Chapter 13 - Inventory Management
CHAPTER 13
INVENTORY MANAGEMENT
Solutions
1.
a. Given:
Determine an A-B-C classification for the following items:
Item
Unit
Cost
Annual
Volume
(00)
1
$100
25
2
$80
30
3
$15
60
4
$50
10
5
$11
70
6
$60
85
Step 1:
Determine the Annual Dollar Value (Unit Cost * Annual Volume) for each item and the sum of
the individual Annual Dollar Values.
Item
Unit
Cost
Annual
Volume
(00)
1
$100
25
2
80
30
3
15
60
4
50
10
5
11
70
6
60
85
Annual
Dollar
Value
$2,500
2,400
900
500
770
5,100
12,170
13-1
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Education.
Chapter 13 - Inventory Management
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
6
1
2
Annual
Dollar
Value
$5,100
Category
A
2,500
2,400
3
900
5
770
4
500
B
C
12,170
Percentage of
Items
Percentage of Annual
Dollar Value
16.67%
41.91%
[(1/6)*100]
[($5,100/$12,170)*100]
33.33%
40.26%
[(2/6)*100]
[($4,900/$12,170)*100]
50.00%
17.83%
[(3/6)*100]
[($2,170/$12,170)*100]
100.00%
100.00%
b. Given:
D = 4,500, S = $36, and H = $10.
Find the EOQ (round to an integer value):
Q0 
2 DS
2(4,500)36

 180 units
10
H
c. Given:
D = 18,000/year, S = $100, H = $40 per unit per year, p = 120 units per day, and u = 90
units/day.
Find the economic production quantity (EPQ) (round to an integer value):
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
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
  
13-2
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Education.
Chapter 13 - Inventory Management
2.
a. Given:
The following table contains figures on the monthly volume and unit costs for a random
sample of 16 items. Develop an A-B-C classification for these items:
Item Unit Cost Usage
K34
$10
200
K35
25
600
K36
36
150
M10
16
25
M20
20
80
Z45
80
200
F14
20
300
F95
30
800
F99
20
60
D45
10
550
D48
12
90
D52
15
110
D57
40
120
N08
30
40
P05
16
500
P09
10
30
13-3
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Education.
Chapter 13 - Inventory Management
Step 1:
Determine the Annual Dollar Value (Unit Cost * Usage) for each item and the sum of the
individual Annual Dollar Values.
Item Unit Cost Usage
K34
$10
200
K35
25
600
K36
36
150
M10
16
25
M20
20
80
Z45
80
200
F14
20
300
F95
30
800
F99
20
60
D45
10
550
D48
12
90
D52
15
110
D57
40
120
N08
30
40
P05
16
500
P09
10
30
Annual
Dollar
Value
$2,000
15,000
5,400
400
1,600
16,000
6,000
24,000
1,200
5,500
1,080
1,650
4,800
1,200
8,000
300
94,130
13-4
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Education.
Chapter 13 - Inventory Management
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).
Annual
Dollar
Percentage of
Item
Value Category
Items
F95 $24,000
18.75%
Z45
16,000
A
[(3/16)*100]
K35
15,000
P05
8,000
F14
6,000
31.25%
D45
5,500
B
[(5/16)*100]
K36
5,400
D57
4,800
K34
2,000
D52
1,650
M20
1,600
F99
1,200
50.00%
C
[(8/16)*100]
N08
1,200
D48
1,080
M10
400
P09
300
94,130
100.00%
Percentage of Annual
Dollar Value
54.83%
[($55,000/$94,130)*100]
31.55%
[($29,700/$94,130)*100]
10.02%
[($9,430/$94,130)*100]
100.00%
b. Given:
Determine an A-B-C classification for the following items:
Item
4021
Usage Unit Cost
90
$1,400
9402
300
12
4066
30
700
6500
150
20
9280
10
1,020
4050
80
140
6850
2,000
10
3010
400
20
4400
5,000
5
13-5
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Education.
Chapter 13 - Inventory Management
Step 1:
Determine the Annual Dollar Value (Usage * Unit Cost) for each item and the sum of the
individual Annual Dollar Values.
Item
4021
Usage Unit Cost
90
$1,400
Annual Dollar
Value
$126,000
9402
300
12
3,600
4066
30
700
21,000
6500
150
20
3,000
9280
10
1,020
10,200
4050
80
140
11,200
6850
2,000
10
20,000
3010
400
20
8,000
4400
5,000
5
25,000
228,000
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).
Annual
Item Dollar Value
4021
$126,000
4400
25,000
4066
21,000
6850
20,000
4050
11,200
9280
10,200
3010
8,000
9402
3,600
6500
3,000
228,000
Percentage
Category
of Items
Percentage of Annual
Dollar Value
A
11.11%
55.26%
[(1/9)*100] [$126,000/$228,000)*100]
B
33.33%
[(3/9)*100]
28.95%
[$66,000/$228,000)*100]
C
55.56%
[(5/9)*100]
15.79%
[$36,000/$228,000)*100]
100.00%
100.00%
c. Determine the percentage of items in each category and the annual dollar value for each
category.
Reference table above.
13-6
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Education.
Chapter 13 - Inventory Management
3.
Given:
D = 1,215 bags per year
S = $10
H = $75
Note: Round the EOQ to an integer value, but round any other values to a maximum of two
decimals.
a. Determine the EOQ:
Q0 
2 DS
2(1,215)10

 18 bags
H
75
b. Determine the average inventory:
Q/2 = 18/2 = 9 bags
c. Determine the number of orders per year:
D
1,215 bags

 67.5 orders
Q 18 bags / order
d. Determine the total cost of ordering and carrying flour:
TC = Carrying cost + Ordering cost


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
  ( )   ( )   ( )   (
)       




e. Assuming that holding cost per bag increases by $9/bag/year, what would happen to total
cost?
New H = $75 + $ 9 = $84.
Q0 
2(1,215)(10)
 17 bags
84




  ( )   ( )   ( )   (
)       




Increase in cost = $1,428.71 – $1,350 = $78.71 per year
13-7
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Education.
Chapter 13 - Inventory Management
4.
Given:
D = 40/day x 260 days/yr. = 10,400 boxes
S = $60. H = $30.
Note: Round the EOQ to an integer value, but round any other values to a maximum of two
decimals.
a. Determine the EOQ:
Q0 
2 DS

H
2(10,400)60
 203.96  204 boxes
30
b. Determine total cost:
TC = Carrying cost + Ordering cost


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
  ( )   ( )   (
)   (
)       




c. Yes, annual ordering and carrying costs always are equal at the EOQ (except when rounding).
d. Determine the total cost for Q = 200 and compare to current total cost:

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  ( )   ( )   (
)   (
)       




$6,120 – $6,118.82 = $1.18 higher per year for Q = 200 (this should be acceptable).
13-8
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Education.
Chapter 13 - Inventory Management
5.
Given:
D = 750 pots/mo. x 12 mo./yr. = 9,000 pots/yr.
C = $2. H = (.30)($2) = $.60/unit/year
S = $20
Note: Round the EOQ to an integer value, but round any other values to a maximum of two
decimals.
a. Determine the additional annual cost for using Q = 1,500:
Step 1:
Determine total cost for Q = 1,500.




  ( )   ( )   (
)    (
)       




Step 2:
Determine EOQ.
Q0 
2 DS

H
2(9,000)20
 774.60  775 pots
.60
Step 3:
Determine total cost for Q = 775.




  ( )   ( )   (
)    (
)       




Step 4:
Determine annual savings from using the EOQ.
$570 – $464.76 = $105.24.
b. The benefit of using the EOQ is that about one half of the storage space would be needed.
13-9
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Education.
Chapter 13 - Inventory Management
6.
Given:
D = 12 * 800 = 9,600
H = .35($10) = $3.50 per crate per year
S = $28
Note: Round the EOQ to an integer value, but round any other values to a maximum of two
decimals.
Step 1:
Determine current total cost for Q = 800 (the manager orders once per month).




