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:
𝑃(0) =
𝑃(1) =
𝑃(2) =
𝑃(3) =
(2.0)0 ⅇ −2.0
0!
(2.0)1 ⅇ −2.0
1!
(2.0)2 ⅇ −2.0
2!
(2.0)3 ⅇ −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 > 3, 2) = 1 - P(0; 2) + P(1; 2) + P(2; 2) + P(3; 2)
P(x > 3, 2) = 1 – (
(2.0)0 ⅇ −2.0
0!
)+(
(2.0)1 ⅇ −1.0
1!
)+(
(2.0)2 ⅇ −2.0
2!
)+(
P(x > 3, 2) = 1 - (0.13533 + 0.27067 + 0.27067 + 0.18045)
P(x > 3, 2) = 0.14288
0.14288 is the probability a delay will occur
(2.0)3 ⅇ −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 − ⅇ −0.60(1.0) = 𝟎. 𝟒𝟓𝟏𝟏𝟗
𝟎. 𝟒𝟓𝟏𝟏𝟗 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 − ⅇ −0.60(2.0) = 𝟎. 𝟔𝟗𝟖𝟖𝟏
𝟎. 𝟔𝟗𝟖𝟖𝟏 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- 𝟎. 𝟔𝟗𝟖𝟖𝟏 = 𝟎. 𝟑𝟎𝟏𝟏𝟗
𝟎. 𝟑𝟎𝟏𝟏𝟗 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
𝜆
0.40
𝑃0 = 1 − 𝑢 = 𝑃0 = 1 − 0.60 = 0.33333
The probability that no units are in the system is 0.33333
b. The average number of customers waiting
𝜆2
0.402
𝐿𝑞 = 𝑢(𝑢−𝜆) = 𝐿𝑞 = 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
𝜆
0.40
𝐿 = 𝐿𝑞 + 𝑢 = 𝐿 = 1.33333 + 0.60 = 𝟐
The average number of customers in the system is 2
d. The average time a customer spends waiting
𝐿𝑞
1.33333
𝑊𝑞 = 𝜆 = 𝑊𝑞 = 0.40 = 𝟑. 𝟑𝟑𝟑𝟑
The average time a customer spends waiting is 𝟑. 𝟑𝟑𝟑𝟑 𝐦𝐢𝐧𝐮𝐭𝐞𝐬
e. The average time a customer spends in the system
1
1
𝑊 = 𝑊𝑞 +
= 𝑊 = 3.33333 +
= 𝟒. 𝟗𝟗𝟗𝟗𝟗
𝑢
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
𝜆
0.40
𝑃𝑤 = 𝑢 = 𝑃𝑤 = 0.60 = 𝟎. 𝟔𝟔𝟔𝟔𝟕
The probability that arriving customers will have to wait for service is 𝟎. 𝟔𝟔𝟔𝟔𝟕
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?
𝜆
1.25
𝑃0 = 1 − 𝑢 = 𝑃0 = 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?
𝜆2
1.252
𝐿𝑞 = 𝑢(𝑢−𝜆) = 𝐿𝑞 = 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?
𝐿𝑞
1.04167
𝑊𝑞 = 𝜆 = 𝑊𝑞 = 1.25 = 𝟎. 𝟖𝟑𝟑𝟑𝟒
The average time a customer spends waiting is 𝟎. 𝟖𝟑𝟑𝟑𝟒 𝐦𝐢𝐧𝐮𝐭𝐞𝐬 or 50 seconds
d. What is the probability that an arriving customer will have to wait for service?
𝜆
1.25
𝑃𝑤 = 𝑢 = 𝑃𝑤 = 2.00 = 𝟎. 𝟔𝟐𝟓
The probability that arriving customers will have to wait for service is 𝟎. 𝟔𝟐𝟓
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 𝑚𝑖𝑛𝑢𝑡ⅇ𝑠
7.5 𝑚𝑖𝑛𝑢𝑡ⅇ𝑠 𝑝ⅇ𝑟 𝑐𝑎𝑙𝑙
=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?
𝜆
5
𝑃0 = 1 − 𝑢 = 𝑃0 = 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?
𝜆2
52
𝐿𝑞 = 𝑢(𝑢−𝜆) = 𝐿𝑞 = 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?
