Problem 7
Bank of San Pedro has only 1 teller. On average, 1 customer comes
every 6 minutes, and it takes the teller an average of 3 minutes to
serve a customer. To improve customer satisfaction, the bank is
going to implement a unique policy called, “We Pay While You
Wait.” Once implemented, the bank will pay each customer $3 per
minute while a customer waits in line. (So the clock starts when a
customer joins the line, and stops when the customer begins to talk
to the teller.) Bank of San Pedro hired you as a consultant and you
are responsible for estimating how much the “We Pay While You
Wait” program will cost. Assume linear cost. If a customer waits for
ten seconds in line, Bank of San Pedro will pay $0.5. Assume that
arrival follows Poisson and service time follows exponential
distribution.
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
1
Problem 7
a) Compute the capacity of the teller
a)
b)
c)
d)
e)
10 customers/hour
3.33 customers/hour
20 customers/hour
30 customers/hour
Cannot be determined
It takes the teller an average of 3 minutes to serve a customer
60/3 = 20 customers per hour
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
2
Problem 7
b) Calculate the proportion of the time the teller is busy.
a) 100%
b) 80%
c) 62%
d)50%
e) 40%
R=10, Rp= 20
R/Rp= 10/20 = 0.5
= 50%
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
3
Problem 7 - Approximation
c) How long, on average, does a customer wait in line?
a) 6 minutes
b) 4.8 minutes
c) 3 minutes
d)2.6 minutes
e) 2 minutes
0.52
11
Ii 

 0.5
(1  0.5) 2
Indeed Ii was even given in the problem.
Ti= Ii/R
Ti= 0.5/10 = .05
0.05(60) = 3 minutes
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
4
Problem 7
d) Calculate the expected “hourly” cost of the “We Pay While
You Wait” program.
a) $9
b) $36
c) $60
d)$90
e) $140
Ii =0.5.
Therefore, a half of a customer is always there.
For each hour one customer gets 60(3) = $180.
Thus 0.5 customer gets $90.
Perhaps you do not believe me.
Each customer waits, on average, 3 minutes.
He or she receives, on average, 3(3) =$9.
There are 10 customers arriving per hour.
The overall hourly cost of this program is 9*10=$90.
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
5
Problem 7
e) Suppose each additional clerk costs X dollars per hour
(including all other clerk related costs such as benefits,
space and equipment hourly costs). Compute the maximum
value of X if it is at our benefit to hire one additional clerk?
If we have two clerks, Rp increases from 20 to 40, and utilization
drops from 0.5 to 10/40 = 0.25
2 ( 2 1)
0.25
Ii 
 0.0447
(1  0.25)
Ii reduced from 0.5 customers to 0.0447 customers = 0.4553
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
6
Problem 7
The number of customers waiting in the line reduced by 0.455.
It means each hour, there are 0.455 less customers waiting in line.
The cost of each hour waiting per 1 customer is $180
The waiting cost of 0.455 customers is 180(0.4553) = 81.95
If the additional clerk costs less than $81.96 per hour it is at our
benefit to hire her.
f) Suppose each additional clerk costs $30 per hour. How many
new clerk should we hire, one or two?
Obviously, it is at our benefit to hire one clerk.
If we hire two clerks (to have 3 clerks), Rp increases to 60, and
utilization drops 10/60 = 0.167
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
7
Problem 7
0.167 2 (31)
Ii 
 0.008
(1  0.167)
Ii is reduced from 0.045 customers to 0.008, a 0.037 customer
reduction.
By adding the third clerk, there are 0.037 less customers waiting
in line (each hour and always)
180(0.037) = about $6-$7
It is not at our benefit to hire the second additional clerk, pay $30
per hour capacity cost to reduce waiting cost by $6-$7 per
hour.
And we will not pay more that $1-$2 to the fourth clerk.
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
8
Problem 7
We did not need to do this much computations for the third
clerk. With two clerks the total number of customers
waiting in line was:
Ii was equal to 0.045 for c =2.
Even if we reduce the number of customers in the waiting line to
0, we have reduced the line by 0.045 customers.
0.045(180) = 8.1
It is not worth the cost $30 to benefit $8
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
9
Exponential Probability Distribution
In a single phase single server service process and exponentially
distributed interarrival time and service times, the actual total
time that a customer spends in the process is also exponentially.
Suppose total time the customers spend in a pharmacy is
exponentially distributed with mean of 15 minutes. The pharmacy
has promised to fill all prescriptions in 30 minutes. What
percentage of the customers cannot be served within this time
limit?
P(x≥30) = EXP(-30/15) = 0.1353
13.53% of customers will wait more than 30 minutes.
= P(x≤30) = EXPON.DIST(30,1/15,1)
= P(x≤30) = 0.864665
P(x ≥ 30) = 1- P(x≤30) = 1- 0.864665 = 0.1353
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
10
Exponential Probability Distribution
90% of customers are served in less than what time interval?
1-e-X0/ = 0.9
Find X0
0
0.064493
0.124827
0.181269
0.234072
0.283469
0.32968
0.372911
0.413354
0.451188
0.486583
0.519695
0.550671
0.57965
0.606759
0.632121
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
15
Chart Title
40
35
30
25
20
15
10
5
0
0
0.1
0.2
0.3
0.4
SOLVER
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
0.5
0.6
0.7
0.8
0.9
1
0
0
15
0.9 34.54
July-2015
11
Exponential & Poisson
The Poisson distribution provides an appropriate
Description of the number of occurrences per interval
The exponential distribution provides an appropriate description
of the length of the interval between occurrences
Compute the probability of arrival of 3 customers in 30
minutes.
One customer arrives per 15 minutes.
The average number of customers arriving in 30 mins is 2.
This is Poisson distribution.
=POISSON.DIST(3,2,1) =0.857123
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
12
Problem 6. Effect of Pooling- M/M/2 vs. 2M/M/1
Tp
R
Queue1
Server 1
2M/M/1
Queue2
R
Server 2
Tp
Server 1
M/M/2
20/hr
Queue
Server 2
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
13
Problem 6. Effect of Pooling- M/M/2 vs. 2M/M/1
10/hr
5 min
Queue1
Server 1
2M/M/1
Queue2
10/hr
Server 2
5 min
Server 1
M/M/2
20/hr
Queue
Server 2
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
14
Problem 6. Effect of Pooling- M/M/2 vs. 2M/M/1
10/hr
Tp= 5 min
Queue1
Server 1
Ip= 0.833
Tp= 5 min
Queue2
10/hr
Server 2
+0.833
+0.833
--------1.667
Ip= 0.833
Design A: R = 10, Tp= 5 min., c =1.
Rp = 1/5 per min. or 12 per hour
U = R/Rp =10/12 = 5/6 
Ip = cU = U = 0.833 or RTp =Ip  10(5/60) = 0.833
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
15
Problem 6. Effect of Pooling- M/M/2 vs. 2M/M/1
U 2 ( c 1) C  C
Ii 