  ( )   ( )   (
)   (
)       




Step 2:
Determine EOQ, total cost for EOQ, and annual savings from using the EOQ.
Q0 
2 DS

H
2(9,600)28
 391.92  392 crates
3.50




  ( )   ( )   (
)   (
)       




Savings per year from using EOQ = $1,736 – $1,371.71 = $364.29
13-10
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Education.
Chapter 13 - Inventory Management
7.
Given:
Demand is projected to be 600 units for the first half of the year and 900 units for the second half.
The monthly holding cost is $2 per unit, and it costs an estimated $55 to process an order.
a. Assuming that monthly demand will be level during each six-month period, determine
an order size that will minimize the sum of ordering and carrying costs for each six-month
period:
Note: We will solve this problem using months, rather than a year, as the period.
First six-month period:
d = monthly demand = 600 / 6 = 100 & H = $2.00 per unit per month.
Q0 
2dS
2(100)55

 74.16  74 units
H
2.00
Second six-month period:
D = monthly demand = 900 / 6 = 150 & H = $2.00 per unit per month.
Q0 
2dS
2(150)55

 90.83  91 units
H
2.00
b. We can use the EOQ only if demand is level (stable).
c. If the vendor is willing to offer a discount of $10 per order for ordering in multiples of 50
units (e.g., 50, 100, 150), would you advise the manager to take advantage of the offer in
either six-month period? If so, what order size would you recommend?
First six-month period:
d = monthly demand = 600 / 6 = 100, H = $2.00 per unit per month, S = $55, & EOQ = 74.
Monthly TC (Q = 74):




( )   ( )   ( )   (
)       




With discount of $10, S = $55 – $10 = $45:
Monthly TC (Q = 50):




( )   ( )   ( )   (
)       




Monthly TC (Q = 100):




( ) ( )  (
)   (
)       




13-11
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Education.
Chapter 13 - Inventory Management
Monthly TC (Q = 150):




( ) ( )  (
)   (
)       




Conclusion: Yes, the manager should take advantage of the offer and order Q = 50 units
during this six-month period.
Second six-month period:
d = monthly demand = 900 / 6 = 150, H = $2.00 per unit per month, S = $55, & EOQ = 91.
Monthly TC (Q = 91):




( )   ( )   ( )   (
)       




With discount of $10, S = $55 – $10 = $45:
Monthly TC (Q = 50):




( )   ( )   ( )   (
)       




Monthly TC (Q = 100):




( ) ( )  (
)   (
)       




Monthly TC (Q = 150):




( ) ( )  (
)   (
)       




Conclusion: Yes, the manager should take advantage of the offer and order Q = 100 units
during this six-month period.
13-12
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Education.
Chapter 13 - Inventory Management
8.
Given:
d = 27,000 jars per month
H = $0.18 per jar per month
S = $60
Company operates 20 days a month
Current Q = 4,000
Note: Round the EOQ to an integer value, but round any other values to a maximum of two
decimals.
a. What penalty is the company incurring by its present order size?
Step 1:
Determine current monthly cost for Q = 4,000.




  ( )   ( )   (
)   (
)       




Step 2:
Determine EOQ, total cost for EOQ, and monthly savings from using the EOQ.
Q0 
2dS
2(27,000)60

 4,242.64  4,243 jars
H
0.18




  ( )   ( )   (
)   (
)       




Savings per month from using EOQ = $765 – $763.68 = $1.32.
Conclusion: Penalty from placing orders of Q = 4,000 = $1.32 per month.
13-13
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Chapter 13 - Inventory Management
b. The manager would prefer ordering 10 times each month (every other day) but would have to
justify any change in order size. One possibility is to simplify order processing to reduce the
ordering cost. What ordering cost would enable the manager to justify ordering every other
day?
Using the current Q = 4,000, total monthly cost = $765.
If the manager orders 10 times per month, Q = 27,000 / 10 = 2,700.
Set TC (Q = 2,700) = $765 and solve for S:




  ( )   ( )   (
)   (
)   




    
    
S = $522 / 10
S = $52.20 (round to two decimals).
This is the order cost that would enable the manager to justify ordering every other day.
13-14
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Education.
Chapter 13 - Inventory Management
9.
Given:
p = 5,000 hotdogs/day
u = 250 hotdogs/day
Factory operates 300 days per year
D = 250 * 300 = 75,000 hotdogs per year
S = $66
H = $0.45 per hotdog per year
Note: Round Qp to an integer value, but round any other values to a maximum of two decimals.
a. Find the optimal run size:
Qp 
2 DS
H
p

p u
2(75,000)66
5,000
 4,812.27  4,812 hotdogs
0.45
5,000  250
b. Number of runs per year:
D / Qp = 75,000 / 4,812 = 15.59 runs per year
c. Days to produce the optimal run quantity:
Qp / p = 4,812 / 5,000 = 0.96 days
13-15
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Education.
Chapter 13 - Inventory Management
10.
Given:
A chemical firm produces 100-pound bags. Demand for the product = 20 tons per day. The
capacity = 50 tons per day. Setup cost = $100, and storage and handling costs = $5 per ton a year.
The firm operates 200 days a year. Note: 1 ton = 2,000 pounds.
p = 50 tons per day * 2,000 pounds per ton = 100,000 pounds per day = 100,000 pounds per day /
100 pounds per bag = 1,000 bags per day
u = 20 tons per day * 2,000 pounds per ton= 40,000 pounds per day = 40,000 pounds per day /
100 pounds per bag = 400 bags per day
D = 400 bags per day * 200 days per year = 80,000 bags per year
S = $100
H = $5 per ton per year = $5 per ton per year / 20 bags per ton = $0.25 per bag per year
Note: Round Qp to an integer value, but round any other values to a maximum of two decimals.
a.
Qp 
b.
I max 
2 DS
H
Qp
p
p

p u
( p  u) 
Average Inventory =
c. Run length =
Qp
d. Runs per year:
p

2(80,000)100
1,000
 10,327.97  10,328 bags
0.25
1,000  400
10,328
(1,000  400)  6,196.8 bags
1,000
I max 6,196.8

 3,098.4 bags
2
2
10,328
 10.33 days
1,000
D 80,000

 7.75 runs per year
Q 10,328
e. S = $25:
Qp 
I max 
2 DS
H
Qp
p
p

p u
( p  u) 
2(80,000)25
1,000
 5,163.98  5,164 bags
0.25
1,000  400
5,164
(1,000  400)  3,098.4 bags
1,000




)  ( )  (
)   (
)  




    
    (




    (
)  ( )  (
)   (
)  




    
Savings when S = $25 = $1,549.19 – $774.60 = $774.59 per year.
13-16
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Education.
Chapter 13 - Inventory Management
11.
Given:
Assembly takes place 5 days a week, 50 weeks a year. It will take a full day to get the machine
ready for a production run of the component for the new product.
S = $300
H = $10.00
p = 200/day
u = 80/day
D = 20,000 (250 days * 80/day)
Note: Round Qp to an integer value, but round any other values to a maximum of two decimals.
a. Optimal run quantity to minimize total annual costs:
2 DS
H
Qp 
p

p u
2(20,000)300
200
 1,414.21  1,414 units
10
200  80
b. Days to produce the optimal run quantity:
Qp
p

1,414
 7.07 days
200
c. Average amount of inventory:
 




     units



  
    units

   
d. The manager would like to run another job between runs of the component for the new
product and needs a minimum of 10 days per cycle (including setup) for the other job:
How much time is available to run the other job? The job must be finished during the pure
consumption time for the component for the new product. The end of the pure consumption
time is when inventory of the component for the new product falls to 0 units. If the other job
takes longer than the pure consumption time, we will run out of inventory of the component
for the new product.
 
  

  


This is the time between starting production runs of the component for the new product.
        


   
       


Plugging in values and solving for Pure Consumption Time:
        
        
         
Conclusion: There will not be enough time to run the other job because the other job requires
10 days, which is .39 days (10 – 9.61) days too many.
  
13-17
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Education.
Chapter 13 - Inventory Management
e. Three options that the manager could consider that will allow this other job to be performed:
1) Try to shorten the setup time of the component for the new product.
2) Increase the run quantity of the component for the new product to allow a longer time
between runs, i.e., run the component less often.
3) Reduce the run size of the other job.
f.
Determine the additional units to produce of the component for the new product and the
increase in total annual cost from this new Q:
We know the following:
The Pure Consumption Time for the component for the new product must equal 10 days to
allow the other job to be run.



  
  

p = 200/day, u = 80/day, and Pure Consumption Time = 10 days.
  
