Chapter 13 – Inventory Management
13–41
Education.
26. Given:
d = 5 boxes/week
d = .5 boxes/week
LT = 2 weeks
S = $2
H= $.20/box/ year
a. Assuming a 52-week year, determine the EOQ:
D = 52 x 5 = 260
b. If ROP = 12, determine risk of a stockout:
LTzdROP d)((LT)
Plugging in values and solving for z:
2(.5)(2)512 z
12 = 10 + .707z
2 = .707z
z = 2 / .707 = 2.83 (round to two decimals)
From Appendix B, Table B, the lead time service level is .9977.
Risk of stockout = 1 – .9977 = .0023 = .23%.
c. OI = 7 weeks. Determine the risk of running out before this order arrives (Q = 36) if the copy
center orders when amount on hand = 12:
Use Formula 13-13 and solve for z:
LTzdROP d)((LT)
Plugging in values and solving for z:
2(.5)(2)512 z
z
707.1012
z
707.2
z = 2 / .707 = 2.83 (round to two decimals)
From Appendix B, Table B, the lead time service level is .9977.
Risk of stockout = 1 – .9977 = .0023 = .23%.
13–42
Education.
27. Given:
d
= 80 lb./day
d = 10 lb./day
LT
= 8 days
LT = 1 day
Determine the ROP that would provide a stockout risk of 10%:
Service Level = 1 – .10 = .90.
28. Given:
d
= 10 rolls/day
d = 2 rolls/day
LT = 3 days
Supermarket is open 360 day a year
S = $1
H = $.40
a. Determine the EOQ:
D = 10 x 360 = 3,600
Chapter 13 – Inventory Management
13–43
Education.
29. Given:
D = 1,200 cases
S = $40 per order
H = $3 per case per year
Service level = 99%
a. Determine the optimal order quantity:
b. Determine the level of safety stock if lead time demand is normally distributed with a mean
of 80 cases and a standard deviation of 6 cases:
30. Given:
ROP = 18 units
Lead time for resupply = 3 days
Usage over the last 10 days:
Day
1
2
3
4
5
6
7
8
9
10
Units
3
4
7
5
5
6
4
3
4
5
Determine the service level achieved by the current ROP. Hint: Use Formula 13-13.
LTzdROP d)((LT)
Step 1:
Calculate the mean and standard deviation of daily demand.
(round to three decimals)
Step 2:
Plug values into Formula 13-13 and solve for z.
13–44
31. Given:
A drugstore uses the fixed-order-interval (FOI) model
Service Level = 98%
OI = 14 days
LT = 2 days
d
= 40 units/day
d = 3 units/day
On-hand inventory in each cycle:
Cycle
On Hand
1
42
2
8
3
103
Using Appendix B, Table B, we look for the z value corresponding to .98.
The closest probability is .9798, which corresponds to z = 2.05.
Cycle 1:
ALTOIzLTOIdQ d
)(
42214)3(05.2)214(40
Q
Q
622.6 = 623 units (round up)
Cycle 2:
8214)3(05.2)214(40
Q
Q
656.6 = 657 units (round up)
Cycle 3:
103214)3(05.2)214(40
Q
Q
561.6 = 562 units (round up)
13–45
32. Given:
Company operates 50 weeks per year
We have the following information on the two items:
P34
P35
d
= 60 units/week
d
= 70 units/week
d = 4 units/week
d = 5 units/week.
LT = 2 weeks
LT = 2 weeks
Unit cost = $15
Unit cost = $20
H = (.30)($15) = $4.50
H = (.30)($20) = 6.00
S = $70
S = $30
Risk = 2.5%
Risk = 2.5%
Can be ordered any time
OI = 4 weeks
a. Determine when to reorder each item:
P34:
Using Appendix B, Table B, we look for the z value corresponding to 1 – .025 = .975:
z = 1.96.
LTzdROP d)((LT)
2(4)96.1(2)60
ROP
13209.13109.11120 ROP
units (round up)
P35: Order every 4 weeks.
b. Compute the EOQ for P34:
13–46
Education.
