15 –
1
Chapter 15
Waiting Line Models
Learning Objectives
1. Be able to identify where waiting line problems occur and realize why it is important to study these
problems.
4. Learn how to use formulas to identify operating characteristics of the following waiting line models:
a. Single-channel model with Poisson arrivals and exponential service times
b. Multiple-channel model with Poisson arrivals and exponential service times
c. Single-channel model with Poisson arrivals and arbitrary service times
5. Know how to incorporate economic considerations to arrive at decisions concerning the operation of a
waiting line.
6. Understand the following terms:
queuing theory steady state
Waiting Line Models
17 – 1
P
(
x
) =
x
e
–
x
!
=
2
x
e
-2
x
!
P
0
= 1
–
=
1
–
0.4
0.6
=
0.3333
L
q
=
2
(
–
)
=
(0.4
)
2
0.6
(0.6
–
0.4)
=
1.3333
L
=
L
q
+
= 1.3333 +
0.4
0.6
= 2
P
n
=
n
P
0
=
0.4
0.6
n
(0.3333)
Solutions:
1. a.
= 5(0.4) = 2 per five minute period
b.
x
P(x)
0
0.1353
1
0.2707
2
0.2707
3
0.1804
2. a. µ = 0.6 customers per minute
P(service time 1) = 1 – e-(0.6)1 = 0.4512
3. a.
b.
c.
4.
n
Pn
0
0.3333
1
0.2222
2
0.1481
3
0.0988
P(n > 3) = 1 – P(n 3) = 1 – 0.8024 = 0.1976
Chapter 15
15 –
2
W
q
=
L
q
= 0.4167 hours (25 minutes)
W
=
W
q
+
1
= .5 hours (30 minutes)
c.
d.
6. a.
0
1.25
1 1 0.375
2
P
= − = − =
b.
22
1.25 1.0417
( ) 2(2 1.25)
q
L
= = =
−−
7. a.
Lq=−=−=
2 2
25
5 5 25 05000
( )
( . )
( . ) .
b.
WL
q
q
= = =
05000
25 020
.
.. hours (12 minutes)
8.
= 1 and
= 1.25
0
1
1 1 0.20
1.25
P
= − = − =
L Lq
= + = + =
05000 25
51. .
15 –
3
3.2 3.2
1
q
q
L
W
= = =
minutes
Even though the services rate is increased to µ = 1.25, this system provides slightly poorer service
due to the fact that arrivals are occurring at a higher rate. The average waiting times are identical,
but there is a higher probability of waiting and the number waiting increases with the new system.
9. a.
0
2.2
1 1 0.56
5
P
= − = − =
b.
10
2.2 (0.56) 0.2464
5
PP
= = =
f.
22
2.2 0.3457
( ) 5(5 2.2)
q
L
= = =
−−
0.157
q
q
L
W
==
hours (9.43 minutes)
10. a.
= 2
µ = 3
µ = 4
Average number waiting (Lq)
1.3333
0.5000
Average number in system (L)
2.0000
1.0000
Average time waiting (Wq)
0.6667
0.2500
Average time in system (W)
1.0000
0.5000
Probability of waiting (Pw)
0.6667
0.5000
Chapter 15
15 –
4
Experienced mechanic = $30(L) + $20
= 30(1) + 20 = $50 per hour
Hire the experienced mechanic
11. a.
= 2.5
= 60/10 = 6 customers per hour
L
q
=
2
(
–
)
=
2.5
2
6
(6
–
2.5)
=
0.2976
L
=
L
q
+
=
0.7143
q
b. No; Wq = 7.14 minutes. Firm should increase the mean service rate (µ) for the consultant or hire a
second consultant.
c.
= 60/8 = 7.5 customers per hour
L
q
=
2
(
–
)
=
2.5
2
7.5
(7.5
–
2.5)
=
0.1667
12.
P
0
= 1
–
=
1
–
15
20
=
0.25
L
q
=
2
(
–
)
=
15
2
20
(20
–
15)
=
2.25
L
=
L
+
=
3
W
=
=
=
+
=
0.20
15 –
5
13. Average waiting time goal: 5 minutes or less.
a. One checkout counter with 2 employees
= 15
= 30 per hour
b. Two channel-two counter system
= 15
= 20 per hour for each
From Table, P0 = 0.4545
14. a.
= =
60
75 8
. customers per hour
b.
0
5
1 1 0.3750
8
P
= − = − =
f. 62.5% of customers have to wait and the average waiting time is 12.5 minutes. Ocala needs to add
more consultants to meet its service guidelines.
15. k = 2,
= 5,
= 8
Using the equation for P0, P0 = 0.5238
𝐿𝑞= (𝜆 𝜇
⁄ )2𝜆𝜇
1!(𝑘𝜇−𝜆)2𝑃0= 0.0676
Chapter 15
P
0
= 1
–
=
1
–
5
10
=
0.50
L
q
=
(
/
)
2
1!
