# Operations Management - Module 5 – Chapter 11

```MODULE 5 – CHAPTER 11
MODULE 5 – ASSIGNMENT CHAPTER 11 – QUESTIONS 1, 2, 3, 6, 14, 15, 18, 19, 24
1. a) What is the mean or expected number of customers that will arrive in a five-minute period?
The mean or expected number of customers to arrive in a 5-minute period is:
0.40 probability of 1 customer per minute x 5 minutes = 5 x 0.40 = 2
2 customers are expected.
b) Assume that the Poisson probability distribution can be used to describe the arrival process.
Use the arrival rate in part (a) and computer the probabilities that exactly 0, 1, 2, and 3 customers
will arrive during a 5-minute period:
๐(0) =
๐(1) =
๐(2) =
๐(3) =
(2.0)0 โ −2.0
0!
(2.0)1 โ −2.0
1!
(2.0)2 โ −2.0
2!
(2.0)3 โ −2.0
3!
= 0.13533
Number of Arrivals Probability
0
0.13533
1
0.27067
2
0.27067
3
0.18045
= 0.27067
= 0.27067
= 0.18045
c. Delays are expected if more than three customers arrive during any 5-minute period. What is
the probability that delays will occur?
P(x &gt; 3, 2) = 1 - P(0; 2) + P(1; 2) + P(2; 2) + P(3; 2)
P(x &gt; 3, 2) = 1 – (
(2.0)0 โ −2.0
0!
)+(
(2.0)1 โ −1.0
1!
)+(
(2.0)2 โ −2.0
2!
)+(
P(x &gt; 3, 2) = 1 - (0.13533 + 0.27067 + 0.27067 + 0.18045)
P(x &gt; 3, 2) = 0.14288
0.14288 is the probability a delay will occur
(2.0)3 โ −2.0
3!
)
MODULE 5 – CHAPTER 11
2. 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. Use the
exponential probability distribution to answer the following questions:
a) What is the probability that the service time is one minute or less?
P(service time ≤ 1.0 min.) = 1 − โ −0.60(1.0) = ๐. ๐๐๐๐๐
๐. ๐๐๐๐๐ is the probability the service time is 1 minute or less
b) What is the probability that the service time is two minutes or less?
P(service time ≤ 2.0 min.) = 1 − โ −0.60(2.0) = ๐. ๐๐๐๐๐
๐. ๐๐๐๐๐ is the probability the service time is 2 minutes or less
c) What is the probability that the service time is more than two minutes?
1- ๐. ๐๐๐๐๐ = ๐. ๐๐๐๐๐
๐. ๐๐๐๐๐ is the probability the service time is more than 2 minutes
3. Determine the following operating characteristics for the system:
a. The probability that no customers are in the system
๐
0.40
๐0 = 1 − ๐ข = ๐0 = 1 − 0.60 = 0.33333
The probability that no units are in the system is 0.33333
b. The average number of customers waiting
๐2
0.402
๐ฟ๐ = ๐ข(๐ข−๐) = ๐ฟ๐ = 0.60(0.60−0.40) = 1.33333
The average number of customers waiting is 1.33333
c. The average number of customers in the system
๐
0.40
๐ฟ = ๐ฟ๐ + ๐ข = ๐ฟ = 1.33333 + 0.60 = ๐
The average number of customers in the system is 2
d. The average time a customer spends waiting
๐ฟ๐
1.33333
๐๐ = ๐ = ๐๐ = 0.40 = ๐. ๐๐๐๐
The average time a customer spends waiting is ๐. ๐๐๐๐ ๐ฆ๐ข๐ง๐ฎ๐ญ๐๐ฌ
e. The average time a customer spends in the system
1
1
๐ = ๐๐ +
= ๐ = 3.33333 +
= ๐. ๐๐๐๐๐
๐ข
0.60
The average time a customer spends in the system is 5 minutes
f. The probability that arriving customers will have to wait for service
๐
0.40
๐๐ค = ๐ข = ๐๐ค = 0.60 = ๐. ๐๐๐๐๐
The probability that arriving customers will have to wait for service is ๐. ๐๐๐๐๐
MODULE 5 – CHAPTER 11
6. During the weeknight evenings, customers arrive at Movies Tonight with an arrival rate of
1.25 customers per minute. The checkout clerk has a service rate of 2 customers per minute.
a. What is the probability that no customers are in the system?
