Exercício - Processos de Poisson - 2024-1

Exercícios processos de poisson quarta-feira, 5 de junho de 2024 00:51 Página 1 de exercícios 4. Consider a post office with two clerks. Three people, A, B, and C, enter simultaneously. A and B go directly to the clerks, and C waits until either A or B leaves before he begins service. What is the probability that A is still in the post office after the other two have left when (a) the service time for each clerk is exactly (nonrandom) ten minutes? (b) the service times are i with probability 1/3, i = 1, 2, 3? (c) the service times are exponential with mean 1/μ? 4. (a) 0 (b) 1/27 (c) 1/4 *5. If X is exponential with rate λ, show that Y = [X] + 1 is geometric with parameter p = 1 – e^–λ, where [x] is the largest integer less than or equal to x. 5. P(Y = n) = P(n – 1 < X < n) = e^–λ(n–1) – e^–λn = (e^–λ)n–1 (1 – e^–λ) 8. If X and Y are independent exponential random variables with respective rates λ and μ, what is the conditional distribution of X given that X < Y? 8. Exponential with rate λ + μ. 9. Machine 1 is currently working. Machine 2 will be put in use at a time t from now. If the lifetime of machine i is exponential with rate λi, i = 1, 2, what is the probability that machine 1 is the first machine to fail? 9. Condition on whether machine 1 is still working at time t, to obtain the answer, 1 – e^–λ1t + e^–λ1t λ1 / λ1 + λ2 Página 2 de exercícios *10. Let X and Y be independent exponential random variables with respective rates λ and μ. Let M = min(X, Y). Find (a) E[M X|M = X], (b) E[M X|M = Y], (c) Cov(X, M). 12. If Xi, i = 1, 2, 3, are independent exponential random variables with rates λi, i = 1, 2, 3, find (a) P{X1 < X2 < X3}, (b) P{X1 < X2|max(X1, X2, X3) = X3}, (c) E[max Xi|X1 < X2 < X3], (d) E[max Xi]. 12. (a) P{X1 < X2 < X3} = P{X1 = min(X1, X2, X3)} P{X2 < X3|X1 = min(X1, X2, X3)} = λ1 / (λ1 + λ2 + λ3)P{X2 < X3|X1 = min(X1, X2, X3)} = (λ1 / (λ1 + λ2 + λ3)) λ2 /(λ2 + λ3) where the final equality follows by the lack of memory property. 14. I am waiting for two friends to arrive at my house. The time until A arrives is exponentially distributed with rate λa, and the time until B arrives is exponentially distributed with rate λb. Once they arrive, both will spend exponentially distributed times, with respective rates μa and μb at my home before departing. The four exponential random variables are independent. (a) What is the probability that A arrives before and departs after B? (b) What is the expected time of the last departure? 14. (a) λa / (λa + λb) λb / (μa + λb μa + λb μa + μb) (b) Let F be the time of the first departure. Write F = T + A where T is the time of the first arrival and A is the additional time from then until the first departure. First take expectations and then condition on who arrives first to obtain E[F] = 1 / λa + λb) + E[A|a] λa / (λa + λb) + E[A|b] λb / (λa + λb) Now use E[A|a] = 1 / (μa + λb) + λb / (μa + λb μa + μb) and E[A|b] = 1 /(μb + λa) + λa / (μb + λa μa + μb) 15. One hundred items are simultaneously put on a life test. Suppose the lifetimes of the individual items are independent exponential random variables with mean 200 hours. The test will end when there have been a total of 5 failures. If T is the time at which the test ends, find E[T] and Var(T). 15. Let Ti denote the time between the (i – 1)th and the ith failure. Then the Ti are independent with Ti being exponential with rate (101 – i)/200. Thus, E[T] = ∑ (i=1 to 5) E[Ti] = ∑ (i=1 to 5) 200 /(101 – i) Var(T) = ∑ (i=1 to 5) Var(Ti) = ∑ (i=1 to 5) (200)^2/(101 – i)^2 Página 4 de exercícios 16. There are three jobs that need to be processed, with the processing time of job j being exponential with rate μj. There are two processors available, so processing on two of the jobs can immediately start, with processing on the final job to start when one of the initial ones is finished. (a) Let Tj denote the time at which the processing of job j is completed. If the objective is to minimize E[T1 + T2 + T3], which jobs should be initially processed if μ1 < μ2 < μ3? (b) Let M, called the makespan, be the time until all three jobs have been processed. With S equal to the time that there is only a single processor working, show that 2E[M] = E[S] + 3 ∑ 1/μi For the rest of this problem, suppose that μ1 = μ2 = μ, μ3 = λ. Also, let P(μ) be the probability that the last job to finish is either job 1 or job 2, and let P(λ) = 1 − P(μ) be the probability that the last job to finish is job 3. (c) Express E[S] in terms of P(μ) and P(λ). Let Pi,j(μ) be the value of P(μ) when i and j are the jobs that are initially started. (d) Show that P1,2(μ) ≤ P1,3(μ). (e) If μ > λ show that E[M] is minimized when job 3 is one of the jobs that is initially started. (f) If μ < λ show that E[M] is minimized when processing is initially started on jobs 1 and 2. (g) Suppose i and j are initially begun, with k waiting for one of them to be