Name (print, please) _______________________________________________ ID ___________________________
Operations Management I 73-331 Winter 2001
Faculty of Business Administration
University of Windsor
Midterm Exam II Solution
Tuesday, March 20, 10:00 – 11:10 am OB B04
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 including 2 pages of Tables
 Please be sure to put your name and student ID number on each page.
 Show your work.
Grading:
Question
Marks:
1
/2
2
/6
3
/4
4
/4
5
/10
6
/4
7
/10
Total:
/40
Name:_________________________________________________
ID:_________________________
Question 1: ( 2 points)
Compute the EOQ for an item with annual holding cost rate 20%, unit cost $5.00, ordering cost $60,
and monthly demand of 1000 units.
Answer
EOQ 
2 K

h
2(60)(100  12)
 1200 units
(0.20)(5.00)
Question 2: (6 points) Montgomery Associates produces switches for scientific equipment and has
gathered information about the production of its 6-13 switches:
Annual production rate 2,000 units
Annual demand rate 1,000 units
Cost per switch
$10
Annual holding cost
Setup cost
$80
20%
a. (2 points) Find the optimal size of each production run.
Answer
EPQ 
2 K
2(80)(1000)

 400 units
 
 1000 
h1  
(0.20)(10)1 

 P
 2000 
b. (2 points) Find the optimal cycle time.
Answer
Cycle time 
Q*


400
 0.4 years
1000
c. (2 points) What is the maximum dollar amount invested in the inventory?
Answer
Maximum Inventory =
 
 1000 
Q * 1    4001 
  200 units
 P
 2000 
Since each unit costs $10, maximum dollar amount invested in inventory =200(10)=$2,000
2
Name:_________________________________________________
ID:_________________________
Question 3: (4 points) Harold Gwynne will start a sandwich-making business. He needs to buy
breads, meats and cheeses, and condiments. He applies the EOQ formula for each item separately
using the same interest rate on every item. The unit costs and the computed EOQ values are as
shown below:
Breads
Meats and Cheeses
Condiments
Cost per unit
$0.5
$4
$2
EOQ (units)
200
100
50
a. (2 points) If Harold wants to use the EOQ order quantities, what maximum dollar investment in
inventory does he need?
Answer
Maximum dollar investment=(200)(0.5)+(100)(4)+(50)(2)=$600
b. (2 points) If Harold has only $300 available to invest in the inventory, find the optimal order
quantities.
Answer
m
Amount available
300

 0.5
Amount needed by the EOQ solution 600
Order quantity
Breads
Meats and Cheeses
Condiments
200(0.5)=100
100(0.5)=50
50(0.5)=25
Question 4: (4 points) Suppose that the annual demand is normally distributed with a mean of 1000
and a standard deviation of 100. Assume that the lead-time is 3 months.
a. (2 points) What are the mean and standard deviation of the lead-time demand?
Answer
3
Lead - time mean,     1000   250 units
 12 
3
Lead - time standard deviation,    annual   100    50 units
 12 
b. (2 points) If a (Q, R) policy is used with the reorder point, R = 335 units, find the corresponding
standardized loss function, L(z). What is the expected number of shortages per cycle?
Answer
R    z  z 
R


335  250
 1.7 . From the Table, L(z) = 0.0183
50
Therefore, expected number of shortages per cycle, n  L( z )  50(0.0183)  0.915 units
3
Name:_________________________________________________
ID:_________________________
Question 5: (10 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 $40 each and sells each fan for $55. Any fans unsold at the end of summer season are
marked down to $30 and sold in a special fall sale.
a. (2 points) What is the underage cost per unit?
Answer
cu  selling price - purchase price  55  40  $15
b. (2 points) What is the overage cost per unit?
Answer
c0  purchase price - salvage value  40  30  $10
c. (2 points) Suppose that probability(demand1000)=0.6. Enrique, the purchasing manager, says
that an optimal order quantity is 1000 units. Is Enrique right? Show your work.
Answer
p
cu
15

