Mathematics > QUESTIONS & ANSWERS > Columbia University - IEOR 41064106-13-Fall-h7- ( ALL SOLUTIONS ARE 100% CORRECT ) (All)

Columbia University - IEOR 41064106-13-Fall-h7- ( ALL SOLUTIONS ARE 100% CORRECT )

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IEOR 4106, HMWK 7, Professor Sigman 1. Cars arrive to a beach parking area (huge, assume of unlimited size) according to a Poisson process at rate ! = 20 per hour. The length of time a car spends pa... rked is an independent random variable with a uniform distribution over the interval (2, 10) hours. Assume that initially (at time t = 0) the parking lot is empty. (a) SOLUTION: Letting X(t) = the number of parked cars at time t, we have an M/G/1 queue with arrival rate ! = 20 and service time distribution the uniform distribution over the interval (2, 10). We know that for each fixed t > 0, X(t) has a Poisson distribution with mean ↵(t) where ↵(t) = ! Z0t P(S > s)ds. Here, P(S > s) = 8><>: 1 0 < s < 2 , (10 " s)/8 2  s < 10, 0 s $ 10 . Thus ↵(1) = ! R01 1ds = ! = 20; ↵(5) = ! Z02 1ds + ! Z25(10 " s)/8ds = 88.75. Since P(S > s) = 0, s $ 10, we observe that for t = 12, ↵(12) = ! R01 P(S > s)ds = !E(S) = ⇢. E(S) = 6, and so the answer is 120. (b) What is the probability that the lot is empty at time t = 3? SOLUTION: ↵(3) = ! Z02 1ds + ! Z23(10 " s)/8ds = 58.75, and we want P(X(3) = 0) = e"↵(3) = 3.06 ⇥ 10"26. (c) What is the long-run average (over all time) number of cars in the lot? SOLUTION: ⇢ = limt!1 ↵(t) = 120 = ↵(12) from (a). (d) Repeat (a)–(c) in the case when the length of time a car spends parked is an independent random variable with an exponential distribution at rate µ = 1/6 per hour. SOLUTION: ↵(t) = ! Z0t e" 6s ds = 120 ⇣1 " e" 6t ⌘ . Therefore, ↵(1) = 18.42, ↵(5) = 67.85, ↵(12) = 103.76. [Show More]

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