Practice Problems: Chapter 12, Inventory Management
Problem 1:
ABC Analysis
Stock Number
Annual $ Volume
Percent of Annual $ Volume
J24
12,500
46.2
R26
9,000
33.3
L02
3,200
11.8
M12
1,550
5.8
P33
620
2.3
T72
65
0.2
S67
53
0.2
Q47
32
0.1
V20
30
0.1
 = 100.0
What are the appropriate ABC groups of inventory items?
Problem 2:
A firm has 1,000 “A” items (which it counts every week, i.e., 5 days), 4,000 “B” items (counted
every 40 days), and 8,000 “C” items (counted every 100 days). How many items should be counted
per day?
Problem 3:
Assume you have a product with the following parameters:
Annual Demand = 360 units
Holding cost per year = $1.00 per unit
Order cost = $100 per order
What is the EOQ for this product?
Problem 4:
Given the data from Problem 3, and assuming a 300-day work year, how many orders should be
processed per year? What is the expected time between orders?
Problem 5:
What is the total cost for the inventory policy used in Problem 3?
Problem 6:
Based on the material from Problems 3 – 5, what would cost be if the demand was actually higher
than estimated (i.e., 500 units instead of 360 units), but the EOQ established in problem 3 above is
used? What will be the actual annual total cost?
Problem 7:
If demand for an item is 3 units per day, and delivery lead-time is 15 days, what should we use for
a simple re-order point?
Problem 8:
Assume that our firm produces Type C fire extinguishers. We make 30,000 of these fire
extinguishers per year. Each extinguisher requires one handle (assume a 300 day work year for
daily usage rate purposes). Assume an annual carrying cost of $1.50 per handle, production setup
cost of $150, and a daily production rate of 300. What is the optimal production order quantity?
Problem 9:
We use 1,000 electric drills per year in our production process. The ordering cost for these is $100
per order and the carrying cost is assumed to be 40% of the per unit cost. In orders of less than 120,
drills cost $78 per unit; for orders of 120 or more the cost drops to $50 per unit.
Should we take advantage of the quantity discount?
Problem 10:
Litely Corp sells 1,350 of its special decorator light switch per year and places orders for 300 of
these switches at a time. Assuming no safety stocks, Litely estimates a 50% chance of no shortages
in each cycle and the probability of shortages of 5, 10, and 15 units as 0.2, 0.15, and 0.15
respectively. The carrying cost per unit per year is calculated as $5 and the stockout cost is
estimated at $6 ($3 lost profit per switch and another $3 loss of goodwill or future sales). What
level of safety stock should Litely use for this product? (Consider safety stock of 0, 5, 10, and 15
units.)
Problem 11:
Presume that Litely carries a modern white kitchen ceiling lamp that is quite popular. The
anticipated demand during lead-time can be approximated by a normal curve having a mean of 180
units and a standard deviation of 40 units. What safety stock should Litely carry to achieve a 95%
service level?
ANSWERS
Problem 1:
ABC Groups
Class
Items
Annual Volume
Percent of $ Volume
A
J24, R26
21,500
79.5
B
L02, M12
4,750
17.6
C
P33, T72, S67, Q47, V20
800
2.9
 = 100.0
Item P33 is a judgment call. It might be considered a B item by some organizations. However, the
modern tendency is to move items to as low a level as possible thereby reducing inventory
management costs.
Problem 2:
Item Class
Quantity
Policy
Number of Items to Count Per Day
A
1,000
Every 5 days
1000/5 = 200/day
B
4,000
Every 40 days
4000/40=100/day
C
8,000
Every 100 days
8000/100=80/day
Total items to count: 380/day
Problem 3:
EOQ 
2* Demand *Order Cost
2*360*100

 72000  268.33 items
Holding cost
1
The EOQ model assumes any real quantity is feasible. The actual quantity ordered may need to be
an integer value and may be affected by packaging or other item characteristics. In the following
Problems an EOQ of 268 is assumed.
Problem 4:
N
Demand 360

 1.34 orders per year
Q
268
T
Working days
 300 /1.34  224 days between orders
Expected number of orders
Problem 5:
Demand *Order Cost (Quantity of Items) *(Holding Cost)

Q
2
360*100 268*1


 134  134  $268
268
2
TC 
Notice that at the EOQ Total Holding Cost and Total Ordering Cost are equal.
Problem 6:
Demand *Order Cost (Quantity of Items) *(Holding Cost)

Q
2
500*100 268*1


 186.57  134  $320.57
268
2
TC 
Note that while demand was underestimated by nearly 50%, annual cost increases by only 20%
(320 / 268  120
. ) an illustration of the degree to which the EOQ model is relatively insensitive to
small errors in estimation of demand.
Problem 7:
ROP = Demand during lead-time = 3 * 15 = 45 units
Problem 8:
The equation used differs from the basic EOQ model by allowing for gradual replenishment, which
affects the average level of inventory.
Q*p 
2* Demand *Order Cost
(2)(30, 000)(150)

 3000 units

 100 
Daily Usage Rate 
1.50 1 
Holding Cost 1 


 300 
 Daily Production Rate 
Problem 9:
Q*p ($78) 
(2)(1000)(100)
 80 units
(0.4)(78)
Q*p ($50) 
(2)(1000)(100)
 100 units  120 to take advantage of quantity discount.
(0.4)(50)
Ordering 100 units at $50 per unit is not possible, the discount does not apply until 120 units are
ordered. We need to compare the total costs for the two alternatives, Q($78) and Q = 120. In this
situation, the Total Cost equation must include the cost of the item since this is not a constant.
Total cos t  Demand * Cost 
Demand * Order Cost (Quantity of Items) * ( Holding cos t )

Q
2
(1000)(100) (80)(0.4)(78)


80
2
$78, 000  $1, 250  $1, 248  $80, 498
Total cost($78)  (1000)(78) 
(1000)(100) (120)(0.4)(50)


120
2
$50, 000  $833  $1, 200  $52, 033
Total cost($50)  (1000)(50) 
Therefore, we should order 120 each time at a unit cost of $50 and a total cost of $52,033. Notice
that Total Holding Cost is not equal to Total Ordering Cost at the lowest cost alternative (Q = 120)
since this is not an EOQ.
Problem 10:
Safety stock  0 units:
Carrying cost equals zero.
Total Stockout Costs = (stockout costs * possible units of shortage * probability of shortage *
number of orders per year)
S0  6*5*0.2*
1350
1350
1350
 6*10*0.15*
 6*15*0.15*
 $128.25
300
300
300
Safety stock  5 units:
Carrying cos t  $5 per unit * 5 units  $25
Stockout cost: S5  6*5*0.15*
1350
1350
 6*10*0.15*
 $60.75
300
300
Total Cost = Carrying cost + Stockout cost  $25  $60.75  $85.75
Safety stock  10 units:
Carrying cos t  10 * 5  $50.00
. *
Stockout cost: S10  6* 5* 015
1350
 $20.25
300
Total Cost = Carrying cost + Stockout cost  $50.00  $20.25  $70.25
Safety stock  15:
Carrying cos t  15* 5  $75.00
Stockout cos ts  0 (There is no shortage if 15 units are maintained)
Total Cost = Carrying cost + Stockout cost  $75.00  $0  $75.00
Therefore: Minimum cost comes from carrying a 10 unit safety stock.
Problem 11:
To find the safety stock for a 95% service level it is necessary to calculate the 95th percentile on
the normal curve. Using the standard Normal table from the text, we find the Z value for 0.95 is
1.65 standard units. The safety stock is then given by:
(165
. * 40)  180  66  180  246 Ceiling Lamps