Name (print, please) _______________________________________________ ID ___________________________
Production Management 73-604 Winter 2001
Faculty of Business Administration
University of Windsor
Midterm Exam 2 Solution
Tuesday, March 27, 4:00 – 6:00 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 two pages of tables
 Please be sure to put your name and student ID number on each page.
 Show your work.
Grading:
Question
Marks:
1
/10
2
/4
3
/8
4
/6
5
/6
6
/6
7
/10
Total:
/50
Name:_________________________________________________
ID:_________________________
Question 1: (10 points)
Multiple Choice Questions
1.1 Assume no other units of production or inventory is available. Which of the following is the
Planned-Order Release using an MRP program if the Gross Requirements are 5,000 and the On
Hand inventory is 1,200?
a. 5,000
b. 4,500
c. 3,800
d. 2,350
e. 1,200
1.2 Which of the following can be used for lot sizing in MRP systems?
a. Low-level coding
b. LUC
c. MRP II
d. Closed-loop MRP
e. None of the above
1.3 Which of the following is a dynamic lot-sizing technique that calculates the order quantity by
comparing the carrying cost and the setup (or ordering) costs for various lot sizes and then
selects the lot in which these are most nearly equal?
a. LTC
b. LUC
c. EOQ
d. L4L
e. None of the above
1.4 Which of the following includes costs of the inspection, testing, and other tasks to ensure that the
product process is acceptable?
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Name:_________________________________________________
ID:_________________________
a. Appraisal cost
b. Prevention costs
c. Internal failure costs
d. External failure costs
e. None of the above
1.5 In the do phase of the PDCA (plan-do-check-act) cycle a plan is
a. improved
b. identified
c. evaluated
d. implemented
e. rejected
1.6 ISO 9000
a. is a series of standards adopted in 1990
b. consists of four primary parts numbered as 9001 through 9004
c. is a well-defined standard
d. and ISO 14000 are the same
e. is intentionally vague
1.7 Which of the following is not a JIT element and does not address elimination of waste?
a. Group technology
b. Kanban production control system
c. Push production
d. Quality at the source
e. Focused factory networks
1.8 All of the following are recognized as wastes in JIT except
a. waste from overproduction
b. waste from flexibility
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Name:_________________________________________________
ID:_________________________
c. transportation waste
d. processing waste
e. waste of motion
1.9 Group Technology is credited with which of the following benefits?
a. Reducing waiting time between process operations
b. Reducing inventory
c. Reducing the number of employees required
d. All of the above
e. None of the above
1.10
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 20 per hour, the safety
stock is 10% of the demand during lead time, the container size is 5, and the lead time to
replenish an order is 5 hours, what is the number of kanban sets?
a. About 5
b. About 20
c. About 22
d. About 27
e. About 110
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Name:_________________________________________________
ID:_________________________
Question 2: (4 points) Joe has been released from his government job. Based on his excellent
performance, he was able to land a job as a production scheduler in a brand-new custom refinishing
auto service shop located near the border. The sequence is customizing first, followed by repainting.
The processing times are as stated below:
Car
Customizing Time (Hours)
Painting (Hours)
1
3.0
1.2
2
2.0
1.0
3
2.5
1.5
Find a schedule that minimizes makespan. Find the optimal makespan.
Answer
Since the minimum processing time is for Job 2 on machine 2, Job 2 is scheduled in the third
position. Among the remaining, the minimum processing time is for Job 1 on machine 2. So, Job 1 is
scheduled near the end, in the second position. Final job sequence is 3, 1, 2 (2 points)
Job 3
Job 1
Starting
time
0
2.5
Machine 1
Processing
time
2.5
3.0
Ending
Time
2.5
5.5
Job 3
5.5
2.0
7.5
5
Starting
time
2.5
5.5
max(4.0,5.5)
7.5
max(6.7,7.5)
Machine 2
Processing
time
1.5
1.2
1.0
Ending
time
4.0
6.7
8.5
makespan
(2 points)
Name:_________________________________________________
ID:_________________________
Question 3: (8 points) A local service station is open 7 days per week, 365 days per year. Sales of
10W40 grade premium oil average 20 cans per day. Inventory holding costs are $0.50 per can per
year. Ordering costs are $10 per order. Lead time is two weeks. Backorders are not practical – the
motorists drive away.
a. (3 points) Based on these data and assuming deterministic demand, calculate the economic order
quantity and reorder point.
Answer
D  (365)( 20)  7300 cans per year
2 DS
2(7300)(10)

 292000  540.37  540 cans (2 points)
H
0.50
R  dL  ( 20)(14)  280 cans (1 point)
Q* 
b. (5 points) The boss is concerned about this model because demand really varies. The standard
deviation of demand was determined from a data sample to be 6.15 cans per day. The manager
wants a 99.5 percent service probability. Determine a new inventory plan based on this
information. Hint: use optimal order quantity from a and find a new reorder point.
Answer
Mean of Lead-time demand,  = (20)(14)=280 cans (1 point)
Standard deviation of lead-time demand,  = 6.15 14 =23.01 cans (1 point)
P(stock-out) = 0.995. To get z, look for area = 0.995-0.5 = 0.495. z  2.57 for area = 0.4949 and
z  2.58 for area = 0.4951. Hence, z  2.575 for area = 0.495 (1 point)
R    z  280  2.575( 23.01)  339.25  339 cans (2 points)
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Name:_________________________________________________
ID:_________________________
Question 4: (6 points) A bearing production shop wants to set up control charts to monitor the
bearing sleeve production process. The following table contains the means and ranges of the
eccentricities (i.e., out-of-roundness) of some samples, measured in thousandths of an inch:
Sample
1
2
3
4
5
Mean
17
14
11
16
12
Range
6
9
5
8
7
(Consider n=5 in Parts a and b.)
Answer
X 
R
X
k
R
k


