Chapter 18 – Management of Waiting Lines
18–21
b. Determine the optimum number of operators:
Machine downtime cost per hour = Average number down x Machine downtime cost per
hour
Average number of machines running =
Average number down = N – J
Operator cost = Number of operators x Operator cost per hour
M = 1:
Machine downtime cost per hour = 2.620 x $70 = $183.40
Operator cost per hour = 1 x $20.00 = $20.00
Machine downtime cost per hour = 1.623 x $70 = $113.61
Operator cost per hour = 2 x $20.00 = $40.00
Total cost per hour = $113.61 + $40.00 = $153.61
M = 3:
Using Table 18.7 with N = 5, X = .280, & M = 3:
F = .993
Chapter 18 – Management of Waiting Lines
18–23
Education.
M = 2:
Using Table 18.7 with N = 10, X = .200, & M = 2:
F = .854
Average number down = N – J = 10 – 6.832 = 3.168
M = 3:
Using Table 18.7 with N = 10, X = .200, & M = 3:
F = .968
Average number down = N – J = 10 – 7.744 = 2.256
M = 4
Using Table 18.7 with N = 10, X = .200, & M = 4:
F = .994
Average number down = N – J = 10 – 7.952 = 2.048
18–24
13. Given:
Trucks arrive at the loading dock at the rate of = 1.2 per hour. A single crew consisting of two
workers can load a truck in an average of about 30 minutes. Crew members receive $10 per hour,
and trucks and drivers reflect an hourly cost of $60. The manager is thinking about adding
another member to the crew. The service rate then would be 2.4 trucks per hour. Assume rates are
a. Would the third member be economical?
Crew size of two ( = 2):
Crew size of three ( = 2.4)μ
Chapter 18 – Management of Waiting Lines
b. Would a fourth member be justifiable if the resulting = 2.6/hour?
Crew size of four ( = 2.6)μ
18–26
14. Given:
Customers are assigned to one of three categories, with Category 1 given the highest priority. An
average of 9 customers arrive per hour and one-third are assigned to each category. There are two
servers, and each can process customers at the rate of 5 customers per hour. Arrival and service
rates can be described by the Poisson distribution.
[Multiple priority model]
All calculated values are carried to four decimals (as is done in the templates).
a. Utilization for the system (ρ)μ
b. Determine the average wait time for units in each class:
Lq using Table 18.4 with / = 1.8 & M = 2 = 7.674
Chapter 18 – Management of Waiting Lines
18–27
Education.
c. Determine the average number of customers in each class that are waiting for service (Lk):
Note: The solutions using the top part of the Multiple Priorities Waiting Line Template are
highlighted in bold below. There could be slight differences in answers due to rounding.
Multiple Priorities Waiting
Line Model
Basic
<Back
Service rate =
5
Increment Δ =
1
Number of
servers
M =
2
Service time 1/ =
0.2000
Class
System
1
2
3
4
Arrival rate
=
9.0000
3
3
3
System Utilization
(Part a)
ρ =
0.9000
Probability system is
empty
P0 =
0.0526
Average number in line
(Part b)
Lq =
7.6737
0.3654
0.9135
6.3947
Average number in
system
Ls =
9.4737
0.9654
1.5135
6.9947
Average time in line
(Part c)
Wq =
0.8526
0.1218
0.3045
2.1316
Average time in system
Ws =
1.0526
0.3218
0.5045
2.3316
Chapter 18 – Management of Waiting Lines
18–28
15. Given:
Customers are assigned to one of two classes when they enter the processing center. The highest-
priority class has an arrival rate of 4 per hour and the other class has an arrival rate of 2 per hour.
Both rates can be described as Poisson. There are 2 servers and each can process customers in an
average of 15 minutes.
1 = 4 customers/hour
2 = 2 customers/hour
= 60 minutes/hour / 15 minutes/customer = 4 customers/hour
M = 2
[Multiple priority model]
The solutions using the top part of the Multiple Priorities Waiting Line Template are highlighted
in bold below:
Multiple Priorities Waiting Line
Model
Basic
<Back
Service rate =
4
Increment Δ =
1
Number
of servers
M =
2
Service time 1/ =
0.2500
Class
System
1
2
3
4
Arrival rate
=
6.0000
4
2
System Utilization
(Part a)
ρ =
0.7500
Probability system is empty
P0 =
0.1429
Average number in line
(Part b)
Lq =
1.9286
0.6429
1.2857
Average number in system
Ls =
3.4286
1.6429
1.7857
Average time in line
(Part c)
Wq =
0.3214
0.1607
0.6429
Average time in system
Ws =
0.5714
0.4107
0.8929
a. What is the system utilization (ρ in the template)?
b. Determine the number of customers of each class that are waiting for service (Lq in the
template):
Chapter 18 – Management of Waiting Lines
Education.
c. Determine the average waiting time for each class (Wq in the template):
Class 1: 0.1607 hours.
Class 2: 0.6429 hours.
d. If the arrival rates of both classes were equal (3/hour), what would the revised average
waiting time (Wq) for each class equal?
The solutions using the top part of the Multiple Priorities Waiting Line Template are
highlighted in bold below:
Multiple Priorities Waiting Line
Model
Basic
<Back
Service rate =
4
Increment Δ =
1
Number of
servers
M =
2
Service time 1/ =
0.2500
Class
System
1
2
3
4
Arrival rate
=
6.0000
3
3
System Utilization
ρ =
0.7500
Probability system is empty
P0 =
0.1429
Average number in line
Lq =
1.9286
0.3857
1.5429
Average number in system
Ls =
3.4286
1.1357
2.2929
Average time in line
(Part d)
Wq =
0.3214
0.1286
0.5143
Average time in system
Ws =
0.5714
0.3786
0.7643
Class 1: 0.1286 hours.
Class 2: 0.5143 hours.
Chapter 18 – Management of Waiting Lines
18–30
16. Given:
A priority waiting system assigns arriving customers to one of four classes. Arrival rates
(Poisson) for the classes are shown below:
Class
Arrivals per Hour
1
2
2
4
3
3
4
2
[Multiple priority model]
The solutions using the top part of the Multiple Priorities Waiting Line Template are highlighted
in bold below:
Multiple Priorities Waiting Line
Model
Basic
<Back
Service rate =
3
Increment Δ =
1
Number of
servers
M =
5
Service time 1/ =
0.3333
Class
System
1
2
3
4
Arrival rate
=
11.0000
2
4
3
2
System Utilization (Part a)
ρ =
0.7333
Probability system is empty
P0 =
0.0209
Average number in line
(Part b)
Lq =
1.1904
0.0666
0.2220
0.3607
0.5411
Average number in system
Ls =
4.8570
0.7333
1.5553
1.3607
1.2077
Average time in line
(Part b)
Wq =
0.1082
0.0333
0.0555
0.1202
0.2705
Average time in system
Ws =
0.4415
0.3666
0.3888
0.4536
0.6039
a. What is the system utilization (ρ in the template)?
ρ = 0.7333