231437624-Heizer-Operation-Management-Solution-PDF.pdf

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Instructor’s Solutions Manual for

Additional Problems

Operations

Management

E I G H T HE I G H T H E D I T I O N E D I T I O N

Principles of

Operations

Management

S I X T HS I X T H E D I T I O N E D I T I O N

Upper Saddle River, New Jersey 07458 Upper Saddle River, New Jersey 07458

JAY HEIZER

Texas Lutheran University

BARRY RENDER

Rollins College

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VP/Editorial Director:

VP/Editorial Director: Jeff Shelstad Jeff Shelstad

Executive Editor:

Executive Editor: Mark Pfaltzgraff Mark Pfaltzgraff

Senior Managing Editor:

Senior Managing Editor: Alana Bradley Alana Bradley

Senior Editorial Assistant:

Senior Editorial Assistant: Jane Avery Jane Avery

Copyright

Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyr

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10 9 8 7 6 5 4 3 2 1 10 9 8 7 6 5 4 3 2 1

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Homework Problem Answers

Chapter

Chapter 1 1 Operations Operations and and Productivity Productivity ...A-1A-1

Chapter

Chapter 3 3 Project Project Management Management ... ... A-3 A-3

Chapter

Chapter 4 4 Forecasting Forecasting ... A-7 A-7

Chapter

Chapter 5 5 Design Design of of Goods Goods and and Services Services ... ... A-11 A-11

Chapter

Chapter 6 6 Managing Managing Quality Quality ... A-15 A-15

Supplement 6:

Supplement 6: Statistical Statistical Process Process Control Control ... A-18 A-18

Chapter

Chapter 7 7 Process Process Strategy Strategy ... A-20 A-20

Supplement 7:

Supplement 7: Capacity Capacity Planning Planning ... A-23 A-23

Chapter

Chapter 8 8 Location Location Strategies Strategies ... A-27 A-27

Chapter

Chapter 9 9 Layout Layout Strategy Strategy ... A-30 A-30

Supplement 10:

Supplement 10: Work Work Measurement Measurement ... A-34 A-34

Chapter

Chapter 12 12 Inventory Inventory Management Management ... ... A-36 A-36

Chapter

Chapter 13 13 Aggregate Aggregate Planning Planning ... ... A-42 A-42

Chapter

Chapter 14 14 Materials Materials Requirements Requirements Planning Planning (MRP) (MRP) & & ERP ERP ... A-46 A-46

Chapter

Chapter 15 15 Short-Term Short-Term Scheduling Scheduling ... A-51 A-51

Chapter

Chapter 16 16 Just-In-Time Just-In-Time and and Lean Lean Production Production Systems Systems ... A-55 A-55

Chapter

Chapter 17 17 Maintenance Maintenance and and Reliability Reliability ... A-57 A-57

Module

Module A: A: Decision Making Decision Making Tools Tools ... A-59 A-59

Module

Module B: B: Linear Linear Programming Programming ... ... A-64 A-64

Module

Module C: C: Transportation Modeling Transportation Modeling ... A-70 A-70

Module

Module D: D: Waiting Line Waiting Line Models Models ... ... A-75 A-75

Module

Module E: E: Learning Learning Curves Curves ... ... A-79 A-79

Module

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1

CHAPTER

Operations and

Productivity

1.1 _a.

(

)(

) (

)(

)

(

)(

) (

)(

)

Last year’s

number of units of output total factor

total dollar value of all inputs used productivity 12,000 units 12, 000 $2.00 14, 000 $10.50 2, 000 $8.00 4, 000 $0.70 $30, 000 12,000 units $219,800

!

"

#

$

₌

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#

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%

&

=

'

+ +

(

)

+ +

_*

= =_{0.0546 units dollar} b.

(

)(

) (

)(

)

(

)(

) (

)(

)

This year’s

number of units of output total factor

total dollar value of all inputs used productivity 14,000 units 14, 000 $2.05 16, 000 $11.00 1,800 $7.50 3,800 $0.75 $26, 000 14,000 units $247,050

!

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$

₌

#

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#

$

%

&

=

'

+ +

(

)

+ +

_*

= =_{0.0567 units dollar} c.

This year’s Last year’s total factor total factor

productivity productivity 0.0567 0.0546 100% 100% Last year’s 0.0546 total factor productivity 3.84% 3.8%

!

"

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"

#

$

₋

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$

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$

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$

#

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₋

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_× ₌ _×

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= + ≈

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1.2

( )( )( )

57,600 0.15

160 12 L

= _{, where}_L =_number_{of laborers employed at the plant.}

So

( )( )(

)

57,600 200 160 12 0.15 L = = Answer : 200 laborers 1.3 _Output=_{28,000 customers}

There are 4 approaches to solving the problem correctly: 1. Input = 7 workers

Then, 28,000 4,000 customers worker 7 =

2. Input =7 40 labor weeks

_{( )}

Then,

( )

28,000

100 customers labor week 7 40

=

3. Input =7 40 50 labor hours

_{( )( )}

Then,

( )( )

28,000

2 customers labor hour 7 40 50 =

4. Input =7 40 $250 dollars of worker wages

_{( )(}

₎

Then,

( )(

)

28,000

0.40 customers per dollar of labor 7 40 $250 = 1.4

( )(

)

6,600 Cadillacs 0.10 labor hours 66,000 labor hours x x = =

There are 300 laborers. So, 66,000 labor hours

220 labor hours laborer 300 laborers = 1.5

( )

52 $90 198 $80 $ output 20,520 $57.00 labor hour 8 45 360 +

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3

CHAPTER

Project

Management

3.1 50 100 150 200 A B C D E F G H I Hours 80 150 200 Gantt Chart 20 120 110 140 170 160 3.2 _{AON Network:} 60 B Purchasing 30 D Sawing 20 A Planning 100 C Excavation 20 E Placement 10 F Assembly 20 G Infill 10 H Outfill 30 I Decoration

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3.3 _{AOA Network:}

Plan

1 2 Purchase 3 Saw 4 Place 5 Assemble 6 Outfill 8 Decorate 9 7 Excavate Infill Dummy A B D E C F G H I

3.4 _Path _{Task Times (Hours)} _{Total Hours}

1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9 1 – 2 – 3 – 4 – 5 – 6 – 8 – 9 1 – 2 – 4 – 5 – 6 – 7 – 8 – 9 1 – 2 – 4 – 5 – 6 – 8 – 9 20 + 60 + 30 + 20 + 10 + 20 + 0 + 30 20 + 60 + 30 + 20 + 10 + 10 + 30 20 + 100 + 20 + 10 + 20 + 0 + 30 20 + 100 + 20 + 10 + 10 + 30 190 180 200 190

The longest path clearly is 1 – 2 – 4 – 5 – 6 – 7 – 8 – 9; hence, this is the critical path, and the project will end after 200 hours.