Plugging in values and solving for Qp:
        


[
 ]  


 

   
 
Using the least common denominator:
 

 
 

 

  
  
 


     
The additional units per run = 1,467 – 1,414 = 53 units per run.
13-18
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Education.
Chapter 13 - Inventory Management
Increase in total cost:
Q = 1,467:


   
     units





    (
)  ( )  (
)   (
) 




     
 

Q = 1,414:




    (
)  ( )  (
)   (
) 




     
Increase = $8,490.98 – $8,485.28 = $5.70 per year
12.
Given:
p = 800 units per day
u = 300 units per day
Q = 2,000 units per batch
Company operates 250 days a year
a. Number of batches of heating elements per year:
D 75,000

 37.5 batches per year
Q 2,000
b. Amount of inventory after 2 days of production:
The number of units produced in 2 days = (2 days)(800 units/day) = 1600 units
The number of units used in 2 days = (2 days) (300 units per day) = 600 units
Inventory build up after the first 2 days of production = 1,600 – 600 = 1,000 units
Current inventory of the heating unit = 0 units
Total inventory after the first 2 days of production = Beginning Inventory + Inventory
Buildup after 2 Days of Production = 0 + 1,000 = 1,000 units.
c. Average Inventory:
I max 
Q
2,000
( p  u) 
(800  300)  1,250 units
p
800
  
 

  


13-19
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Education.
Chapter 13 - Inventory Management
d. The other component requires 4 days (including setup). Setup time for the heating element =
0.5 days. Is there enough time to run the other component between batches of heating
elements?
How much time is available to run the other component? The other component must be
finished during the pure consumption time for the heating element. The end of the pure
consumption time is when inventory of the heating element falls to 0 units. If the other
component takes longer than the pure consumption time, we will run out of inventory of the
heating element.
 
   
  


This is the time between starting production runs of the heating element.
        


   
        


Plugging in values and solving for Pure Consumption Time:
        
        
         
Conclusion: There will not be enough time to run the other component because the other
component requires 4 days, which is .33 (4 – 3.67) days too many.
  
13-20
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 13 - Inventory Management
13.
Given:
D = 18,000 boxes/year
S = $96
H = $.60/box/year
Price Schedule:
Number of Boxes
Price per Box (P)
1,000-1,999
$1.25
2,000-4,999
$1.20
5,000-9,999
$1.15
10,000+
$1.10
a. Determine the optimal order quantity (round to an integer value):
Step 1:
Compute the common minimum point.
Q
2 DS
2(18,000)96

 2,400 boxes
H
.60
This quantity is feasible in the range 2000-4,999.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit
costs.


  ( )   ( )   


TC2,400 =
2,400
18,000
(.60) 
($96)  $1.20(18,000)  $23,040
2
2,400
TC5,000 =
5,000
18,000
(.60) 
($96)  $1.15(18,000)  $22,545.60
2
5,000
10,000
18,000
(.60) 
($96)  $1.10(18,000)  $22,972.80
2
10,000
Conclusion: Optimal order quantity = 5,000 boxes.
TC10,000 =
b. Determine number of orders per year:
D 18,000

 3.6 orders per year (round to a maximum of two decimals)
Q 5,000
13-21
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Education.
Chapter 13 - Inventory Management
14.
Given:
D = 25 stones/day * 200 days/year = 5,000 stones/year
S = $48
Price Schedule:
Number of Stones Price per Stone (P)
1-399
$10
400-599
$9
600+
$8
a. H = $2. Determine the optimal order quantity:
Step 1:
Compute the common minimum point.
Q
2 DS
2(5,000)48

 489.90  490 stones
2
H
This quantity is feasible in the range 400-599.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit
costs.


  ( )   ( )   




  (
)  (
)     




  (
)  (
)     


Conclusion: Optimal order quantity = 600 stones.
b. H = 30% of unit cost
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you
find a feasible minimum point.
Minimum point P = $8:
2 DS
2(5,000)48

 447.21  447 Not feasible
H
.30(8)
13-22
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Education.
Chapter 13 - Inventory Management
Minimum point P = $9:
2 DS
2(5,000)48

 421.64  422 Feasible
H
.30(9)
Step 2:
Compare the total cost at Q = 422 to Q = 600.


  ( )   ( )   




  (
)     (
)     




  (
)     (
)     


Conclusion: Optimal order quantity = 600 stones.
c. Lead time = 6 working days. Determine ROP:
ROP = 25 stones/day * 6 days = 150 stones.
13-23
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Education.
Chapter 13 - Inventory Management
15.
Given:
D = 4,900
S = $50
H = 40% of purchase cost
Price Schedule:
Range
Price per Unit (P)
1-999
$5.00
1,000-3,999
$4.95
4,000-5,999
$4.90
6,000+
$4.85
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you find
a feasible minimum point.
Minimum point P = $4.85:
2 DS
2(4,900)50

 502.57  503 Not feasible
H
.40(4.85)
Minimum point P = $4.90:
2 DS
2(4,900)50

 500 Not feasible
H
.40(4.90)
Minimum point P = $4.95:
2 DS
2(4,900)50

 497.47  497 Not feasible
H
.40(4.95)
Minimum point P = $5.00:
2 DS
2(4,900)50

 494.97  495 Feasible
H
.40(5.00)
13-24
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 13 - Inventory Management
Step 2:
Compare the total cost at Q = 495 to Q = 1,000, 4,000, & 6,000.


  ( )   ( )   




  (
)     (
)     




  (
)     (
)     


  (


)     (
)     


  (


)     (
)     


Conclusion: Optimal order quantity = 495 units. Note: The total cost for 1,000 units is only $.05
different.
13-25
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 13 - Inventory Management
16.
Given:
D = 800 * 12 = 9,600
S = $40
H = 25% of purchase cost
Price Schedule Supplier A:
Range
Price per Unit (P)
1-199
$14.00
200-499
$13.80
500+
$13.60
Price Schedule Supplier B:
Range
Price per Unit (P)
1-149
$14.10
150-349
$13.90
350+
$13.70
We need to find the optimal quantity for each supplier and select the supplier with the minimum
cost.
Supplier A:
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you find
a feasible minimum point.
Minimum point P = $13.60:
2 DS
2(9,600)40

 475.27  475 Not feasible
H
.25(13.60)
Minimum point P = $13.80:
2 DS
2(9,600)40

 471.81  472 Feasible
H
.25(13.80)
Step 2:
Compare the total cost at Q = 472 to Q = 500.


  ( )   ( )   




  (
)     (
)     




  (
)     (
)     


Conclusion: Optimal order quantity from Supplier A: 500 units with TC = $132,718.
13-26
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 13 - Inventory Management
Supplier B:
Step 1:
Beginning with the lowest unit price, compute minimum points for each price range until you find
a feasible minimum point.
Minimum point P = $13.70:
2 DS
2(9,600)40

 473.53  474 Feasible
H
.25(13.70)
Step 2:
Compute total cost for Q = 474.


  (
)     (
)     


Conclusion: Optimal order quantity from Supplier B: 474 units with TC = $133,141.86.
Compare total cost for Q = 500 from Supplier A to total cost for Q = 474 from Supplier B:
Conclusion: Optimal order quantity = 500 units from Supplier A.
13-27
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 13 - Inventory Management
17.
Given:
D = 3,600 boxes per year
Q = 800 boxes (recommended)
S = $80/order
H = $10/box/year
Price Schedule:
Range
Price per Unit (P)
1-199
$1.20
200-800
$1.10
801+
$1.00
If the firm decides to order 800 boxes, the total cost is computed as follows:


  ( )   ( )   




  (
)   (
)     


If the firm decides to order 801 boxes, the total cost is computed as follows:


  (
)   (
)     


Even though the inventory total cost curve is fairly flat around its minimum, when there are
quantity discounts, there are multiple U shaped total inventory cost curves. Therefore, when the
quantity changes from 800 to 801, we shift to a different total cost curve.
Conclusion: The order quantity of 801 is preferred to the order quantity of 800 because the total
cost for Q = 801 is lower.
Determine the optimal Q:
Step 1:
Compute the common minimum point.
Q
2 DS
2(3,600)80

 240 boxes
10
H
This quantity is feasible in the range 200-800.
13-28
Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill
Education.
Chapter 13 - Inventory Management
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit
costs.


