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

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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

Category

$5,100

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%

A

2,500 B

2,400

3

900

5

770

4

500

C

12,170

b. Given: D = 4,500, S = $36, and H = $10. Find the EOQ (round to an integer value):

2 DS  H

Q0 

2(4,500)36  180 units 10

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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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 K34

Unit Cost Usage $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

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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 K34 K35

Unit Cost Usage $10 200 25 600

Annual Dollar Value $2,000 15,000

K36

36

150

5,400

M10

16

25

400

M20

20

80

1,600

Z45

80

200

16,000

F14

20

300

6,000

F95

30

800

24,000

F99

20

60

1,200

D45

10

550

5,500

D48

12

90

1,080

D52

15

110

1,650

D57

40

120

4,800

N08

30

40

1,200

P05

16

500

8,000

P09

10

30

300 94,130

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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 F95 Z45 K35 P05

Annual Dollar Percentage of Percentage of Annual Value Category Items Dollar Value $24,000 18.75% 54.83% 16,000 A [(3/16)*100] [($55,000/$94,130)*100] 15,000 8,000

F14

6,000

D45

5,500

K36

5,400

D57

4,800

K34

2,000

D52

1,650

M20

1,600

F99

1,200

N08

1,200

D48

1,080

M10

400

P09

300

B

31.25% [(5/16)*100]

31.55% [($29,700/$94,130)*100]

C

50.00% [(8/16)*100]

10.02% [($9,430/$94,130)*100]

100.00%

100.00%

94,130

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

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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.

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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  H

2(1,215)10  18 bags 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 ( )

( )

(

)

(

)

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

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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 ( )

( )

(

)

(

)

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: ( )

( )

(

)

(

)

$6,120 – $6,118.82 = $1.18 higher per year for Q = 200 (this should be acceptable).

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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.

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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

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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): ( )

( )

(

)

(

)

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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.

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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.

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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.

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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

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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 p

d. Runs per year:



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)  )

(

)

5,164 (1,000  400)  3,098.4 bags 1,000 )

(

2(80,000)25 1,000  5,163.98  5,164 bags 0.25 1,000  400

( )

)

( )

(

)

(

(

)

)

(

)

Savings when S = $25 = $1,549.19 – $774.60 = $774.59 per year.

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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:

Qp 

2 DS H

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 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

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.

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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

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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.

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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

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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 H 2

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)

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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.

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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)

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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 H 10

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 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

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 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

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 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

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 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

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 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

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 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

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 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

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 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

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. ̅

̅̅̅̅

(

̅

√̅̅̅̅ )

√( )

√(

28.

) (

(

) ) (round up)

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 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

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 = 30.

(round up)

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 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

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 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

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 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

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 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: 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 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

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 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

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 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

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. ( 37.

)(

)

pounds (assuming that fractional values are possible)

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 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

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 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

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 Demand Frequency Frequency 0 .30 .30 1 .20 .50 2 .20 .70 3 .15 .85 4 .10 .95 5 .05 1.00 1.00 a. Determine the implied range of excess cost per machine: Cs = $10 Ce = unknown 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.

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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.

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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. 13-50 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

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. 13-52

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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:

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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 

(Q  B) 2 D 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

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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 t1    27.22 days 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)     H  CB 10 20 

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 H 10 13-63

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Chapter 13 - Inventory Management

Total cost planned shortage model:

(Q  B) 2 (190  63) 2 H (10)  $424.45 Annual carrying cost = 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:

Q  155  H   (10)  $775.00 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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