 0.6
cu  co 15  10
Since probability(demand1000) = 0.6 = p, an optimal order quantity is 1000 units.
So, yes, Enrique is right.
d. (2 points) If the demand is uniformly distributed between 400 to 1400 units, find the optimal order
quantity.
Answer
Q *  a  p(b  a)  500  0.6(1400  400)  1000 units
e. (2 points) If the demand is normally distributed with a mean of 900 and a standard deviation of
120, find the optimal order quantity.
Answer
Area=0.10
Hence, the area between mean and z = 0.6-0.5 = 0.1 (Area (2))
From the Table, z = 0.25 for area = 0.0987 (nearest to 0.1)
Q *    z  900  (0.25)(120)  930 units
Probability
First, find the z for which area on the left = p = 0.6 (Area (1)+(2))
 100
(2)
(1)
(3)
=900 z = 0.25
Demand
4
Name:_________________________________________________
ID:_________________________
Question 6: (4 points) Riverside Furniture produces two furniture pieces J-123 and H-234 using a
single lathe. The company uses a rotation cycle policy with a cycle time of 0.1 year.
J-123
H-234
Annual production rate
30,000
40,000
Annual demand rate
12,000
18,000
a. (2 points) Compute the production quantities for each product in each cycle.
Answer
Production quantity in each cycle, Qi  i T
Since, T  0.1 year, we get the following production quantities in each cycle:
Production
cycle
quantity
in
each
J-123
H-234
12,000(0.1)=1,200 units
18,000(0.1)=1,800 units
b. (2 points) Compute the proportion of time the lathe is idle.
Answer
Production time in each cycle, T1,i 
Qi
Pi
Hence, we get the following production times in each cycle:
Production times in each cycle
J-123
H-234
1,200/30,000=0.04 year
1,800/40,000=0.045 year
Thus, idle time in each cycle
= cycle time – production time for J-123 – production time for H-234
= 0.1-0.04-0.045
= 0.015 year
Hence the proportion of time the lathe is idle = idle time/cycle time = 0.015/0.1 = 0.15 = 15%
5
Name:_________________________________________________
ID:_________________________
Question 7: ( 10 points)
A supplier has quoted the following discount schedule.
Category
Order Size
Unit Cost
1
0-499
$5.00
2
500-1999
$4.05
3
2000 and over
$3.20
Inventory holding costs are 20% per year, it costs $50 to place an order, and annual demand is
20736 units. What order quantity should be selected?
Answer
(1.5 points) Price level: c3 = $3.2 per unit (cheapest)
2 K
2(50)( 20736)

 1800 units
Optimal order quantity (EOQ), Q * 
h
0.20(3.20)
The quantity is not feasible, because the price level is available for order quantity of 2000+.
Hence, a candidate for optimal order quantity is 2000 units and the next price level must be considered.
(1.5 points) Price level: c2 = $4.05 per unit
2 K
2(50)( 20736)

 1600 units
Optimal order quantity (EOQ), Q * 
h
0.20( 4.05)
The quantity is feasible, because the price level is available for order quantity of 500-1999.
Hence, a candidate for optimal order quantity is 1600 units and it is not necessary to consider any other price
levels.
(3 points) Order quantity 2000 units
hQ (0.2)(3.20)( 2000)

 $640
Annual holding cost =
2
2
K 50( 20736)
Annual ordering cost =

 $518.4
Q
2000
Annual cost of item = c3  20,736(3.20)  $66,355.2
Total annual cost = 640+518.4+66,355.2=$67,513.6
(3 points) Order quantity 1600 units
hQ (0.2)( 4.05)(1600)

 $648
Annual holding cost =
2
2
K 50( 20736)
Annual ordering cost =

 $648
Q
1600
Annual cost of item = c2  20736(4.05)  $83,980.8
Total annual cost = 648+648+83,980.8=$85276.8
(1 point) Conclusion: The minimum total annual cost is obtained from an order quantity of 2000 units. Hence
the optimal order quantity is 2000 units with a total annual cost of $67,513.6
6