  ( )   ( )   




  (
)   (
)     




  (
)   (
)     


Conclusion: Optimal order quantity = 240 boxes.
13-29
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Education.
Chapter 13 - Inventory Management
18.
Given:
Daily usage = 800 feet/day & lead time = 6 days.
Service level desired: 95%.
Stockout risk for various levels of safety stock:
.10 for 1,500 feet; .05 for 1,800 feet; .02 for 2,100 feet; and .01 for 2,400 feet.
Stockout risk should = 1.00 – .95 = .05. This requires a safety stock of 1,800 feet.
ROP = Expected demand during lead time + Safety stock = EDDLT + SS =
(800 feet/day x 6 days) + 1,800 feet = 6,600 feet.
19.
Given:
EDDLT = 300 units
dLT = 30 units
a. Determine ROP for 1% risk of stockout:
Using Appendix B, Table B, we look for the z value corresponding to 1.00 – .01 = 0.99.
The closest probability is .9901, which corresponds to z = 2.33.
ROP =                (round up)
b. SS = 69.9 = 70 units (round up)
c. Stockout risk of 2% is > 1%. Greater stockout risk = smaller z = less safety stock &
smaller ROP.
20.
Given:
EDDLT = 600 lb.
dLT = 52 lb.
Stockout risk = 4%
a. Determine SS for 4% risk of stockout.
Using Appendix B, Table B, we look for the z value corresponding to 1.00 – .04 = 0.96.
The closest probability is .9599, which corresponds to z = 1.75.
SS =      units
b. ROP = EDDLT + SS = 600 + 91 = 691 units
c. With no safety stock, stockout risk is 50% (z = 0.00).
13-30
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Education.
Chapter 13 - Inventory Management
21.
Given:

d = 21 gallons/week
d = 3.5 gallons/week
LT = 2 days & the dairy is open 7 days a week
Service level= 90%
Hint: Work in terms of weeks
a. Determine ROP and days of supply on hand:
Using Appendix B, Table B, we look for the z value corresponding to .90.
The closest probability is .8997, which corresponds to z = 1.28.
ROP  d (LT)  z( d ) LT  21(2/7)  1.28(3.5) (2/7)  8.39  9 gallons (round up)
Days of Supply = 9 / (21/7) = 9 / 3 = 3 days of supply on hand at the ROP
b. OI = 10 days & 8 gallons are on hand at the order time:
 10 2 
Q  d (OI  LT )  z d OI  LT  A  21    1.28(3.5) 12 / 7  8  33.87  34
 7 7
(round up)
Determine the probability of experiencing a stockout before this order arrives:
Risk of a stockout at the end of the initial lead time:
Using Formula 13-13, set the ROP equal to the quantity on hand when the order is placed and
solve for z:
ROP  d (LT)  z ( d ) LT

   ( )  √

    
2 = 1.871z
z = 2 / 1.871
z = 1.07 (round to two decimals)
From Appendix B, Table B, the lead time service level is .8577.
Risk of stockout before this order arrives = 1 - .8577 = .1423 = 14.23%.
13-31
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Education.
Chapter 13 - Inventory Management
c. The manager is using the ROP model described in part a. One day after placing an order with
the supplier, the manager receives a call that the order will be delayed and will arrive 3 days
from the initial order date. Two gallons have been sold since the order was placed (one day
ago).
Determine the probability of a stockout:
ROP = 9 gallons.
After one day, quantity on hand = 9 – 2 = 7 gallons.
Determine the probability of experiencing a stockout before this order arrives (in 2 days):
Risk of a stockout at the end of the initial lead time:
Using Formula 13-13, set the ROP equal to the quantity on hand 1 day after the order was
placed and solve for z:
ROP  d (LT)  z ( d ) LT

   ( )  √

    
1 = 1.871z
z = 1 / 1.871
z = 0.53 (round to two decimals)
From Appendix B, Table B, the lead time service level is .7019.
Risk of stockout before this order arrives = 1 - .7019 = .2981 = 29.81%.
13-32
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Education.
Chapter 13 - Inventory Management
22.
Given:

d = 30 gallons/day
ROP = 170 gallons
SS = 50 gallons and provides a stockout risk of 9%
Step 1:
Solve for the standard deviation of demand during the lead time.
We know that SS = 50.
Using Appendix B, Table B, we look for the z value corresponding to 1.00 – .09 = .91.
The closest probability is .9099, which corresponds to z = 1.34.
  
Plug in values and solve for  :
   

 

   gallons (round to three decimals)
Step 2:
Determine the SS.
Stockout risk = 3%.
  
Using Appendix B, Table B, we look for the z value corresponding to 1.00 – .03 = .97.
The closest probability is .9699, which corresponds to z = 1.88.
       (round up)
13-33
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Education.
Chapter 13 - Inventory Management
23.
Given:
d = 85 boards/day
ROP = 625 boards
LT = 6 days
LT = 1.1 days
Determine the probability of a stockout:
  
    
625 = (85 x 6) + z (85) (1.1)
625 = 510 + 93.5z
115 = 93.5z
z = 115 / 93.5
z = 1.23 (round to two decimals)
Using Appendix B, Table B, we find a probability of .8907.
The risk of a stockout = 1 - .8907 = .1093 = 10.93%.
24.
Given:
Service level = 96%
d = 12 units/day
d = 2 units/day
LT = 4 days
LT = 1 day
a. Determine the ROP:
Using Appendix B, Table B, we look for the z value corresponding to .96.
The closest probability is .9599, which corresponds to z = 1.75.

  
  ̅ 
  ̅  
      √        
ROP = 48 + 22.14
ROP = 70.14 = 71 units (round up)
b. The model might not be appropriate if seasonality were present because during the busy times
of the year, the ROP would be set too low (causing stockouts) and during the slow times of
the year, the ROP would be set too high (causing excess inventory).
13-34
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Education.
Chapter 13 - Inventory Management
25.
Given:
LT = 4 x (1 – 0.25) = 4 x 0.75 = 3 days
S = $30
D = 4,500 gallons
H = $3
360 days/year
d = 2 gallons/day
Price List:
Quantity
Unit Price
1-399
$2.00
400-799
$1.70
800+
$1.62
a. Determine the optimal order quantity:
Step 1:
Compute the common minimum point.
Q
2 DS
2(4,500)30

 300 gallons
H
3
This quantity is feasible in the range 1-399.
Step 2:
Determine total cost for the common minimum point and for the price breaks of all lower unit
costs.


  ( )   ( )   




  (
)  (
)     




  (
)  (
)     




  (
)  (
)     


Conclusion: Optimal order quantity = 400 gallons.
b. Acceptable stockout risk = 1.5%. Determine ROP:
d
4 ,500
 12.5 / day
360
Using Appendix B, Table B, we look for the z value corresponding to 1 - .015 = .985:
z = 2.17.
ROP  d (LT)  z ( d ) LT
ROP  12.5(3)  2.17(2) 3
ROP  12.5(3)  2.17(2) 3
ROP  45.02 = 46 gallons (round up).
13-35
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Education.
Chapter 13 - Inventory Management
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:
ROP  d (LT)  z ( d ) LT
Plugging in values and solving for z:
12  5(2)  z (.5) 2
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:
ROP  d (LT)  z ( d ) LT
Plugging in values and solving for z:
12  5(2)  z (.5) 2
12  10  .707 z
2  .707 z
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%.
13-36
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Education.
Chapter 13 - Inventory Management
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.
Using Appendix B, Table B, we look for the z value corresponding to .90.
The closest probability is .8997, which corresponds to z = 1.28.

  
  ̅ 
  ̅  
      √        
    √        
         (round up)
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


  

   

 
b. Determine the ROP that will provide a service level of 96%:
Using Appendix B, Table B, we look for the z value corresponding to .96.
The closest probability is .9599, which corresponds to z = 1.75.
ROP  d (LT)  z ( d ) LT
ROP  10(3)  1.75(2) 3
ROP  30  6.06  36.06  37 (round up)
13-37
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Education.
Chapter 13 - Inventory Management
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:


  



    (round to an integer value)
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:
EDDLT = 80
dLT = 6
Using Appendix B, Table B, we look for the z value corresponding to 0.99.
The closest probability is .9901, which corresponds to z = 2.33.
SS =        (round up)
30.
Given:
ROP = 18 units
Lead time for resupply = 3 days
Usage over the last 10 days:
1
3
Day
Units
2
4
3
7
4
5
5
5
6
6
7
4
8
3
9
4
10
5
Determine the service level achieved by the current ROP. Hint: Use Formula 13-13.
ROP  d (LT)  z ( 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.
18  4.6(3)  z(1.265) 3
18 = 13.8 + 2.191z
4.2 = 2.191z
z = 4.2 / 2.191 = 1.92 (round to two decimals)
From Appendix B, Table B, the lead time service level is .9726 = 97.26%.
13-38
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Education.
Chapter 13 - Inventory Management
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:
Q  d (OI  LT )  z d OI  LT  A
Q  40(14  2)  2.05(3) 14  2  42
Q  622.6 = 623 units (round up)
Cycle 2:
Q  40(14  2)  2.05(3) 14  2  8
Q  656.6 = 657 units (round up)
Cycle 3:
Q  40(14  2)  2.05(3) 14  2  103
Q  561.6 = 562 units (round up)
13-39
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Education.
Chapter 13 - Inventory Management
32.
Given:
Company operates 50 weeks per year
We have the following information on the two items:
P34
d = 60 units/week
d = 4 units/week
LT = 2 weeks
Unit cost = $15
H = (.30)($15) = $4.50
S = $70
Risk = 2.5%
Can be ordered any time
P35
d = 70 units/week
d = 5 units/week.
LT = 2 weeks
Unit cost = $20
H = (.30)($20) = 6.00
S = $30
Risk = 2.5%
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.
ROP  d (LT)  z ( d ) LT
ROP  60(2)  1.96(4) 2
ROP  120  11.09  131.09  132 units (round up)
P35: Order every 4 weeks.
b. Compute the EOQ for P34:
D = 60 x 50 = 3,000 units/year


