ISyE 6650 Probabilistic Models Fall Homework 5 Solution

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1 ISyE 6650 Probabilistic Models Fall 2007 Homework 5 Solution 1. (Ross 5.3) Due to the memoryless property of the exponential distribution, the conditional distribution of X, given that X > 1, is the same as the unconditional distribution of 1 + X. Hence, the correct answer is (a). 2. (Ross 5.9) We have that: P {M1 is first failing machine} P {M1 is first failing machine M1 failed before starting M2}P {M1 failed before starting M2} + P {M1 is first failing machine M1 failed after starting M2}P {M1 failed after starting M2} 1 (1 e λ1t ) + λ 1 λ 1 + λ 2 e λ1t 3. (Ross 5.15) Let T i denote the time between the (i 1) th and the i th failure. Then the T i are independent with T i being exponential with rate (101 i)/200. Thus, E[T] E[T i ] 101 i V ar(t) i1 5 V ar(t i ) i1 i1 5 i1 (200) 2 (101 i) 2 4. (Ross 5.20) (a) Obviously, your time in server 1 is exponential with rate µ 1. Furthermore, due to the memoryless property of the exponential distribution, the remaining time of A in server 2 is exponential with rate µ 2. Hence, for this part, essentially we are seeking the probability that among these two exponential random variables, the first one has the minimum value. This is equal to P A µ1 µ 1+µ 2 (b) A quick answer to this part is obtained by noticing that B will not be in the system when you proceed to the second station, only when both A and B complete their service on server 2 while you still are processed by server 1. From part (a), the probability that A completes his service in server 2

2 µ before you complete service in server 1 is equal to 2 µ 1+µ 2. A similar argument to that in part (a) establishes that the probability that B completes service in server 2 before you complete service in server 1 is also equal to µ 2 µ µ 1+µ 2. Hence, the requested probability is equal to P B 1 ( 2 µ 1+µ 2 ) 2 (c) Following the book suggestion, and using the results derived in parts (a) and (b), we get that E[T] 1/µ 1 + 1/µ 2 + P A /µ 2 + P B /µ 2 5. (Ross 5.34) A key observation for solving this problem is that the inter-event times of a Poisson process with rate λ are exponentially distributed with the same rate λ. Then, we reason as follows: (a) Obviously, patient A will obtain a new kidney only if the next kidney arrives before she dies. Recognizing that all the involved times are exponentially distributed and reasoning as in problem 5.20, we obtain that the λ λ+µ A relevant probability is equal to (b) For patient B to obtain a new kidney, first patient A must die or obtain a new kidney, and then, the next kidney must arrive before B λ+µ dies. The probability for the first event is equal to A λ+µ A+µ B. Due to the memoryless property of the exponential distribution, the probability λ of the second event is equal to λ+µ B. Hence, the requested probability is equal to λ+µ A λ+µ A+µ B λ λ+µ B 6. (Ross 5.50) Let T denote the time until the next train arrives; and so T is uniform on (0,1). Note that, conditional on T, X is Poisson with, both, mean and variance equal to 7T. (a) E[X] E[E[X T]] E[7T] 7/2 (b) By the conditional variance formula: V ar[x] E[V ar[x T]]+V ar[e[x T]] E[7T]+V ar[7t] 7(1/2)+49(1/12) 91/ (Ross 5.61) One way to reason about this problem is as follows: Consider any single flaw in this system. Then, the probability that this flaw will have been identified by time t is equal to G(t). Since the identification of each flaw by time t is independent from the identification of the other flaws, by time t, we can classify each flaw in the system into two categories, detected and undetected, with corresponding classification probabilities G(t) and 1 G(t). Since the original distribution of flaws is Poisson with rate c, this classification generates two other Poisson distributions, independent from each other, with rates cg(t) and c(1 G(t)). Hence, the responses to the problem questions are: 2