1U
2
2
a
2 (11)
2
p
10/hr
Queue1
Ii = 4.167
1 1


1U
2
U

0.833
2
Ti = 25 min.
2
Ti = 25 min.
Tp= 5 min
Server 1
Ip= 0.833
Tp= 5 min
2
1  0.833
Ii = 4.167
Queue2
10/hr
Ii = 4.167
Server 2
Ip= 0.833
RTi =Ii
10Ti=4.167 
Ti = .4167 hr or 25 min.
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
16
Problem 6. Effect of Pooling- M/M/2 vs. 2M/M/1
10
Ti = 25 min.
Queue1
Tp=5
Server 1
2M/M/1
Queue2
10
Ii = 2*4.167 = 8. 334
Server 2
Tp
Server 1
M/M/2
20/hr
Queue
Server 2
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
17
Problem 6. Effect of Pooling- M/M/2 vs. 2M/M/1
Server 1
M/M/2
20/hr
Queue
Server 2
C C
U
Ii 

1U
2
2 ( c 1)
2
a
2
p
2 ( 2 1)
12  12


1U
2
U
Ii = 3.839
RTi =Ii
20Ti=3.839 
Ti = 0.1919 hr or 11.52.
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
18
Problem 6. Effect of Pooling- M/M/2 vs. 2M/M/1
Server 1
M/M/2
20/hr
Queue
Server 2
C C
U
Ii 

1U
2
2 ( c 1)
2
a
2
p
2 ( 2 1)
12  12


1U
2
U
Ii = 3.839
RTi =Ii
20Ti=3.839 
Ti = 0.1919 hr or 11.52.
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
19
Effect of Pooling

Rp
12
12
12
12
Tp
5
5
5
5
R
10
10.5
11
11.5
U
0.833
0.875
0.917
0.958
Ii
4.167
6.125
10.083
22.042
Ti
25
35
55
115
Rp
24
24
24
24
Tp
5
5
5
5
R
20
21
22
23
U
0.833
0.875
0.917
0.958
Ii
3.839
5.768
9.697
21.624
Ti
11.516
16.481
26.445
56.410
2Ii
8.333
12.250
20.167
44.083
Why is Design B better than A?