  

    


c. Compute the order quantity for P35 if 110 units are on hand at the time the order is placed:
Using Appendix B, Table B, we look for the z value corresponding to 1 - .025 = .975:
z = 1.96.
Q  d (OI  LT )  z d OI  LT  A
Q  70(4  2)  1.96(5) 4  2  110
Q  334.01 = 335 units (round up)
13-40
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Education.
Chapter 13 - Inventory Management
33.
Given:
We have the following list of items:
Item
H4-010
H5-201
P6-400
P6-401
P7-100
P9-103
TS-300
TS-400
TS-041
V1-001
Estimated
Annual Demand
20,000
60,200
9,800
14,500
6,250
7,500
21,000
45,000
800
33,100
Ordering
Cost
50
60
80
50
50
50
40
40
40
25
Holding
Cost (%)
20
20
30
30
30
40
25
25
25
35
Unit
Price
2.50
4.00
28.50
12.00
9.00
22.00
45.00
40.00
20.00
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
H4-010
H5-201
P6-400
P6-401
P7-100
P9-103
TS-300
TS-400
TS-041
V1-001
Unit
Price
2.50
4.00
28.50
12.00
9.00
22.00
45.00
40.00
20.00
4.00
Estimated Annual
Demand
20,000
60,200
9,800
14,500
6,250
7,500
21,000
45,000
800
33,100
Annual
Dollar
Value
50,000
240,800
279,300
174,000
56,250
165,000
945,000
1,800,000
16,000
132,400
3,858,750
13-41
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Education.
Chapter 13 - Inventory Management
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).
Annual
Percentage of
Dollar
Percentage of
Annual Dollar
Item
Value
Category
Items
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
P9-103
165,000
V1-001 132,400
C
60%
15.38%
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
H4-010
H5-201
P6-400
P6-401
P7-100
P9-103
TS-300
TS-400
TS-041
V1-001
Estimated Annual
Demand
20,000
60,200
9,800
14,500
6,250
7,500
21,000
45,000
800
33,100
Ordering
Cost
50
60
80
50
50
50
40
40
40
25
Unit
Holding
Cost ($)
.50
.80
8.55
3.60
2.70
8.80
11.25
10.00
5.00
1.40
EOQ
2,000
3,005
428
635
481
292
386
600
113
1,087
13-42
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Education.
Chapter 13 - Inventory Management
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)
19
20
21
22
23
24
25
26
27
28
29
Relative
Frequency
.01
.05
.12
.18
.13
.14
.10
.11
.10
.04
.02
Cs = Rev – Cost = $4.80 – $3.20 = $1.60
Ce = Cost – Salvage = $3.20 – $2.40 = $.80
 



 
     
Demand
(dozens)
19
20
21
22
23
24
25
26
27
28
29
Relative
Frequency
.01
.05
.12
.18
.13
.14
.10
.11
.10
.04
.02
Cumulative
Frequency
.01
.06
.18
.36
.49
.63
.73
.84
.94
.98
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%.
13-43
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Education.
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:
Cs = $88,000
Ce = $100 + 1.45($100) = $245


 

 
     
[From Poisson Table with  = 3.2]
Cumulative
x
Probability
0
.041
1
.171
2
.380
3
.603
4
.781
5
.895
6
.955
7
.983
8
.994
9
.998
10
1.000
Because .997 falls between the cumulative probabilities of .994 and .998, the optimal number
of spare parts to order = 9. The resulting service level will be .998 = 99.8%.
b. Determine the range of shortage cost for which carrying 0 spare parts would be the best
strategy:
Determine the value of Cs for which the service level = the service level of stocking 0 spare
part and solve for Cs:
Service Level for 0 Spare Parts = .041

 
 
  

 
 
  
    
    
    
   
   (round to two decimals)
Conclusion: Carrying 0 spare parts is the best strategy if the shortage cost is less than or
equal to $10.47.
13-44
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Education.
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.


 

 
     
    
    
    
   
   per quart (round to two decimals)
b. This might be a reasonable figure because it probably is close to the lost profit per quart
during strawberry season.
13-45
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Education.
Chapter 13 - Inventory Management
38.
Given:
Demand can be approximated with a Poisson distribution with a mean of 6 per day. It costs $9 to
prepare each cake. Fresh cakes sell for $12 each. Day-old cakes sell for $9 each. One half of the
day-old cakes are sold and the rest thrown out.
Cs = Rev – Cost = $12 – $9 = $3.00 per cake
Ce = Cost – Salvage = $9 – (1/2)($9.00) = $4.50/cake
 



 
     
[From Poisson Table with  = 6]
Cumulative
x
Probability
0
.003
1
.017
2
.062
3
.151
4
.285
5
.446
6
.606
Because .4 falls between the cumulative probabilities of .285 and .446, the optimal number of
cakes to prepare = 5. The resulting service level will be .446 = 44.6%.
39.
Given:
Purchase price = $1.00 per pound. Salvage value = $.80 per pound. Burgers sell for $.60 each.
Four hamburgers can be prepared from each pound of beef. Labor, overhead, meat, buns, and
condiments cost $.50 per burger. Demand is normally distributed with a mean of 400 pounds per
day and a standard deviation of 50 pounds per day. Hint: Shortage costs must be in dollars per
pound.
d = 400 pounds/day
d = 50 pounds/day
Determine the optimal daily order quantity:
Cs = $.10/burger x 4 burgers/pound = $.40/pound
Ce = Cost – Salvage = $1.00 – $.80 = $.20/pound
 

 

 
      
Using Appendix B, Table B, we find that .6667 falls closest to .6664:
z = 0.43.
           pounds (assuming that fractional values are possible)
13-46
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Education.
Chapter 13 - Inventory Management
40.
Given:
Demand for rug cleaning machines is shown in the table below. Machines are rented by the day
only. Profit on rug cleaners = $10/day. Clyde has 4 rug-cleaning machines.
Demand
0
1
2
3
4
5
Frequency
.30
.20
.20
.15
.10
.05
1.00
Cumulative
Frequency Frequency
.30
.30
.20
.50
.20
.70
.15
.85
.10
.95
.05
1.00
1.00
a. Determine the implied range of excess cost per machine:
  
Cs = $10
Ce = unknown
Demand
0
1
2
3
4
5
For 4 machines to be optimal, the SL ratio must be ≥ .85 and ≤ .95.
Step 1:
Set SL = .85 and solve for Ce:


 

 
     
     
    
    
  
   (round to two decimals)
Step 2:
Set SL = .95 and solve for Ce:


 

 
     
     
    
    
  
    (round to two decimals)
Conclusion: Implied range of excess cost: $.53 ≤  ≤ $1.76.
13-47
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Chapter 13 - Inventory Management
b. If Clyde protests that the answer from part a is too low, does this suggest an increase or a
decrease in the number of machines he stocks?
If the excess cost is supposed to be higher, then the number of machines should be decreased.
When excess cost increases, SL decreases along with the optimum stocking level.
c. Suppose now that excess cost per day = $10 and the shortage cost per day is unknown.
Assuming that the optimal number of machines is 4, what is the implied range of shortage
cost?
  
Cs = unknown
Ce = $10
For 4 machines to be optimal, the SL ratio must be ≥ .85 and ≤ .95.
Step 1:
Set SL = .85 and solve for Cs:


 

 
     
    
    
    
   
   (round to two decimals)
Step 2:
Set SL = .95 and solve for Cs:


 

 
     
    
    
    
   
   (round to two decimals)
Conclusion: Implied range of shortage cost: $56.67 ≤  ≤ $190.00.
13-48
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Chapter 13 - Inventory Management
41.
Given:
Spares cost $200 each. Unused spares have a salvage value of $50 each. If a part fails and a spare
is not available, 2 days will be needed to obtain a replacement and install it. The cost for idle
equipment is $500/day.
Probability of usage:
0
1
2
3
Number
.10
.50
.25
.15
Probability
a. Use the ratio method to determine the number of spares to order:
# of Spares
0
1
2
3
Probability of
Demand
.10
.50
.25
.15
Cumulative
Probability
.10
.60
.85
1.00
Cs = Cost of stockout = ($500 per day) (2 days) = $1000
Ce = Cost of excess inventory = Unit Cost – Salvage Value = $200 – $50 = $150
SL 
Cs
1,000

