#### Transcript Module 8B Inventory Management Problems

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Inventory Management
(solved problems and exercises)
For Operations Management, 9e by
Krajewski/Ritzman/Malhotra
© 2010 Pearson Education
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 1
Solved Problem 1
Booker’s Book Bindery divides SKUs into three classes,
according to their dollar usage. Calculate the usage values of
the following SKUs and determine which is most likely to be
classified as class A.
SKU Number
Description
1
Boxes
2
Quantity Used
per Year
Unit Value
(\$)
500
3.00
Cardboard
(square feet)
18,000
0.02
3
Cover stock
10,000
0.75
4
Glue (gallons)
75
40.00
5
Inside covers
20,000
0.05
6
Reinforcing tape
(meters)
3,000
0.15
7
Signatures
150,000
0.45
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 2
Solved Problem 1
SOLUTION
The annual dollar usage for each item is determined by
multiplying the annual usage quantity by the value per unit. As
shown in Figure 12.11, the SKUs are then sorted by annual
dollar usage, in declining order. Finally, A–B and B–C class
lines are drawn roughly, according to the guidelines presented
in the text. Here, class A includes only one SKU (signatures),
which represents only 1/7, or 14 percent, of the SKUs but
accounts for 83 percent of annual dollar usage. Class B
includes the next two SKUs, which taken together represent 28
percent of the SKUs and account for 13 percent of annual dollar
usage. The final four SKUs, class C, represent over half the
number of SKUs but only 4 percent of total annual dollar usage.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 3
Solved Problem 1
SKU
Number
Description
Quantity
Used per
Year
Unit Value
(\$)
Annual Dollar
Usage (\$)
500