Planning 1 Excavate Purchasing 2 LF = 90 LS = 30 EF = 80 ES = 20 Sawing 3 LF = 120 LS = 90 EF = 110 ES = 80 Placement 4 LF = 140 LS = 120 EF = 140 ES = 120 LF = 20 LS = 0 EF = 20 ES = 0 Assembly 5 LF = 150 LS = 140 EF = 150 ES = 140 Outfill 6 8 7 LF = 170 LS = 160 EF = 160 ES = 150 Infill Dummy LF = 170 LS = 150 EF = 170 ES = 150 Decoration 9 LF = 200 LS = 170 EF = 200 ES = 170 LF = 120 LS = 20 EF = 120 ES = 20 A B C D E F _G _H I

Answer : The critical path therefore is A – C – E – F – G – I (200 hours). The activities that can be delayed include ones with slack times > 0. Thus, B (10 hours), D (10 hours), and H (10 hours) can be delayed.

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3.5

( )

(

)

( )

2 2 2 2 2 4

Mean: Variance Standard Deviation

6 6 6 20 120 20 A : 20 A : 11.11 A : 3.33 6 36 6 60 360 60 B : 60 B : 100.00 B : 10.00 6 36 6 120 600 120 C : 100 C : 400.00 C : 20.00 6 36 6 10 180 10 D : 30 D : 2.78 D : 1.67 6 36 6 a+ m b+ b a− b a−

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%

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%

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= = = = = = = = = = = =

( )

2 2 2 2 2 10 120 10 E : 20 E : 2.78 E : 1.67 6 36 6 0 60 0 F : 10 F : 0.00 F : 0.00 6 36 6 40 120 40 G : 20 G : 44.44 G : 6.67 6 36 6 4 60 4 H : 10 H : 0.44 H : 0.67 6 36 6 40 180 40 I : 30 I : 44.44 I : 6.67 6 36 6 = = = = = = = = = = = = = = =

3.6 _{Since the critical path is A – C – E – F – G – I, only those variances are along the critical}

path are used.

Therefore, the variances along critical path are 11.11, 400, 2.78, 0, 44.44, and 44.44 . So the sum of these variances 502.77= .

Thus, the project completion standard deviation = 502.77≅ 22.4 .

µ = mean time of critical path 200 hrs=

22.4 hrs

σ =

The z value 240 200 40 1.8 22.4 22.4

−

= = = _{. Using the cumulative normal distribution table in} Appendix I of the text, we observe that 96.4 percent of the distribution lies to the left of 1.8 standard deviations. Hence, there is a 100 96.4− = 3.6% chance that it will take more than 240 hrs to build the garden/picnic area.

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3.7 _{The critical path is A – C – E – F – G – I. Hence, the project completion variance}

11.11 400 2.78 0 44.44 44.44 502.77. = + + + + + =

So, the project completion standard deviation = 502.77≅ 22.4.

The cumulative normal distribution tells us that 90% of the area lies to the left of 1.29 standard deviations. Therefore, amount of time to build the garden/picnic area should be

(

)

200 22.4 1.29+ = 200 29 + = 229 hours .

3.8 _a. _{Activity on Nodes Diagram of the project.}

A 1 B 1 C 4 E 2 F 2

b. The critical path, listing all critical activities in chronological order:

(

)

A B E F 1 1 2 2 6 A C F 1 4 2 7. This is the CP. not CP → → → + + + = → → + + = c. The project duration (in weeks):

7 (This is the length of CP.)

d. The slack (in weeks) associated with any and all non-critical paths through the project: Look at the paths that aren’t critical—only 1 here—so from above:

A→ → →B E F 7 6 1− = week slack.

3.9 _{We have only 1 activity with probabilistic duration.}

(

)

8 1 4 2 Duedate 1 2 0.5 0.5 Z µ σ − + + −

= = = _{= (length of entire path is 7, not 4). For a}_z = ₂_, this means P

₍

Due date 8<

₎

=97.72% (table lookup) for the path so chance of being OVER 8 weeks is 2.28% (and we know non-CP path will be only 6 weeks)

3.10 _{Helps to modify the AON with the lowest costs to crash}

1. CP is A→ → ; C is cheapest to crash, so take it to 3 wks at $200.C F (and $200 < $250)

2. Now both paths through are critical. We would need to shorten A or F, or shorten C and either B/E. This is not worth it, so we would not bother to crash any further.

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4

CHAPTER

Forecasting

4.1 _Present =_{period week 6.}

(

)

So: ₇ 1 ₆ 1 ₅ 1 ₄ 1 ₃ 1

_{( ) ( ) ( ) ( )}

52 1 63 1 48 1 70 3 4 4 6 3 4 4 6 F = A + A + A + A = + + + =56.75 patients 4.2 1 120 — 2 136 — 3 114 128 4 116 125 t t t A F 120 136 256 128 2 2 _{Checking Data} 136 114 250 125 2 2 +

₊

= =

,,

-+

_,

= =

,.

