Name (print, please) _______________________________________________ ID ___________________________
Production Management 73-604 Winter 2002
Odette School of Business
University of Windsor
Midterm Exam 2 Solution
Tuesday, March 26, 7:00 – 9:00 pm
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and a two-sided formula sheet.
Time available: 2 hours
Instructions:
 This exam has 12 pages including this cover page and one page of table
 Please be sure to put your name and student ID number on each page.
 Show your work.
Grading:
Question
Marks:
1
/10
2
/7
3
/6
4
/4
5
/8
6
/5
7
/9
8
/6
9
/10
Total:
/65
Name:_________________________________________________
ID:_________________________
Question 1: (10 points)
Multiple Choice Questions
1.1 Suppose that we want to set up a kanban control system and want to determine the number of
kanban card sets needed. If the expected demand during the lead time is 30 per hour, the safety
stock is 20% of the demand during lead time, the container size is 10, and the lead time to
replenish an order is 5 hours, what is the number of kanban sets?
a. About 3
b. About 4
c. About 18
d. About 30
e. About 36
1.2 Which of the following are related to JIT?
a. A philosophy of waste reduction
b. Pull production
c. Multi-skilled workers
d. Strong supplier relations
e. All of the above
1.3 The costs of quality include which of the following?
a. Appraisal costs
b. Prevention costs
c. Internal failure costs
d. External failure costs
e. All of the above
1.4 Lot for lot minimizes
a. carrying cost
b. ordering cost
2
Name:_________________________________________________
ID:_________________________
1.5 Which of the following is the ISO 9000 form of certification that requires that a “qualified” national
or international standards or certifying agency serve as an auditor?
a. First party
b. Second party
c. Third party
d. All of the above
e. None of the above
1.6 Silver-Meal heuristic and least unit cost heuristic perform better if the costs
a. change over time
b. do not change over time
1.7 ______________ reduce setup time
a. Cellular layouts
b. Poka-yoke
1.8 The fact that the EOQ cost curve is flat near the optimal order quantity implies that
a. if there are some managerial reasons to order Q units such that Q  EOQ, but Q is near
EOQ, then one may order Q units without causing a large increase in inventory cost
b. inventory cost is not sensitive to the cost of buying items
1.9 Reduced material handling cost and setup time are advantages of
a. cellular layouts
b. job shops
1.10 Following charts are suitable for variable data:
a. X and R charts
b. p and c charts
3
Name:_________________________________________________
ID:_________________________
Question 2: (7 points) Three jobs must be processed on a single machine that starts at 8:30 am.
The processing times and due dates are given below:
Job
Processing Time (Hours)
Due Date
J1
2
11:30 am
J2
3
12:30 pm
J3
1
3:30 pm
Assuming that the jobs are processed in the sequence J1, J2, J3, compute makespan, total
completion time, maximum lateness, and average tardiness.
Job
Start Time
Processing
Time
Completion
Time
Due Date
Lateness
Tardiness
(1 point)
(1 point)
(1 point)
J1
0
2
2
3
-1
0
J2
2
3
5
4
1
1
J3
5
1
6
7
-1
0
Makespan = 6 hours (1 point)
Total completion time = 2+5+6 = 13 hours (1 point)
Maximum lateness = max(-1,1,-1) = 1 hour (1 point)
Average tardiness = (0+1+0)/3 = 1/3 hour (1 point)
4
Name:_________________________________________________
ID:_________________________
Question 3: (6 points) Three jobs are to be scheduled on two machines M1 and M2. Assume that
every job is first processed on M1 and then on M2. The processing times are as follows:
Job
M1
M2
J1
7
9
J2
8
4
J3
5
6
a. (3 points) Using Johnson’s algorithm for two-machine flow-shop, find a sequence of jobs in order
to minimize makespan. In each iteration, show the job you assign and position to which the job is
assigned.
Iteration 1: J2 has min processing time, 4 on M2. Assign J2 in the end, Position 3. Delete J2.
Positions
1
2
3
J2
Iteration 2: J3 has min processing time, 5 on M1. Assign J3 in the beginning, Position 1. Delete
J3.
Positions
1
2
3
J3
J2
Iteration 3: Assign J1 in the remaining position.
Positions
1
2
3
J3
J1
J2
b. (3 points) Compute makespan given by the sequence obtained in part a.
M1
Job
M2
Start
Process
Finish
Start
Process
Finish
J3
0
5
5
5
6
11
J1
5
7
12
12
9
21
12
8
20
21
4
25
J2
Makespan = 25
5
Name:_________________________________________________
ID:_________________________
Question 4: (4 points) Jumbo’s restaurant is trying to create a consecutive days off schedule that
uses the fewest workers. Use the following information to create a consecutive days off schedule:
Day
M
Tu
W
Th
F
S
Su
Requirements
2
3
2
3
4
4
2
Days off for worker 1: Su, M
M
Worker 1
2
Tu
3
W
2
Th
3
F
4
S
4
Su
2
Days off for worker 2: Tu, W
M
Worker 2
2
Tu
2
W
1
Th
2
F
3
S
3
Su
2
Days off for worker 3: W, Th
M
Worker 3
1
Tu
2
W
1
Th
1
F
2
S
2
Su
1
Days off for worker 4: Su, M
M
Worker 3
0
Tu
1
W
1
Th
1
F
1
S
1
Su
0
Hence, the solution is as follows:
Two workers get days off on Su and M
One worker gets days off on Tu and W
One worker gets days off on W and Th
6
Name:_________________________________________________
ID:_________________________
Question 5: (8 points) The following matrix contains the material handling costs (in thousand
dollars) associated with assigning Machines 1, 2 and 3 to Locations A, B and C. Assign machines to
locations to minimize material handling costs. State the optimal assignment and the associated cost.
Machines
Locations
A
B
C
1
$50
$90
$90
2
60
90
80
3
80
70
50
a. (2 points) Show the matrix obtained after row reduction
Machines
Locations
A
B
1
0
40
2
0
30
3
30
20
C
40
20
0
b. (2 points) Continue from part a and show the matrix obtained after column reduction
Machines
Locations
A
B
1
0
20
2
0
10
3
30
0
C
40
20
0
c. (2 points) Continue from part b, and show the optimal solution
Machines
Locations
A
B
1
0
10
2
0
0
3
40
0
C
30
10
0
Hence, the optimal solution is:
Machine 1 is assigned to location A
Machine 2 is assigned to location B
Machine 3 is assigned to location C
d. (2 points) What is the cost associated with the optimal assignment obtained in part c?
Assignment
Machine 1 is assigned to location A
Machine 2 is assigned to location B
Machine 3 is assigned to location C
Total
Cost
$50
$90
$50
$140
7
Name:_________________________________________________
ID:_________________________
Question 6: (5 points) Irwin sells a particular model of fan, with most of the sales being made in the
summer months. Irwin makes a one-time purchase of the fans prior to each summer season at a cost
of $30 each and sells each fan for $55. Any fans unsold at the end of summer season are marked
down to $15 and sold in a special fall sale.
a. (1 point) What is the marginal loss per unit?
ML = Purchase price – salvage value = $30 – $15 = $15
b. (1 point) What is the marginal profit per unit?
MP = Selling price – purchase price = $55 – $30 = $25
c. (3 points) If the demand is uniformly distributed between 400 to 1400 units, find the optimal order
quantity.
ML
15