3.00
=
1,500
Cardboard
(square feet)
18,000

0.02
=
360
3
Cover stock
10,000

0.75
=
7,500
4
Glue (gallons)
75

40.00
=
3,000
5
Inside covers
20,000

0.05
=
1,000
6
Reinforcing tape
(meters)
3,000

0.15
=
450
7
Signatures
150,000

0.45
=
67,500
Total
81,310
1
Boxes
2
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12 – 4
Solved Problem 1
Percentage of Dollar Value
100 –
Class B
Class C
90 –Class
80 –
A
70 –
60 –
50 –
40 –
Figure 12.11 – Annual Dollar Usage for
Class A, B, and C SKUs
Using Tutor 12.2
30 –
20 –
10 –
0–
10
20 30 40 50 60 70 80 90 100
Percentage of SKUs
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12 – 5
Solved Problem 2
Nelson’s Hardware Store stocks a 19.2 volt cordless drill that is
a popular seller. Annual demand is 5,000 units, the ordering
cost is \$15, and the inventory holding cost is \$4/unit/year.
a. What is the economic order quantity?
b. What is the total annual cost for this inventory item?
SOLUTION
a. The order quantity is
EOQ =
2DS
H
=
2(5,000)(\$15)
\$4
=
37,500 = 193.65 or 194 drills
b. The total annual cost is
Q
194
5,000
D
(\$4) +
(\$15) = \$774.60
C = 2 (H) +
(S) =
Q
2
194
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12 – 6
Solved Problem 3
A regional distributor purchases discontinued appliances from
various suppliers and then sells them on demand to retailers in
the region. The distributor operates 5 days per week, 52 weeks
per year. Only when it is open for business can orders be
received. Management wants to reevaluate its current inventory
policy, which calls for order quantities of 440 counter-top
mixers. The following data are estimated for the mixer:
Average daily demand (d) = 100 mixers
Standard deviation of daily demand (σd) = 30 mixers
Lead time (L) = 3 days
Holding cost (H) = \$9.40/unit/year
Ordering cost (S) = \$35/order
Cycle-service level = 92 percent
The distributor uses a continuous review (Q) system
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12 – 7
Solved Problem 3
a. What order quantity Q, and reorder point, R, should be used?
b. What is the total annual cost of the system?
c. If on-hand inventory is 40 units, one open order for 440
mixers is pending, and no backorders exist, should a new
order be placed?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 8
Solved Problem 3
SOLUTION
a. Annual demand is
D = (5 days/week)(52 weeks/year)(100 mixers/day)
= 26,000 mixers/year
The order quantity is
EOQ =
2DS
=
H
=
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
2(26,000)(\$35)
\$9.40
193,167 = 440.02 or 440 mixers
12 – 9
Solved Problem 3
The standard deviation of the demand during lead time
distribution is
σdLT = σd L = 30 3 = 51.96
A 92 percent cycle-service level corresponds to z = 1.41
Safety stock = zσdLT = 1.41(51.96 mixers) = 73.26 or 73 mixers
Average demand during lead time = dL = 100(3) = 300 mixers
Reorder point (R) = Average demand during lead time + Safety stock
= 300 mixers + 73 mixers = 373 mixers
With a continuous review system, Q = 440 and R = 373
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 10
Solved Problem 3
b. The total annual cost for the Q systems is
Q
D
C = 2 (H) + Q (S) + (H)(Safety stock)
26,000
440
C=
(\$9.40) +
(\$35) + (\$9.40)(73) = \$4,822.38
440
2
c. Inventory position = On-hand inventory + Scheduled receipts
– Backorders
IP = OH + SR – BO = 40 + 440 – 0 = 480 mixers
Because IP (480) exceeds R (373), do not place a new order
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12 – 11
Solved Problem 4
Suppose that a periodic review (P) system is used at the
distributor in Solved Problem 3, but otherwise the data are the
same.
a. Calculate the P (in workdays, rounded to the nearest day)
that gives approximately the same number of orders per
year as the EOQ.
b. What is the target inventory level, T? Compare the P system
to the Q system in Solved Problem 3.
c. What is the total annual cost of the P system?
d. It is time to review the item. On-hand inventory is 40 mixers;
receipt of 440 mixers is scheduled, and no backorders exist.
How much should be reordered?
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12 – 12
Solved Problem 4
SOLUTION
a. The time between orders is
EOQ
440
P=
(260 days/year) =
(260) = 4.4 or 4 days
D
26,000
b. Figure 12.12 shows that T = 812 and safety stock
= (1.41)(79.37) = 111.91 or about 112 mixers. The
corresponding Q system for the counter-top mixer requires
less safety stock.
Figure 12.12 –
OM Explorer Solver
for Inventory
Systems
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12 – 13
Solved Problem 4
c. The total annual cost of the P system is
dP
D
C = 2 (H) + dP (S) + (H)(Safety stock)
100(4)
26,000
C=
(\$9.40) +
(\$35) + (\$9.40)(1.41)(79.37)
100(4)
2
= \$5,207.80
d. Inventory position is the amount on hand plus scheduled
receipts minus backorders, or
IP = OH + SR – BO = 40 + 440 – 0 = 480 mixers
The order quantity is the target inventory level minus the
inventory position, or
Q = T – IP = 812 mixers – 480 mixers = 332 mixers
An order for 332 mixers should be placed.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 14
Solved Problem 5
Grey Wolf Lodge is a popular 500-room hotel in the North
Woods. Managers need to keep close tabs on all room service
items, including a special pine-scented bar soap. The daily
demand for the soap is 275 bars, with a standard deviation of 30
bars. Ordering cost is \$10 and the inventory holding cost is
\$0.30/bar/year. The lead time from the supplier is 5 days, with a
standard deviation of 1 day. The lodge is open 365 days a year.
a. What is the economic order quantity for the bar of soap?
b. What should the reorder point be for the bar of soap if
management wants to have a 99 percent cycle-service level?
c. What is the total annual cost for the bar of soap, assuming a
Q system will be used?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 15
Solved Problem 5
SOLUTION
a. We have D = (275)(365) = 100,375 bars of soap; S = \$10; and
H = \$0.30. The EOQ for the bar of soap is
EOQ =
2DS
=
H
=
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
2(100,375)(\$10)
\$0.30
6,691,666.7 = 2,586.83 or 2,587 bars
12 – 16
Solved Problem 5
b. We have d = 275 bars/day, σd = 30 bars, L = 5 days,
and σLT = 1 day.
σdLT =
Lσd2 + d2σLT2 =
(5)(30)2 + (275)2(1)2 = 283.06 bars
Consult the body of the Normal Distribution appendix for
0.9900. The closest value is 0.9901, which corresponds to
a z value of 2.33. We calculate the safety stock and reorder
point as follows:
Safety stock = zσdLT = (2.33)(283.06) = 659.53 or 660 bars
Reorder point = dL + Safety stock = (275)(5) + 660 = 2,035 bars
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 17
Solved Problem 5
c. The total annual cost for the Q system is
Q
D
C = 2 (H) + Q (S) + (H)(Safety stock)
2,587
100,375
C=
(\$0.30) +
(\$10) + (\$0.30)(660) = \$974.05
2
2,587
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12 – 18
Solved Problem 6
Zeke’s Hardware Store sells furnace filters. The cost to place an
order to the distributor is \$25 and the annual cost to hold a filter
in stock is \$2. The average demand per week for the filters is 32
units, and the store operates 50 weeks per year. The weekly
demand for filters has the probability distribution shown on the
left below.
The delivery lead time from the distributor is uncertain and has
the probability distribution shown on the right below.
Suppose Zeke wants to use a P system with P = 6 weeks and a
cycle-service level of 90 percent. What is the appropriate value
for T and the associated annual cost of the system?
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12 – 19
Solved Problem 6
Demand
Probability
Probability
24
0.15
1
0.05
28
0.20
2
0.25
32
0.30
3
0.40
36
0.20
4
0.25
40
0.15
5
0.05
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12 – 20
Solved Problem 6
SOLUTION
Figure 12.13 contains output from the Demand During the
Protection Interval Simulator from OM Explorer.
Figure 12.13 – OM Explorer Solver for Demand during the Protection Interval
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 21
Solved Problem 6
Given the desired cycle-service level of 90 percent, the
appropriate T value is 322 units. The simulation estimated the
average demand during the protection interval to be 289 units,
consequently the safety stock is 322 – 289 = 33 units.
The annual cost of this P system is
6(32)
50(32)
C=
(\$2) +
(\$25) + (33)(\$2)
2
6(32)
= \$192.00 + \$208.33 + \$66.00 = \$466.33
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 22
Solved Problem 7
Consider Zeke’s inventory in Solved Problem 6. Suppose that
he wants to use a continuous review (Q) system for the filters,
with an order quantity of 200 and a reorder point of 140. Initial
inventory is 170 units. If the stockout cost is \$5 per unit, and all
of the other data in Solved Problem 6 are the same, what is the
expected cost per week of using the Q system?
SOLUTION
Figure 12.14 shows output from the Q System Simulator in OM
Explorer. Only weeks 1 through 13 and weeks 41 through 50
are shown in the figure. The average total cost per week is
\$305.62. Notice that no stockouts occurred in this simulation.
These results are dependent on Zeke’s choices for the reorder
point and lot size. It is possible that stockouts would occur if
the simulation were run for more than 50 weeks.
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12 – 23
Solved Problem 7
Figure 12.14 – OM Explorer Q System Simulator
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12 – 24
Solved Problem D1
Peachy Keen, Inc., makes mohair sweaters, blouses with Peter
Pan collars, pedal pushers, poodle skirts, and other popular
clothing styles of the 1950s. The average demand for mohair
sweaters is 100 per week. Peachy’s production facility has the
capacity to sew 400 sweaters per week. Setup cost is \$351. The
value of finished goods inventory is \$40 per sweater. The
annual per-unit inventory holding cost is 20 percent of the
item’s value.
a. What is the economic production lot size (ELS)?
b. What is the average time between orders (TBO)?
c. What is the total of the annual holding cost and setup cost?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 25
Solved Problem D1
SOLUTION
a. The production lot size that minimizes total cost is
ELS 
2 DS
H
2100  52 \$351
400
p