 .870
Cs  Ce 1,000  150
Because .870 is between the cumulative probabilities of .85 and 1.00, we need to order 3
spares.
b. Use the tabular method to determine the number of spares to order:
Stocking
Level
0
1
2
3
Demand = 0
Prob. = 0.10
$0
.10(1)($150)=$15
.10(2)($150)=$30
.10(3)($150)=$45
Demand = 1
Prob. = 0.50
.50(1)($1000)=$500
$0
.50(1)($150)=$75
.50(2)($150)=$150
Demand = 2
Prob. = 0.25
.25(2)($1000)=$500
.25(1)($1000)=$250
$0
.25(1)($150)=$37.50
Demand = 3
Prob. = 0.15
.15(3)($1000)=$450
.15(2)($1000)=$300
.15(1)($1000)=$150
$0
Expected
Cost
$1,450
$565
$255
$232.50
We should order 3 spares.
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Chapter 13 - Inventory Management
42.
Given:
Cakes cost $33 each, and they sell for $60 each. Unsold cakes are reduced to half price on
Monday, and typically one-third of those are sold. Any that remain are donated.
Probability of demand:
0
1
2
3
Number
.15
.35
.30
.20
Probability
a. Use the ratio method to determine the number of cakes to prepare to maximize expected
profit:
Probability of
Cumulative
# of Cakes
Demand
Probability
0
.15
.15
1
.35
.50
2
.30
.80
3
.20
1.00
Cs = Selling Price – Unit Cost = $60 – $33 = $27
Ce = Unit Cost – Salvage Value = $33 – [(1/3)(1/2)($60)] = $23
Cs
27

 .54
Cs  Ce 27  23
Because the service level of .54 falls between the cumulative probabilities of .50 and .80, the
supermarket should stock 2 cases of wedding cakes.
SL 
b. Use the ratio method to determine the number of cakes to prepare to maximize expected
payoff.
Expected Payoff = Expected Profit – Expected Cost
Expected Profit = Probability of Demand * Expected Profit per Cake Sold at Regular Price
($27) * Number of Cakes Sold at Regular Price.
Expected Cost = Probability of Demand * Expected Cost per Cake Left Over (Ce = $23) *
Number of Cakes Left Over.
Stocking
Level
0
1
2
3
Demand = 0
Prob. = .15
[Sell 0, Over 0]
(.15 * 0 * $27) –
(.15 * 0 * $23) =
$0
[Sell 0, Over 1]
(.15 * 0 * $27) –
(.15 * 1 * $23) =
-$3.45
[Sell 0, Over 2]
(.15 * 0 * $27) –
(.15 * 2 * $23) =
-$6.90
[Sell 0, Over 3]
(.15 * 0 * $27) –
(.15 * 3 * $23) =
-$10.35
Demand = 1
Prob. = .35
[Sell 0, Over 0]
(.35 * 0 * $27) –
(.35 * 0 * $23) =
$0
[Sell 1, Over 0]
(.35 * 1 * $27) –
(.35 * 0 * $23) =
$9.45
[Sell 1, Over 1]
(.35 * 1 * $27) –
(.35 * 1 * $23) =
$1.40
[Sell 1, Over 2]
(.35 * 1 * $27) –
(.35 * 2 * $23) =
-$6.65
Demand = 2
Prob. = .30
[Sell 0, Over 0]
(.30 * 0 * $27) –
(.30 * 0 * $23) =
$0
[Sell 1, Over 0]
(.30 * 1 * $27) –
(.30 * 0 * $23) =
$8.10
[Sell 2, Over 0]
(.30 * 2 * $27) –
(.30 * 0 * $23) =
$16.20
[Sell 2, Over 1]
(.30 * 2 * $27) –
(.30 * 1 * $23) =
$9.30
Demand = 3
Prob. = .20
[Sell 0, Over 0]
(.20 * 0 * $27) –
(.20 * 0 * $23) =
$0
[Sell 1, Over 0]
(.20 * 1 * $27) –
(.20 * 0 * $23) =
$5.40
[Sell 2, Over 0]
(.20 * 2 * $27) –
(.20 * 0 * $23) =
$10.80
[Sell 3, Over 0]
(.20 * 3 * $27) –
(.20 * 0 * $23) =
$16.20
Expected
Payoff
$0
$19.50
$21.50
$8.50
Conclusion: The supermarket should stock 2 cases of wedding cakes. This number of cakes
will maximize the expected payoff.
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Chapter 13 - Inventory Management
43.
Given:
On average, 18 ticket holders cancel their reservations, so the company intentionally overbooks
the flight. Cancellations can be described by a normal distribution with a mean equal to 18 and a
standard deviation of 4.55. Profit per passenger = $99. If a passenger is bumped, the company
pays that passenger $200.
Determine the number of tickets to overbook:
Cs = $99, Ce = $200
SL 
Cs
99

 .3311
Cs  Ce 99  200
Using Appendix B, Table B, we find that .3311 falls closest to .3300:
z = -0.44.
             tickets to overbook
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Chapter 13 - Inventory Management
Case: UPD Manufacturing
Given:
OI = 6 weeks
S = $32
H = $.08/unit/week
d = 89 units/week
LT = 5 working days = 1 week
1.
Students must recognize that without demand variability, the fixed order interval order quantity
equation reduces to:
      
UPD places an order every 6 weeks and the lead-time is 1 week. Therefore, when the order is
placed, there will be 89 units on hand (d x LT = 1 week * 89 units/week).
Because A = 89 = d x LT, the fixed order interval order quantity equation further reduces to the
following:
        
         units
Therefore, ordering at six-week intervals requires an order quantity of 534 units.
Optimal order quantity as determined by using the basic EOQ equation:
Q
2dS
2(89)(32)