๐
1.25
๐0 = 1 − ๐ข = ๐0 = 1 − 2.00 = 0.375
The probability that no customers are in the system is 0.375
b. What is the average number of customers waiting for service?
๐2
1.252
๐ฟ๐ = ๐ข(๐ข−๐) = ๐ฟ๐ = 2.00(2.00−1.25) = 1.04167
The average number of customers waiting is 1.04167
c. What is the average time a customer waits for service to begin?
๐ฟ๐
1.04167
๐๐ = ๐ = ๐๐ = 1.25 = ๐. ๐๐๐๐๐
The average time a customer spends waiting is ๐. ๐๐๐๐๐ ๐ฆ๐ข๐ง๐ฎ๐ญ๐๐ฌ or 50 seconds
d. What is the probability that an arriving customer will have to wait for service?
๐
1.25
๐๐ค = ๐ข = ๐๐ค = 2.00 = ๐. ๐๐๐
The probability that arriving customers will have to wait for service is ๐. ๐๐๐
e. Do the operating characteristics indicate that the one-clerk checkout system provides an
acceptable level of service?
A customer having a 62.5% chance of having to wait approximately 50 seconds is an acceptable
level of service in my opinion.
14. The customer calls follow a Poisson probability distribution, with an arrival rate of five calls
per hour. Qn average, it takes 7.5 minutes for a consultant to answer a customer s questions.
a. What is the service rate in terms of customers an hour?
60 ๐๐๐๐ข๐กโ๐
7.5 ๐๐๐๐ข๐กโ๐  ๐โ๐ ๐๐๐๐
=8
The service rate is 8 customers per hour
b. What is the probability that no customers are in the system and the consultant is idle?
๐
5
๐0 = 1 − ๐ข = ๐0 = 1 − 8 = 0.375
The probability that no customers are in the system is 0.375
MODULE 5 – CHAPTER 11
c) What is the average number of customers waiting for a consultant?
๐2
52
๐ฟ๐ = ๐ข(๐ข−๐) = ๐ฟ๐ = 8(8−5) = 1.04167
The average number of customers waiting is 1.04167
d) What is the average time a customer waits for a consultant?
๐ฟ๐
1.04167
๐๐ = ๐ = ๐๐ =
= ๐. ๐๐๐๐๐๐
5
The average time a customer spends waiting is ๐. ๐๐๐๐๐๐ ๐ก๐จ๐ฎ๐ซ๐ฌ or 12 minutes and 30
seconds
e) What is the probability that a customer will have to wait for a consultant?
๐
5
๐๐ค = ๐ข = ๐๐ค = 8 = ๐. ๐๐๐
The probability that arriving customers will have to wait for service is ๐. ๐๐๐
f) If Ocala customer service guidelines state that no more than 35% of customers should have to
wait for technical support and the waiting time should be 2 minutes or less, Ocala is not meeting
its customer service guidelines as 62.5% of customer have to wait and the average wait time is
over 12 minutes. Ocala needs to either hire more consultants or make the system more efficient.
They could consider adding self-service prompts to receive prerecorded instructions for common
technical problems with the option to escalate to a consultant if required.
15. What effect would adding a second consultant have on customer service?
Would two consultants allow Ocala to meet its customer service guidelines of 35% or less
customers wait and the wait time is 2 minutes or less?
๐
๐0 = 0.52395 ๐๐  ๐โ๐ ๐ก๐๐๐โ 11.4: ๐ข =
5
8
= ๐. ๐๐๐
Average table values for 0.60 and 0.65 = (0.5385+0.5094)/2 = 0.52395
๐ท๐ = ๐. ๐๐๐๐๐
The probability that an arriving unit has to wait for service:
1
๐ ๐
๐๐ข
1
5 2
2(8)
๐๐ค = ๐! (๐ข) (๐๐ข−๐) ๐0 = 2! (8) (2(8)−5) 0.52395 = ๐. ๐๐๐๐๐๐
The probability that a customer would have to wait with 2 servers would decrease to 0.148849,
which is within the Ocala customer service guidelines of 35% or less.
