Last Name _________________________ First Name _________________________ ID _______________________
Production Management 73-604 Fall 2004
Odette School of Business
University of Windsor
Exam 1
Thursday, October 14, 5:30 – 6:50 pm
Instructor : Mohammed Fazle Baki
Aids Permitted : Calculator, straightedge, and a one-page formula sheet.
Time available : 1 hour 20 minute
Instructions:

This exam has 10 pages including this cover page.

Please be sure to put your name and student ID number on each odd numbered page.

Show your work.

Show your answers upto four decimal places.
Grading:
Question
1
Marks:
/10
2
3
/10
/10
4
5
/22
/13
Total: /65

Last Name _________________________ First Name _________________________ ID _______________________
Question 1: (10 points) Circle the most appropriate answer
1.1 If learning is faster, the rate of learning, L is a. more and the learning curve is steeper b. more and learning curve is flatter c. less and the learning curve is steeper d. less and learning curve is flatter
1.2 The following is a disadvantage of too frequent capacity addition: a. Cost of capital tied to build and maintain excess capacity b. Installation c. Unused capacity d. Both b and c
1.3 The exponential smoothing method with
 
0 .
25 yields the same distribution of forecast errors as the simple moving average method with a. N = 7 b. c.
N
N


5
3 d. none of the above
1.4 The demand for lawn mower in spring is more than average quarterly demand. Which of the following is not a possible multiplicative seasonal factor for demand of lawn mower in spring? a. 1.75 b. 0.75 c. 4.75 d. Both b and c
1.5 Which of the following forecasting method yields the smoothest series? a. Exponential smoothing method with b. Exponential smoothing method with c. Simple moving average method with

N


1


.
00
1
0 .
10 d. Both b and c
1.6 Which of the following is an advantage of level strategy? a. Less inventory b. Efficiency of machines an workers c. High worker morale d. Both b and c
1.7 An aggregate production plan is developed using a. an aggregate unit of production after maser production schedule is developed in terms of actual units b. an aggregate unit of production and master production schedule translates the plan in terms of actual units c. actual units of production after maser production schedule is developed in terms of an aggregate unit d. actual units unit of production and master production schedule translates the plan in terms of an aggregate unit
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Last Name _________________________ First Name _________________________ ID _______________________
1.8 It’s possible to find an optimal aggregate production plan without considering the following: a. cost of labor b. cost of material c. cost of capital assets such as plants and machines d. Both b and c
1.9 Which of the following is true? a. Efficient supply chain is better than responsive supply chain b. Responsive supply chain is better than efficient supply chain c. Responsive supply chain is better for the innovative products and efficient supply chain for the functional products d. Efficient supply chain is better for the innovative products and responsive supply chain for the functional products
1.10 Sales for the CNC machines for the past two years were as follows:
Quarter, 2002 Quantity (units) Quarter, 2003 Quantity (units)
Winter 22 Winter 42
Spring
Summer
Fall
35
30
38
Spring
Summer
Fall
45
40
48
Which of the following is a most appropriate multiplicative seasonality index for the fall quarter for the sales of the CNC machines? a. 0.3040 b. 1.2160 c. 1.1467 d. 1.0971
Question 2: (10 points)
The following table reproduces some parts of Exhibits TN2.5 and TN2.6.
Unit
1
2
3
4
5
6
7
8
20
Unit Improvement Factor Cumulative Improvement Factor
80% 85% 90% 95% 80% 85% 90% 95%
1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000 1.0000
0.8000 0.8500 0.9000 0.9500 1.8000 1.8500 1.9000 1.9500
0.7021 0.7729 0.8462 0.9219 2.5021 2.6229 2.7462 2.8719
0.6400 0.7225 0.8100 0.9025 3.1421 3.3454 3.5562 3.7744
0.5956 0.6857 0.7830 0.8877 3.7377 4.0311 4.3392 4.6621
0.5617 0.6570 0.7616 0.8758 4.2994 4.6881 5.1008 5.5380
0.5345 0.6337 0.7439 0.8659 4.8339 5.3217 5.8447 6.4039
0.5120 0.6141 0.7290 0.8574 5.3459 5.9358 6.5737 7.2612
0.3812 0.49540 0.6342 0.8012 10.4800 12.4000 14.6100 17.1300
A time standard was set as 0.70 hour per unit based on the 3 rd unit produced. If the task had an 80 percent learning curve,
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Last Name _________________________ First Name _________________________ ID _______________________ a. (5 points) what would be the expected time for the 4 th unit?
Y
3

K * UIF
80
%,
3
So, K

Y
3
UIF
80 %, 3

0
.
70
0
.
7021
 0
.
9970
(2 points)
Y
4

K * UIF
80
%,
4
 0
.
9970
*
0
.
6400  0
.
6381 hours (3 points including 1 for unit) b. (5 points) what would be the total expected time for 17 units starting from the 4 th units, i.e., from the 4 th unit to the 20 th unit?
20
 i
 4
Y i
 i
20

 1
Y i
 i
3

 1
Y i


K * CIF
80
%,
20
0
.
9970
*
10
.