 266.83  267 (round to an integer value)
h
.08
The weekly total cost based on the EOQ is given below:
 d  Q
 89 
 267 
TCEOQ   S    H  
32  
.08
 267 
 2 
Q  2 
TCEOQ  10.67  10.68  $21.35 / week
The weekly total cost based on six-week fixed order interval (FOI) order quantity is given below:
 d  Q
 89 
 534 
TCFOI   S    H  
32  
.08
 534 
 2 
Q  2 
TCFOI  5.33  21.36  $26.69 / week
Weekly savings of using EOQ rather than 6-week FOI = $26.69 – $21.35 = $5.34
The annual savings = (52 weeks) ($5.34/week) = $277.68
2.
The total annual savings as a result of switching from the six-week FOI to EOQ are relatively
small and switching to the optimal order quantity may not be warranted. However, if the FOI
approach is used with other parts or components as well, the total potential loss may be
significant.
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Chapter 13 - Inventory Management
Case: Harvey Industries
To improve the current inventory control system, the new president may want to consider the following:
1.
Computerize the inventory control system. Rationale: There are too many parts for the current
manual system.
2.
Currently, no paperwork is used when items are withdrawn from the stockroom when they are
needed on the shop floor. Harvey Industries may either want to establish a procedure for
recording the transactions in the stockroom or invest in a bar coding system. If a bar coding
system is purchased, it has to be coordinated with the new computerized inventory control
system. Establishing a cycle counting procedure may be very helpful also. Rationale: As a result
of these actions, the inventory accuracy should improve substantially.
3.
It appears that utilization of A-B-C inventory classification system is needed. Rationale: Harvey
Industries rarely should experience stockouts in those “A” items that account for $220,684 of
$314,673 in annual purchases or for any “A” items for which a stockout leads to significant
downtime costs. ABC analysis will allow Harvey Industries to establish an appropriate degree of
control over items in terms of order quantity and ordering frequency.
Case: Grill Rite
Recommendations:
1.
The president’s stance on steady output conflicts with seasonal demand. However, it is unlikely
that this will change. One alternative might be to identify a complementary product that would
offset seasonal demand for electric grills.
2.
The main problem is inventory management. Therefore, we recommend the following:
a. Having a single, centralized warehouse: This will lower the need for safety stock due to the
canceling effect of random variability in orders from the various regions. Conversely, with
separate warehouses, each warehouse needs a relatively larger safety stock to guard against
variations in demand.
b. What is needed is overall control of the system that would take into account seasonal
variations in demand and achieve a better match between regional demand and supply. This
might involve making or improving regional forecasts. In any case, improved system
visibility is essential: direct access to regional warehouse data by the main warehouse is
needed to be able to coordinate and set priorities on inventory shipments to regional
warehouses. That should take care of most of the problem.
c. It also may pay to examine the feasibility of shipping from one warehouse to another when a
shortage occurs. Relevant costs would include transaction costs and transportation costs
versus the potential increase in profit by avoiding the shortage.
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Chapter 13 - Inventory Management
Case: Farmers Restaurant
1.
Inventory management is crucial not only to Farmers Restaurant, but to businesses in general.
Customer satisfaction and customer return is contingent upon proper inventory management. If
customers visit the Farmers Restaurant and are unable to receive the food they desire due to a
stockout, the customer may be dissatisfied and may not return to Farmers Restaurant. In addition
to customer satisfaction, total food costs are important to businesses also. In the restaurant
industry, if too much of a product is ordered and not used; it could result in product waste due to
the items expiring. This would result in an increase in total food costs when the goal is to keep
food costs low! Overstocking products also can negatively affect Farmers Restaurant and other
businesses. By having more products on-hand than needed, Farmers Restaurant is tying up funds
that might be more productive elsewhere. Overall, it is imperative that Farmers Restaurant try to
manage inventory levels successfully due to these potential/existing problems.
2.
A fixed-interval ordering system is appropriate given that the manager reviews inventory and
places orders once a week from the supplier.
3.
Given:
SL = 95%
̅   units/week
   units/week
Gravy mix comes in packs of 2
There are currently 3 packs in inventory = 6 units
LT = 2 days = 2/7 weeks
OI = 1 week
Using Appendix B, Table B, we look for the z value corresponding to .95:
.95 falls midway between .9495 (z = 1.64) and .9505 (z = 1.65).
Using z = 1.64:
Q  d (OI  LT )  z d OI  LT  A
Q  35(1  2 / 7)  1.64(3.5) 1  2 / 7  6
Q  45.51 = 46 units (round up) = 23 of the 2-packs.
Using z = 1.65:
Q  d (OI  LT )  z d OI  LT  A
Q  35(1  2 / 7)  1.65(3.5) 1  2 / 7  6
Q  45.55 = 46 units (round up) = 23 of the 2-packs.
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Chapter 13 - Inventory Management
4.
Given:
A = 12
Determine the risk of a stockout at the end of the initial lead time and at the end of the second
lead time:
Use Formula 13-13 to determine the risk of stockout at the end of the initial lead time:
ROP  d (LT)  z ( d ) LT
12  35(2/7)  z (3.5) 2 / 7
12  10  1.871z
2  1.871z
z  1.07 (round to two decimals)
Using Appendix B, Table B, we look for the corresponding service level: .8577.
The risk of a stockout = 1 - .8577 = .1423 = 14.23%.
Determine the risk of a stockout at the end of the second lead time:
Use Formula 13-16 to determine the risk of stockout at the end of the second lead time:
Q  d (OI  LT )  z d OI  LT  A
46  35(1  2 / 7)  z(3.5) 1  2 / 7  12
46  45  3.969 z  12
46  33  3.969 z
13  3.969 z
   (round to two decimals)
Using Appendix B, Table B, we look for the corresponding service level: .9995.
The risk of a stockout = 1 - .9995 = .0005 = .05%.
5.
Kristin may want to consider dealing with a nearby supplier to be able to order more frequently or
to reduce transportation costs. In addition, if she ordered from a nearby supplier, she could have
the option of sending an employee to the supplier’s facility to pick up emergency orders. On the
other hand, she may want to keep her current supplier due to competitive prices and/or
exceptional customer service.
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Chapter 13 - Inventory Management
Operations Tour: Bruegger’s Bagel Bakery
1.
If too little inventory is maintained, there is a risk of a stockout and potential lost sales. In
addition, if there is not sufficient work-in-process inventory, the production process may become
too inefficient, raising the cost of production. On the other hand, if too much inventory is
maintained, the carrying cost may become excessively high.
2.
a. Customers judge the quality of bagels by their appearance (size, shape, and shine), taste, and
consistency. Customers are also very interested in receiving high service quality.
b. Bruegger’s checks quality at every stage of operation, from choosing suppliers of ingredients,
careful monitoring of ingredients, and keeping equipment in good operating condition to
monitoring output at each step of the production process. At the stores, employees watch for
deformed bagels and remove them.
c. Steps for Bruegger’s Bagel Bakery Operations:
1) Purchase ingredients from suppliers
2) Receive ingredients from suppliers
3) Mix basic ingredients into the dough
4) Shape the dough into individual bagels
5) Ship bagels to stores in refrigerated trucks
6) Unload and store the bagels
7) Boil bagels in kettle of water and malt
8) Bake bagels for 15 minutes
9) Sell bagels to customers
The company can improve quality at each step by monitoring output more carefully and with
training and education of the employees.
3.
The basic ingredients can be purchased using either fixed order interval or fixed order quantity
models, e.g., EOQ. The EPQ model is most appropriate for deciding the size of the production
quantity.
4.
If there were a bagel-making machine at each store, the company would have to invest in more
machinery, more space for production and storage, and more worker training for the production
of bagels. However, the lead time to make the bagels would be shortened. The shorter lead time
would provide faster, more flexible response to customer demands and fresher bagels.
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Chapter 13 - Inventory Management
Enrichment Module: EPQ Problem
This enrichment module consists of an EPQ problem to solidify the concepts associated with the
Economic Production Quantity model.
Problem
A company produces plastic powder in lots of 2,000 pounds, at the rate of 250 pounds per hour. The
company uses powder in an injection molding process at the steady rate of 50 pounds per hour for an
eight-hour day, five days a week. The manager has indicated that the setup cost is $100 for this product,
but “We really have not determined what the holding cost is.”
a.
b.
What weekly holding cost per pound does the lot size imply, assuming the lot size is optimal?
Suppose the figure you compute for holding cost has been shown to the manager, and the
manager says that it is not that high. Would that mean the lot size is too large or too small?
Explain.
Solution to Enrichment Module Problem
a.
Q  2,000 lbs.
p  250 lbs. / hr.
u  50 lbs. / hr.
S  $100
d  (50 lbs. / hr.) x (8 hrs. / day ) x (5 days / week )  2,000 lbs. / week
Q
2dS  p 

x
H  p  u 
2,000 
2(2,000)(100)  250 
x

H
 250  50 
2,000 
400,000
x(1.25)
H
500,000
H
500,000
(2,000) 2 
H
500,000
H
 $.125 / lb. / week
4,000,000
Decreasing the value of carrying cost (H) will result in an increase in the lot size.
Because holding inventory is not as expensive, the firm could afford to carry more
inventory and therefore produce a larger batch.
2,000 
b.
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Chapter 13 - Inventory Management
Enrichment Module 2: Inventory Model with Planned Shortages
In most cases, shortages are undesirable and should be avoided. However, in certain circumstances, it
may be desirable to plan and allow for shortages. Planned shortages are implemented for high dollar
volume items where the inventory carrying cost is very high. The model discussed in this section refers to
the specific type of shortages called backorders. When a customer attempts to purchase an out-of-stock
item, the firm does not lose the sale. The customer waits until the purchased order arrives from the
supplier. If there were no additional cost associated with backordering, there would be no incentive for
the firm to maintain any inventory. However, there are costs associated with backordering. The tangible
part of the backorder cost involves the cost of expediting the delivery (special delivery) and production of
the backordered item. The intangible part of the backorder cost involves the loss of goodwill due to the
fact that the customers are forced to wait for their orders. The longer the waiting period, the higher the
backorder cost due to loss of goodwill will be.
There is a direct trade-off between the inventory carrying cost and the cost of a planned shortage in the
form of backorders. In many cases, the cost of backorders can be offset easily by the reduction in carrying
costs. The model discussed in this section will not be valid if a customer decides not to wait for the
backorder.
The fixed order quantity inventory model with planned shortages (backorders) is very similar to the basic
EOQ model. When the reorder point is reached, a new economic order quantity (Q) is placed. Figure 1
shows the schematic representation of this model. The size of the backorder is B units and the maximum
inventory is Q – B units. The average size of the backorder is B/2 for each order cycle. T is defined as the
amount of time between two successive orders (a complete order cycle). t1 is the part of the order cycle
where the customer orders are met from stock. In other words, during t1 there is positive inventory level.
On the other hand, t2 is the period of time in the order cycle where the inventory is depleted and all the
customer orders are placed on backorder (stockout period).
Symbol definitions used to explain various concepts are listed below.
H = carrying cost per unit per year
S = ordering cost per batch (lot)
D = annual demand
Q* = optimal order quantity
B = size of the backorder
CB = backorder cost per unit per year
B* = optimal planned backorder quantity
T = Q/D (length of the complete order cycle in years) or
T = Q/d (length of the complete order cycle in days)
t1 = (Q – B)/D or (Q – B)/d (time period during which inventory is positive)
t2 = B/D or B/d (time period during which there is no inventory)
In this model, the average inventory is not Q/2 or not even (Q – B)/2 because during the shortage period
there are no units in inventory. The average inventory calculation for this model can be explained with the
following example:
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Chapter 13 - Inventory Management
A large local car dealership orders a certain brand of automobiles from a car manufacturer located in
Detroit. Order quantity (Q) is 500 units, annual demand (D) is 7,500 and the firm operates 300 working
days per year. Due to the high holding costs, the company plans to backorder (B) 200 cars per order cycle.
Determine the average inventory.
d = (D/number of operational days) = 7,500/300 = 25 units (daily consumption)
T = Q/d = 500/25 = 20 days (time between orders is 20 days)
t1 = (Q – B)/d = (500 – 200)/25 = 300/25 = 12 days (time period during which there is no shortage)
t2 = B/d = 200/25 = 8 days (time period during which there is no inventory)
The dealership will carry an average of (Q – B)/2 units during t1 and no units during t2. Therefore, total
number of unit days during the inventory cycle can be computed by multiplying t1 by (Q – B)/2
Number of unit days of inventory/cycle  t1 *
Number of unit days of inventory/cycle 
QB
2
(Q  B) (Q  B)
d
2
(Q  B) 2
2d
In other words, an average of 150 units are carried in inventory for 12 days and zero units are carried for 8
days (shortage period). Therefore, total number of unit days of inventory during the complete order cycle
is (150)(12) = 1800.
Because there are a total of 20 days in the complete order cycle, the average inventory can be computed
by dividing the total number of unit days of inventory by the number of days in the inventory cycle. In
this example, the average inventory is equal to 1,800/20 or 90 units. Therefore, the average inventory can
be computed by using the following formula:

