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ISyE 3044 – Spring 2009
Homework #3 Solutions
1. Banks, Carson, Nelson, and Nicol, Chapter 6 (Queueing): Do most of #1–10.
6–1. 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
ρ
L = .
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
c−1
(cρ)n (cρ)c −1
P0 = + = 0.4545,
n=0 n! (c!)(1 − ρ)
and then
(cρ)c+1 P0
L = cρ + = 0.8727.
c(c!)(1 − ρ)2
Therefore, the expected cost is 33.09, so we take c = 2. 2
6–2. 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

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assumed to occur at random, what arrival rate can be tolerated if the average wait
in the sky is not to exceed 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
3µ2
λ ≤ = 4/9. 2
1 + 3µ
6–3. The Port of Trop can service only one ship at a time. However, there is moor-
ing 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 exponen-
tially 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 = µ)
a[1 − (N + 1)aN + N aN +1 ]
L = = 1.735. 2
(1 − aN +1 )(1 − a)
6–4. 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 ser-
vice 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.

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(a) P0 = 1/7.
(b) LQ = 1.9286.
(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
6–5. 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 partic-
ular 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
LQ = = ≤ 5
21−ρ 2 µ(µ − λ)
if and only if (after some algebra)
λ2 + 40λ − 160 ≤ 0.
The roots for this are
−40 ± (40)2 + 4(160)
λ = .
2
We’ll take the larger root, λ = 3.664 cakes/hr. 2
6–6. 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 in-
dependently 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.

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Solution: Model the exam as an M/Ek /1 queue with k = 3, λ = 1/60 pa-
tient/min., and µ = 1/(15 + 15 + 15) = 1/45. Then ρ = λ/µ = 3/4, and the
M/Ek /1 table gives
1 + k ρ2
LQ = = 1.5. 2
2k 1 − ρ
6–7. 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 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,
ρ2 (1 + σ 2 µ2 )
LQ = = 0.833 mechanics.
2(1 − ρ)
This implies that the avg cost per hr is $15LQ = 12.50. 2
6–8. 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
LQ = = 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
Also, the average delay cost is $1500wQ = $56.25. 2
6–9. 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

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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,
ρ2 (1 + σ 2 µ2 )
L = ρ+ = 2.055
2(1 − ρ)
and so w = L/λ = 6.85 hrs. 2
By the way, note that the expected waiting time is
L 1
wQ = − = 4.35.
λ µ
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
ρ ρ2 (1 + σ 2 µ2 )
w = L/λ = + ≤ 6.5.
λ 2λ(1 − ρ)
After algebra, we take σ 2 ≤ 0.417 hrs2 . 2
6–10. 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 1 minutes. What’s the expected number of vehicles in the
3
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,
ρ2 (1 + σ 2 µ2 )
L = ρ+ = 2.577. 2
2(1 − ρ)
In what follows, we’ll work with some of the Arena models we intro-
duced during lecture. The models referenced below are from my webpage,
www.isye.gatech.edu/∼sman/courses/3044/.

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2. Arena Baby Model 3-1. Just run and play around with this model. You don’t
have to turn anything in.
3. Arena Arrivals Model 4-1. Let’s add Part C’s to the model. Specifically, sup-
pose that Part C’s arrive one-at-a-time according to a Poisson process (that is,
i.i.d. exponential interarrivals) with a mean time between arrivals of 20 minutes.
Suppose these parts go to the Prep C Process station (with one server) and that
the time to prep these parts is normal with mean 4 minutes and standard deviation
1 minute. After this, they go the Sealer Process (just like everyone else), where
they experience sealer times that are exponentially distributed with mean 3. Fi-
nally, Part C’s go through the same inspection process as everyone else. Your job is
to alter the model to accommodate the above changes. You may have to add some
capacity to the sealer and rework resources in order to avoid gigantic lines. Run
the model for 32 hours and let me know how many parts are shipped. salvaged,
and scrapped. Also let me know any other info you find interesting, e.g., how many
resources you needed at each station.
4. Arena Call Center Model 5-1.
(a) Run the model for 10 replications of one day, and see how may people are still
left in the system when everything completely shuts down at 7 p.m.
(b) Let’s suppose that all of the tech support personnel have undergone additional
training and can now handle all three of the products. What is the effect on
the overall model performance? Suggestion: Run 10 independent replications
of the model before and after their magical new abilities appear, and see if the
customer waiting times change at all.