5 116 114 230 115 Answer 2 2 F = + = = =

4.3 _{Method 1:} _{MAD : 0.20 0.05 0.05 0.20}+ + + = _0.5000← _better

MSE : 0.04 0.0025 0.0025 0.04+ + + = 0.0850 Method 2: MAD : 0.1 0.20 0.10 0.11 0.5100+ + + =

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4.4 _y = +_{a bx} 4 1 2 1 58,538 75.75 191.5 23,209 i i i n i i x y x y x = = = = = =

/

(

)(

)

(

)

(

)

2 58, 538 4 75.75 191.5 513.50 2 256.75 23, 209 4 75.75 191.5 2 75.75 40 40 2 85 210 b a y x x y − = = = − = − = ≈ + =

0

≈ 4.5 t Day Actual Demand Forecast Demand 1 Monday 88 88 2 Tuesday 72 88 3 Wednesday 68 84 4 Thursday 48 80 5 Friday 72 ←Answer

(

)

1 1 t t t F+ = α A + −α F . Let 1 4

α = . Let Monday forecast demand = 88

( ) ( )

2 3 4 5 1 3 88 88 88 4 4 1 3 72 88 18 66 84 4 4 1 3 68 84 17 63 80 4 4 1 3 48 80 12 60 72 4 4 F F F F = + = = + = + = = + = + = = + = + =

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4.6 _{Winter Spring Summer} _Fall 2001 1, 400 1,500 1, 000 600 2002 1, 200 1, 400 2,100 750 2003 1, 000 1,600 2, 000 650 2004 900 1,500 1,900 500 4, 500 6,000 7,000 2,500 Average over all seasons: 20,000 1,250

16 = Average over spring: 6,000 1,500

4 = Spring index: 1,500 1.2 1, 250 =

( )

5,600 Answer : 1.2 1, 680 4

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sailboats

4.7 _{We need to find the smoothing constant}_α_{. We know in general that}

(

)

1 1

t t t

F+ = α A + −α F , 1, 2, 3

t = . Choose either t = or 2 t = (3 t = won’t let us find1 α because

( ) (

)

2 50 50 1 50

F = = α + −α holds for anyα ). Let’s pick, e.g., t = . Then2

( ) (

)

3 48 42 1 50 F = = α + −α . So 48 42 50 50 2 8 1 . 4 α α α α = + − − = − =

Now we can find F :₅ F ₅ = α

( ) (

46 + −1 α

)

50, with 1 4 α = . So

( ) ( )

5 1 3 46 50 49 Answer 4 4 F = + = ←

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4.8 _Let

1, 2, , 6

X X ! X be the prices; Y Y₁, , ₂ !, Y ₆ be the number sold.

6 1 _{Average price} _3.25833 6 i i X X = = = =

/

(1) 6

1 _{Average number sold 550.00}

6 i i Y Y = = = =

/

(2)

All calculations to the 1 nearest th 100,000

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6 1 9,783.00 i i i X Y = =

/

(3) 6 2 1 67.1925 i i X = =

/

(4)

Then y≈ +a bx, where y = number sold , x = price, and

( )( )

( )

(

) (

)( )

(

)

( ) ( )

6 1 6 2 2 2 1 9, 783 6 3.25833 550 969.489 277.61395 3.49222 67.1925 6 3.25833 1,454.5578 i i i i i X Y n X Y b X n X a Y b X = = − − − = = = =− − − = − =

/

So at x =1.80, y = 1,454.5578 277.61395 1.80−

₍

₎

= 954.85270. Now round to the nearest integer: Answer : 955 dinners

4.9 _{Tracking Signal}

(

)

1 MAD n t t t A F = − =

/

Month At F t At −F t

(

At −F t

)

May 100 100 0 0 June 80 104 24 –24 July 110 99 11 11 August 115 101 14 14 September 105 104 1 1 October 110 104 6 6 November 125 105 20 20 December 120 109 11 11 Sum: 87 Sum: 39 So: MAD : 87 10.875 8 = 39 1

Answer : 3.586 to the nearest th 10.875 1,000

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5

CHAPTER

Design of Goods

and Services

5.1 $27,500 $27,500 Use K1 (0.80) (0.20) 90 of 100 non-defect 70 of 100 non-defect $42,500 –$32,500 $4,062.50 (0.85) (0.15) 90 of 100 non-defect 75 of 100 non-defect $12,500 –$43,750 Use K2 $24,375 (0.90) (0.10) 95 of 100 non-defect 80 of 100 non-defect –$18,750 –$75,000 Use K3 Answer: $27,500—use K1 Outcome Calculations

( )( )(

)

( )( )(

)

90 10 $100, 000 500 300 $1.20 500 300 $1.30 100 100 $100, 000 $162, 000 $19, 500 $42, 500

1

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,

2 ,

₋ ₊ ₋ ₌

3 (

)(

)

(

)(

)

70 30 $100, 000 150, 000 $1.20 150, 000 $1.30 100 100 $100, 000 $126, 000 $58, 500 $32, 500

1

₋ ₊ ₋ ₌

,

2 ,

₋ ₊ ₋ ₌₋

3

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(

)(

)

(

)(

)

90 10 $130, 000 150, 000 $1.20 150, 000 $1.30 100 100 $130, 000 $162, 000 $19, 500 $12, 500

1

₋ ₊ ₋ ₌

,

2 ,

₋ ₊ ₋ ₌

3 (

)(

)

(

)(

)

75 25 $130, 000 150, 000 $1.20 150, 000 $1.30 100 100 $130, 000 $35, 000 $48, 750 $43, 750

1

₋ ₊ ₋ ₌

,

2 ,

₋ ₊ ₋ ₌₋

3 (

)(

)

(

)(

)

95 5 $180, 000 150, 000 $1.20 150, 000 $1.30 100 100 $180, 000 $171, 000 $9, 750 $18, 750

1

₋ ₊ ₋ ₌

,

2 ,

₋ ₊ ₋ ₌₋

3 (

)(

)

(

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)

80 20 $180, 000 150, 000 $1.20 150, 000 $1.30 100 100 $180, 000 $144, 000 $39, 000 $75, 000

1

₋ ₊ ₋ ₌

,

2 ,

₋ ₊ ₋ ₌₋

3

5.2 84.0 Use D1 84.0 (0.4) F market 99.0 (0.6) U market 74.0 66.0 (0.3) F market 80.0 (0.7) U market 60.0 80.2 (0.6) F market 89.2 (0.4) U market 66.7 Use D0 Use D2

(All $ figures in millions in tree)

(

)

(

)

(

)

(

)

(

)

(

)

$ Profits : D0 F : 1,000 80, 000 $80, 000, 000 D0 U : 750 80, 000 $60, 000, 000 D1 F : 1, 000 100, 000 1, 000, 000 $99, 000, 000 D1 U : 750 100, 000 1, 000, 000 $74, 000, 000 D2 F : 1, 000 90, 000 800, 000 $89, 200, 000 D2 U : 750 90, 000 800 − = − = − − = − − = − − = − − _{, 000} =_{$66, 700, 000} Answer : Answer: Design D1 has an expected profit of $84,000,000.