17  14  11  16  12 70

 14
5
5
6  9  5  8  7 35

7
5
5
a. (2 points) Construct a three-sigma X chart
Answer
UCL X  X  A2 R  14  (0.58)( 7)  18.06
LCL X  X  A2 R  14  (0.58)( 7)  9.94
b. (2 points) Construct a three-sigma R chart
Answer
UCL R  D4 R  ( 2.11)( 7 )  14.77
LCL R  D3 R  (0)( 7 )  0
c. (2 points) The bearing production shop undertakes a process capability study. The lower and
upper specification limits (in thousandths of an inch) for the eccentricity of the bearing sleeves are
5 and 25. If a sample has a mean of 12 and standard deviation of 4, what is Cpk.
Answer
 X  LTL UTL - X 
12  5 25  12 
C pk  min 
,
,
  min 
  min0.5833,1.0833  0.5833 ,
3

3

3(4)
3(4)




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Name:_________________________________________________
ID:_________________________
Question 5: ( 6 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 $20 each and sells each fan for $40. Any fans unsold at the end of summer season are marked
down to $15 and sold in a special fall sale.
a. (1 points) What is the marginal loss per unit?
Answer
ML= Purchase price - Salvage value = 20 - 10 = $5
b. (1 points) What is the marginal profit per unit?
Answer
MP= Selling price - Purchase price = 40 – 20 = $20
c. (4 points) If the demand is uniformly distributed between 500 to 1500 units, find the optimal order
quantity.
Answer
ML
5
P

 0.20 (2 points)
MP  ML 20  5
Now, find the Q so that P(demand Q) = 0.20
Probability
Q* = b-0.25(b-a) = 1500 - 0.20 (1500-500) = 1300 units (2 points)
Area
= 0.20
Area = 0.80
500
Demand
Q*
8
1500
Name:_________________________________________________
ID:_________________________
Question 6: ( 6 points) UA Hamburger Hamlet (UAHH) places a daily order for its high-volume
items (Hamburger patties, buns, milk, and so on). UAHH counts its current inventory on hand
once per day and phones in its order for delivery 24 hours later. Determine the number of
hamburgers UAHH should order for the following conditions:
Average daily demand:
600
Standard deviation of daily demand:
100
Desired service probability:
99%
Hamburger inventory:
800
Answer
T  1 day (1 point)
L  1 day (1 point)
 T  L   T  L  100 1  1  141.42 hamburgers (1 point)
Since service level P=0.99, the z value corresponds to area = 0.99-0.50 = 0.49
Hence, z  2.33 (1 point)
q  d (T  L )  z T  L  I  600(1  1)  2.33(141.42)  800  729.5  729 (2 points)
Probability
 141.42
Area=0.01
Area=0.49
z = 2.33
=600
Demand
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Name:_________________________________________________
ID:_________________________
Question 7: ( 10 points) A supplier has quoted the following discount schedule.
Category
Order Size
Unit Cost
1
0-599
$4.84
2
600-1799
$4.00
3
1800 and over
$3.24
Inventory holding costs are 25% per year, it costs $50 to place an order, and annual demand is 9,801
units. What order quantity should be selected?
Answer
(1.5 points) Price level: c3 = $3.24 per unit (cheapest)
2 DS
2(9801)(50)

 1100 units
Optimal order quantity (EOQ), Q * 
h
0.25(3.24)
The quantity is not feasible, because the price level is available for order quantity of 1800+.
Hence, a candidate for optimal order quantity is 1800 units and the next price level must be considered.
(1.5 points) Price level: c2 = $4.00 per unit
2 DS
2(9801)(50)

 990 units
Optimal order quantity (EOQ), Q * 
h
0.25( 4.00)
The quantity is feasible, because the price level is available for order quantity of 600-1799.
Hence, a candidate for optimal order quantity is 990 units and it is not necessary to consider any other price
levels.
(3 points) Order quantity 1800 units
hQ (0.25)(3.24)(1800)

 $729
Annual holding cost =
2
2
DS (9801)(50)
Annual ordering cost =

 $272.25
Q
1800
Annual cost of item = Dc 3  9801(3.24)  $31,755.24
Total annual cost = 729+272.25+31,755.24=$32,756.49
(3 points) Order quantity 990 units
hQ (0.25)( 4.00)(990)

 $495
Annual holding cost =
2
2
DS (9801)(50)
Annual ordering cost =

 $495
Q
990
Annual cost of item = Dc 2  9801(4.00)  $39,204
Total annual cost = 495+495+39,204=$40,194
(1 point) Conclusion: The minimum total annual cost is obtained from an order quantity of 1800 units. Hence
the optimal order quantity is 1800 units with a total annual cost of $32,756.49
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