The average number of units in the waiting line:
(๐โ๐ข)๐ ๐๐ข
(5โ8)2 5(8)
๐ฟ๐ = (๐−1!)(๐๐ข−๐)2 ๐0 = (2−1!)(2(8)−5)2 0.52395 = 0.06766
MODULE 5 – CHAPTER 11
The average time a customer spends waiting:
๐๐ =
๐ฟ๐
๐
= 0.013532 = or 0.81191 minutes or approximately 49 seconds.
A 49 second wait time is within the Ocala customer service guidelines of 2 minutes or less.
18. The airport has three screening stations available. The service rate for processing passengers
at each screening station is 3 passengers per minute. The arrival rate is 5.4 passengers per
minute. Assume that processing times at each screening station follow an exponential
distribution and that arrivals follow a Poisson distribution.
a. Suppose two of the three screening stations are open on Monday morning. Compute the
operating characteristics for the screening facility.
๐ = 5.4 ๐๐๐ ๐ โ๐๐โ๐๐  ๐โ๐ ๐๐๐๐ข๐กโ
๐ข = 3 ๐๐๐ ๐ โ๐๐โ๐๐  ๐โ๐ ๐๐๐๐ข๐กโ ๐โ๐ ๐๐๐๐ ๐
๐ = 2
1. ๐โโ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐กโโ๐โ ๐๐โ ๐๐ ๐๐ข๐ ๐ก๐๐โ๐๐  ๐ค๐๐๐ก๐๐๐ ๐๐๐ ๐ โ๐๐ข๐๐๐ก๐ฆ ๐ ๐๐โโ๐๐๐๐.
๐0 = 0.0526 (5.4/3 = 1.80 ๐/ ๐ข ๐๐๐ก๐๐ − ๐โ๐ ๐ก๐๐๐โ 11.4 ๐๐ ๐กโโ ๐กโ๐ฅ๐ก:
= ๐. ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐กโโ๐โ ๐๐โ ๐๐ ๐๐ข๐ ๐ก๐๐โ๐๐  ๐ค๐๐๐ก๐๐๐ ๐๐๐ ๐ โ๐๐ข๐๐๐ก๐ฆ ๐ ๐๐โโ๐๐๐๐.
2. the average number of customers in line:
(๐โ๐ข)๐ ๐๐ข
(5.4โ3)2 5.4(3)
๐ฟ๐ = (๐−1!)(๐๐ข−๐)2 ๐0 = (2−1!)(2(3)−5.4)2 0.0526 = ๐. ๐๐๐๐
There is an average of 7.6691 customers in line.
3. The average number of customers in the system:
๐
๐ฟ = ๐ฟ๐ + ๐ข = 7.6691 + 1.8 = 9.4691 average number of customers in the system
There is an 9.4691 average number of customers in the system
4. The average time a customer spends in the waiting line:
๐๐ =
๐ฟ๐
๐
=
7.6691
5.4
= 1.4202
The average time a customer spends in the waiting line is 1.4202 minutes
5. The average time a customer spends in the system is:
๐ = ๐๐ +
1
๐ข
= 1.4202 + 0.33333 = 1.75353
The average time a customer spends in the system is: 1.75353 minutes
MODULE 5 – CHAPTER 11
b. Because of space considerations, the facility manager’s goal is to limit the average number of
passengers waiting in line to 10 or fewer. Will the two-screening-station system be able to meet
the manager’s goal? Yes, the average number of passengers in line with 2 screening stations
is 7.6991.
c. What is the average time required for a passenger to pass through security screening?
1.75353 minutes or 1 minute, 45 seconds.
19. When security level is raised to high, the service rate for processing passengers is reduced to
2 per minute at each screening station. The arrival rate is 5.4 passengers per minute.