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5.3 $14,000 $10,000 (0.3) Demand rises $30,000 –$20,000 Purchase overhead hoist (0.5) Demand stays same (0.2) Demand falls $10,000 $14,000 (0.4) Demand rises $20,000 (0.6) Demand stays same $10,000 Purchase forklift $0 Do nothing

Answer : Maximum expected payoff $14,000=

5.4

Low demand (0.4)

$380,000 Upgrade to D

160K –$50,000

Use A Low demand (0.4)

$300,000 High demand (0.6) 180K $0 $300,000 High demand (0.6) Use B 302K $250,000 High demand (0.6) Use C 380K $0 No upgrade to D Low demand (0.4) $0 Do nothing Note: K = $1,000’s

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5.5

Bread & Rolls Pies & Cakes

Support

Support No support

No support Bread & Rolls

Support

No support Full Service

5.6

Bread & Rolls Pies & Cakes

$15,000

$10,000 Bread & Rolls

Full Service

Support ( = 0.40)

No support ( = 0.60)EMV = $12,000

$25,000

$5,000 Support ( = 0.40)

No support ( = 0.60)EMV = $13,000

$35,000

–$10,000 Support ( = 0.40)

No support ( = 0.60)EMV = $7,500 p p p p p p

Based upon this decision tree, Jeff should consider most seriously the medium-sized shop carrying bread, rolls, pies, and cakes.

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6

CHAPTER

Managing

Quality

6.1 _1. _{Appearance of food} 2. Portion size 3. Lighting 4. Speed of service 5. Knowledge of server 6. Quality of service 7. Appearance of room

8. Appropriate amount of space 9. View of stage and audio

Item Overall Grade

Rated A B C D E 1. 20 28 1 1 0 2. 4 2 30 14 0 3. 19 20 3 8 0 4. 4 5 25 5 11 5. 0 0 27 18 7 6. 9 30 7 0 4 7. 19 18 13 0 0 8. 0 26 24 0 0 9. 0 0 0 20 30 Item Weights

Rated 4 3 2 1 0 Total Average 1. 80 84 2 1 0 167 2.61 2. 16 6 60 14 0 96 1.50 3. 76 60 6 8 0 150 2.34 4. 16 15 50 5 0 86 1.34 5. 0 0 54 18 0 72 1.13 6. 36 90 14 0 0 140 2.19 7. 76 54 26 0 0 156 2.44 8. 0 78 48 0 0 126 1.97 9. 0 0 0 20 0 20 0.31

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c. A check sheet will help categorize the comment cards Check Sheet Positive Negative Appearance of food Portion size ! ! ! ! ! Lighting ! Speed of service ! ! Knowledge of server ! Quality of service ! ! Appearance of room

Appropriate amount of space !

View of stage and audio ! ! ! ! ! ! Other ! ! ! ! ! chilly

d. The written comments are not always consistent: Portion size is highly rated in comments, but 5th in overall grade. View/audio is lowest rated in both.

6.2 _a. 8 9 10 11 12 13 14 1 2 3 4 5 6 x y minutes Trips 0

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6.3 _a. 5 10 15 20 25 30 35 W R I M O 40 0 2 4 6 10 14 24 30 34 36

b. 39% of complaints are W, demeaning towards women.

6.4 mislabeled Manpower Incorrect measurement Operator misreads display Inadequate cleanup Technician calculation off Machines Temperature controls off Variability Antiquated scales Inadequate flow controls Equipment in disrepair Incorrect Formulation Materials Jars Incorrect weights Damaged raw material instructions Methods Lack of clear Priority miscommunication Incorrect maintenance Inadequate instructions

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6

SUPPLEMENT

Statistical Process

Control

S6.1 _{We are given a target of}_{X =} ₄₂₀_{. So} _LCL _{420 4} 25 ₄₀₀

25 X Z n σ

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= −

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= −

_#

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=

%

&

%

&

. 25 UCL 420 4 440 25 X Z n σ

!

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= +

_#

_$

= +

_#

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=

%

&

%

&

. Thus,

Answer : LCL 400 calories UCL 440 calories = = S6.2 7 5 9 250 250 250 7 5 9 300 _0.040 30 7,500 7,500 p = + + "+ = + + "+ = =

(

)

(

)

(

)

(

)

1 UCL 0.040 3 0.01239 0.077 1 LCL 0.040 3 0.01239 0.003 p p p Z n p p p Z n − = + = + = − = − = − =

S6.3 _{We want}_{Z = , since}₂

(

_{1 0.0455}−

)

= _0.9545 _{which implies}_{Z = from the Normal Table.}₂

UCL = +c 2 c, where c =average number of breaks= 3 : 3 2 3+ = 6.46.

S6.4 _Z =₃ _{for -chart}_x _{. Here,}_{n = so}₄

2 0.729

A = (from Table S6.1). x =2.0, R = 0.1,

( )

2 UCL_x = x + A R =2.0 0.729 0.1 + = 2.07 S6.5 _{C chart} 0.0027 1.0000 0.0027 0.9973 0.49865 3 2 2 Z −

= =

0

_{= (see Normal Table)}

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S6.6 _x _Z _answers n σ

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±

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=

%

&

384 16 lbs. 24 0.12 2 0.08 3 16.00 0.08 16.08 UCL 16.00 0.08 15.92 LCL x x x Z n σ = =

!

"

₌

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₌

#

$

#

$

%

&

%

&

+ = = − = = S6.7 _{x =}_1.00_,_R = _0.10_, 2 0.483

A = (from Table S6.1),

(

)( )

2

LCL= x −A R =1− 0.483 0.10 = 0.9517 weeks

S6.8 _R = _3.25 _mph,_{Z = , with}₃ _{n = , from Table S6.1,}₈

UCL 1.864 6.058 LCL 0.136 0.442 R R = = = = S6.9 30 Number of defects 250 1 300 _0.04 30 7,500 i p= = = =

/

, n =250

( )( )

(

)

( )( )

(

)

0.04 0.96 UCL 2 0.04 2 0.0124 0.0648 250 0.04 0.96 LCL 2 0.04 2 0.0124 0.0152 250 p p p p = + = + = = − = − =

S6.10 _a. _{We are counting attributes and we have no idea how many total observations there are}

(the proportion of drivers who weren’t offended enough to call!) This is a C-chart. b. Use mean of 6 weeks of observations 36 6

6 = for c , as true c is unknown.

