03q2sol

# 03q2sol - Urban OR Quiz 2 Solutions(2003 Problem 1(a Wo = 2...

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Urban OR: Quiz 2 Solutions (2003) Problem 1: (a) W o = ƒ 2 λ i E [ S i 2 ] = 12 1 + 12 9 = 1 min 2 60 2 60 2 i = 1 W q 1 = W o = 1 = 1.25 min (1 ρ 1 ) (1 12 ) 60 W q 2 = (1 ρ 1 )(1 W o ρ 1 ρ 2 ) = (1 12 )(1 1 12 12 ) = 6.25 min 60 60 20 W q , priority = (1.25 + 6.25) / 2 = 3.75 min λ [ 1 2 + σ S 2 ] 24 [2 2 + 1] W q , FCFS = W q , M / G /1 = µ = 60 = 5 min 2(1 ρ ) 2 (1 24 ) 30 The expected waiting time for the FCFS system is greater than the expected waiting time for the priority system, as expected. When we assign priority to the customers with the shortest expected service time, we minimize the expected waiting time (see Corollary on page 239 of the text). (b) Use expression (4.107a) – note that ρ 1 =20/60=1/3 and ρ 2 =20/20=1 and ρ 1 + ρ 2 = 4/3 (>1). E [ S 1 2 ] E [ S 2 2 ] 1 1 2 9 ρ 1 + (1 ρ 1 ) + 2 E [ S 1 ] 2 E [ S 2 ] 3 2 1 3 2 3 W q ,1 = (1 ρ 1 ) = ( ) = 1.75 min 2 3 and W q ,2 = ∞ (c) Now, since ρ 1 =1, neither class of customers attains steady state. Both classes experience infinite expected waiting time in the long run.

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Problem 2. (a) The ambulances are almost always available. The system is either in state 000 or (very rarely) in state 001 or 010 or 100. Thus, whenever an ambulance is dispatched, it is dispatched to a random location in its own district or sector (i.e., the set of points within one mile of the ambulance's home location). Thus the response distance is a r.v. uniform over [0,1]. The mean response distance is 1/2 mile. (b) The workload or utilization factor is ρ = λ /(3 µ ) < 1. Since µ = 1, λ max = 3 calls/hour. (c) Here all three servers are almost always busy. That is, with probability 1 - epsilon , there is always a queue of waiting 'customers.' FCFS means that the location of the next queued customer is uniform over the entire triangle. The triangle circumference is 3 miles. The ambulance takes a shortest route from its home location to the customer. The customer is in one of two 3-mile-length segments emanating from the ambulance’s home location, and conditioned on the segment he or she is in, is at a location that is uniformly distributed over that segment. Thus the mean travel distance is 3/2 = 1.5 mi. (d) The ambulance will travel more than one mile if it is serving an inter-district dispatch. Let P j = steady state probability the system is in state j , where state j corresponds to j customers in the system. A random customer arrives and finds the system is state j with probability P j . If the system is in state 0, there is no chance of
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