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01q2sol

# 01q2sol - Urban Operations Research Compiled by James S...

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± Urban Operations Research Compiled by James S. Kang Fall 2001 Quiz 2 Solutions 12/5/2001 Problem 1 (Larson, 2001) λ 2 1 (a) One is tempted to say yes by setting ρ = = 2 × 2 = 2 .Bu t λ = 2 is not the rate at which customers are accepted into the system because we have a loss system. Thus the answer is no, and we must derive the correct ﬁgure. We can use the following aggregate birth-death process (state transition diagram for an M/M/2 queueing system with no waiting space) to compute the workloads: 2 1 2 2 0 2 4 The balance equations and the normalization equation are 2 P 0 =2 P 1 2 P 1 =4 P 2 P 0 + P 1 + P 2 =1 Solving the equations, we obtain 2 2 1 P 0 = ,P 1 = 2 = . 5 5 5 The workloads of server 1 and server 2 are then given by 1 2 1 2 ρ 1 = P 1 + P 2 = 2 = P 1 + P 2 = . 2 5 2 5 (b) The 2-dimensional hypercube state transition diagram is given below. From the steady-state probabilities computed in part (a) and the symmetry of the system, we have 2 1 1 1 P 00 = P 0 = 11 = P 2 = 10 = P 01 = P 1 = . 5 5 2 5 The fraction of dispatches that take server 1 to sector 2 is λ 2 1 1 1 f 12 = P 10 = = . (1 P 11 ) λ (1 1 )2 5 8 5 1

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± 2 0,0 1,0 1,1 0,1 2 1 2 2 2 2 1 ¯ (c) The mean travel time to a random served customer, T , is obtained by ¯ T = f 11 T 1 (sector 1) + f 22 T 2 (sector 2) + f 12 T 1 (sector 2) + f 21 T 2 (sector 1) . Since the travel speed is constant, let us ﬁrst compute the mean travel distance to a random ¯ customer, D . ¯ D = f 11 D 1 (sector 1) + f 22 D 2 (sector 2) + f 12 D 1 (sector 2) + f 21 D 2 (sector 1) . Using the knowledge of Chapter 3, we have 1 1 2 1 1 2 D 1 (sector 1) = + = ,D 2 (sector 2) = + = , 3 3 3 3 3 3 1 4 1 4 D 1 (sector 2) = 1 + = 2 (sector 1) = 1 + = . 3 3 3 3 We compute f 11 as follows: λ 1 1 2 1 3 f 11 = ( P 00 + P 10 )= + = . (1 P 11 ) λ (1 1 5 )2 5 5 8 Invoking the symmetries, we know 1 3 f 21 = f 12 = ,f 22 = f 11 = . 8 8 Putting all together, D ¯ = 3 · 2 + 3 · 2 + 1 · 4 + 1 · 4 = 5 mile . 8 3 8 3 8 3 8 3 6 ¯ D ¯ Hence the mean travel time to a random served customer is T = hr = 3 . 0 sec. This 1000 2
means that changes in total service time due to changes in travel time are insigniﬁcant and ¯ therefore the Markov models applies. Note that another way to compute D is P 00 ( 2 )+( P 01 + P 10 )( 1 · 2 + 1 · 4 ) 2 ( 2 )+ 2 ( 1 · 4 + 1 · 2 ) 5 D ¯ = 3 2 3 2 3 = 5 3 5 2 3 2 3 = .

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01q2sol - Urban Operations Research Compiled by James S...

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