๐ = 5.4 ๐๐๐ ๐ โ๐๐โ๐๐  ๐โ๐ ๐๐๐๐ข๐กโ
๐ข = 2 ๐๐๐ ๐ โ๐๐โ๐๐  ๐โ๐ ๐๐๐๐ข๐กโ ๐โ๐ ๐๐๐๐ ๐
๐ = 3
1. ๐โโ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐กโโ๐โ ๐๐โ ๐๐ ๐๐ข๐ ๐ก๐๐โ๐๐  ๐ค๐๐๐ก๐๐๐ ๐๐๐ ๐ โ๐๐ข๐๐๐ก๐ฆ ๐ ๐๐โโ๐๐๐๐.
๐0 = 0.0526 (5.4/3 = 2.70 ๐/ ๐ข ๐๐๐ก๐๐ − ๐โ๐ ๐ก๐๐๐โ 11.4 ๐๐ ๐กโโ ๐กโ๐ฅ๐ก:
= ๐. ๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ก๐ฆ ๐กโโ๐โ ๐๐โ ๐๐ ๐๐ข๐ ๐ก๐๐โ๐๐  ๐ค๐๐๐ก๐๐๐ ๐๐๐ ๐ โ๐๐ข๐๐๐ก๐ฆ ๐ ๐๐โโ๐๐๐๐.
2. the average number of customers in line:
(๐โ๐ข)๐ ๐๐ข
(5.4โ2)3 5.4(2)
๐ฟ๐ = (๐−1!)(๐๐ข−๐)2 ๐0 = (3−1!)(3(2)−5.4)2 0.02525 = ๐. ๐๐๐๐๐
There is an average of 7.45494 customers in line.
3. The average number of customers in the system:
๐
๐ฟ = ๐ฟ๐ + ๐ข = 7.45494 + 2.70 = 10.15494 average number of customers in the system
There is an 10.15494 average number of customers in the system
4. The average time a customer spends in the waiting line:
๐๐ =
๐ฟ๐
๐
=
๐.๐๐๐๐๐
5.4
= 1.38054
The average time a customer spends in the waiting line is 1.38054 minutes
5. The average time a customer spends in the system is:
๐ = ๐๐ +
1
๐ข
= 1.38054 + 0.50 = 1.75353
The average time a customer spends in the system is: 1.88054 minutes
MODULE 5 – CHAPTER 11
a. The facility manager’s goal is to limit the average number of passengers line to 10 or fewer.
How many screening stations must be open in order to manager's goal?
3 Screen stations must be opened with an average of ๐. ๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐ ๐๐๐๐.
b. What is the average time required for a passenger to pass through security screening?
1.88054 minutes or 1 minute and 53 seconds.
24. The average time between the arrival of a customer at the food-service counter and his or her
departure with a filled order is 10 minutes. During the game, customers arrive at the rate of four
per minute. The food-service operation requires an average of 2 minutes per customer order.
๐ = 4 ๐๐ข๐ ๐ก๐๐โ๐๐  ๐โ๐ ๐๐๐๐ข๐กโ
๐ข = 0.50 ๐๐ข๐ ๐ก๐๐โ๐๐  ๐โ๐ ๐๐๐๐ข๐กโ
a. What is the service rate per server in terms of customers per minute?
๐ข = 1 ๐๐๐๐ข๐กโ /2 ๐๐๐๐ข๐กโ๐  ๐โ๐ ๐๐ข๐ ๐ก๐๐โ๐ = 0.50 ๐๐ข๐ ๐ก๐๐โ๐๐  ๐โ๐ ๐๐๐๐ข๐กโ
b. What is the average waiting time in the line prior to placing an order?
W = 10 ๐๐๐๐ข๐กโ๐  = ๐๐ +
1
0.50
=
Wq = 10 minutes – 2 minutes = 8
the total time waiting in line is 8 minutes.
c. On average, how many customers are in the food-service system?
Wq=Lq / λ
8= Lq / 4
Lq = 32
L= Lq + λ/ μ
L= 32 + 4/0.5
L=40 (therefore 40 is the average number of units in the system)
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