Solution to Queueing Problems1
1. Customers arrive at a local 7-11 at the rate of  = 40 per hour (and follow a Poisson
process). The only employee in the store can check them out at a rate of  = 60 per hour
(following an exponential distribution). Compute the following:
a. The percentage of time that the employee is busy with the checkouts.



S
40
60 *1
 0.6667

b. The average length of the queue.
Lq
2

1 
2

0.6667 

0.3333
 1.3333
c. The average number of customers in the system.
L
 Lq  Ls
 1.3333  0.6667
2
These problems appear in Operations Management, 6th Edition, by Jay Heizer and Barry Render (PrenticeHall, 2001, ISBN 0-536-62524-7) and are used here with permission. The solutions were written by David
Juran (2001).
1
d. The average time spent waiting in the queue.


Wq
 1   
0.6667
600.3333
 0.03333 hours
 2 minutes

e. The average time in the system.
1
 
1

60  40
 0.05 hours
 3 minutes

W
2. There is only one copying machine in the student lounge of the business school.
Students arrive at a rate of  = 40 per hour (according to a Poisson distribution).
Copying takes an average of 40 seconds, or  = 90 per hour (according to an
exponential distribution). Compute the following:
a. The percentage of time that the machine is used.



S
40
90 *1
 0.4444

b. The average length of the queue.
Lq
2

1 
2

0.4444

0.5556
 0.3556
Operations Management
2
Profs. Juran and Pinedo
c. The average number of students in the system.
 Lq  Ls
L
 0.3556  0.4444
 0.8
d. The average time spent waiting in the queue.

Wq

 1   
0.4444
900.5556
 0.0089 hours
 0.5333 minutes

e. The average time in the system.
1
 
1

90  40
 0.02 hours
 1.2 minutes

W
3. Due to a recent increase in business, a law firm secretary must now word-process an
average of 20 letters a day (assume a Poisson distribution). It takes him approximately
20 minutes to type each letter (assume an exponential distribution). Assuming the
secretary works 8 hours a day:
a. What is the secretary’s utilization rate?



S
 20 


8 * 60 


 1 
  *1
 20 
0.04167

0.05
 0.8333
Operations Management
3
Profs. Juran and Pinedo
b. What is the average waiting time before the secretary word-processes a letter?
Wq


 1   
0.8333
0.050.1667 
 100 minutes

c. What is the average number of letters waiting to be done?
Lq
2

1 
2

0.8333

0.1667
 4.1667
d. What is the probability that the secretary has more than 5 letters to do?
PN  5
Operations Management
k 1

  

6
 0.8333
 0.3349
4
Profs. Juran and Pinedo
4. Sam the Vet is running a rabies-vaccination clinic for dogs at the local grade school.
Sam can “shoot” a dog every three minutes. It is estimated that the dogs will arrive
independently and randomly throughout the day at a rate of one dog every 6 minutes
according to a Poisson distribution. Also assume that Sam’s shooting times are
exponentially distributed. Compute the following:
a. The probability that Sam is idle.
1 
 1

S
1
 
6
 1  
1
  *1
3
 1 0.5
 0.5
b. The proportion of time that Sam is busy.



S
1
 
6
  
1
  *1
 3
 0.5
c. The average number of dogs being vaccinated and waiting to be vaccinated.
L
 Lq  Ls
2

0 .5 

 0 .5
0 .5
 0.5  0.5
1
Operations Management
5
Profs. Juran and Pinedo
d. The average number of dogs waiting to be vaccinated.

Lq
2
1 
2

0 .5 

0.5
 0.5
e. The average time a dog waits before getting vaccinated.
Wq


 1   
0.5
1
 0.5
 3
 3 minutes

f. The average amount of time a dog spends waiting in line and being vaccinated.
W
Operations Management
1
 
1

1 1

3 6
 6 minutes

6
Profs. Juran and Pinedo