OPIM 201 – Operations Management
Homework 1 Solutions
Question 1
The cafeteria can be divided into 4 subsystems:
1. Queue for tray gathering and food/drinks ordering
2. Gathering trays and ordering food/drinks
3. Queue for cashier
4. Cashier
Queue for cash
cashier
20%
I3cash = 3
Queue for tray
gathering &
food ordering
I1=15
Gathering trays
and ordering
food/drinks
40%
T4cash = 2 min
Queue for CC
cashier
CC Cashier
I3CC = 10
T4CC = 3 min
40%
T2 = 6 min
Cash Cashier
Queue for
Student ID
cashier
Student ID
Cashier
I3ID=4
T4ID = 2 min
We know that the flow rate to all subsystems (R1, R2, R3 = R3cash+ R3CC + R3ID and R4 = R4cash+ R4CC +
R4ID) is equal to 5 customers per minute as the system is in steady state. Thus, the flow rate to cash
cashier (and its queue) is R3cash =20%*5 = 1 customer per min, the flow rate to credit card cashier
is R3CC = 40%*5 = 2 customers per min and the flow rate to student ID cashier is R3ID = 40%*5 = 2
customers per min.
1/8
Now using Little’s formula for every subsystem we get
T1 = I1/R1 = 15/ 5 = 3 minutes
I2 = T2*R2 = 6min* 5 per min = 30 customers
T3cash = I3cash / R3cash = 3 / 1 = 3 minutes
T3CC = I3CC / R3CC = 10 / 2 = 5 minutes
T3ID = I3ID / R3ID = 4 / 2 = 2 minutes
I4cash = T4cash*R4cash = 2 min* 1 per min = 2 customers
I4CC = T4CC*R4CC = 3 min* 2 per min = 6 customers
I4ID = T4ID*R4ID = 2 min* 2 per min = 4 customers
A. Payment with Student ID results in the shortest waiting time a T3ID + T4ID = 2 + 2 = 4 minutes
is less than T3cash + T4cash = 3 + 2 = 5 min and T3CC + T4CC = 5 + 3 = 8 min.
B. There are on average I1 + I2 + I3cash + I4cash + I3CC + I4CC + I3ID + I4ID = 15 + 30 + 3 + 2 + 10 + 6 +
4 + 4 = 74 students
C. The average time a student spends in the cafeteria is equal to T1 + T2 + 20%*( T3cash + T4cash)
+ 40%*( T3CC + T4CC) + 40%*( T3ID + T4ID) = 3 + 6 + 20%*5 + 40%*8 + 40%*4 = 14.8 minutes.
We can also find the average flow time using Little’s law, i.e., 74 students / 5 students per
min = 14.8 minutes.
D. (i) When the percentage of students paying with credit card decreases from 40% to 30%,
R3CC becomes 30%*5 = 1.5 customers per min and R3ID becomes 50%*5 = 2.5 customers
per min. Thus, I4CC = T4CC*R4CC = 3 min* 1.5 per min = 4.5 customers and I4ID = T4ID*R4ID = 2
min* 2.5 per min = 5 customers. We can calculate the average number of students as I1 +
I2 + I3cash + I4cash + I3CC + I4CC + I3ID + I4ID = 15 + 30 + 3 + 2 + 10 + 4.5 + 4 + 5 = 73.5. Using
Little’s law, the average flow time is then 73.5 students / 5 students per min = 14.7
minutes.
(ii) When there is a reduction of 2 customers in the credit card checkout lane, the average
number of students in the cafeteria becomes average I1 + I2 + I3cash + I4cash + I3CC + I4CC + I3ID
+ I4ID = 15 + 30 + 3 + 2 + 8 + 6 + 4 + 4 = 72. . Using Little’s law, the average flow time in then
72 students / 5 students per min = 14.4 minutes.
Option (ii) results in the greatest reduction in average flow time.
2/8
Question 2
A. Process flow diagram is as follows:
Cutting
1
10 min
Stamping
Labelling
4 min
5 min
Cutting
2
10 min
B. Gantt chart for the production of four jobs:
Cutting Machine 1
1
3
Cutting Machine 2
2
4
Buffer 1
2
Stamping Machine
1
4
2
Buffer 2
Labelling Machine
3
4
2
1
4
2
3
4
C. A rush order of one product can be satisfied in 10 + 4 + 5 = 19 minutes.
D. Cycle time of the process is 5 minutes, so the capacity of the process is 60/5 =12 units per
hour.
E. Average WIP can be calculated by finding the average flow time and flow rate in the
process. Average Flow time is equal to (19+24)/2 = 21.5 minutes and average flow rate is
1/5 per minute. Thus Average WIP is 21.5 minutes * 1/5 per minute = 4.3 units.
3/8
F. We have two stages that are bottleneck in this process, i.e., cutting stage and labelling
stage. With a budget of $10000, we can buy 2 additional machines. If we want to increase
process capacity as much as possible, we need to buy one cutting machine and one
labelling machine. After this investment, cutting stage has a total capacity of 3*60/10 = 18
per hour, stamping stage has a capacity of 60/4=15 per hour, and labelling stage has a
capacity of 2*60/5 = 24 per hour. Stamping stage becomes the bottleneck and the new
capacity of the process after investment is 15 units per hour.
Question 3
The following table shows capacity calculations per week when the order size is 50.
