Test 1 Solutions - prestage wait for uniform 5, 1 min move...

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5 ISyE 3044/Spring 2009: Solutions to Test #1 1. (a) Using the transformation t =4 u , we write μ = ± 1 0 4 1+ u du . The Monte Carlo estimate is ˆ μ = 1 9 9 ² i =1 X i ; X i = 4 1+ U i . (b) We have ˆ μ =2 . 45 and S 2 = 1 8 9 i =1 ( X i - ˆ μ ) 2 =0 . 38 2 . Using t 8 , 0 . 05 =1 . 86, we get the conFdence interval 2 . 45 ± 1 . 86 0 . 38 9 =2 . 45 ± 0 . 24 = (2 . 21 , 2 . 69) . (c) No. 2. The car wash is modeled by an M / G /1 queueing system with λ =0 . 1/min, μ =0 . 125/min, and σ 2 = (12 - 4) 2 / 12 = 5 . 33. (a) The number of arrivals in a 30-minute period, N (30), is Poisson(3). The answer is Pr[ N (30) 3] = 1 - Pr[ N (30 2] = 1 - e - 3 (1 + 3 + 3 2 ) = 1 - 8 . 5 e - 3 =0 . 58 . (c) ρ =0 . 8. (d) Using the P-K formula we have W Q = 0 . 1(8 2 +5 . 33) 2(1 - 0 . 8) = 17 . 33 minutes. (e) By Little’s law, the average number of cars in the system is L = λ ( W Q + 8) = 2 . 53 . 3. Notice that a truck cannot leave the pre-stage area before the staging area has available space. begin P inspect arriving procedure move into Q
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Unformatted text preview: prestage wait for uniform 5, 1 min move into Q stage move into Q scan use R scan for uniform 8, 4 min send to die end 4. (a) i time in system( i ) / 6 = 218 / 6 = 36 . 33. (b) Let X ( t ) be the number of entities in the system at time t . Area under sample path of X ( t ) Time 6th customer leaves = 218 100 = 2 . 18 . (c) 1. (d) The queue length at time t is Q ( t ) = max { X ( t )-1 , } . The answer is 1 100 100 Q ( t ) dt = 1 . 18. (e) Entity i starts service when entity i-1 leaves the system. Since we have the departure times, we can obtain the service times by subtracting the service initiation times from the departure times. The answer is 102 / 6 = 16 . 67....
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This note was uploaded on 09/17/2009 for the course ISYE 3044 taught by Professor Alexopoulos during the Spring '08 term at Georgia Institute of Technology.

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