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ISyE 3044 – Fall 2002 Homework #3 Solutions A. The following problems are transcribed from Banks, Carson, Nelson, and Nicol, Chapter 6. You don’t have to turn these problems in. 1. A tool crib has exponential interarrival and service times, and it serves a very large group of mechanics. The mean time between arrivals is 4 minutes. It takes 3 minutes on the average for a tool-crib attendant to service a mechanic. The attendant is paid $10 per hour and the mechanic is paid $15 per hour. Would it be advisable to have a second tool-crib attendant? Solution: Model the crib as an M/M/c queue with λ = 1/4, µ = 1/3, and c = 1 or 2. The expected cost per hour to run this system is 10c + 15L, where L is the long-run average number of customers in the system. For the case c = 1, we have ρ = λ/µ = 3/4, and so the the M/M/1 queueing table gives L =

ρ . 1−ρ

Therefore, the expected cost is 55. For the case c = 2, we have ρ = λ/cµ = 0.375, and so the the M/M/2 queueing table gives ½· c−1 ¸¾−1 X (cρ)n ¸ · (cρ)c P0 = + = 0.4545, (c!)(1 − ρ) n=0 n! and then L = cρ +

(cρ)c+1 P0 = 0.8727. c(c!)(1 − ρ)2

Therefore, the expected cost is 33.09, so we take c = 2.

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2. A two-runway (one runway for landing, one for taking off) airport is being designed for propeller-driven aircraft. The time to land an airplane is known to be exponentially distributed with a mean of 1.5 minutes. If airplane arrivals are assumed to occur at random, what arrival rate can be tolerated if the average wait in the sky is not to exceed

2 3 minutes? Solution: Model this as a single M/M/1 landing strip with µ = 2/3 (assume service times are exponential). We want the maximum arrival rate λ such that wQ ≤ 3. By the M/M/1 table, ρ λ wQ = = ≤ 3 µ(1 − ρ) µ(µ − λ) if and only if λ ≤

3µ2 = 4/9. 1 + 3µ

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3. The Port of Trop can service only one ship at a time. However, there is mooring space for three more ships. Trop is a favorite port of call, but if no mooring space is available, the ships have to go to the Port of Poop. An average of seven ships arrive each week, according to a Poisson process. The Port of Trop has the capacity to handle an average of eight ships a week, with service times exponentially distributed. What’s the expected number of ships waiting or in service at the Port of Trop? Solution: Model this as an M/M/1/N queue with λ = 7, µ = 8, and N = 4. This yields a = λ/µ = 7/8. Then from the M/M/1/N table, we get (since λ 6= µ) a[1 − (N + 1)aN + N aN +1 ] L = = 1.735. (1 − aN +1 )(1 − a)

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4. At Metropolis City Hall, two workers “pull strings” every day. Strings arrive to be pulled on the average of one every ten minutes throughout the day. It takes an average of 15 minutes to pull a string. Both times between arrivals and service times are exponentially distributed. What is the probability that there are no strings to be pulled in the system at a random point in time? What’s the expected number of strings waiting to be pulled? What’s the probability that both string pullers are busy? What’s the effect on performance if a third string puller, working at the same speed as the first two, is added to the system? Solution: Model string pulling as an M/M/2 queue with λ = 1/10 and µ = 1/15. This gives ρ = λ/(cµ) = 0.75. Then from the M/M/c table, we have the following. (a) P0 = 1/7. (b) LQ = 1.9286.

3 (c) Pr(L(∞) ≥ 2) = 0.6429. (d) If we add a third worker, so that c = 3, we now get P0 = 0.2105, LQ = 0.2368, and Pr(L(∞) ≥ 3) = 0.2368. 2 5. At Tony and Cleo’s bakery, one kind of birthday cake is offered. It takes 15 minutes to decorate this particular cake, and the job is performed by one particular baker. In fact, this is all the baker does. What mean time between arrivals (exponentially distributed) can be accepted if the mean length of the queue for decorating is not to exceed 5 cakes? Solution: Model the bakery as an M/D/1 queue with µ = 4/hr. Want the maximum λ such that LQ ≤ 5. In other words, by the M/D/1 table, 1 ρ2 1 λ2 = ≤ 5 21−ρ 2 µ(µ − λ)

