MODULE D
DISCUSSION QUESTIONS
2.
The assumptions underlying the standard waiting line or queuing model are:

Arrivals are served on a “first come, first served” (FCFS, or FIFO) basis; and every arrival
waits to be served regardless of the length of the line or queue.

All arrivals are independent of preceding arrivals, and the average number of arrivals per unit
time (arrival rate) does not change over time.

Arrival rates are described by a Poisson probability distribution, and arrivals come from an
infinite or very large source.

Service times vary from one customer to another and are independent of one another, but their
average rate is known.

Service times are described by a negative exponential probability distribution.

The effective service rate is faster than the arrival rate.
4.
If the service rate is not greater than the arrival rate, the line will increase in length indefinitely.
12.
This deals with the interesting issue of the value of waiting time. Some service organizations place a
very low value on your time, leading to a good classroom discussion.
15.
Supermarkets simply do not have enough space to set up a common line, so they use a separate queue
for each checkout station.
16.
The cost trade-offs in queuing are waiting cost versus service cost.
END-OF-MODULE PROBLEMS
D.1
(a)
(b)
(c)
(d)
(e)
D.2
(a)
(b)
(c)
(d)
 40 2


 60 3
 40  2
2
4
Lq 


    6060  40  3

40
Ls 

2
   60  40

40
Wq 

 0.033 hours  2 minutes
    6060  40

1
1
1


hour  3 minutes
   60  40 20
  40 /hour,   60 /hour
Ws 
40
 0.44  44%
90
 40  2
Lq 
 0.356
9090  40 
40
Ls 
 0.8
90  40
40
Wq 
 0.0089 hours  0.533 minutes  32 seconds

90 90  40

Quantitative Module D: Waiting Line Models
1
D.4
Ws 
(a)
P0  1    0.5
(b)
(c)
(d)
(e)
D.5
(a)
P0  1    0.25
(d)
(e)
(f)
(a)
(b)
(c)
(a)
(b)
(c)
D.8
1
Ws 
(c)
D.7

 0.5


Ls 
1

2
Lq 
 0.5
    

Wq 
 0.05 hours
    

(f)
(b)
D.6
1
1

hour  12
. minutes  72 seconds
90  40 50
  40 /hour,   90 /hour
(e)
(a)

 01
. hours
where:   20 /hour,   10 /hour

 0.75


Ls 
3

2
Lq 
 2.25
    

Wq 
 015
. hours
   

Ws 
1
 0.2 hours

where:   20 /hour,   15 /hour

 0.667


Wq 
 0.667 time units  minutes
    
2
where:   180 /hour,   120 /hour
Lq 
 1333
.
    

2
 3.2
    
1
1
Ws 
 hour (10 minutes)
 6

where   30 /hour,   24 /hour
   0.8

Lq 
The utilization rate,  , is given by:

(b)
2
 3
  0.375
 8
The average down time, Ws , is the time the machine waits to be serviced plus the time taken to
perform the service.
Instructor’s Solutions Manual t/a Operations Management
Ws 
(c)
The number of machines waiting to be served, Lq , is, on average:
Lq 
(d)
1
1

 0.2 days, or 1.6 hours
    8  3
2
32

 0.225 machines waiting
    88  3
Probability that more than one machine is in the system:
3
P
 
n  k 8
k 1
2
9
3
P
  
 0.141
n 1 8
64
Probability that more than two machines are in the system:
Pn>2 = 0.053
Pn>3 = 0.020
Pn>4 = 0.007
D.11   4 students/minute,   60 12  5 students/minute
(a)
The probability of more than two students in the system, Pn2 , is given by:
Pn>2 = (4/5)3 = 0.512
The probability of more than three students in the system, Pn3 , is given by:
Pn>3 = (4/5)4 = 0.410
The probability of more than four students in the system, Pn4 , is given by:
Pn>4 = (4/5)5 = 0.328
(b)
The probability that the system is empty, P0 , is given by:
P0  1 
(c)
The average waiting time, Wq , is given by:
Wq 
(d)