(

)

UCL= + c z c = + 6 3 2.45 =13.3

(

)

LCL= − c z c = − 6 3 2.45 = −1.3, or 0.

c. It is in control because all weeks’ calls fall within interval of

[

0, 13 .

]

d. Instead of using 36 6

6 = , we now use c = .4 UCL 4 3 4 4 3 2

( )

10 = + = + = _.

( )

(25)

7

CHAPTER

Process

Strategy

7.1 _a. _{Find breakeven points,}

p X . Mass Customization: 1, 260, 000 60+ X =120X → X _p = 21, 000 Intermittent: 1, 000, 000 70+ X =120X → X _p = 20, 000 Repetitive: 1, 625, 000 55+ X =120X → X _p = 25, 000 Continuous: 1,960, 000 50+ X =120X → X _p = 28, 000 b. Find least-cost process at X =24, 000 units.

Fixed cost VC Units

Mass Customization: 1, 260, 000 60 24, 000+

₍

₎

= 2, 700, 000 Intermittent: 1, 000, 000 70 24, 000+

₍

₎

= 2, 680, 000 Repetitive: 1, 625, 000 55 24, 000+

₍

₎

= 2, 945, 000 Continuous: 1, 960, 000 50 24, 000+

(

)

= 3,160, 000 The least-cost process: Intermittent Process.

c.

Anticipated Intermittent Production Process

Volu me Breakeven Point

24,000> 20,000 ? yes!

#$% #$%

Annual Profit Using Intermittent Process: $ 120 24, 000

'

₍

₎

− 2, 680, 000

)

_*

= $200, 000 Answer : The intermittent process will maximize annual profit.

(26)

7.2 _{Use a crossover chart. First graph. Then solve for breakpoint(s).} 5 1 2 3 10 15 20 25 V R MC 0 0 1,000’s of Ovens I P2 P1 Cost (Millions of dollars) Finding value of P2: 1, 250, 000 50 P2+

(

)

= 2, 000, 000 5 P2+

(

)

. So 2 3 P2 16,666 = units. (Note: P1 12,500= ).

Answer : For volumes of production V such that 2 3 16,666 ≤ ≤V 25,000. 7.3 2 4 6 8 10 12 14 5,000 15,000 V Volume 10,000 20,000 I R C I R C Cost (millions) 7,500 0 0

(

)

1, 000, 000 1, 650 3, 000, 000 1, 250 400 2, 000, 000 I&R 5,000 Intersect 1, 000, 000 1, 650 5, 000 $9, 250, 000 x x x x + = +

1 ,

₌

,

2

₌

,

₊ ₌

3 (

)

3, 000, 000 1, 250 7, 500, 000 650 600 4, 500, 000 R&C 7,500 Intersect 3, 000, 000 1, 250 7, 500 $12, 375, 000 x x x x + = +

1 ,

₌

,

2

₌

,

₊ ₌

3

(27)

7.4 _{Breakeven points} a. R: 21, 000, 000 450+ x= 750x

0

x= 70, 000 : 26, 250, 000 400 750 75, 000 C + x= x

0

x= : 15, 000, 000 500 750 60, 000 M + x= x

0

x=

b. Least cost process at x = 65,000 Cost

R: $50,250,000 C : $52,250,000

M : $47,500,000 ← lowest cost with Mass Customization c. 65,000 demand > 60,000 breakeven for M

7.5 _{Breakeven points}

a. Continuous : 2, 400, 000 20+ x= 80x

0

x= 40, 000 Repetitive : 1, 950, 000 30+ x= 80x

0

x= 39, 000

Mass Customization : 1, 480, 000 40+ x =80x

0

x= 37, 000 Intermittent : 1,800, 000 40+ x=80x

0

x= 45, 000

b. Least cost process at x = 48,000 Continuous: $3,360,000 ← least cost Repetitive: $3,390,000

Mass Customization: $3,400,000 Intermittent: $3,720,000

c. Is 48,000 > 40,000? Yes, so we use continuous process. Annual profit= $480,000

7.6 4,000 11,000 15,000Volume I R C $ 2,000 M (11,000; 1,350,000) (4,000; 860,000) (2,000; 300,000) widest

Repetitive has the widest production volume range over which it is a least-cost process.

7.7 _{Total profit now:}

Profit =40, 000 2.00 20, 000 40, 000 0.75 80, 000 20, 000 30, 000× − − × = − − = 30, 000 Total profit with new machine:

Profit =50, 000 2.00 2, 000 50, 000 1.25 100, 000 25, 000 62, 500 12, 500× − − × = − − = Since profit decreases with the new piece of equipment added to the line, purchase of the machine probably would not be a good investment.

(28)

7

SUPPLEMENT

Capacity

Planning

S7.1 _{Problem is under risk and has two decisions, so use a decision tree:}

109 High demand 70 135 (0.6) Medium demand (0.4)

No additional expansion ₉₀

135 Additional minor expansion

148 High demand (0.6) Medium demand (0.4) 40 220 148 Small expansion Large expansion

(Payoffs and Expected Payoffs are in $1,000’s)

Answer : Ralph should undertake a large expansion. Then the annual expected profit will equal $148,000.

(29)

S7.2 High demand 140,000 (0.3) No additional $90,000 $140,000 70,000 $70,000 Small expansion Large expansion minor expansion Additional minor expansion Medium demand (0.7) $40,000 High demand (0.3) 14,000 Medium demand (0.7) $105,000 –$25,000 Maximum value = $70,000 S7.3 $18,000 Small expand Demand up small (0.4) 16,000 Demand up medium (0.6) $10,000 $20,000 $0 No expand Demand up medium (0.3) 18,000 Demand up large (0.7) –$10,000 $34,000 Large expand Answer : $18,000

(30)

S7.4 (1) 50,000 100,000 200,000 300,000 400 x 1,000 2,000 (1) 250 (2) (1) (2) (3) (3) (2)

Cap level (2) is lowest for all x so 1, 000≤ ≤ x 2, 000

S7.5 _{Actual (or expected)}_{Output =} Effective Capacity

Efficiency (text Equation S7-3)

4.8 cars = 5.5 cars × 0.880. Therefore in one 8 hour day one bay accommodates 38.4 cars =

(

8 hrs 4.8 cars per hr ×

)

and to do 200 cars per day requires 5.25 bays or 6 bays = 200 cars

38.4 cars per bay

!