(Q  B) 2
2d
Average inventory 
Q
d
(Q - B) 2
Average inventory 
2Q
Using a similar logic, we can also develop the average backlog formula. The dealership will experience
shortage (backorders) for 8 days during the order cycle. The average amount of backorder on a given
shortage day is B/2. Based on this information, the total number of backorder unit days can be computed
using the following equation: (t2) (B/2) = (B/D)(B/2) = B2 /2D.
In our example, there are 8 days of a planned shortage period. During this period, an average of 200/2 =
100 units of backorders are realized. Therefore, the total number of backorder unit days during the order
cycle is (8)(100) = 800 units. Because there are a total of 20 days in the order cycle, the average
backorder quantity for the complete order cycle can be determined by dividing the total number of
backorder unit days by the number of days in the complete inventory cycle. In this example, using the
above equation, we obtain an average backorder quantity of 800/20 = 40 units. The general equation for
the average backorder quantity is:
13-59
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Chapter 13 - Inventory Management
B2
Averagebackorder  2d
Q
d
Averagebackorder 
B2
2Q
Annual inventory carrying cost still is calculated by multiplying the average inventory by the inventory
carrying cost per unit per year. The formula for the annual ordering cost is the same as it was for the basic
EOQ model. The annual backorder cost is determined by multiplying the average backorder quantity by
the backorder cost per unit per year.
The annual inventory carrying cost is given by:
(Q  B) 2
H
2Q
The annual ordering and backordering costs are given by the following respective formulas:
D
S
Q
B2
CB
2Q
Therefore, the total annual inventory cost (TC) can be expressed by summing the annual inventory
carrying cost, the annual ordering cost, and the annual backordering cost as shown in the following
formula:
TC 
D
(Q  B) 2
B2
H S
CB
2Q
Q
2Q
Taking the first total derivative of the above total cost formula with respect to Q, setting the resulting
equation to zero, and solving for Q will result in the following optimal quantity (Q*) and optimal
backorders (planned shortages) (B*) formulas:
Q* 
2 DS  H  C B 


H  C B 
 H 
B*  Q * 

H B
13-60
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Chapter 13 - Inventory Management
Figure 1
An inventory situation with planned shortages
Inventory
Maximum Inventory Level
Q–B
Q
Stockout B
Time
t2
T = Q/d
Example:
XYZ Company distributes a major part for the F–15 fighter jets. Due to the very high holding cost, the
company wants to implement a model with planned shortages. The annual demand is 81,000 and the
company operates 300 days per year. The annual carrying cost rate is 10% of the unit cost and the unit
cost of this item is $1,000. The ordering cost per batch is estimated at $500.
a.
Determine the optimal order quantity and total annual inventory cost (ordering cost + carrying
cost) using the basic EOQ model with no backorders.
b.
If each unit backordered costs the company $200 per unit per year, what would be the optimal
order quantity and the optimal size of the planned backorder?
c.
Determine the annual carrying cost, the annual ordering cost, the annual backordering cost, and
the annual total inventory cost for the planned shortage model used in part b.
d.
Determine the values of t1, t2 and T in days.
e.
Should the company adopt the planned backorder model of part b or the basic EOQ model of part
a, which does not allow backorders?
D = 81,000 units
S = $500
d = 81,000/300 days = 27 units per day
H = ($1,000) (.10) = $100
CB = $200
a.
Q* 
2DS
H
Q* 
2(81,000)(500)
 900
100
D
 81,000 
Annual ordering cost =  S  
(500)  45,000
 900 
Q
Q
 900 
H  
(100)  45,000
2
 2 
Annual carrying cost = 
Total annual cost = $45,000 + $45,000 = $90,000
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Chapter 13 - Inventory Management
b.
Q* 
2 DS  ( H  C B ) 


H  C B

Q* 
2(81,000)(500)  100  200 


100
 200 
Q * 1,215,000  1,102.3  1,102
 H 

B  Q * 
 H  CB 
 100 
B  (1,102)
  367.33  367
 100  200 
c.
Annual carrying cost 
(Q  B) 2
H
2Q
(1,102  367) 2
(100)
2(1,102)
 24,511.12

D
 81,000 
Annual ordering cost   ( S )  
(500)
 1,102 
Q
 36,751.36
 B2 
C B
Annual backorderi ng cost  
 2Q 
367 2

(200)  12,222.23
2(1,102)
Let TC = Total annual inventory cost
TC = $24,511.12 + $36,751.36 + $12,222.23 = $73,484.71
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Chapter 13 - Inventory Management
d.
81,000
 27
300
Q 1,102
T 
 40.81 days
d
27
Q  B 1,102  367

 27.22 days
t1 
d
27
B * 367
t2 

 13.59 days
d
27
d
e.
The model with planned backorders is preferred because the total annual inventory cost of the
basic EOQ inventory model is substantially higher than the total annual inventory cost of the
planned backorder model.
TCbasic EOQ = $90,000
TCbackorder = $73,484.71
$90,000 – $73,484.71 = $16,515.29 difference
Problems
The manager of an inventory system believes that inventory models are important decision-making aids.
Although the manager often uses an EOQ policy, he has never considered a backorder model because of
his assumption that backorders are “bad” and should be avoided. However, with upper management’s
continued pressure for cost reduction, you have been asked to analyze the economics of a backordering
policy for some products that possibly can be backordered. For a specific product with D = 800 units per
year, S = $150, H = $10, and CB = $20, what is the cost difference in the EOQ and the planned shortage or
backorder model? If the manager adds constraints that no more than 35% of the units may be backordered
and that no customer will have to wait more than 20 days for an order, should the backorder inventory
policy be adopted? Assume 250 working days per year.
Solution to Problem
D = 800 units/year
S = $150
H = $10/unit/year
CB = $20/unit/year
Planned shortage model:
Q* 
2 DS  ( H  C B ) 
2(800)(150) (10  20)

 

20
H  CB
10

Q*  36,000  189.74  190 units
 H
B*  Q * 
 H  CB

10
  (190)   63.33  63 units
 30 

EOQ model:
Q* 
2 DS
2(800)(150)

 154.92  155 units
10
H
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Chapter 13 - Inventory Management
Total cost planned shortage model:
Annual carrying cost =
(Q  B) 2
(190  63) 2
H
(10)  $424.45
2Q
2(190)
D
 800 
S  
(150)  $631.58
 190 
Q
Annual ordering cost = 
 B2 
 (63) 2 
CB  
20  $208.89
 2Q 
 2(190) 
Annual backordering cost = 
TC = $424.45 + $631.58 + $208.89 = $1,264.92
Total cost regular EOQ model:
 155 
Q
(10)  $775.00
H  
 2 
2
Annual carrying cost = 
D
 800 
S  
(150)  $774.19
 155 
Q
Annual ordering cost = 
TC = $775.00 + $774.19 = $1,549.19
TCDifference = $1,549.19 - $1,264.92 = $284.27
Using the planned shortage model will result in annual savings of $284.27.
Number of orders =
D 800

 4.21 orders
Q 190
D

Q
Expected annual number of units short = (B) 
Expected annual number of units short = (63)(4.21) = 265.23
D 800

 3.2 units/day
250 250
63
t2 =
 19.69 days
3.2
265.23
Because 19.69 < 20 and
= .3315 < .35, the backorder inventory policy should be adopted.
800
d=
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