0.20\$40 
400  100
pd
 456,300
4
 780 sweaters
3
b. The average time between orders is
TBOELS 
780
ELS
 0.15 year

5,200
D
Converting to weeks, we get
TBOELS  0.15 year 52 weeks/yea r   7.8 weeks
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12 – 26
Solved Problem D1
c. The minimum total of setup and holding costs is
C

Q pd 
D

 H    S 
2 p 
Q
780  400  100 
5,200
\$351

0.20  \$40 
2  400 
780
 \$2,340/year  \$2,340/year  \$4,680/year
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12 – 27
Solved Problem D2
A hospital buys disposable surgical packages from Pfisher, Inc.
Pfisher’s price schedule is \$50.25 per package on orders of 1 to
199 packages and \$49.00 per package on orders of 200 or more
packages. Ordering cost is \$64 per order, and annual holding
cost is 20 percent of the per unit purchase price. Annual
demand is 490 packages. What is the best purchase quantity?
SOLUTION
We first calculate the EOQ at the lowest price:
EOQ 49.00 
2 DS

H
2490\$64.00 
 6,400  80 packages
0.20\$49.00 
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12 – 28
Solved Problem D2
This solution is infeasible because, according to the price
schedule, we cannot purchase 80 packages at a price of \$49.00
each. Therefore, we calculate the EOQ at the next lowest price
(\$50.25):
EOQ 50.25
2 DS


