Name (print, please) _______________________________________________ ID ___________________________
Operations Management I 73-331 Winter 2001
Faculty of Business Administration
University of Windsor
Midterm Exam I Solution
Tuesday, February 13, 10:00 – 11:10 am
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and a one-sided formula sheet.
Time available: 1 hour 10 min
Instructions:
 This exam has 8 pages.
 Please be sure to put your name and student ID number on each page.
 Show your work.
Grading:
Question
Marks:
1
/5
2
/5
3
/8
4
/6
5
/6
6
/10
Total:
/40
Name:_________________________________________________
ID:_________________________
Question 1: ( 5 points)
An analyst predicts that an 80 percent experience curve should be an accurate predictor of the cost
of producing a new product. Suppose that the cost of the first unit is $2,000. What would he predict
is the cost of producing 50th unit?
Answer
Y (u )  au  b
 au
 ln(L ) 

 
 ln(2 ) 
 2000(50)
(1 point)
 ln(0.8 ) 

 
 ln(2 ) 
 2000(50) 0.321928
 2000(0.283827)
 $567.6542
2
(1 point)
(2 points)
(1 point)
Name:_________________________________________________
ID:_________________________
Question 2: ( 5 points)
An oil company believes that the cost of construction of new refineries obeys a relationship of the
type f ( y)  ky a , where y is measured in units of barrels per day and f ( y ) in millions of dollars. From
the past experience, each doubling of the size of a refinery at a single location results in an increase
in the construction costs of about 80 percent. Furthermore, a plant size of 20,000 barrels per day
costs $10 million. Find the values of k and a.
Answer
f (2 y )  1.8 f ( y )
or, k (2 y ) a  1.8k ( y ) a
or, 2 a  1.8
(2 points)
or, a ln( 2)  ln( 1.8)
or, a 
ln( 1.8)
 0.847997
ln( 2)
f (20,000)  10
(1 point)
(1 point)
or, k (20,000) 0.847997  1.8k ( y ) a
or, k 
10
 0.00225
(20,000) 0.847997
3
(1 point)
Name:_________________________________________________
ID:_________________________
Question 3: ( 8 points)
Observations of the demand for a certain part stocked at a part supply depot during the calendar
year of 2000 were:
Month
January
February
March
Demand
89
57
144
Month
April
May
June
Demand
221
177
280
a. (3 points) Using a three-month moving average method, determine the forecasts for April through
June 2000.
Answer
89  57  144
 96.7
3
57  144  221

 141
3
144  221  177

 181
3
FApril 
(1 point)
FMay
(1 point)
FJune
(1 point)
b. (3 points) Using an exponential smoothing method with   0.1 and a March forecast of 150,
determine the forecasts for April through June 2000.
Answer
FApril  0.1(144)  0.9(150)  149.4
(1 point)
FMay  0.1(221)  0.9(149.4)  156.6
(1 point)
FJune  0.1(177)  0.9(156.6)  158.6
(1 point)
c. (2 points) For each method in (a) and (b), compute MAD. Based on the MAD values, comment on
which method to use.
Answer
MAD Exp Sm
221  96.7  177  141  280  181
124.33  36.33  99.33 260