 0.375
ML  MP 15  25
Q  b  pb  a  1400  0.3751400  400  1025
p
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Name:_________________________________________________
ID:_________________________
Question 7: (9 points) The annual demand for a product is 27,040 units. The weekly demand is 520
units with a standard deviation of 100 units. The cost to place an order is $50, and the time from
ordering to receipt is four weeks. The annual inventory carrying cost is $2.5 per unit.
a. (4 points) Compute the optimal order quantity.
Q
2 DS
227,04050

 1,040 units
H
2.50
b. (5 points) Compute the reorder point necessary to provide a 98 percent service probability.
For a 98 percent service probability, the area under the normal distribution curve on the left side
of z , F z   0.98
Hence, look for area = 0.98-0.50 = 0.48 in the standard normal table
From the standard normal table, z  2.055 for area = 0.48.
Hence, R  d L  z L  520  4  2.055  200  2,491 units
9
Name:_________________________________________________
ID:_________________________
Question 8: (6 points) Charlie’s Pizza orders all of its pepperoni, olives, anchovies, and mozzarella
cheese to be shipped directly from Italy. An American distributor stops by every five weeks to take
orders. Because the orders are shipped directly from Italy, they take four weeks to arrive. Charlie’s
Pizza uses an average of 200 pounds of pepperoni each week, with a standard deviation of 40
pounds. Charlie’s prides itself on offering only the best-quality ingredients and a high level of service,
so it wants to ensure a 98 percent probability of not stocking out on pepperoni. Assume that the
sales representative just walked in the door and there are currently 1000 pounds of pepperoni in the
walk-in cooler. How many pounds of pepperoni must be ordered?
T  5 weeks
L  4 weeks
For a 98 percent probability of not stocking out, the area under the normal distribution curve on
the left side of z , F z   0.98
Hence, look for area = 0.98-0.50 = 0.48 in the standard normal table
From the standard normal table, z  2.055 for area = 0.48.
Q  d T  L   z T  L  I  2005  4  2.055120  1000  1,046.6 units
10
Name:_________________________________________________
ID:_________________________
Question 9: (10 points) Comptek Computers wants to reduce a large stock of personal computers it
is discontinuing. It has offered the University Bookstore a quantity discount pricing schedule if the
store will purchase the personal computers in volume, as follows:
Quantity
Price
1-9
$950
10-49
925
50+
900
The annual inventory holding cost is 25%, the ordering cost is $160, and annual demand for this
particular model is estimated to be 125 units. Compute the optimal order size.

First, consider the cheapest price level of c3  $900 per unit. (1 point)
h3  Ic3  0.25  900  $225 /unit/year
EOQ3 
2 K
2125160

 13.33 units (1 point)
h3
0.25900
Since the price level of c3  $900 is not available for an order quantity Q  EOQ3 = 13.33 units,
EOQ3 is infeasible and a candidate for optimal order quantity is Q3  50 , because 50 is the
minimum order quantity for the price level of c3  $900. (1 point)

Now, consider the next price level, c2  $925 per unit. (1 point)
h2  Ic2  0.25  925  $231.25 /unit/year
EOQ2 
2 K

h2
2125160
 13.15 (1 point)
0.25925
Since the price level of c2  $925 is available for an order quantity Q  EOQ2 = 13.15 units, EOQ2
is feasible and a candidate for optimal order quantity is Q2  13.15  13. (1 point)

It’s not necessary to consider the other price level.

Now, compute total cost for each candidate for optimal order quantity:
(1 point)
Holding cost
Ordering cost
Cost of item
Total cost
Qj
h jQ j
c j
2
K
Qj
= Holding cost +
Ordering cost +
Cost of item
0.25  900  50
2
 5,625
160  125
50
 400
900  125
$118,525
 112,500
(1 point)
0.25  925  13
2
 1,503.125
160  125
13
 1538.46
925  125
$118,667
 115,625
(1 point)
j
3
2

Q3  50
Q2  13
Conclusion: The total cost is minimum, $118,525 for Q3  50 . Therefore, an optimal order
quantity is Q3  50 . (1 point)
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