"

#

$

%

&

S7.6 _a.

_{( )}

(

)

450 BEP $ 878.05 0.5125 1 _i _i _i F V P W = = =

'

−

)

(

*

/

Breakeven

_{( )}

$ =$878.05

b. Number of pizzas required at breakeven:

Whole pizzas =

(

878.05 0.30 5.00×

)

= 52.7 → 53 Slices =

(

878.05 0.05 0.75 58.5×

)

= →59

Whole pizzas to make slices 59 6= = 9.8→10

Therefore, he needs a total of 63 pizzas. He does not have sufficient capacity.

S7.7 _a. _{Remember that Yr 0 has no discounting.}

Initial coat $1,000,0000 yearly maint 75,000 members dues/member Salvage cost $50,000 yearly dues $300,000 500 $600 Discount rate 0.100

Year Cost Revenues Profit PV Mult PV Profit 0 $1,075,000 $300,000 $–775,000 1 –$775,000 1 75,000 300,000 225,000 0.9 $202,500 2 75,000 300,000 225,000 0.81 $182,250 3 75,000 300,000 225,000 0.729 $164,025 4 75,000 300,000 225,000 0.6561 $147,623

(31)

Assume dues are collected at the beginning of each year. This is a simplification—in reality, people are likely to join throughout the year. (Technically, if equipment is sold at the end of year 5, it should probably appear as a final revenue stream in year 6 but the difference is only $2,952.45.

b. Special deal comparison: $3,000 for all 6 years. Compare the PV cash stream of yearly dues from one member to that of the deal. Since we specified the club will always be full, we can make the assumption that the member (or her replacement) will always be paying the annual fee.

Initial cost $0 yearly maint $0 Salvage cost $0 yearly dues $600 Discount rate 0.100

(Membership fee)

Year Cost Revenues Profit PV Mult PV Profit 0 $0 $600 $600 1 $600 1 0 600 600 0.9 $540 2 0 600 600 0.81 $486 3 0 600 600 0.729 $437 4 0 600 600 0.6561 $394 5 0 600 600 0.59049 $354 undisc. Profit 3,600 PV Profit $2,811 Since this is less than $3K, the special deal is worth more to the Health Club. Note also: If Health Club member is using same discount rates, it’s better for her to pay yearly.

S7.8_Breakeven:

Costs = Revenues

500 0.50+ × = ×b b 0.75 where b = number of units at breakeven or b

₍

0.75 0.50−

₎

=500, and 500 2,000 units

0.25 b = =

a. breakeven in units = 2,000 units

(32)

8

CHAPTER

Location

Strategies

8.1

(

)( ) (

)( )

(

)( ) (

)( )

2, 000 2.5 5, 000 2.5 10, 000 5.5 7, 000 5.0 10, 000 8.0 20, 000 7.0 14, 000 9.0 6.67 2, 000 5, 000 10, 000 7, 000 10, 000 20, 000 14, 000 x C + + + + + + = = + + + + + +

(

)( ) (

)( )

(

)( ) (

)( )

2, 000 4.5 5, 000 2.5 10, 000 4.5 7, 000 2.0 10, 000 5.0 20, 000 2.0 14, 000 2.5 3.02 68,000 y C + + + + + + = =

8.2 _Site _{Total Weighted Score}

A 174 B 185 C 187 D 165 8.3

/

_{Population weights}=_{5, 000}

( )(

0 2, 050

) ( )( ) ( )(

1 550 2 1, 025

) ( )( ) ( )( ) ( )( )

3 775 2 250 2 350 0.525 5,000 x C = + − + + + − + − =

( )(

0 2, 050

) ( )( ) ( )(

1.5 550 1 1, 025

) ( )( ) ( )( ) ( )( )

3 775 3 250 1 350 0.205 5,000 y C = + + − + + − + − = Coordinates:

₍

0.525, 0.205

₎

8.4 25 50 75 100 125 150 1 2 3 4 5 6 V $ cost (millions) (2, 60) (1, 35) (1/2, 30) (0, 25) (0, 20) (0, 10) 10,000’s of Autos = V 0 A B C

(33)

8.5 _Site _Score

A 5w +320 B 4w +330 C 3w +370 D 5w +255

Find all w from 1–30 so that:

3 370 5 320 50 2 25 3 370 4 330 40 40 3 370 5 255 115 2 57.5 w w w w w w w w w w w w + ≥ + ⇔ ≥ ⇔ ≤ + ≥ + ⇔ ≥ ⇔ ≤ + ≥ + ⇔ ≥ ⇔ ≤ Answer : For all w such that 1.0≤ ≤w 25.0

8.6 2 4 6 8 10 12 5 10 15 20 25 30 V TC (millions $) (1, 35) (30, 9.5) (20, 7) 35 (thousands) L.A. K.C. Charlotte Char K.C. L.A. cutoff 0 4,100, 000 180 1, 000, 000 300 For all so: 4,100, 000 180 2, 000, 000 250

0 35, 000 3,100,000 120 2,100, 000 70 0 35, 000 1 25,833 Answer : 30, 000 35, 000 3 30,000 V V V V V V V V V V V V + ≤ +

1 ,

₊ _≤ ₊

2 ,

_{≤ ≤}

3

≤

1 ,

⇔

₂

≤

,

_{≤ ≤}

3

1

_≤

,

⇔

₂

0

≤ ≤

,

_≤

3

(34)

8.7

/

_{weights 14,000}= _{(or 14 for calculations below)}

( )( ) ( )( ) ( )( )

2 3.5 8 7 6 3.5 6.0 14 x C = + + =

( )( ) ( )( ) ( )( )

6 3.5 1 7 2 3.5 2.5 14 y C = + + =

(35)

9

CHAPTER

Layout

Strategy

9.1 _a. _Cycle_time

( )(

60 60 sec

)

3,600 ₂₀_sec_per_PLA

180 PLAs 180 = = =

b. Theoretical number of work stations task time 60 3 cycle time 20 =

/

= = c. Yes, it is feasible. 9.2 1 2 3 4 D A C B Department pair

( )

( )( )

1 2 1 2 1 2 1 2 1 2 1 2 Weekly $ Cost : 8 800 3, 200 : 6 700 2,100 : 4 400 800 : 10 300 1,500 : 7 200 700 : 9 600 2, 700 $11,000 AB AC AD BC BD CD = = = = = =

(36)

9.3 _a.

