MSC-600_Module_09_Practice Problems_9_Solution
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Dec 6, 2023
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MSC 600 Quantitative Methods
Practice Problems_9 Solutions
1)
Willow Brook National Bank operates a drive-up teller window that allows
customers to complete bank transactions without getting out of their cars. On
weekday mornings, arrivals to the drive-up teller window occur at random,
with an arrival rate of 24 customers per hour or 0.4 customers per minute.
Assume that the service times for the drive-up teller follow an exponential
probability distribution with a service rate of 36 customers per hour, or 0.6
customers per minute. (Single-server drive-up bank teller operation)
a)
What is the mean or expected number of customers that will arrive in a
five-minute period?
b)
Assume that the Poisson probability distribution can be used to describe
the arrival process. Use the arrival rate in part (a) and compute the
probabilities that exactly 0, 1, 2 and 3 customers will arrive during a five
minute period.
c)
Delays are expected if more than three customers arrive during any five-
minute period. What is the probability that delays will occur?
d)
What is the probability that the service time is one minute or less?
e)
What is the probability that service time is more than two minutes?
f)
What is the probability that there are no customers in the system?
g)
What is the average number of customers waiting?
h)
What is the average number of customers in the system?
i)
What is the average time a customer spends waiting?
j)
What is the average time a customer spends in the system?
k)
What is the probability that arriving customers will have to wait for
service?
Solution:
a.
= 5(0.4) = 2 per five minute period
b.
x
P(x)
0
0.1353
1
0.2707
2
0.2707
3
0.1804
c.
P
(Delay Problems) =
P
(
x
> 3) = 1 -
P
(
x
3) = 1 - 0.8571 = 0.1429
P
(
x
)
=
x
e
-
x
!
=
2
x
e
-2
x
!
MSC 600 Quantitative Methods
d.
µ
= 0.6 customers per minute
P
(service time
1) = 1 -
e
-(0.6)1
= 0.4512
e.
P
(service time
2) = 1 -
e
-(0.6)2
= 0.6988
P
(service time > 2) = 1 - 0.6988 = 0.3012
3.
f.
g.
h.
i.
j.
k.
2)
Pete’s Market is a small local grocery store with only one checkout counter.
Assume that shoppers arrive at the checkout lane according to a Poisson
probability distribution, with an arrival rate of 15 customers per hour. The
checkout service times follow an exponential probability distribution, with a
service rate of 20 customers per hour.
a)
Compute the operating characteristics for this waiting line.
b)
If the manager’s service goal is limit the waiting time prior to beginning to
check-out process to no more than five minutes, what recommendation
would you provide regarding the current checkout system?
Solution:
P
0
=
1 -
= 1 -
0.4
0.6
= 0.3333
L
q
=
2
(
-
)
=
(0.4)
2
0.6 (0.6 - 0.4)
= 1.3333
L
=
L
q
+
=
1.3333 +
0.4
0.6
=
2
W
q
=
L
q
=
1.3333
0.4
=
3.3333 min.
W
=
W
q
+
1
=
3.3333 +
1
0.6
=
5 min.
P
w
=
=
0.4
0.6
= 0.6667
MSC 600 Quantitative Methods
P
0
=
1 -
= 1 -
15
20
= 0.25
L
q
=
2
(
-
)
=
15
2
20 (20 - 15)
= 2.25
L
=
L
+
= 3
W
q
=
L
q
= 0.15 hours
(9 minutes)
W
=
W
q
+
1
= 0.20 hours
(12 minutes)
P
w
=
=
15
20
= 0.75
With
W
q
= 9 minutes, the checkout service needs improvements.
3)
After reviewing the waiting line of the previous problem, the manager of
Pete’s Market wants to consider one of the following alternatives for
improving service. What alternative would you recommend? Justify your
recommendation.
a)
Hire a second person to bag the groceries while cash register operator is
entering the cost data and collecting money form the customer. With this
improved single-server operation, the service rate could be increased to
30 customers per hour.
b)
Hire a second person to operate a second checkout counter. The two-
server operation would have a service rate of 20 customers per hour for
each server.
Solution:
Average waiting time goal: 5 minutes or less.
a.
One checkout counter with 2 employees
=
15
=
30 per hour
L
q
=
2
(
-
)
=
15
2
30 (30 - 15)
= 0.50
W
q
=
L
q
= 0.0333 hours
(2 minutes)
b.
Two channel-two counter system
=
15
=
20
per hour for each
P
0
= 0.4545
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MSC 600 Quantitative Methods
L
q
=
(
/
)
2
1! (2 (20) - 15)
2
P
0
=
(15 / 20)
2
(15) (20)
(40 - 15)
2
(0.4545) = 0.1227
W
q
=
L
q
= 0.0082 hours
(0.492 minutes)
Recommend one checkout counter with two people.
This meets the service goal with
W
q
=
2 minutes.
The two counter system has better service, but has the added cost of installing a
new counter.
4)
Jobs arrive randomly at a particular assembly plant: assume that the arrival
rate is five jobs per hour. Service times (in minutes per job) do not follow the
exponential probability distribution. Two proposed designs for the plant’s
assembly operation are shown.
Service Times
Design
Mean
Standard
Deviation
A
6.0
3.0
B
6.25
0.6
a)
What is the service rate in jobs per hour for each design?
b)
For the service rates in part (a), what design appears to provide the best
or fastest service rate?
c)
What are the standard deviations of the service times in hours?
d)
Use the M/G/1 model to compute the operating characteristics for each
design.
e)
Which design provides the best operating characteristics? Why?
Solution
a.
Design
µ
A
60/6 = 10
B
60/6.25 = 9.6
b.
Design A with
µ
= 10 jobs per hour.
c.
3/60 = 0.05 for A
0.6/60 = 0.01 for B
d.
Characteristic
Design A
Design B
MSC 600 Quantitative Methods
P
0
0.5000
0.4792
L
q
0.3125
0.2857
L
0.8125
0.8065
W
q
0.0625
0.0571
W
0.1625
0.1613
P
w
0.5000
0.5208
e.
Design B is slightly better due to the lower variability of service times.
System A:
W
= 0.1625 hrs
(9.75 minutes)
System B:
W
= 0.1613 hrs
(9.68 minutes)