H
2490 \$64.00 
 6,241  79 packages
0.20\$50.25 
This EOQ is feasible, but \$50.25 per package is not the lowest price.
Hence, we have to determine whether total costs can be reduced by
purchasing 200 units and thereby obtaining a quantity discount.
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12 – 29
Solved Problem 2
C
C 79 
Q
D
 H    S   PD
2
Q
79
490
0.20  \$50.25  
\$64.00  \$50.25490
2
79
 \$396.98/year  \$396.68/year  \$24,622.50  \$25,416.44/year
C 200 
200
490
0.20  \$49.00 
\$64.00  \$49.00490
2
200
 \$980.00/year  \$156.80/year  \$24,010.00  \$25,146.80/year
Purchasing 200 units per order will save \$269.64/year, compared to
buying 79 units at a time.
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12 – 30
Solved Problem D3
Swell Productions is sponsoring an outdoor conclave for
owners of collectible and classic Fords. The concession stand
in the T-Bird area will sell clothing such as T-shirts and official
Thunderbird racing jerseys. Jerseys are purchased from
Columbia Products for \$40 each and are sold during the event
for \$75 each. If any jerseys are left over, they can be returned to
Columbia for a refund of \$30 each. Jersey sales depend on the
weather, attendance, and other variables. The following table
shows the probability of various sales quantities. How many
jerseys should Swell Productions order from Columbia for this
one-time event?
Sales Quantity
Probability
Quantity Sales
Probability
100
0.05
400
0.34
200
0.11
500
0.11
300
0.34
600
0.05
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12 – 31
Solved Problem D3
SOLUTION
Table D.1 is the payoff table that describes this one-period
inventory decision. The upper right portion of the table shows
the payoffs when the demand, D, is greater than or equal to the
order quantity, Q. The payoff is equal to the per-unit profit (the
difference between price and cost) multiplied by the order
quantity. For example, when the order quantity is 100 and the
demand is 200,
Payoff = (p – c)Q =
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
(\$75 - \$40)100 = \$3,500
12 – 32
Solved Problem D3
TABLE D.1
|
PAYOFFS
Demand, D
100
200
300
400
500
600
Expected
Payoff
100
\$3,500
\$3,500
\$3,500
\$3,500
\$3,500
\$3,500
\$3,500
200
\$2,500
\$7,000
\$7,000
\$7,000
\$7,000
\$7,000
\$6,775
300
\$1,500
\$6,000
\$10,500
\$10,500
\$10,500
\$10,500
\$9,555
400
\$500
\$5,000
\$9,500
\$14,000
\$14,000
\$14,000
\$10,805
500
(\$500)
\$4,000
\$8,500
\$13,000
\$17,500
\$17,500
\$10,525
600
(\$1,500)
\$3,000
\$7,000
\$12,000
\$16,500
\$21,000
\$9,750
Q
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12 – 33
Solved Problem D3
The lower-left portion of the payoff table shows the payoffs
when the order quantity exceeds the demand. Here the payoff is
the profit from sales, pD, minus the loss associated with
returning overstock, l(Q – D), where l is the difference between
the cost and the amount refunded for each jersey returned and
Q – D is the number of jerseys returned. For example, when the
order quantity is 500 and the demand is 200,
Payoff = pD – l(Q – D) =
(\$75 - \$40)200 – (\$40 – \$30)(500 – 200)
= \$4,000
The highest expected payoff occurs when 400 jerseys are ordered:
Expected payoff400 =
(\$500  0.05) + (\$5,000  0.11)
+ (\$9,500  0.34) + (\$14,000  0.34)
+ (\$14,000  0.11) + (\$14,000  0.05)
= \$10,805
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12 – 34
Application 12.1
Suppose that you are reviewing the inventory policies on an
\$80 item stocked at a hardware store. The current policy is to
replenish inventory by ordering in lots of 360 units. Additional
information is:
D = 60 units per week, or 3,120 units per year
S = \$30 per order
H = 25% of selling price, or \$20 per unit per year
What is the EOQ?
SOLUTION
EOQ =
2DS
H
=
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
2(3,120)(30)
= 97 units
20
12 – 35
Application 12.1
What is the total annual cost of the current policy (Q = 360), and
how does it compare with the cost with using the EOQ?
Current Policy
EOQ Policy
Q = 360 units
Q = 97 units
C = (360/2)(20) + (3,120/360)(30)
C = (97/2)(20) + (3,120/97)(30)
C = 3,600 + 260
C = 970 + 965
C = \$3,860
C = \$1,935
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12 – 36
Application 12.1
What is the time between orders (TBO) for the current policy
and the EOQ policy, expressed in weeks?
SOLUTION
360
TBO360 =
(52 weeks per year) = 6 weeks
3,120
TBOEOQ =
97
(52 weeks per year) = 1.6 weeks
3,120
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12 – 37
Application 12.2
The on-hand inventory is only 10 units, and the reorder point
R is 100. There are no backorders and one open order for
200 units. Should a new order be placed?
SOLUTION
IP = OH + SR – BO = 10 + 200 – 0 = 210
R = 100
Decision: Place no new order
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12 – 38
Application 12.3
Suppose that the demand during lead time is normally
distributed with an average of 85 and σdLT = 40. Find the safety
stock, and reorder point R, for a 95 percent cycle-service level.
SOLUTION
Safety stock = zσdLT = 1.645(40) = 65.8 or 66 units
R = Average demand during lead time + Safety stock