K
4800
* CIF
80
%,
3
 0
.
9970
*
2
.
5021
 7
.
9540 hours
Question 3: (10 points)
A firm must decide whether to construct a small, medium, or large stamping plant. A consultant’s report indicates a 0.20 probability that demand will be low and a 0.80 probability that demand will ve high.
If the firm builds a small facility and demand turns out to be low, the net present value will be $42 million; if demand turns out to be high, the firm can either subcontract and realize the net present value of $42 million or expand for a net present value of $48 million.
The firm could build a medium-size facility as a hedge: If demand turns out to be low, its net present value is estimated at $22 million; if demand turns out to be high, the firm could do nothing and realize a net present value of $46 million, or it could expand and realize a net present value of $50 million.
If the firm builds a large facility and demand is low, the net present value will be -$20 million, whereas high demand will result in a net present value of $72 million. a. (5 points) Construct a decision tree to analyze this problem. b. (5 points) What is the best decision and what is the expected payoff?
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Last Name _________________________ First Name _________________________ ID _______________________
Decision tree is shown above. The expected values are also shown on nodes.
EV(5) = max(42,48) = $48m
EV(6) = max(46,50) = $50m
EV(2) = 0.20(42) + 0.80(48) = $46.8m
EV(3) = 0.20(22) + 0.80(50) = $44.4m
EV(4) = 0.20(-20) + 0.80(72) = $53.6m
EV(1) = max(46.8, 44.4, 53.6) = $53.6m
Hence, the best decision is to construct a large plant and the expected payoff is $53.6m.
5

Last Name _________________________ First Name _________________________ ID _______________________
Question 4: (22 points)
Historical demand for a product is: x Month Demand
1
2
January
February
41
45
3
4
March
April
46
51
5 May 52 a. (5 points) Using a simple three-month moving average, find April and May forecast. Compute
MAD.
F
F
April
May
E
April


41 
45 
45  46
3
46  51
 44 
3
51  


7
,
44
47
E
.
33
May
 47
.
33  52   4
.
67
MAD

E
April

E
May 
 7 
2
 4
.
67
 5
.
8334
2 b. (3 points) Using a single exponential smoothing with
 forecast
F
June

F
May
 0
.
20

A
May

F
May

 50 

0
.
20

52
0 .
20 and a May forecast = 50, find June
 50

 50
.
4 c. (6 points) Using a double exponential smoothing method with
T
1

3 , find F
2
, T
2
and FIT
2
.
 
0 .
25 ,
 
0 .
20 , F
1

37 , and
FIT
1
F
2
T
2


FIT
2

F
1

T
1
FIT
1
T
1
 



F
2


A
1

37

 3
FIT
1
FIT
1




3
40


40  0
.
25

41
0
.
20

40
.
25 
 40
40



 40
.
25
3
.
05

F
2

T
2
 40
.
25  3
.
05  43
.
30
6

Last Name _________________________ First Name _________________________ ID _______________________ d. (8 points) Find June forecast using linear regression
Month x y xy x
2
January
February
1
2
41
45
41
90
1
4
March
April
May
Sum
Average
(2 points)
3
4
5
15
3
46
51
52
235
47
138
204
260
733
------
9
16
25
55
------
Slope, b


 x xy
2 
 n n x
 
2 y

733
55


5
5
  
  2
47

28
10
 2
.
80
(3 points)
Intercept, a
 y
 b x

Forecast for June, Y
6
47  2
.
8
 
 38
.
6
(2 points)
 a
 bx
 38
.
6  2
.
8
 
 55
.
4 (1 point)
Question 5: (13 points)
Mr. Meadows Cookie Company makes a variety of chocolate chip cookies in the plant in Albion,
Michigan. Based on orders received and forecasts of buying habits, it is assumed that the demand for the next three months is 1200, 1300 and 1100, expressed in thousands of cookies. During a 50day period when there were 100 workers, the company produced 2 million cookies. There are currently 143 workers employed, and there is no starting inventory of cookies. a. (2 points) Compute productivity in number of cookies per worker per day
Productivity

2
,
000
,
000
50  100
 400 cookies per worker per day b. (3 points) Assume that the numbers of workdays over the three months are respectively 19, 21 and 20. The management plans to hire 15 workers at the beginning of Month 1, and 5 at the beginning of Month 2. The management also plans to lay off 34 workers at the beginning of
Month 3. Show the number of units produced and ending inventory carried in each of the next three months
Month Demand
000s
Beginning
Inventory
000s
Number of
Workers
Hired
Number of
Workers
Fired
Total
Workers
Number of Days of Work
Production
000s
Ending
Inventory
000s
1
2
1200
1300
3 1100
Total -------
0
0.800
70.000
-------
15
5
0
20
0
0
34
34
158
163
129
-------
19
21
20
-------
1200.800
1369.200
1032.000
-------
0.800
70.000
2.000
72.800
7

Last Name _________________________ First Name _________________________ ID _______________________ c. (3 points) Assume that the inventory holding cost is 10 cents per cookie per month, backorder cost 20 cents per cookie, hiring cost $300 per worker, and firing cost $500 per worker. Compute total cost of the plan.
Hiring cost = 300(20) =
Firing cost = 500(34) =
$6,000
$17,000
Inventory holding cost = 72.8(1000)(0.10) = $7,280
Total cost = $30,280 d. (5 points) Ignore parts b and c .
Assume that the numbers of workdays over each of the three months is 20. The management plans to use a constant level of workforce. What is the minimum numbers of workers required in order to avoid shortages?
Units produced per worker per month = 20(400) = 8,000
Month Demand
000s
Cumulative
Demand
000s
Cumulative Units
Produced per
Worker
000s
Number of
Workers Required
1 1200 1200 8  1200
/
8   150
2 1300 2500 16
24
 2500
/
16   157
 3600
/
24   150 3 1100 3600
Minimum number of workers required = max(150, 157, 150) = 157.
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