3 (a) Poisson with mean cg(t). (b) Poisson with mean c[1 G(t)]. (c) Independent. An alternative, longer way to get to the above results, is by conditioning on the number of flaws in the system; for instance, for part (a), we would have: P {failures by time t k} P {flaws detected by time t k} P {flaws detected by time t k total flaws i}p {total flaws i} i0 But c ci P {total flaws i} e i! and { P {flaws detected by time t k total flaws i} 0 i < k i k i! k!(i k)! (G(t))k (1 G(t)) i k Hence, P {failures by time t k} P {flaws detected by time t k} c ci i! e i! k!(i k)! (G(t))k (1 G(t)) i k ik c [cg(t)]k e k! c [cg(t)]k e k! [c(1 G(t))] i k (i k)! ik [c(1 G(t))] j j! j0 c [cg(t)]k e e c cg(t) k! cg(t) [cg(t)]k e k! But the last expression implies a Poisson distribution with rate cg(t). 8. (Ross 6.1) Let us assume that the state is (n,m). Male i mates at a rate λ with female j, and therefore it mates at a rate λm. Since there are n males, matings occur at a rate λnm. Therefore, v (n,m) λnm. 3

4 Since any mating is equally likely to result in a female as in a male, we have P (n,m);(n+1,m) P (n,m);(n,m+1) (Ross 6.5) (a) Yes. (b) It is a pure birth process (since the only state accessed from state X(t) is X(t) + 1). (c) Consider a state X(t) i for some i 1,...,N; i.e., there are i infected individuals in that state. Then, in this state, any single contact between two individuals will lead to a new infection with probability i(n i) ( N 2 ) p. Hence, it follows that the birth rate in this state is λ i λpi(n i)/( N 2 ). But then, the expected time to transition from state i to state i+1 is equal to 1/λ i N(N 1) 1 2λp i(n i), and E[time all infected X(0) 1] N(N 1) 2λp N i1 1 i(n i) 10. (Ross 6.13) According to the Kendall notation introduced in class, this is an M/M/1/2 system, i.e., a single server queueing system, with exponential inter-arrival and processing times, and a total buffering capacity equal to two (including the single-unit capacity of the server). So, it can be modeled by a CTMC with a state space S {0,1,2}, where each state declares the corresponding number of customers in the system. Furthermore, the instantaneous transition rates for transitioning from state i to state i+1 are λ 0 λ 1 3, while the rates for transitioning from state i to state i 1 are µ 1 µ 2 4. Then, the limiting probabilities of this chain satisfy the following balance equations: 3P 0 4P 1 P P 0 (3 + 4)P 1 3P 0 + 4P 2 P P 1 ( 3 4 )2 P 0. And since Σ 2 0P i 1, we get P 0 (1 + 3/4 + (3/4) 2 ) 1 16/37. 4

5 (a)the average number of customers in the shop is P 1 + 2P 2 (3/4 + 2(3/4) 2 )P 0 30/16(1 + 3/4 + (3/4) 2 ) 1 30/37. (b)the proportion of customers that enter the shop is λ(1 P 2 ) λ (c) Now µ 8, and so 1 P P 0 (1 + 3/8 + (3/8) 2 ) 1 64/97. So the proportion of customers who now enter the shop is 1 P 2 1 (3/8) 2 (264/97) 1 9/97 88/97. The rate of added customers is therefore λ(88/97) λ(28/37) 3(88/97 28/37) The business he does would improve by 0.45 customers per hour. 11. (Ross 6.17) Say the state is 0 if the machine is up, and that it is i when it is down due to a type i failure, i 1,2. The balance equations for the limiting probabilities are as follows. λp 0 µ 1 P 1 + µ 2 P 2 µ 1 P 1 λpp 0 µ 2 P 2 λ(1 p)p 0 P 0 + P 1 + P 2 1. These equations are easily solved to give the results P 0 (1 + λp/µ 1 + λ(1 p)/µ 2 ) 1 P 1 λpp 0 /µ 1,P 2 λ(1 p)p 0 /µ (Extra Credit: 5.12) (a) P {X 1 < X 2 < X 3 } P {X 1 min(x 1,X 2,X 3 )}P {X 2 < X 3 X 1 min(x 1,X 2,X 3 )} λ 1 P {X 2 < X 3 X 1 min(x 1,X 2,X 3 )} λ 1 + λ 2 + λ 3 λ 1 λ 2 λ 1 + λ 2 + λ 3 λ 2 + λ 3 5