33. Given:
We have the following list of items:
Item
Estimated
Annual Demand
Ordering
Cost
Holding
Cost (%)
Unit
Price
H4-010
20,000
50
20
2.50
H5-201
60,200
60
20
4.00
P6-400
9,800
80
30
28.50
P6-401
14,500
50
30
12.00
P7-100
6,250
50
30
9.00
P9-103
7,500
50
40
22.00
TS–300
21,000
40
25
45.00
TS–400
45,000
40
25
40.00
TS–041
800
40
25
20.00
V1–001
33,100
25
35
4.00
a. Classify the items as A, B, or C:
Step 1:
Determine the Annual Dollar Value (Unit Price x Estimated Annual Demand) for each item
and the sum of the individual Annual Dollar Values:
Item
Unit
Price
Estimated Annual
Demand
Annual
Dollar
Value
H4-010
2.50
20,000
50,000
H5-201
4.00
60,200
240,800
P6-400
28.50
9,800
279,300
P6-401
12.00
14,500
174,000
P7-100
9.00
6,250
56,250
P9-103
22.00
7,500
165,000
TS–300
45.00
21,000
945,000
TS–400
40.00
45,000
1,800,000
TS–041
20.00
800
16,000
V1-001
4.00
33,100
132,400
3,858,750
Chapter 13 – Inventory Management
13–47
Education.
Step 2:
Arrange the items in descending order based on Annual Dollar Values. Determine the A, B,
and C items. Then, determine the percentage of items and the percentage of Annual Dollar
Value for each category (round to two decimals).
Item
Annual
Dollar
Value
Category
Percentage of
Items
Percentage of
Annual Dollar
Value
TS–400
1,800,000
A
20%
71.14%
TS–300
945,000
P6-400
279,300
B
20%
13.48%
H5-201
240,800
P6-401
174,000
C
60%
15.38%
P9-103
165,000
V1–001
132,400
P7-100
56,250
H4-010
50,000
TS–041
16,000
3,858,750
100.00%
100.00%
Note: An alternate solution could be to include P6-400 through V1-001 in the B category.
b. Determine the EOQ for each item (round to nearest integer):
Item
Estimated Annual
Demand
Ordering
Cost
Unit
Holding
Cost ($)
EOQ
H4-010
20,000
50
.50
2,000
H5-201
60,200
60
.80
3,005
P6-400
9,800
80
8.55
428
P6-401
14,500
50
3.60
635
P7-100
6,250
50
2.70
481
P9-103
7,500
50
8.80
292
TS–300
21,000
40
11.25
386
TS–400
45,000
40
10.00
600
TS–041
800
40
5.00
113
V1-001
33,100
25
1.40
1,087
13–48
Education.
34. Given:
Demand for jelly doughnuts is shown in the table below. Labor, materials, and overhead are
estimated to be $3.30 per dozen, doughnuts are sold for $4.80 per dozen, and leftover doughnuts
are sold at half price.
Demand
(dozens)
Relative
Frequency
19
.01
20
.05
21
.12
22
.18
23
.13
24
.14
25
.10
26
.11
27
.10
28
.04
29
.02
Demand
(dozens)
Relative
Frequency
Cumulative
Frequency
19
.01
.01
20
.05
.06
21
.12
.18
22
.18
.36
23
.13
.49
24
.14
.63
25
.10
.73
26
.11
.84
27
.10
.94
28
.04
.98
29
.02
1.00
Because .67 falls between the cumulative frequencies of .63 and .73, Don should stock 25 dozen
to attain a service level of at least .67. The resulting service level will be .73 = 73.00%.
Chapter 13 – Inventory Management
35. Given:
Purchase price for spare part X135 = $100 each. Carrying and disposal costs = 145% of the
purchase price. Stockout cost = $88,000. Demand for parts will approximate a Poisson
distribution with a mean of 3.2 parts.
a. Determine the optimal number of spare parts to order:
Chapter 13 – Inventory Management
36. Given:
Purchase price = $4.20 per pound. Selling price = $5.70 per pound. Salvage price = $2.40 per
pound. Daily demand can be approximated by a normal distribution with a mean of 80 pounds
and a standard deviation of 10 pounds.
d
= 80 pounds/day
d = 10 pounds/day
Cs = Rev – Cost = $5.70 – $4.20 = $1.50 per pound
Ce = Cost – Salvage = $4.20 – $2.40 = $1.80 per pound
Using Appendix B, Table B, we find that .4545 falls closest to .4562:
z = -0.11.
pounds (assuming that fractional values are possible)
37. Given:
Daily demand can be approximated by a normal distribution with a mean of 40 quarts per day and
a standard deviation of 6 quarts per day. Excess cost = $.35 per quart. The grocer orders 49 quarts
per day.
d
= 40 quarts/day
d = 6 quarts/day
a. Determine the implied shortage cost per quart:
Cs = Rev – Cost = unknown
Ce = $.35
Step 1:
Determine z value.
Using Appendix B, Table B, we find that z = 1.50 corresponds to a service level = .9332 =
93.32%.
Step 2:
Plug in .9332 and solve for Cs.