lot size units 10 Warehouse order filling cost unit 10 Average production time days 21 Production time standard deviation days 1000 Purchase cost unit 228 Transportation Transport between Distributor and Retailer 25 Transport cost unit 1 Average time intransit days 0 Transit time standard deviation days Transport between Warehouse and Distributor 70 Transport cost unit 5 Average time intransit days 1 Transit time standard deviation days Transport between Factory and Warehouse 78 Transport cost unit 9 Average time intransit days 3 Transit time standard deviation days 229 CHAPTER 15 LOGISTICSSUPPLY CHAIN ORGANIZATION All questions in this chapter require individual judgment and response No answers are offered 230 CHAPTER 16 LOGISTICSSUPPLY CHAIN AND CONTROL All questions in this chapter require individual judgment and response No answers are offered