hmwk12sol - ISyE 3232 Stochastic Manufacturing and Service...

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ISyE 3232 Stochastic Manufacturing and Service Systems Fall 2011 H. Ayhan Solutions to Homework 12 1. The throughput in each case is equal to the e±ective arrival rate λ ef = λ (1 P(Customer arrives when all three lines are busy)) To calculate the probability of a customer arriving when all three lines are busy, I will use a Continuous Time Markov Chain (CTMC) X K ( t ) to represent the number of people in the system at time t when there are K servers. Because the system has only three phone lines, the state space is S = { 0 , 1 , 2 , 3 } .G i v e n K , the stationary distribution can be calculated using the cut method. The average number of people in the queue can also be calculated from the stationary distribution of the DTMC. With the throughput and average number of people in the queue, Little’s Law allows us to calculate the average waiting time. (a) The rate diagram for the case with one server ( K = 1) is ± ²³ ´ ±±± ±²³ ±²³ ±²³ Making a cut between 0 and 1, 1 and 2, and 2 and 3 gives the equations: η 0 = 1 2 η 1 η 1 = 1 2 η 2 η 2 = 1 2 η 3 Setting η 0 = 1 gives the invariant measure η = { 1 , 2 , 4 , 8 } ,w ith ° 3 i =0 η i = 15. Using the relation between stationary distributions and invariant measures, π n = η n ± ° 3 i =0 η i ² results in the stationary distribution π = { 1 / 15 , 2 / 15 , 4 / 15 , 8 / 15 } . The probability of a caller arriving to see all three lines busy equals π 3 ,so λ ef = λ (1 π 3 )=1 ³ 1 8 15 ´ = 7 15 per minute Because there is one server, the number in the queue is one less than the state of the Markov chain. The average number of people in the queue is
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hmwk12sol - ISyE 3232 Stochastic Manufacturing and Service...

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