𝐿𝑞
1.04167
𝑊𝑞 = 𝜆 = 𝑊𝑞 =
= 𝟎. 𝟐𝟎𝟖𝟑𝟑𝟒
5
The average time a customer spends waiting is 𝟎. 𝟐𝟎𝟖𝟑𝟑𝟒 𝐡𝐨𝐮𝐫𝐬 or 12 minutes and 30
seconds
e) What is the probability that a customer will have to wait for a consultant?
𝜆
5
𝑃𝑤 = 𝑢 = 𝑃𝑤 = 8 = 𝟎. 𝟔𝟐𝟓
The probability that arriving customers will have to wait for service is 𝟎. 𝟔𝟐𝟓
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?
𝜆
𝑃0 = 0.52395 𝑎𝑠 𝑝ⅇ𝑟 𝑡𝑎𝑏𝑙ⅇ 11.4: 𝑢 =
5
8
= 𝟎. 𝟔𝟐𝟓
Average table values for 0.60 and 0.65 = (0.5385+0.5094)/2 = 0.52395
𝑷𝟎 = 𝟎. 𝟓𝟐𝟑𝟗𝟓
The probability that an arriving unit has to wait for service:
1
𝜆 𝑘
𝑘𝑢
1
5 2
2(8)
𝑝𝑤 = 𝑘! (𝑢) (𝑘𝑢−𝜆) 𝑝0 = 2! (8) (2(8)−5) 0.52395 = 𝟎. 𝟏𝟒𝟖𝟖𝟒𝟗
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:
(𝜆∕𝑢)𝑘 𝜆𝑢
(5∕8)2 5(8)
𝐿𝑞 = (𝑘−1!)(𝑘𝑢−𝜆)2 𝑝0 = (2−1!)(2(8)−5)2 0.52395 = 0.06766
MODULE 5 – CHAPTER 11
The average time a customer spends waiting:
𝑊𝑞 =
𝐿𝑞
𝜆
= 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.
𝜆 = 5.4 𝑝𝑎𝑠𝑠ⅇ𝑛𝑔ⅇ𝑟𝑠 𝑝ⅇ𝑟 𝑚𝑖𝑛𝑢𝑡ⅇ
𝑢 = 3 𝑝𝑎𝑠𝑠ⅇ𝑛𝑔ⅇ𝑟𝑠 𝑝ⅇ𝑟 𝑚𝑖𝑛𝑢𝑡ⅇ 𝑝ⅇ𝑟 𝑘𝑖𝑜𝑠𝑘
𝑘 = 2
1. 𝑇ℎⅇ 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎⅇ𝑟ⅇ 𝑎𝑟ⅇ 𝑛𝑜 𝑐𝑢𝑠𝑡𝑜𝑚ⅇ𝑟𝑠 𝑤𝑎𝑖𝑡𝑖𝑛𝑔 𝑓𝑜𝑟 𝑠ⅇ𝑐𝑢𝑟𝑖𝑡𝑦 𝑠𝑐𝑟ⅇⅇ𝑛𝑖𝑛𝑔.
𝑃0 = 0.0526 (5.4/3 = 1.80 𝜆/ 𝑢 𝑟𝑎𝑡𝑖𝑜 − 𝑝ⅇ𝑟 𝑡𝑎𝑏𝑙ⅇ 11.4 𝑖𝑛 𝑡ℎⅇ 𝑡ⅇ𝑥𝑡:
= 𝟎. 𝟎𝟓𝟐𝟔 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎⅇ𝑟ⅇ 𝑎𝑟ⅇ 𝑛𝑜 𝑐𝑢𝑠𝑡𝑜𝑚ⅇ𝑟𝑠 𝑤𝑎𝑖𝑡𝑖𝑛𝑔 𝑓𝑜𝑟 𝑠ⅇ𝑐𝑢𝑟𝑖𝑡𝑦 𝑠𝑐𝑟ⅇⅇ𝑛𝑖𝑛𝑔.
2. the average number of customers in line:
(𝜆∕𝑢)𝑘 𝜆𝑢
(5.4∕3)2 5.4(3)
𝐿𝑞 = (𝑘−1!)(𝑘𝑢−𝜆)2 𝑝0 = (2−1!)(2(3)−5.4)2 0.0526 = 𝟕. 𝟔𝟗𝟗𝟏
There is an average of 7.6691 customers in line.
