spring2011-final-tuesday-A

# spring2011-final-tuesday-A - ISyE 3232 Stochastic...

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Unformatted text preview: ISyE 3232 Stochastic Manufacturing and Service Systems Spring 2011 Section A J. G. Dai Final (May 3, 2011) This is a closed book test. No calculator is allowed. 1. (18 points) Consider a production system that consists of three single-server stations. Each station has an infinite size waiting buffer. When a job to a station finds the server busy, the job waits in the buffer; when the server is free, the server picks the next job from the buffer. Assume that jobs arrive at station 1 following a Poisson process with rate α = 1 job per minute. After being processed at station 1, a job is sent to station 2 without any delay. After being processed at station 2, it will be sent to station 3 with probability 50% or to station 1 with probability 50% to be reworked. After being processed at station 3, it is shipped to a customer with probability 50% or is sent to station 2 with probability 50% to be reworked. The job processing times at station i are iid exponentially distributed with mean m i minutes, i = 1 , 2 , 3. Assume that m 1 = . 3 minutes , m 2 = . 2 minutes , m 3 = . 45 minutes . (a) (5 points) What is long-run average utilization for station i , i = 1 , 2 , 3. (b) (4 points) What is the long-run fraction of time that station 1 has 2 jobs, station 2 has 3 jobs, and station 3 has one job? (Leaving your answer in a numerical expression is OK. ) (c) (3 points) What is the long-run average number of jobs at station 2, including those in the buffer and possibly the one being served? (d) (3 points) What is the throughput of the system? Here, throughput is the rate at which the jobs are being shipped to customers. (e) (3 points) What is the long-run average time in system per job? Solution. (a) Solve the following equations for traffic density, λ 1 = α + 0 . 5 λ 2 λ 2 = λ 1 + 0 . 5 λ 3 λ 3 = 0 . 5 λ 2 Then λ 1 = 3 ,λ 2 = 4 ,λ 3 = 2, the long run average utilization for 3 stations are: ρ 1 = λ 1 m 1 = 0 . 9 , ρ 2 = λ 2 m 2 = 0 . 8 , ρ 3 = λ 3 m 3 = 0 . 9 . 1 (b) (1- ρ 1 ) ρ 2 1 (1- ρ 2 ) ρ 3 2 (1- ρ 3 ) ρ 3 3 = 0 . 0006 (c) ρ 2 1- ρ 2 = 0 . 8 / . 2 = 4 . (d) Throughput is equal to . 5 λ 3 = 1 . (e) The long run average number of customers in the system is equal to L = ρ 1 1- ρ 1 + ρ 2 1- ρ 2 + ρ 3 1- ρ 3 = . 9 . 1 + . 8 . 2 + . 9 . 1 = 22 . By Little’s law, the long-run average time in system is equal to L/α = 22 minutes. 2. (15 points; 3 points each) Let X = { X ( t ) : t ≥ } be a continuous time Markov chain with state space { 1 , 2 , 3 } with generator G = - 4 ? 1 1 ? 2 1 ?- 2 . Using matlab , one can compute e G via command expm ( G ) = . 20539 0 . 36211 0 . 43250 . 19865 0 . 35727 0 . 44407 . 19865 0 . 33896 0 . 46239 ....
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## This note was uploaded on 01/16/2012 for the course ISYE 3232 taught by Professor Billings during the Spring '07 term at Georgia Institute of Technology.

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spring2011-final-tuesday-A - ISyE 3232 Stochastic...

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