(

)

(

)

274 seconds Cycle time Cycle time seconds

i t n =

/

=

( )

(

)

60 60 seconds

C.T. 60 seconds per truck 60 trucks

= = _so 274 _4.5667 ₅ 60

n = = → = .n Answer : 5

b. Steps 1 and 2: Sample Answer 60 seconds

From (a) number of stations is at least 5

c =

+

-.

Precedence diagram: A 40 B 30 D 40 E 6 H 20 C 50 F 25 G 15 I 18 J 30 Step 3 Task Number of Successors Task Number of Successors A 9 F 2 B 4 G 2 C 4 H 1 D 2 I 1 E 2 J 0

(37)

Step 4

Available

and Fit Assigned Station 1 A A A

B, C — —

Station 2 B, C B, C C (Broke a tie) B, F, G — —

Station 3 B, F, G B, F, G B

D, E, F, G E, F, G F (Broke ties) D, E, G — —

Station 4 D, E, G D, E, G D (Broke ties) E, G E, G G (Broke a tie) E, I — — Station 5 E, I E, I E I, H I, H H (Broke a tie) I I I J — — Station 6 J J J Answer : Station Tasks

(Other answers possible, 1 A depending upon how ties 2 C are broken in above 3 B, F

procedure) 4 D, G 5 E, H, I 6 J c. n = 6 work stations are in our answer.

( ) ( )

274 Efficiency 0.7611 C.T. 6 60 i t n =

/

= =

9.4 _a. _{First assignment costs}

(

_{8, 000 7, 200 1, 600 4,800 8, 000 800}

)

1

2 $15,200

= + + + + + × =

b. New layout costs

₍

8, 000 9, 600 6, 400 3, 600 2, 000 800

₎

1 $15, 200 2

= + + + + + × = No improvement—both yield the same cost.

(38)

9.5 _{Cost of 3 attempts:} Work Area Attempt 1 2 3 Cost a. 1 S D M $1,088 b. 2 D S M $1,142 c. 3 M D S $1,100 ↑

( )( ) ( )( ) ( )( )

2 23 10

'

₍

+ 32 5 + 20 8

)

_*

= $1,100

9.6 _a. _{Theoretical minimum number of stations} task times

cycle time =

/

Cycle time 60 12 minutes 5

= = _{. So minimum number of stations} 48 _{4 stations} 12 = = b. A 10 min WS #1 B 12 min WS #2 C 8 min WS #3 D 6 min WS #4 E F 12 min WS #5

This requires 5 stations—it cannot be done with 4. c. Efficiency 48 48 80%

5 12 60 = = =

× for 5 stations.

9.7 _{There are three alternatives:}

Station Alternative 1 Tasks Alternative 2 Tasks Alternative 3 Tasks 1 A, B, F A, B A, F, G

2 C, D C, D H, B

3 E G, H C, D

4 G, H E E

5 I I I

(39)

10

SUPPLEMENT

Work

Measurement

S10.1 _{Required sample size}

2 Zs n hx

!

"

= =

_#

_$

%

&

where s = 0.15, x = 0.4, z =1.96 (for 95% confidence),

10% h = accuracy level

( )( )

2 1.96 0.15 54.0225 54 0.10 0.4 n=

'

4 )

5

= ≈

4

5 (

*

S10.2 _a. 2 Zs n hx

!

"

=

_#

_$

%

&

Thus,

( )( )

0.10 0.40

( )

12 0.924 0.15 hx n Z s = = =

Referring to Appendix I (Standard Normal Table), Area= 0.64= 64%. The confidence level when n =12 is 64%, as opposed to 95% when n = 54 (in Problem S10.1)

b. Average observed time 0.331 0.243 0.484 0.4484 minutes 12

+ + +

= " =

Normal time = Average time × perf. rating = 0.4484 0.90× = 0.4036 minutes Standard time Normal time 0.4036 0.429 minutes

1 allowance factor 1 0.06

= = =

− −

S10.3 _{Average observed time} 100 hours 60 minutes 0.75 _{22.5 minutes}

200 units × ×

= =

Normal time = 22.5 minutes 1.1 24.75 minutes× =

Standard time for job Normal time for process 24.75 29.12 minute unit 1 Allowance fraction 1 0.15

= = =

− −

S10.4 _a. _{Observed time} sum of times 1.74 _{0.10875 minutes} _{6.525 seconds}

number of cycles 16

= = = =

b. Normal time

(

Observed time

)

(

Performance rating factor

)

6.525 95% 6.2 seconds

= × = ×

=

c. Standard time normal time 6.2 6.2 6.739 seconds 1 allowance factor 1 8% 92%

= = = =

(40)

S10.5 _{Normal time 10 minutes 0.90}= × = _{9 minutes}

Allowance fraction Personal Fatigue Delay 5 3 1 9 0.15 60 minutes 60 60

+ + + +

= = = =

Normal time 9

Standard time 10.59 minutes 1 Allowance fraction 1 0.15

= = =

− −

S10.6 _{Observation (Minutes Per Cycle)}

Element Rating 1 2 3 4 5 Average Time Normal Time 1 100% 1.5 1.6 1.4 1.5 1.5 1.5 1.50 2 90% 2.3 2.5 2.1 2.2 2.4 2.3 2.07 3 120% 1.7 1.9 1.9 1.4 1.6 1.7 2.04 4 100% 3.5 3.6 3.6 3.6 3.2 3.5 3.50 Normal time for lab test = 9.11

Standard time for lab test Normal time for process 9.11 11.1 minutes 1 Allowance fraction 1 0.18

= = =

(41)

12

CHAPTER

Inventory

Management

12.1 _{An ABC system classifies the top 70% of dollar volume items as A, the next 20% as B, and}

the remaining 10% as C items. Similarly, A items constitute 20% of total number of items, B items are 30%; and C items are 50%.