3. The average number of customers in the system:
𝜆
𝐿 = 𝐿𝑞 + 𝑢 = 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:
𝑊𝑞 =
𝐿𝑞
𝜆
=
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:
𝑊 = 𝑊𝑞 +
1
𝑢
= 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.
𝜆 = 5.4 𝑝𝑎𝑠𝑠ⅇ𝑛𝑔ⅇ𝑟𝑠 𝑝ⅇ𝑟 𝑚𝑖𝑛𝑢𝑡ⅇ
𝑢 = 2 𝑝𝑎𝑠𝑠ⅇ𝑛𝑔ⅇ𝑟𝑠 𝑝ⅇ𝑟 𝑚𝑖𝑛𝑢𝑡ⅇ 𝑝ⅇ𝑟 𝑘𝑖𝑜𝑠𝑘
𝑘 = 3
1. 𝑇ℎⅇ 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎⅇ𝑟ⅇ 𝑎𝑟ⅇ 𝑛𝑜 𝑐𝑢𝑠𝑡𝑜𝑚ⅇ𝑟𝑠 𝑤𝑎𝑖𝑡𝑖𝑛𝑔 𝑓𝑜𝑟 𝑠ⅇ𝑐𝑢𝑟𝑖𝑡𝑦 𝑠𝑐𝑟ⅇⅇ𝑛𝑖𝑛𝑔.
𝑃0 = 0.0526 (5.4/3 = 2.70 𝜆/ 𝑢 𝑟𝑎𝑡𝑖𝑜 − 𝑝ⅇ𝑟 𝑡𝑎𝑏𝑙ⅇ 11.4 𝑖𝑛 𝑡ℎⅇ 𝑡ⅇ𝑥𝑡:
= 𝟎. 𝟎𝟐𝟓𝟐𝟓 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎⅇ𝑟ⅇ 𝑎𝑟ⅇ 𝑛𝑜 𝑐𝑢𝑠𝑡𝑜𝑚ⅇ𝑟𝑠 𝑤𝑎𝑖𝑡𝑖𝑛𝑔 𝑓𝑜𝑟 𝑠ⅇ𝑐𝑢𝑟𝑖𝑡𝑦 𝑠𝑐𝑟ⅇⅇ𝑛𝑖𝑛𝑔.
2. the average number of customers in line:
(𝜆∕𝑢)𝑘 𝜆𝑢
(5.4∕2)3 5.4(2)
𝐿𝑞 = (𝑘−1!)(𝑘𝑢−𝜆)2 𝑝0 = (3−1!)(3(2)−5.4)2 0.02525 = 𝟕. 𝟒𝟓𝟒𝟗𝟒
There is an average of 7.45494 customers in line.
3. The average number of customers in the system:
𝜆
𝐿 = 𝐿𝑞 + 𝑢 = 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:
𝑊𝑞 =
𝐿𝑞
𝜆
=
𝟕.𝟒𝟓𝟒𝟗𝟒
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:
𝑊 = 𝑊𝑞 +
1
𝑢
= 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 𝟕. 𝟒𝟓𝟒𝟗𝟒 𝒑𝒂𝒔𝒔𝒆𝒏𝒈𝒆𝒓𝒔 𝒊𝒏 𝒍𝒊𝒏𝒆.
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.
𝜆 = 4 𝑐𝑢𝑠𝑡𝑜𝑚ⅇ𝑟𝑠 𝑝ⅇ𝑟 𝑚𝑖𝑛𝑢𝑡ⅇ
𝑢 = 0.50 𝑐𝑢𝑠𝑡𝑜𝑚ⅇ𝑟𝑠 𝑝ⅇ𝑟 𝑚𝑖𝑛𝑢𝑡ⅇ
a. What is the service rate per server in terms of customers per minute?
𝑢 = 1 𝑚𝑖𝑛𝑢𝑡ⅇ /2 𝑚𝑖𝑛𝑢𝑡ⅇ𝑠 𝑝ⅇ𝑟 𝑐𝑢𝑠𝑡𝑜𝑚ⅇ𝑟 = 0.50 𝑐𝑢𝑠𝑡𝑜𝑚ⅇ𝑟𝑠 𝑝ⅇ𝑟 𝑚𝑖𝑛𝑢𝑡ⅇ
b. What is the average waiting time in the line prior to placing an order?
W = 10 𝑚𝑖𝑛𝑢𝑡ⅇ𝑠 = 𝑊𝑞 +
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)