Download soln_mt_2_f02_331 - University of Windsor

Last Name ______________________ First Name _________________________ ID ___________________________
Operations Management I 73-331 Fall 2002
Odette School of Business
University of Windsor
Midterm Exam II Solution
Wednesday, November 20, 10:00 – 11:20 am
Education Building Room ED 1101
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and a one-sided formula sheet.
Time available: 1 hour 20 min
Instructions:
 This exam has 22 pages including this cover page and 8 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
/10
3
/10
4
/10
5
/10
6
/15
Total:
/65
Name:_________________________________________________
ID:_________________________
Question 1: (10 points) Circle the most appropriate answer
1.1 Obsolescence is considered when estimating
a. opportunity cost of capital
b. holding cost
c. ordering/setup cost
d. stock-out/penalty cost
e. all of the above
f. none of the above
1.2 In the finite production rate model, the inventory is accumulated during the uptime at the rate of
a. P
b. 
c.  
d. P  
e. P  
f. 1 / 
1.3 In the context of the all-units quantity discount model, an EOQ is feasible if
a. it’s not a breakpoint
b. it’s the cheapest
c. it gives the minimum total cost
d. it falls within the interval that corresponds to the unit cost used to compute it
e. a, b and c
f. none of the above
1.4 In the space constrained problem, the optimal solution requires the use of
a. the available space if space available  space required by the EOQ units
b. the space required by the EOQ units if space required by the EOQ units  space available
c. the available space if space available > space required by the EOQ units
d. the space required by the EOQ units if space required by the EOQ units > space available
e. none of the above
f. a and b
1.5 The fund constrained model assumes
a. quantity discount
b. limited space
c. finite production rate
d. multiple products
e. uncertain demand
f. shortages
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Name:_________________________________________________
ID:_________________________
1.6 If multiple items are produced using a single production facility, we must not plan the production
of the items separately in order to avoid
a. infeasible/non-optimal solution or shortages
b. machine breakdown
c. damages
d. long uptime
e. long downtime
f. none of the above
1.7 The following sequence of production of items A, B, C and D is an example of the rotation cycle
policy
a. A, C, A, B, D, A, C, A, B, D, A, C, A, B, D
b. A, B, C, D, B, C, D, A, C, D, A, B
c. D, C, A, B, D, C, A, B, D, C, A, B
d. a and b
e. b and c
f. c and a
1.8 If the demand is restricted to the following, then the demand is discrete:
a. 500 gm to 800 gm
b. 500 gm or more
c. 800 gm or less
d. Multiple of 100 bottles up to a maximum of 1000 bottles
e. All of the above
f. None of the above
1.9 The single period model is best suited for
a. newspapers
b. perishable items
c. seasonal items
d. automobiles
e. a, b, c
f. none of the above
1.10
a.
b.
c.
d.
e.
f.
The expected annual number of units purchased
is a function of the order size Q
is not a function of the order size Q
is a function of the safety stock
is not a function of the safety stock
a and c
b and d
3
Name:_________________________________________________
ID:_________________________
Question 2: (10 points)
Suppose that item A has a unit cost of $4.00, an ordering cost of $100, and a quarterly demand of
450 units. It is estimated that cost of capital is approximately 20 percent per year. Annual storage
cost amounts to 3 percent and breakage to 2 percent of the value of the each item.
a. (2 points) Compute holding cost per unit per year.
I  0.20  0.03  0.02  0.25
h  Ic  0.254  $1 per unit per year
b. (2 points) Compute annual demand.
  4504  1800 units per year
c. (3 points) Compute EOQ of item A.
EOQ=
2K
21001800

 600 units
h
1
d. (3 points) Suppose that both items A and B should be purchased and there is only $640 available
for buying items A and B. The unit cost of item B is $8 and the EOQ of item B is 500 units. What
is the optimal order quantity of item A?
Fund required by the EOQ order quantity of Item A = 600(4) = $2,400
Fund required by the EOQ order quantity of Item B = 500(8) = $4,000
Total fund required by the EOQ order quantities of Items A and B
= 2,400+4,000 = $6,400 (1 point)
Fund available = $640
Hence, m =
fund available
640