6 where the final equality follows by the lack of memory property. (b) P {X 1 < X 2 X 3 max(x 1,X 2,X 3 )} P {X 1 < X 2 < X 3 } P {X 1 < X 2 < X 3 } + P {X 2 < X 1 < X 3 } λ 1 λ 1+λ 2+λ 3 λ 2 λ 1 λ 2 λ 1+λ 2+λ 3 λ 2+λ 3 λ 2+λ 3 + λ 2 λ 1+λ 2+λ 3 λ 1 λ 1+λ 3 1/(λ 2 + λ 3 ) 1/(λ 2 + λ 3 ) + 1/(λ 1 + λ 3 ) (c) Recognizing that max X i X [1] + (X [2] X [1] ) + (X [3] X [2] ) where X [i] denotes the random variable with the i-th smallest value, and taking into consideration the ordering suggested by the conditioning, and the memoryless property of the exponential distribution, we get that: E[max X i X 1 < X 2 < X 3 ] 1 λ 1 + λ 2 + λ λ 2 + λ λ 3 (d) We can compute this expectation by conditioning as in part (c) above. Then, we get: E[max X i ] i j k λ j λ i 1 1 [ ], λ 1 + λ 2 + λ 3 λ j + λ k λ 1 + λ 2 + λ 3 λ j + λ k λ k where the sum is over all 6 permutations of 1,2, (Extra Credit: Ross 5.47) (a) 1/(2µ) + 1/λ i.e., the expected time until one server becomes free plus the expected time until a customer appears after the release of the server. (b) Let T i denote the time until both servers are busy when you start with i busy servers, i 0.1. Then, E[T 0 ] 1/λ + E[T 1 ] Now, starting with 1 server busy, let T be the time until the first event(arrival or departure); Let X 1 if the first event is an arrival and let it be 0 if it is a departure; Let Y be the additional time after the first event until both servers are busy. Then, E[T 1 ] E[T] + E[Y ] 1/(λ + µ) + E[Y X 1]λ/(λ + µ) + E[Y X 0]µ/(λ + µ) 1/(λ + µ) + E[T 0 ]µ/(λ + µ) 6

7 where in the last equation we have taken into consideration that E[Y X 1] 0. Thus, or Also, E[T 0 ] 1/λ 1/(λ + µ) + E[T 0 ]µ/(λ + µ) E[T 0 ] (2λ + µ)/λ 2 E[T 1 ] (λ + µ)/λ 2 (c) Let L i denote the time until a customer is lost when you start with i busy servers, and notice that, due to the memoryless property of the exponential distribution, L 2 characterizes also the time between two successive lost customers. To compute E[L 2 ], we reason as in part (b), and we get that E[L 2 ] 1/(λ + µ) + E[L 1 ]µ/(λ + µ) 1/(λ + µ) + (E[T 1 ] + E[L 2 ])µ/(λ + µ) 1/(λ + µ) + µ/λ 2 + E[L 2 ]µ/(λ + µ) Thus, E[L 2 ] 1/λ + µ(λ + µ)/λ (Extra Credit: Ross 5.55) Notice that the current arrival will be the n-th departure from the system, and over the time interval until that departure, the occurring events will be either arrivals or departures with corresponding probabilities λ/(λ+µ) and µ/(λ+µ). Considering each arrival as a failure and each departure as a success, it follows that the number of arrivals until the n-th departure is distributed according to a negative binomial distribution with success probability equal to µ/(λ + µ). Hence, P {X m} ( n + m 1 n 1 µ λ )( λ + µ )n ( λ + µ )m 15. (Extra Credit: Ross 6.20) Letting the state be the number of failing machines, we obtain a birth and death process with parameters λ i λ, i 0,1 µ i µ, i 1,2 Based on the material of pages in your textbook, we have that E[time to go from 0 to 2] E[T 0 ] + E[T 1 ] 2/λ + µ/λ 2 7

8 and V ar(time to go from 0 to 2) V ar(t 0 ) + V ar(t 1 ) 1 λ λ(λ + µ) + µ λ 3 + µ µ + λ (2/λ + µ/λ2 ) 2 Computing the limiting probabilities of this process, we can express the requested probability as P 0 + P λ/µ 1 + λ/µ + (λ/µ) (Extra Credit: Ross 6.21) Now we have a birth and death process with the same state space as before, and parameters λ i λ, i 0,1 Therefore, P 0 + P 1 µ i iµ, i 1,2 1 + λ/µ 1 + λ/µ + (λ/µ) 2 /2 and so the probability that at least one machine is up is higher in this case. 8