4106-08-h4-solu

4106-08-h4-solu - HMWK 4 Solutions Problems from the Text...

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HMWK 4 Solutions Problems from the Text Ch. 5, Page 346: 4, 37, 44 (Hint: Wald’s Equation) 4. (a) Here, A and B leave together after 10 minutes, and then C enters service. Therefore the answer is zero. (b) The only way A can still be in service after B and C leave is if A’s service time is 3, and B and C’s both are 1. The probability of this event is (1 / 3) 3 = 1 / 27 . (c) First of all, A’s service time S A must be greater than B’s , S B , and this happens with probability 1 / 2 = P ( S A > S B ), and then this must be followed by A’s remaining service time being greater than C’s service time, which by the memoryless property also has probability 1 / 2, and is independent of the past. Therefore (by independence) the answer is the product, 1 / 4 . 37. She is uninjured during time of length s if and only if N ( s ) = 0; P ( N ( s ) = 0) = e - λs , with λ = 3. For s = 2 60 , 5 60 , 10 60 , 20 60 minutes, we see that e - 3 s = e - 1 10 ,e - 1 4 ,e - 1 2 ,e - 1 44. (METHOD I:) Let X n denote the iid exponential λ interarrival times for the Poisson process. Let τ = min { n 1 : X n > T } , a stopping time. Then the waiting time W can be written as W = τ - 1 X n =1 X n = ± τ X n =1 X n ² - X τ , where X τ has the conditional distribution of ( X | X > T ). Clearly, τ has a geometric distribution with “success” probability p = P ( X > T ) = e - λT ; and E ( τ ) = 1 /p = e λT . W = 0 if and only if τ = 1; P ( W = 0) = p = e - λT . From Wald’s equation E ( W ) = E ( τ ) E ( X ) - E ( X | X > T ) = e
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4106-08-h4-solu - HMWK 4 Solutions Problems from the Text...

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