(
k
–
)
2
P
0
=
0.0333
W
q
=
L
q
=
0.0067
hours (24.12 seconds)
WL
q
q
= = =
00676
500135
.. hours (0.81 minutes)
Two consultants meet service goals with only 14.88% of customers waiting with an average waiting
time of 0.81 minutes (49 seconds).
16. a.
b.
22
50.50
( ) 10(10 5)
q
L
= = =
−−
c.
0.1
q
q
L
W
==
hours (6 minutes)
17. a. From Table, P0 = 0.60
b.
c.
18. Arrival rate:
= 5.4 per minute
Service rate:
= 3 per minute for each station
a. Using the table of values of P0,
/
= 1.8, k = 2, and P0 = 0.0526
Waiting Line Models
15 –
7
7.67 1.8 9.47
q
LL
= + = + =
7.67 1.42
5.4
q
q
L
W
= = =
minutes
b. The average number of passengers in the waiting line is 7.67. Two screening stations will be able to
meet the manager’s goal.
c. The average time for a passenger to move through security screening is 1.75 minutes.
19. a. For the system to be able to handle the arrivals, we must have k
>
, where k is the number of
channels. With
= 2 and
= 5.4, we must have at least k = 3 channels.
To see if 3 screening stations are adequate, we must compute Lq.
Using the table of values of P0,
/
= 2.7, k = 3, and P0 = 0.02525 (halfway between
/
= 2.6 and
/
= 2.8)
b. The average time required for a passenger to pass through security screening is
7.45 1.38
5.4
q
q
L
W
= = =
11.38 0.5 1.88
q
WW
= + = + =
minutes
Note: The above results are based on using the tables of P0 and an approximate value for P0. If a
computer program is used, we obtain exact results as follows:
Chapter 15
15 –
8
20. a. Note
1.2 1.60 1.
0.75
= =
Thus, one postal clerk cannot handle the arrival rate.
Try k = 2 postal clerks
2.3704
q
q
L
W
==
minutes
13.7037
q
WW
= + =
minutes
Pw = 0.7111
b. Try k = 3 postal clerks.
From Table with
2.1 2.80
.75
==
and k = 3, P0 = 0.0160
3
0
2
( / ) 12.2735
2(3 )
q
LP
==
−
Pw = 0.8767
Three postal clerks will not be enough in two years. Average time in system of 7.1778 minutes and
an average of 15.0735 customers in the system are unacceptable levels of service. Post office
expansion to allow at least four postal clerks should be considered.
Waiting Line Models
15 –
9
21. From question 11, a service time of 8 minutes has µ = 60/8 = 7.5
L
q
=
2
(
–
)
=
(2.5)
2
7.5
(7.5
–
2.5)
=
0.1667
L
=
L
q
+
=
0.50
Total Cost = $25L + $16
= 25(0.50) + 16 = $28.50
Total Cost = 25(0.4356) + 2(16) = $42.89
Use the one consultant with an 8 minute service time.
22.
= 24
Characteristic
System A
(k = 1, µ = 30)
System B
(k = 1, µ = 48)
System C
(k = 2, µ = 30)
a. P0
0.2000
0.5000
0.4286
b. Lq
3.2000
0.5000
0.1524
c. Wq
0.1333
0.0200
0.0063
f. Pw
0.8000
0.5000
0.2286
System C provides the best service.
23. Service Cost per Channel
System A:
6.50
+
20.00
=
$26.50/hour
System B:
2(6.50)
+
20.00
=
$33.00/hour
System C:
6.50
+
20.00
=
$26.50/hour
Total Cost = cwL + csk
System A:
+
26.50(1)
=
$126.50
System B:
+
33.00(1)
=
System C:
25(0.9524)
+
26.50(2)
=
Chapter 15
15 –
10
24.
= 4, W = 10 minutes
a. µ = 1/2 = 0.5
25. A two-server system with a queue dedicated to each server effectively operates as two one-server waiting
line systems each with λ = 0.75 / 2 = 0.375 customers arriving per hour and µ = 1 customer processed per
hour.
Thus for each server:
𝐿𝑞=𝜆2
𝜇(𝜇−𝜆) = 0.3752
1(1−0.375)= 0.225 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠
𝐿 = 𝐿𝑞+𝜆
𝜇= 0.225+0.375
1= 0.6 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠
For the entire system (both servers and their respective dedicated queue) on average there are 2 × 0.225 =
0.45 customers waiting and 2 × 0.6 = 1.2 customers in the system. On average, each customer experiences
0.6 minutes of waiting time and is in the system 1.6 minutes.
As shown in Section 15.3, in the two-server system with a single shared queue, there are 0.1227 customers
waiting and 0.8737 customers in the system on average. Furthermore, each customer experiences 0.1636
minutes of waiting time and is in the system 1.1636 minutes.