Step
1
2
Setup time
(minutes/order)
0
200
Processing time
(minutes/board)
8
10
3
50
12
40*60*(50+50∗12) =184.6
4
100
6
40*60*(100+50∗6) =300
5
200
2
40*60*(
6
60
4
Capacity per week
40*60/8 = 300
50
40*60*(200+50∗10) =171.4
50
50
50
) =400
200+50∗2
50
40*60*(60+50∗4) =461.5
When the order size is 50, the capacity of the process is 171.4 /50 = 3.428 orders per week, which
generates a profit of 3.428* $1000 per order = $3428 per week.
The following table shows capacity calculations per week when the order size is 500.
Step
1
2
Setup time
(minutes/order)
0
200
Processing time
(minutes/board)
8
10
Capacity per week
3
50
12
40*60*(50+500∗12) = 198.3
4
100
6
40*60*(100+500∗6) = 387.1
5
200
2
40*60*(200+500∗2) = 1000
6
60
4
40*60*(60+500∗4) = 582.5
40*60/8 = 300
500
40*60*(200+500∗10) = 230.8
500
500
500
500
4/8
When the order size is 500, the capacity of the process is 198.3 /500 = 0.3966 orders per week,
which generates a profit of 0.3966* $9000 per order = $3569.4 per week.
Circulon should offer option ii (order size must be 500).
Question 4
As demand rate is now given in this question, we can pick an arbitrary demand rate to find the
resource with the highest implied utilization. Assume that demand rate is 1 unit per hour. Next,
we will calculate capacity requested by demand and available capacity per hour.
Step
Processing time
(min)
1
2
3
4
6
8
4
9
Available
capacity
(min/hr)
60
60
60
60
Requested Capacity
(min/hr)
Implied
Utilization
1*6 + 0.2 *6 = 7.2
1*8 + 0.2 *8 = 9.6
1*4 + 0.2 *4 = 4.8
1*9 = 9
12%
16%
8%
15%
A. Second resource is the bottleneck as it has the highest implied utilization.
B. The implied utilization of the bottleneck is 16% with a demand rate of 1 unit per hour. The
capacity of the process is then 1/0.16 = 6.25 units per hour.
C. After a training program is implemented, the rework rate is reduced to 10%. The
bottleneck is now resource 4 and the capacity of the process is 1/0.15 = 6.67 units per
hour.
Step
Processing time
(min)
1
2
3
4
6
8
4
9
Available
capacity
(min/hr)
60
60
60
60
Requested Capacity
(min/hr)
Implied
Utilization
1*6 + 0.1 *6 = 6.6
1*8 + 0.1 *8 = 8.8
1*4 + 0.1 *4 = 4.4
1*9 = 9
11%
14.7%
7.3%
15%
5/8
Question 5
A.
30
18
27
16
23
13
20
19
25
22
18
8∗60∗60
B. Cycle Time = 600 = 48 seconds
The desired cycle time is <= 48 seconds such that at least 600 units per shift can be produced.
C. As Total Task Time/Desired Cycle Time = 231 / 48 = 4.81, the minimum number of
workstations required to meet the production requirement is 5.
D.
Task
RPW
A
231
B
127
C
121
D
93
E
81
F
81
G
79
G H
I
H
63
I
61
J
36
K
13
Tasks ordered in decreasing value of RPW:
A
B
C
Station
1
2
3
4
5
6
D
E
Tasks
A, B
C, E
D, G
F, H
I, J
K
F
J
K
Station Time
19 + 16 = 35
22 + 18 = 40
30 + 18 = 48
20 + 27 = 47
25 + 23 = 48
13
6/8
The actual cycle time of the assignment is 48 seconds.
Total Task Time
231
Line Efficiency = Actual No of Stations x Actual CT = 6 X 48 = 80.2%
Question 6
In this waiting line problem, a = 1/40 hours = 1.5 minutes. p = 6 minutes, CVA = 1 and CVP = 1.
A. When m = 5, utilization of checkout clerks is equal to p/ma = 6/(5*1.5) = 80%.
B. Using the wait time formula, we obtain
Tq= [6 / 5] * [0.80√12-1/(1-0.80)] * [(12 + 12)/2] = 3.46 minutes.
Thus, the average customer waiting time is 3.46 minutes and the total time a customer spends
on waiting and checkout is 3.46 + 6 = 9.46 minutes.
C. The average number of customers waiting in the checkout line is Iq = Tq/a = 3.46 / 1.5 = 2.31
customers.
D. If the manager wants to keep the average customer waiting time below 1 minute, he needs
to hire more clerks. We will start adding more checkout clerks and stop when Tq < 1 min.
With m = 5, Tq = 3.46 minutes.
With m = 6, u = 66.7% and Tq= [6 / 6] * [0.667√14-1/(1-0.667)] * [(12 + 12)/2] = 0.987 minutes.
This means hiring one more checkout clear is enough to keep the average customer waiting
time below 1 minute.
E. Total cost can be expressed as the sum of customer waiting cost and staffing cost.
Staffing cost is $20 X m per hour.
Customer waiting cost can be expressed as average waiting time per customer in hours x
number of customers served per hour x waiting cost per hour = (Tq /a) x $15.
The following table shows the calculations:
7/8
m
Staffing cost per hour
Waiting cost per hour
5
6
7
5 x 20 = $100
6 x 20 = $120
7 x 20 = $140
(3.46/60) x 40 x 15 = 34.6
(0.987/60) x 40 x 15 = 9.87
(0.37/60) x 40 x 15 = 3.7
Total cost
per hour
134.6
129.87
143.7
As the total cost is minimized with m=6, 6 checkout cashiers should be opened to minimize
the total cost per hour.
8/8