 0.10 (1 point)
fund required
6,400
Therefore, the optimal order quantity of Item A = m  EOQA = 0.10(600) = 60 units (1 point)
4
Name:_________________________________________________
ID:_________________________
Question 3: (10 points)
This problem uses the same data from problem 2, re-written in the following: Suppose that item A
has a unit cost of $4.00, an ordering cost of $100, and a quarterly demand of 450 units. It is
estimated that cost of capital is approximately 20 percent per year. Annual storage cost amounts to 3
percent and breakage to 2 percent of the value of the each item.
Assume that item A has a production rate of 2400 items per year.
a. (2 points) Compute EPQ of item A.
EOQ=
2 K

h'
2 K

 
h1  
 P
21001,800
 1200 units
 1,800 
11 

 2,400 
b. (2 points) What is the cycle time of item A if only item A is produced?
Q * EPQ 1,200
T


 0.6667 years


1,800
Item C has a production rate of 1440 items per year, a unit cost of $48.00, an ordering cost of $150,
and a monthly demand of 12 units.
c. (4 points) What is the cycle time if both items A and C are produced in a single facility? Assume
negligible setup times for both items A and C.
T* 


2 K j
 h'
j
j

2K1  K 2 

h'1 1  h' 2  2
2100  150
 
 1,800 
11 
1,800  Ic 2 1  2
P2
 2,400 

2250

0.25  1,800  12  0.90  144

 2

2100  150
  
  
h1 1  1 1  h2 1  2  2
P1 
P2 



2100  150
 1,800 
 12  12 
11 
1,800  0.25  481 
144
1,440 
 2,400 

500

450  1,555.2
500
= 0.4994 years
2005.2
d. (2 point) What is the optimal order quantity of item A?
Q *  T *  1,8000.4994  898.92 units
5
Name:_________________________________________________
ID:_________________________
Question 4: (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 $60 each and sells
each fan for $70. Any fans unsold at the end of summer season are marked down to $20 and sold in
a special fall sale.
a. (2 points) What is the underage cost per unit?
cu  Selling price – purchase price = 70-60 = $10/unit
b. (2 points) What is the overage cost per unit?
co  Purchase price – salvage value = 60-20 = $40/unit
c. (3 points) If the demand is uniformly distributed between 350 and 850 units, find the optimal order
quantity.
For the optimal order quantity Q , Probability(demand  Q ), p 
cu
10

 0.20 (2 points)
cu  co 40  10
Hence, Q *  a  pb  a   350  0.20850  350  450 units (1 point)
d. (3 points) If the demand is normally distributed with a mean of 600 and a standard deviation of
75, find the optimal order quantity.
For the optimal order quantity Q , Probability(demand  Q ), p 
cu
10

 0.20
cu  co 10  40
Find the standard normal z -value for which cumulative area on the left, F z   p  0.20 .
Since Table A-1 gives area between z  0 and positive z -values, find z -value corresponding to
Table A-1 area = 0.50-0.20 = 0.30
Hence, z  0.845 (2 points for the correct value of z) Note: z is negative because F z   p  0
f(x)
Hence, Q *    z  600   0.84575  536.625 units (1 point)
Table A-1 gives
area from the
center (z = 0)
=0.50-0.20 =0.30
Area, F(z)
=p
= 0.20
=75
z=? =600
z=0
6
Name:_________________________________________________
ID:_________________________
Question 5: (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
1-12
$1000
13-25
950
26+
945
The annual inventory holding cost is 20%, the ordering cost is $80, 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  $940 per unit. h3  Ic3  0.20  945  $189 /unit/year
EOQ3 
2 K
280125

 10.29 units (1 point)
h3
189
Since the price level of c3  $945 is not available for an order quantity Q  EOQ3 = 10.3 units,
EOQ3 is infeasible and a candidate for optimal order quantity is Q3  26 , because 26 is the
minimum order quantity for the price level of c3  $945.