LQ = if and only if (after some algebra)

λ2 + 40λ − 160 ≤ 0. The roots for this are λ =

−40 ±

q

(40)2 + 4(160)

2 We’ll take the larger root, λ = 3.664 cakes/hr.

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6. Patients arrive for a physical examination according to a Poisson process at the rate of one per hour. The physical exam requires three stages, each one independently and exponentially distributed with a service time of 15 minutes. A patient must go thru all three stages before the next patient is admitted to the treatment facility. Determine the average number of delayed patient, LQ , for this system. Solution: Model the exam as an M/Ek /1 queue with k = 3, λ = 1/60 patient/min., and µ = 1/(15 + 15 + 15) = 1/45. Then ρ = λ/µ = 3/4, and the M/Ek /1 table gives LQ =

1 + k ρ2 = 1.5. 2k 1 − ρ

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7. Suppose that mechanics arrive randomly at a tool crib according to a Poisson process with rate λ = 10 per hour. It is known that a single tool clerk serves a mechanic in 4 minutes on the average, with a standard deviation of approximately 2 minutes. Suppose

4 that mechanics make $15 per hour. Find the steady-state average cost per hour of mechanics waiting for tools. Solution: Model the process as an M/G/1 queue with λ = 10/hr, µ = 15/hr, and σ = (1/30) hr. Thus, we have ρ = 2/3, and from the M/G/1 table, LQ =

ρ2 (1 + σ 2 µ2 ) = 0.833 mechanics. 2(1 − ρ)

This implies that the avg cost per hr is $15LQ = 12.50.

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8. Arrivals to an airport are all directed to the same runway. At a certain time of the day, these arrivals form a Poisson process with rate 30 per hour. The time to land an aircraft is a constant 90 seconds. Determine LQ , wQ , L, and w for this airport. If a delayed aircraft burns $5000 worth of fuel per hour on the average, determine the average cost per aircraft of delay waiting to land. Solution: Model the process as an M/D/1 queue with λ = 30/hr and µ = 40/hr, so that ρ = 3/4. Then from the M/D/1 table, ρ2 = 1.125 planes 2(1 − ρ) wQ = LQ /λ = 0.0375 hr ρ2 L = ρ+ = 1.875 2(1 − ρ) w = L/λ = 0.0625 hr 2 LQ =

Also, the average delay cost is $1500wQ = $56.25.

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9. A machine shop repairs small electric motors which arrive according to a Poisson process at a rate of 12 per week (5-day, 40-hour workweek). An analysis of past data indicates that engines can be repaired, on the average, in 2.5 hours, with a variance of 1 hour2 . How many working hours should a customer expect to leave a motor at the repair shop (not knowing the status of the system)? If the variance of the repair time could be controlled, what variance would reduce the expected waiting time to 6.5 hours? Solution: Model the pump as an M/G/1 queue with λ = 12/40 = 0.3/hr, µ = 1/2.5 = 0.4/hr, and σ = 1 hr. Thus, ρ = 0.75. Then from the M/G/1 table, L = ρ+

ρ2 (1 + σ 2 µ2 ) = 2.055 2(1 − ρ)

5 and so w = L/λ = 6.85 hrs.

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By the way, note that the expected waiting time is L 1 − = 4.35. λ µ

wQ =

This suggests that what the second part of the problem really wants is to find the variance that reduces w (not wQ ) to 6.5 hrs. In this case, it turns out that we need w = L/λ =

ρ ρ2 (1 + σ 2 µ2 ) + ≤ 6.5. λ 2λ(1 − ρ)

After algebra, we take σ 2 ≤ 0.417 hrs2 .

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10. Arrivals to a self-service gasoline pump occur in a Poisson fashion at the rate of 12 per hour. Service time has a distribution which averages 4 minutes with a standard deviation of 1 13 minutes. What’s the expected number of vehicles in the system? Solution: Model the pump as an M/G/1 queue with λ = 12/hr, µ = 15/hr, and σ = 0.02222 hr. Thus, ρ = 4/5. Then from the M/G/1 table, L = ρ+

ρ2 (1 + σ 2 µ2 ) = 2.577. 2(1 − ρ)

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