Design A the waiting time of customer is dependent on the
processing time of those ahead in the queue
Design B, the waiting time of customer is only partially dependent
on each preceding customer’s processing time
Combining queues reduces variability and leads to reduce waiting
times
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
20
Problem 7 - Exact
M/M/s
Arrival rate
Service rate
Number of servers
Utilization
P(0), probability that the system is empty
Ii, expected queue length
I, expected number in system
Ti, expected time in queue
T, expected total time in system
Probability that a customer waits
Operations Management: Waiting Lines 2
0.16667
0.33333
1
0.16667
0.33333
2
0.16667
0.33333
3
50.00%
0.5000
0.5000
1.0000
3.0000
6.0000
0.5000
25.00%
0.6000
0.0333
0.5333
0.2000
3.2000
0.1000
16.67%
0.6061
0.0030
0.5030
0.0182
3.0182
0.0152
Ardavan Asef-Vaziri
July-2015
21
Problem 8.a. Exponential Probability Distribution
Average trade time in Ameritrade is one second. Ameritrade has
promised its customers if trade time exceeds 5 second it is free (a
$10.99 cost saving. The same promises have been practiced by
Damion Pizza (A free regular pizza) and Wells Fargo ($5 if
waiting time exceeds 5 minutes). There are 150,000 average daily
trade. What is the cost to Ameritrade”
P(x≥ X0) = e-X0/= e-5/1 = EXP(-5) = 0.006738
Probability of not meeting the promise is 0.6738%
0.006738*150,000* = 1011 orders
@10.99 per order = 10.99*1011 = $11111 per day
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
22
Problem 8.b. Exponential Probability Distribution
What was the cost if they had improved their service level by
50%? that is to make it free for transactions exceeding 2.5 secs.
P(x≥ X0) = e-X0/= e-2.5/1 = EXP(-2.5) = 0.082085
8.2085%*150,000*10.99 = $135317 per day
We cut the promised time by half, our cost increased 12 times.
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
23
Problem 8.c
Bank of San Pedro has only 1 teller. On average, 1 customer comes
every 6 minutes, and it takes the teller an average of 3 minutes to
serve a customer. To improve customer satisfaction, the bank is
going to implement a unique policy called, “We Pay If You Wait
Long.” Once implemented, the bank will pay $10 to any customer
being in the bank for more than 10 minutes.
Ta and Tp both have exponential distribution.
Compute Ti.
0.52
11
Ti= Ii/R
Ii 
(1  0.5)

2
 0.5
Ti= 0.5/10 = .05
Ti = 0.05(60) = 3 minutes
Tp = 3 minutes
T = Ti +Tp = 6 min
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
24
Problem 8.c. Exponential Probability Distribution
P(x≥ X0) = e-X0/= e-10/6 = EXP(-10/6) = 0.188876
Probability of not meeting the promise is = 18.89%
0.180*10 = $1.89
Each hour 10 customers arrive
Total cost = 1.89*10 = $19.9 per hour.
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
25
Problem 9
American Vending Inc. (AVI) supplies vending food to a large
university. Because students often kick the machines out of
anger and frustration, management has a constant repair
problem. The machines break down on an average of 3/hr, and
the breakdowns are distributed in a Poisson manner. Downtime
costs the company $25/hr/machine, and each maintenance
worker gets $4 per hr. One worker can service machines at an
average rate of 5/hr, distributed exponentially; 2 workers
working together can service 7/hr, distributed exponentially;
and a team of 3 workers can do 8/hr, distributed exponentially.
What is the optimal maintenance crew size for servicing the
machines?
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
26
Problem 9
Members in a team
1
2
3
c=????
c=1
U=??
Demand
3
3
3
Capacity
5
7
8
Ii
U
0.600 0.90
0.429 0.32
0.375 0.23
U 2( c 1)
Ii 
1U
Down Time Cost Capacity Cost
4
22.5
8
8.0
12
5.6
Total Cost
26.5
16.0
17.6
U2

1U
Down time cost = 25Ii
Capacity Cost= 4(# of team members)
Total Cost
Operations Management: Waiting Lines 2
Ardavan Asef-Vaziri
July-2015
27
Problem 9
Have I made any mistakes?
Downtime costs the company $25 /hr/machine.
When the machine is down?
Until it is up.
In the waiting line it is down. In the processor until the end of
the process it is down.
There fore, besides Ii, I also need Ip
Members
Waiting
in a team Demand Capacity U
Ii
Cost
1
3
5
0.600 0.90 22.50
2
3
7
0.429 0.32 8.04
3
3
8
0.375 0.23 5.63
Operations Management: Waiting Lines 2
In Process
Cost
15.00
10.71
9.38
Labor
Cost
4
8
12
Ardavan Asef-Vaziri
Total Cost
41.50
26.75
27.00
July-2015
28
Problem 9
Lets check by using Ti and Tp instead of Ii and Ip
Members
in a team Demand Capacity U
Ii
Ti
1
3
5
0.600 0.90 0.30
2
3
7
0.429 0.32 0.11
3
3
8
0.375 0.23 0.08
Processing
time (hr) /
Machine
0.20
0.14
0.13
Operations Management: Waiting Lines 2
In
Waiting
Process Waiting
Cost
Cost
Cost
/Machine /Machine /Hour
7.50
5.00
22.50
2.68
3.57
8.04
1.88
3.13
5.63
Ardavan Asef-Vaziri
In
Process
Cost
/hour
15.00
10.71
9.38
July-2015
Labor
Cost
4.00
8.00
12.00
Total
Cost
41.50
26.75
27.00
29