completed. Then E[Tj] + E[Tj] + E[Tk] = 1/μi + 1/μj + 1/μi + μj + 1/μk = 3 ∑ 1/μi + 1/μi + μj Hence, the preceding is minimized when μi + μj is as large as possible, showing that it is optimal to begin processing on jobs 2 and 3. Consequently, to minimize the expected sum of the completion times the jobs having largest rates should be initiated first. (b) Letting Xi be the processing time of job i, this follows from the identity 2(M − S) + S = 3 ∑ Xi which follows because if we interpret Xi as the work of job i then the total amount of work is 3 ∑ Xi, whereas work is processed at rate 2 per unit time when both servers are busy and at rate 1 per unit time when only a single processor is working. (c) E[S] = 1/μ P(μ) + 1/λ P(λ) = μ/μ + λ/μ + λ/μ + μ/λ P1,3(μ) = P1,3(μ) (e) If μ > λ then E[S] is minimized when P(μ) is as large as possible. Hence, because minimizing E[S] is equivalent to minimizing E[M], it follows that E[M] is minimized when jobs 1 and 3 are initially processed. (f) In this case E[M] is minimized when jobs 1 and 2 are initially processed. In all cases E[M] is minimized when the jobs having smallest rates are initiated first. 17. A set of n cities is to be connected via communication links. The cost to construct a link between cities i and j is Ci,j, i ≠ j. Enough links should be constructed so that for each pair of cities there is a path of links that connects them. As a result, only n − 1 links need be constructed. A minimal cost algo- rithm for solving this problem (known as the minimal spanning tree problem) first constructs the cheapest of all ( n 2) links. Then, at each additional stage it chooses the cheapest link that connects a city without any links to one with links. That is, if the first link is between cities 1 and 2, then the second link will either be between 1 and one of the links 3, . . ., n or between 2 and one of the links 3, . . ., n. Suppose that all of the ( n 2) costs Ci,j are independent expo- nential random variables with mean 1. Find the expected cost of the preceding algorithm if (a) n = 3, (b) n = 4. 17. Let Ci denote the cost of the ith link to be constructed, i = 1, . . . , n − 1. Note that the first link can be any of the ( n 2) possible links. Given the first one, the second link must connect one of the 2 cities joined by the first link with one of the n − 2 cities without any links. Thus, given the first constructed link, the next link constructed will be one of 2(n − 2) possible links. Similarly, given the first two links that are constructed, the next one to be constructed will be one of 3(n − 3) possible links, and so on. Since the cost of the first link to be built is the minimum of ( n 2) exponentials with rate 1, it follows that E[C1] = 1/ ( n 2) Introduction to Probability Models 90 By the lack of memory property of the exponential it follows that the amounts by which the costs of the other links exceed C1 are independent exponentials with rate 1. Therefore, C2 is equal to C1 plus the minimum of 2(n − 2) independent exponentials with rate 1, and so E[C2] = E[C1] + 1/2(n − 2) Similar reasoning then gives E[C3] = E[C2] + 1/3(n − 3) and so on. *18. Let X1 and X2 be independent exponential random variables, each having rate μ. Let X(1) = minimum(X1, X2) and X(2) = maximum(X1, X2) Find (a) E[X(1)]. (b) Var[X(1)]. (c) E[X(2)]. (d) Var[X(2)]. *19. In a mile race between A and B, the time it takes A to complete the mile is an exponential random variable with rate λa and is independent of the time it takes B to complete the mile, which is an exponential random variable with rate λb. The one who finishes earliest is declared the winner and receives Re−art if the winning time is t, where R and a are constants. If the loser receives 0, find the expected amount that runner A wins. 19. Using that the winning time is exponential with rate r = λa + λb, independent of who wins, gives that with X equal to amount that runner A wins E[X] = λa/λa + λb R ∫ e−at re−rt/(λa + λb/λa + λb r re−at dt = R λa/λa + λb r r + a Introduction to Probability Models 20. Consider a two-server system in which a customer is served first by server I, then by server 2, and then departs. The service times at server i are exponential random variables with rates μi, i = 1, 2. When you arrive, you find server 1 free and two customers at server 2—customer A in service and customer B waiting in line. (a) Find PA, the probability that A is still in service when you move over to server 2. (b) Find PB, the probability that B is still in the system when you move over to server 2. (c) Find E[T ], where T is the time that you spend in the system. Hint: Write T = S1 + S2 + WA + WB where Si is your service time at server i, WA is the amount of time you wait in queue while A is being served, and WB is the amount of time you wait in queue while B is being served. 20. (a) PA = μ1/μ1 + μ2 (b) PB = 1 − (μ2/μ1 + μ2)2 (c) E[T ] = 1/μ1 + 1/μ2 + PA/μ2 + PB/μ2