R = 85 + 66 = 151 units
Find the safety stock, and reorder point R, for an 85 percent
cycle-service level.
Safety stock = zσdLT = 1.04(40) = 41.6 or 42 units
R = Average demand during lead time + Safety stock
R = 85 + 42 = 127 units
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12 – 39
Application 12.4
Grey Wolf lodge is a popular 500-room hotel in the North
Woods. Managers need to keep close tabs on all of the room
service items, including a special pint-scented bar soap. The
daily demand for the soap is 275 bars, with a standard deviation
of 30 bars. Ordering cost is \$10 and the inventory holding cost
is \$0.30/bar/year. The lead time from the supplier is 5 days, with
a standard deviation of 1 day. The lodge is open 365 days a
year.
What should the reorder point be for the bar of soap if
management wants to have a 99 percent cycle-service?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 40
Application 12.4
SOLUTION
d = 275 bars
L = 5 days
σd = 30 bars
σLT = 1 day
σdLT =
Lσd2 + d2σLT2 = 283.06 bars
From the Normal Distribution appendix for 0.9900, z = 2.33.
We calculate the safety stock and reorder point as follows;
Safety stock = zσdLT = (2.33)(283.06) = 659.53 or 660 bars
Reorder point + safety stock = dL + safety stock
= (275)(5) + 660 = 2,035 bars
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12 – 41
Application 12.5
The Discount Appliance Store uses a continuous review system
(Q system). One of the company’s items has the following
characteristics:
Demand = 10 units/wk (assume 52 weeks per year)
Ordering and setup cost (S) = \$45/order
Holding cost (H) = \$12/unit/year
Lead time (L) = 3 weeks (constant)
Standard deviation in weekly demand = 8 units
Cycle-service level = 70%
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12 – 42
Application 12.5
SOLUTION
What is the EOQ for this item?
D = 10/wk  52 wks/yr = 520 units
EOQ =
2DS
H
=
2(520)(45)
= 62 units
12
What is the desired safety stock?
σdLT = σd L = 8 3 = 14 units
Safety stock = zσdLT = 0.525(14) = 8 units
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12 – 43
Application 12.5
What is the desired reorder point R?
R = Average demand during lead time + Safety stock
R = 3(10) + 8 = 38 units
What is the total annual cost?
62
520
C = 2 (\$12) + 62 (\$45) + 8(\$12) = \$845.42
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12 – 44
Application 12.5
Suppose that the current policy is Q = 80 and R = 150. What will
be the changes in average cycle inventory and safety stock if
your EOQ and R values are implemented?
Reducing Q from 80 to 62
Cycle inventory reduction = 40 – 31 = 9 units
Safety stock reduction = 120 – 8 = 112 units
Reducing R from 150 to 38
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12 – 45
Application 12.6
The on-hand inventory is 10 units, and T is 400. There are no
back orders, but one scheduled receipt of 200 units. Now is the
time to review. How much should be reordered?
SOLUTION
IP = OH + SR – BO
= 10 + 200 – 0 = 210
T – IP = 400 – 210 = 190
The decision is to order 190 units
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
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Application 12.7
Return to Discount Appliance Store (Application 12.4), but now
use the P system for the item.
Previous information
Demand = 10 units/wk (assume 52 weeks per year) = 520
EOQ = 62 units (with reorder point system)
Lead time (L) = 3 weeks
Standard deviation in weekly demand = 8 units
z = 0.525 (for cycle-service level of 70%)
Reorder interval P, if you make the average lot size using the
Periodic Review System approximate the EOQ.
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 47
Application 12.7
SOLUTION
Reorder interval P, if you make the average lot size using the
Periodic Review System approximate the EOQ.
P = (EOQ/D)(52) = (62/529)(52) = 6.2 or 6 weeks
Safety stock
Safety stock =  d P  L
 0.5258 6  3  12.6 or 13 units
Target inventory
T = d(P + L) + safety stock for protection interval
T = 10(6 + 3) + 13 = 103 units
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 48
Application 12.7
Total cost
dP
D
C = 2 (H) + dP (S) + HzσP + L
10(6)
520
=
(\$12) +
(\$45) + (13)(\$12) = \$906.00
10(6)
2
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 49
Application D.1
A domestic automobile manufacturer schedules 12 two-person
teams to assemble 4.6 liter DOHC V-8 engines per work day.
Each team can assemble 5 engines per day. The automobile
final assembly line creates an annual demand for the DOHC
engine at 10,080 units per year. The engine and automobile
assembly plants operate 6 days per week, 48 weeks per year.
The engine assembly line also produces SOHC V-8 engines.
The cost to switch the production line from one type of engine
to the other is \$100,000. It costs \$2,000 to store one DOHC V-8
for one year.
a. What is the economic lot size?
b. How long is the production run?
c. What is the average quantity in inventory?
d. What is the total annual cost?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 50
Application D.1
SOLUTION
a. Demand per day = d = 10,080/[(48)(6)] = 35
2 DS
ELS 
H
210,080 100,000 
60
p

