The document contains 5 practice problems related to waiting line and queuing models. Problem 1 involves calculating metrics like idle time, busy time, and average wait times for a single-server information desk. Problem 2 calculates total expected costs over an 8-hour day based on the results of Problem 1. Problem 3 examines adding a second server and calculates related metrics. Problem 4 describes the parameters of a coffee machine system. Problem 5 describes the characteristics of a repair shop system with exponential service times and Poisson arrivals.

1. Practice Problems: Module D, Waiting-Line Models
Problem 1:
A new shopping mall is considering setting up an information desk manned by one employee.
Based upon information obtained from similar information desks, it is believed that people will
arrive at the desk at a rate of 20 per hour. It takes an average of 2 minutes to answer a question. It is
assumed that the arrivals follow a Poisson distribution and answer times are exponentially
distributed.
(a) Find the probability that the employee is idle.
(b) Find the proportion of the time that the employee is busy.
(c) Find the average number of people receiving and waiting to receive some information.
(d) Find the average number of people waiting in line to get some information.
(e) Find the average time a person seeking information spends in the system.
(f) Find the expected time a person spends just waiting in line to have a question answered (time
in the queue).
Problem 2:
Assume that the information desk employee in Problem 1 earns $10 per hour. The cost of waiting
time, in terms of customer unhappiness with the mall, is $12 per hour of time spent waiting in line.
Find the total expected costs over an 8-hour day.
Problem 3:
The shopping mall has decided to investigate the use of two employees on the information desk.
(a) Find the probability of no people in the system.
(b) Find the average number of people waiting in this system.
(c) Find the expected time a person spends waiting in this system.
(d) Assuming the same salary level and waiting costs as in Problem 2, find the total expected costs
over an 8-hour day.
1

2. Problem 4:
Three students arrive per minute at a coffee machine that dispenses exactly four cups per minute at
a constant rate. Describe the system parameters.
Problem 5:
A repairman at a local metal working shop services their five drill presses. Service time averages
10 minutes and is exponentially distributed. Machines breakdown after an average of 70 minutes
operation (following a Poisson distribution). Describe the major system characteristics.
2

3. ANSWERS
Problem 1:
(a) P0  1 
(b) p 

20
 1
 0.33  33%

30

 0.66

(c) Ls 



20
 2 people
30  20
2
202

 1.33 people
(d) Lq 
 (    ) 30(30  20)
(e) Ws 
1
1

 0.10 hours
   30  20
(f) Wq 

20

 0.0667 hours
 (    ) 30(30  20)
3

4. Problem 2:
From the solution to Problem 1:
The average person waits 0.0667 hours and there are 160(20 arrivals * 8 hours) arrivals per day.
Therefore: Total waiting time = 160 x 0.0667 = 10.67 hours
Total cost for waiting = Total waiting time * Cost per hour = 10.67 * $12 = $128 per day.
Salary cost = 8 hours * $10 = $80
Total cost = Salary cost + Waiting cost = $80 + $128 = $208 per day.
Problem 3:
  20 per hour   30 per hour M  2 open channels (servers)
(a)
P0 
1
1  20  1  20  1  20   2(30) 
       
0!  30  1!  30  2  30   2(30)  20 


0
1

1
2 1  4  60 
1    

3 2  9  60  20 

2
1
1
  50%
2 1 2
1 
3 3
4

5. (b)
2

 20  
 (20)(30)   
 30    1   20
Ls  
 (1)[(2)(30)  20]2   2  30
 






  800  
 3   1 2
 
 
 1600  2 3





1 8
  0.75 people
12 12
(c)
Ws 
Ls


0.75
 0.0375 hours
20
Problem 4:
Lq 
2
 1.125 people in the queue on average
2 (   )
Wq 

 0.375 minutes in the queue waiting
2 (    )
Ls  Lq 

 1.87 people in the system

Ws  Wq 
1

 0.625 minutes in the system
5

6. Problem 5:
N=5
T = 10 minutes
U = 70 minutes
M = 1 server
X
T
10

 0.125
T  U 10  70
From Table D.8: where X = .125 and number of service channels = 1, D = 0.473, F = 0.920
Average number waiting = L = N(1 – F) = 5(1 – 0.920) = 0.4
Average number of machines running = J = NF(1 – X) = 5(0.920)(1 – 0.125) = 4.025 machines
Average number of machines being serviced = H = FNX = (0.920)(5)(0.125) = 0.575 machines
Probability of no wait = 1 – D = 1 – 0.473 = 0.527
6