 86.67
3
3
3
221  149.4  177  156.6  280  158.6 71.60  20.44  121.40 213.44



 71.15
3
3
3
MAD 3-Month MA 

(1 point)
Use Exponential smoothing method, which has a MAD 71.15 < 86.67 (1 point)
4
Name:_________________________________________________
ID:_________________________
Question 4: ( 6 points)
Here are the data for the past 4 months of actual sales of a particular product:
Month
Actual Demand
1
220
2
227
3
229
4
234
a. (5 points) Calculate the double exponential smoothing forecasts for months 1-4 using S0=210,
G0=10, =0.10, and  = 0.20.
Answer
Month
t
1
2
3
4
b.
Actual
St
demand
(2 points)
220
0.1(220)+0.9(220)
=220
227
0.1(227)+0.9(230)
=229.7
229
0.1(229)+0.9(239.64)
=238.576
234
0.1(234)+0.9(248.3032)
=246.87
Gt
(2 points)
0.2(220-210)
+0.8(10)=10
0.2(229.7-220)
+0.8(10)=9.94
0.2(238.576-229.7)
+0.8(9.94)=9.7272
0.2(246.87-238.576)
+0.8(9.7272)=9.4411
(1 point) What is F4,6?
Answer
F4,6 = S4 +2G4 =246.87+2(9.4411) = 265.76
5
Ft-1,t
(1 point)
210+10
=220
220+10
=230
229.7+9.94
=239.64
238.576+9.7272
=248.3032
Name:_________________________________________________
ID:_________________________
Question 5: ( 6 points)
Windsor Swiss Milk Products manufactures and distributes ice cream in Ontario. The company
wants to expand operations by locating another plant in northern Sarnia. The size of the new plant
will be a function of the expected demand for ice cream within the area served by the plant. A market
survey is currently under way to determine that demand. Windsor Swiss wants to estimate the
relationship between the manufacturing cost per gallon and the number of gallons sold in a year to
determine the demand for ice cream and thus the size of the new plant. The following data have
been collected:
Plant
Thousands of Gallons Sold, X
Cost per Thousand Gallons, Y
1
416.9
$1,015
2
472.5
973
3
250.0
1,046
4
372.1
1,006
a. (5 points) Develop a regression equation to forecast the cost per gallon as a function of the
number of gallons produced.
Answer
x
416.9
472.5
250
372
1511.4
377.85
Sum
Average
y
1015
973
1046
1006
4040
1010
xy
423153.5
459742.5
261500
374232
1518628
x^2
173805.61
223256.25
62500
138384
597945.86
(3 points)
Slope 
4(1518628)  1511.44040
4(597945.86)  1511.4
2
 0.29356
Intercept  y  slope( x)  1010  (0.29356)(377.85)  1120.9215
(1 point)
(1 point)
b. (1 point) Suppose that the market survey indicates a demand of 325,000 gallons in Sarnia.
Estimate the manufacturing cost per gallon for a plant producing 325,000 gallons per year.
Answer
y  a  bx  1120.9215  (0.29356)(325)  $1025 / thousand gallons sold
 $1.025/gal lon sold
6
Name:_________________________________________________
ID:_________________________
Question 6: ( 10 points)
Windsor Cookie Company makes a variety of chocolate chip cookies. Based on orders received and
forecasts of buying habits, it is estimated that the demand for the next three months is 700, 1000,
and 900, expressed in thousands of cookies. Historically, each worker produces 250 cookies per
working day. Assume that the numbers of workdays over the three months are respectively 25, 20
and 16. There are currently 75 workers employed, and there are 200 thousand cookies in the
inventory.
a. (5 points) What is the minimum constant workforce required to meet demand over the next three
months?
Answer
Month
Production
Required
1
700-200=500
Cumulativ
e Net
Production
Required
(2 point)
500
Number
of
Working
Days
Cumulative
Cumulative
Number of Number of Units
Working
Produced Per
Days
Worker
25
25
25250=6250
2
1000
1500
20
45
45250=11250
3
900
2400
16
61
61250=15250
Number of
Workers Needed
(2 points)
500,000/6250
=80
1,500,000/11250
=133.33=134
2,400,000/15250
=157.38=158
Minimum number of workers needed to meet the demand = 158 (1 point)
(Continued…)
7
Name:_________________________________________________
ID:_________________________
b. (5 points) Assume that the inventory holding cost is 20 cents per cookie per month, cost of hiring
is $200 per worker hired and cost of firing is $300 per worker fired. Evaluate the cost of the plan
derived in (a).
Answer
Number of workers hired = 158-75=83
Hiring cost = 83  200 = $16,600 (1 point)
Month
Production
Required
1
700,000
Beginnin
g
Inventory
200,000
2
1000,000
487,500
3
900,000
277,500
Actual Production
(1 point)
Ending Inventory
(158)(25)(250)=987,50
0
(158)(20)(250)=790,00
0
(158)(16)(250)=632,00
0
487,500
Total
277,500
9,500
774,500
Inventory holding cost = 774,500(0.2)=$154,900 (1 point)
Total cost = $16,600 + $154,900 = $171,500 (2 points)
8