4
 1   1  0.8  0.2

5

4

 0.8 minutes
    55  4
The expected number of students in the queue, Lq , is given by:
Quantitative Module D: Waiting Line Models
3
Lq 
(e)
2
42

 3.2 students
    55  4
The average number of students in the system, Ls , is given as:
Ls 
(f)

4

 4 students
  5 4
Adding a second channel, we have:
  4 students minute
60
 5 students minute
12
M2

(b)
The probability that the two channel system is empty, P0 , is given by:
P0 
2   10  4 6


 0.429
2   10  4 14
Thus, the probability of an empty system when using the second channel, is 0.429.
(c)
The average waiting time, Wq , for the two channel system is given by:
Wq 
(d)
2
16

 0.038 minutes
 (2   )( 2   ) 5(14)(6)
The average number of students in the queue for the two channel system, Lq , is given by:
Lq  Wq  4 / minute( 0.038 minutes)  0.152 students
(e)
The average number of students in the two channel system, Ls , is given by:
LS  Lq 
4

 0.152  0.8  0.952 students

Instructor’s Solutions Manual t/a Operations Management
D.12   30 trucks/hour,   35 trucks/hour
(a)
The average number of trucks in the system, Ls , is given by:
Ls 
(b)


30
30

 6 trucks in the system
35  30 5
The average time spent by a truck in the system, Ws , is given by:
Ws 
(c)

1
1
1

  0.2 hours  12 minutes
   35  30 5
The utilization rate for the bin area,  , is given by:

(d)
 30 6

  0.857
 35 7
The probability that there are more than three trucks in the system, Pn3 , is given by:
Pn>3 = (30/35)4 = 0.540
Thus, the probability that there are more than three trucks in the system is 0.540.
(e)
Unloading cost:
Cu  16
(f)
hours
trucks
hours
$
 30
 0.2
 18
 16  30  0.2  18  $1728 day
day
hour
truck
hour
or $12,096 per week
Enlarging the bin will cut waiting costs by 50% next year. First, we must compute annual
waiting costs:
Annual waiting costs  2
weeks
days
$
7
 1728
 $24,192
year
week
day
Enlarging the bin will cut waiting costs by 50% next year, resulting in a savings of $12,096.
Because the cost of enlarging the bin is only $9,000, the cooperative should proceed to
enlarge the bin. The net savings is $3,096 (= 12,096 – 9000).
D.13   12 calls/hour,   60 4  15 calls/hour
(a)
The average time the catalogue customer must wait, Wq , is given by:
Wq 
(b)
The average number of callers waiting to place an order, Lq , is given by:
Lq 
(c)

12
12
12



 0.267 hours
    1515  12 15  3 45
2
122
144
144



 3.2 customers
    1515  12 15  3 45
To decide whether or not to add the second clerk, we must

Compute present total cost

Compute total cost with the second clerk
Quantitative Module D: Waiting Line Models
5

Compare the two
Present total cost:
Ct hour  service cost  waiting cost
 $10 per hour  12 calls per hour  0.267 hours waiting per call  $25 per hour 
 10  12  0.267  25  10  80.1 hour  $90.10 hour
To determine total cost using the second clerk (a second channel):
Wq 
2
144

 0.0127 hours
 (2   )( 2   ) 15(42)(18)
Cost with two clerks:
Ct hour  service cost  waiting cost
calls
hours
$
 0.0127
 25
 20  12  0.0127  25
hour
call
hour
 20  3.81  $23.81 hour
 20  12
There is a saving of $90.10 – $23.81 = $66.29/hour
Thus, a second clerk should certainly be added!
(d)
With three clerks the cost goes to $30.47. So the costs are:
1 clerk
2 clerks
3 clerks
$90.10
$23.81
$30.47
For 3 clerks Wq  0.00158 (from Excel OM)
$30  12  0.0015  $25  $30  0.47  $30.47
Therefore, optimum number of clerks is 2.
6
Instructor’s Solutions Manual t/a Operations Management