Now, consider the next price level, c2  $950 per unit. h2  Ic2  0.20  950  $190 /unit/year
EOQ2 
2 K
280125

 10.26 units (1 point)
h2
190
Since the price level of c2  $950 is available for an order quantity Q  EOQ2 = 10.3 units, EOQ2
is feasible and a candidate for optimal order quantity is Q2  13 , because 13 is the minimum order
quantity for the price level of c2  $950.

Now, consider the next price level, c1  $1,000 per unit. h1  Ic1  0.20 1,000  $200 /unit/year
EOQ1 
2 K
280125

 10 units (1 point)
h1
200
Since the price level of c1  $1,000 is available for an order quantity Q  EOQ1 = 10 units, EOQ1
is feasible and a candidate for optimal order quantity is Q1  10 .
(Continued…)
7
Name:_________________________________________________

Now, compute total cost for each candidate for optimal order quantity:
Candidate
Qj
j
3
Q3  26
(1 point)
2
Q2  13
(1 point)
1
Q2  10
(1 point)

ID:_________________________
(2 points)
(1 point)
Holding cost
Ordering cost
Cost of item
Total cost
hjQ j
c j
2
K
Qj
Holding cost +
Ordering cost +
Cost of item
189  26
 2,457
2
80  125
 384.62
26
945  125  118,125
$120,966.50
190  13
 1,235
2
80  125
 769.23
13
950  125  118,750
$120,754.23
200  10
 1,000
2
80  125
 1,000
10
1000  125  125,000
$127,000
Conclusion: The total cost is minimum, $120,754.23 for Q3  13 . Therefore, an optimal order
quantity is Q3  13 . (1 point)
8
Name:_________________________________________________
ID:_________________________
Question 6: (15 points)
The home appliance department of a large department store is using a lot size-reorder point system
to control the replenishment of a particular model of FM table radio. The store sells an average of
1,000 radios each year. The annual demand follows a normal distribution with a standard deviation
of 60. The store pays $20 for each radio, which it sells for $65. The holding cost is 25 percent per
year. Fixed costs of replenishment amount to $450. If a customer demands the radio when it is out of
stock, the customer will generally go elsewhere. Loss-of-goodwill costs are estimated to be about
$30 per radio. Replenishment lead time is three months. Currently, the store is using Q  450 and
R  300 . Compute
a. (2 points) the mean and standard deviation of the lead time demand

3
3
 0.25 years,     1,000  250 units, (1 point)
12
12
   y   60
3
 30 units (1 point)
12
b. (1 point) the annual holding cost per unit
h  Ic  0.25  20  $5 per unit per year
c. (1 point) the stock-out cost per unit
p = loss of profit + good will = (65-20) +30 =$75 per unit
d. (1 point) the safety stock
R    300  250  50 units
e. (1 point) the expected number of units stock-out per cycle
z
R   300  250

 1.67

30
Lz   0.0197
n  Lz   300.0197  0.5910 units per cycle
(Continued…)
9
Name:_________________________________________________
ID:_________________________
f. (2 points) the annual holding cost
hQ
5  450
 h R    
 5300  250  1,125  250  $1,375 (1 point for each part)
2
2
g. (2 points) the annual ordering cost
K 450  1,000

 $1,000
Q
450
h. (2 points) the annual stock-out cost
np 0.591  75  1,000

 $98.50
Q
450
i.
(1 point) the total annual holding, ordering and stock-out cost
1,375  1,000  98.50  $2,473.50
j.
(1 point) the probability of not stocking out during the lead time
z
R   300  250

 1.67

30
Table A-4: The probability of not stocking out during the lead time = F z  1.67
 0.9525
Table A-1: The probability of not stocking out during the lead time
= the area on the left of z  1.67
= P   z  1.67  P   z  0  P0  z  1.67
= 0.50  P0  z  1.67  0.5  0.4525  0.9525
k. (1 point) the fill rate, up to four decimal places
  1
n
0.5910
 1
 0.9987  99.87%
Q
450
10