26. a. Express
and µ in mechanics per minute
= 4/60 = 0.0667 mechanics per minute
µ = 1/6 = 0.1667 mechanics per minute
Lq =
Wq = 0.0667(4) = 0.2668
15 –
11
L
q
=
2
2
+
(
/
)
2
2
(1
–
/
)
=
(0.25
)
2
(2
)
2
+
(0.25
/
0.3125
)
2
2
(1
–
0.25
/
0.3125)
=
2.225
b. Lq = 0.0667(1) = 0.0667
W = 1 + 1/0.1667 = 7 minutes
L =
W = (0.0667)(7) = 0.4669
27. a. 2/8 hours = 0.25 per hour
b. 1/3.2 hours = 0.3125 per hour
c.
28.
= 5
a.
Design
µ
A
60/6 = 10
B
60/6.25 = 9.6
c. 3/60 = 0.05 for A 0.6/60 = 0.01 for B
d.
Characteristic
Design A
Design B
P0
0.5000
0.4792
Lq
0.3125
0.2857
0.5000
0.5208
Chapter 15
15 –
12
e. Design B is slightly better due to the lower variability of service times.
System A:
W = 0.1625 hrs
(9.75 minutes)
System B:
W = 0.1613 hrs
(9.68 minutes)
29. a.
= 3/8 = .375
µ = 1/2 = .5
c.
Current System (
= 1.5)
New System (
= 0)
Lq
=
1.7578
Lq
=
1.125
TC = cwL + csk = 35 (1.875) + 32 (1) = $97.63
d. Yes; Savings = 40 ($115.77 – $97.63) = $725.60
Note: Even with the advantages of the new system, Wq = 3 shows an average waiting time of 3
hours. The company should consider a second channel or other ways of improving the emergency
repair service.
30. a.
= 42 µ = 20
i
(
/µ)i / i !
0
1.0000
1
2.1000
2
2.2050
3
1.5435
j
Pj
0
1/6.8485
=
0.1460
1
2.1/6.8485
=
0.3066
2
2.2050/6.8485
=
0.3220
3
1.5435/6.8485
=
0.2254
b. 0.2254
c. L =
/µ(1 – Pk) = 42/20 (1 – 0.2254) = 1.6267
31. a.
= 20 µ = 12
i
(
/µ)i / i !
0
1.0000
1
1.6667
2
1.3889
j
Pj
0
1/4.0556
=
0.2466
1
1.6667/4.0556
=
0.4110
2
1.3889/4.0556
=
0.3425
P2 = 0.3425 34.25%
b. k = 3 P3 = 0.1598
32. a.
= 40 µ = 30
i
(
/µ)i / i !
0
1.0000
1
1.3333
2
0.8888
3.2221
i
(
/µ)i / i !
3
0.3951
4
0.1317
P2 = 0.2758
Chapter 15
15 –
14
33. a.
= 0.05 µ = 0.50
/µ = 0.10 N = 8
n
N
!
(
N
–
n
) !
n
0
1.0000
1
0.8000
2
0.5600
3
0.3360
4
0.1680
5
0.0672
6
0.0202
7
0.0040
8
0.0004
P0 = 1/2.9558 = 0.3383
𝐿𝑞= 𝑁 − (𝜆+𝜇
𝜆)(1−𝑃0)= 8 − (0.55
0.05)(1−0.3383)= 0.7215
L = Lq + (1 – P0) = 0.7213 + (1 – 0.3383) = 1.3832
b. P0 = 0.4566
Lq = 0.0646
c. One Employee
Cost = 80L + 20
= 80(1.3832) + 20 = $130.65
34. N = 5
= 0.025 µ = 0.20
/µ = 0.125
a.
n
N
!
(
N
–
n
) !
n
0
1.0000
1
0.6250
2
0.3125
3
0.1172
4
0.0293
5
0.0037
P0 = 1/2.0877 = 0.4790
b.
0
0.225
(1 ) 5 (1 0.4790) 0.3110
0.025
q
L N P
+
= − − = − − =
c. L = Lq + (1 – P0) = 0.3110 + (1 – 0.4790) = 0.8321
d.
WL
N L
q
q
=−=−=
( )
.
( . )( . ) .
03110
5 08321 0025 29854 min
Time at Copier: 12 x 7.9854 = 95.8 minutes/day
Wait Time at Copier: 12 x 2.9854 = 35.8 minutes/day
g. Yes. Five administrative assistants x 35.8 = 179 min. (3 hours/day)
3 hours per day are lost to waiting.
Chapter 15
15 –
16
35. N = 10
= 0. 25 µ = 4
/µ = 0.0625
a.
n
N
!
(
N
–
n
) !
n
0
1.0000
1
0.6250
2
0.3516
3
0.1758
4
0.0769
5
0.0288
P0 = 1/2.2698 = 0.4406
b. 𝐿𝑞= 𝑁 − (𝜆+𝜇
𝜆)(1−𝑃0)=10− (4.25
0.25)(1−0.4406)= 0.4895
f. TC = cw L + cs k
= 50 (1.0490) + 30 (1) = $82.45
g. k = 2
TC = cw L + cs k
= 50L + 30(2) = $82.45
50L = 22.45
L = 0.4490 or less.
15 –
17