 1,555.38
2,000
60  35
pd
or 1,555 engines
b. The production run
Q 1,555

 25.91 or 26 production days
p
60
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12 – 51
Application D.1
c. Average inventory
I max Q  p  d  1,555  60  35 
 


  324 engines
2
2 p 
2  60 
d. Total annual cost
C

I max
D
Q pd 
D
 H    S   
 H    S 
2
Q
2 p 
Q
1,555  60  35 
10,080
\$100,000

\$2,000 
2  60 
1,555
 \$647,917  \$648,231
 \$1,296,148
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 52
Application D.2
A supplier’s price schedule is:
Order Quantity
Price per Unit
0–99
\$50
100 or more
\$45
If ordering cost is \$16 per order, annual holding cost is 20 percent of the
purchase price, and annual demand is 1,800 items, what is the best order
quantity?
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 53
Application D.2
SOLUTION
Step 1:
EOQ 45.00 
2 DS

H
21,800 16 
 80 units (infeasibl e)
45 0.2 
EOQ 50.00 
2 DS

H
21,800 16 
 76 units (feasible)
500.2 
Step 2:
C 76 
76
1,800
50  0.2 
16  501,800  \$90,759
2
76
C100 
100
1,800
45  0.2 
16  451,800  \$81,738
2
100
The best order quantity is 100 units
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
12 – 54
Application D.3
For one item, p = \$10 and l = \$5. The probability distribution for
the season’s demand is:
Demand
Demand
(D)
Probability
10
0.2
20
0.3
30
0.3
40
0.1
50
0.1
Complete the following payoff matrix, as well as the column on the right
showing expected payoff. (Students complete highlighted cells) What is the
best choice for Q?
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12 – 55
Application D.3
D
Expected
Payoff
Q
10
20
30
40
50
10
\$100
\$100
\$100
\$100
\$100
\$100
20
50
200
200
200
200
170
30
0
40
–50
100
250
400
400
175
50
–100
50
200
350
500
140
300
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall.
300
12 – 56
Application D.3
D
Q
10
20
30
40
50
10
\$100
\$100
\$100
\$100
\$100
\$100
20
50
200
200
200
200
170
30
0
150
300
300
300
195
40
–50
100
250
400
400
175
50
–100
50
200
350
500
140
Payoff
if Q = 30 and D = 20:
pD – l(Q – D) =
Payoff
Expected
Payoff
10(20) – 5(30 – 20) = \$150
if Q = 30 and D = 40:
Expected payoff
pD = 10(30) = \$300
if Q = 30:
0(0.2) + 150(0.3) + 300(0.3 + 0.1 + 0.1) = \$195
Q = 30 has the highest payoff at \$195.00
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