Item Code Number Average Dollar Volume Percent of Total $ Volume 1289 400 3.75 2347 300 4.00 2349 120 2.50 2363 75 1.50 2394 60 1.75 2395 30 2.00 6782 20 1.15 7844 12 2.05 8210 8 1.80 8310 7 2.00 9111 6 → × = → × = → × = → × = → × = → × = → × = → × = → × = → × = → × 1, 500.00 44.0% 1, 200.00 36.0% 300.00 9.0% 112.50 3.3% 105.00 3.1% 60.00 1.8% 23.00 0.7% 24.60 0.7% 14.40 0.4% 14.00 0.4% 3.00 18.00 0.5% $3, 371.50 100% =

Answer : The company can make the following classification: A: 1289, 2347

B: 2349, 2363, 2394, 2395

C: 6782, 7844, 8210, 8310, 9111

12.2 _{D (Annual Demand) = 4,800 units, P (Purchase Price/Unit) = $27, H (Holding Cost) = $2}

S (Ordering Cost) = $30. So,

(

Order Quantity

)

240 2 2 4,800 30

(

)( )

2 DS Q H ∗ ₌ ₌ ₌ .

(

)

(

)

2 240 30 4,800

Thus, Total Annual Cost 4,800 27

2 2 2 240 129, 600 240 600 $130, 440 HQ SD TC = PD + + = × +

!

#

×

"

$

+

!

#

×

"

$

%

&

%

&

= + + =

(42)

12.3 _{D (Annual Demand) = 14,558, P (Purchase Price/Unit) = $5, H (Holding Cost/Unit) = $4,} S (Ordering Cost/Order) = $22, 2 2 14, 558 22 400 4 DS Q H × ×

= = = _{tons per order}

(

5 14,558

)

4 400 22 14,558 72, 790 800 800.69 2 2 2 400 $74,390.69 HQ SD TC = PD + + = × +

!

_#

×

"

_$

+

!

_#

×

"

_$

= + +

%

&

%

&

=

Answer : The optimal order quantity

_{( )}

Q =400 tons order ; total annual inventory cost

( )

TC =$74,391.

12.4 _{D (Annual Demand) = 400 × 12 = 4,800, P (Purchase Price/Unit) $350 unit}= _,

H (Holding Cost/Unit) $35 unit year = , S (Ordering Cost/Order) $120 order = . So, 2 2 4,800 120 181.42 181 35 DS Q H × ×

= = = = _{units (rounded off).}

(

)

(

)

35 181 120 4,800

Thus, Total Cost 4,800 325

2 2 181 1, 560, 000 3,168 3,182 $1, 566, 350 HQ SD TC PD Q × ×

!

"

!

"

= + + = × +

_#

_$

+

_#

_$

%

&

%

&

= + + = However, if Bell Computers orders 200 units,

(

4,800 325

)

35 200 120 4,800 1, 440, 000 3,500 2,880 $1, 446,380 2 200

TC = × +

!

_#

×

"

_$

+

!

_#

×

"

_$

= + + =

%

&

%

&

Answer : Bell Computers should order 200 units for a minimum total cost of $1,446,380.

12.5 1 2 2 4,800 120 181 units 35 DS Q H × × = = = 2 2 2 4,800 120 188 units 32.5 DS Q H × × = = = 3 2 2 4,800 120 196 units 30 DS Q H × × = = =

181 units cannot be bought at $350, hence that isn’t feasible. 196 units cannot be bought at $300, hence that isn’t possible either. So, EOQ =188 units.

(

)

(

)

32.5 188 120 4,800 Thus, 188 units 325 4,800 2 2 188 1, 560, 000 3, 055 3, 064 $1, 566,119 HQ SD TC PD Q × ×

!

"

!

"

= + + = × +

_#

_$

+

_#

_$

%

&

%

&

= + + =

(

200 units

)

(

300 4,800

)

30 200 120 4,800 2 2 200 1, 440, 000 3, 000 2, 880 $1, 445, 880 HQ SD TC PD Q × ×

!

"

!

"

= + + = × +

_#

_$

+

_#

_$

%

&

%

&

= + + =

(43)

12.6 _D =_{12,500 year}_{, so}_d =_{50 day}_,_p =_{300 day}_,_S =_{$30 order}_,_H =_{$2 unit year} a. 2 2 12, 500 30 300 612.37 1.095 671 2 300 50 DS p Q H p d × × = × = × = × = − −

b. Number of production runs

_{( )}

12,500 18.63 671

D N

Q

= = =

c. Maximum inventory level

(

_max

)

1 671 1 50 671 1 1 559 300 6 d I Q p

!

"

_!

_"

_!

_"

= −

_#

=

_$

_#

−

_$

=

_#

−

_$

=

%

&

%

&

%

&

d. Days of demand satisfied by each production run 250 13.42 18.63

= = _{days in demand} only mode

Time in production for each order 671 2.24 300

Q p

= = = _{days in production for each} order. Total time =13.42 days per cycle.

Thus, percent of time in production 2.24 16.7% 13.42

= = _.

12.7 _{H =}_{$2 unit year}_,_S = _{$10 order}_, _{4, 000} 4,000 ₁₆

250 D=

!

#

d = =

"

$

%

&

. So, 2 2 4, 000 10 200 2 DS Q H × ×

= = = _{. ROP (reorder point)} _{= × , l (lead time = 5 days).}_{l d} Thus, ROP= × 5 16=80 units.

Answer : Saveola, Inc. should place an order for 200 frames every time the inventory of frames falls to 80 units. This will be their inventory policy.

12.8 ₃₀₀ 2 5, 400 34

(

)

H

= _{; Square both sides} _90,000 367,200 H = _, 367,200 _$4.08 90,000 H = = 12.9 _a. 2 2 6, 000 30

(

)( )

_{189.74 units} 10 DS EOQ H = = = b. Average inventory = 94.87

c. Optimal number of orders/year = 31.62

d. Optimal days between orders 250 7.91 31.62

= =

e. Total annual inventory cost 601,897.37= (including the $600,000 cost of goods)

12.10 _a. _{Holding cost = $530.33}

b. Set up cost = $530.33 c. Unit costs = $56,250.00 d. Total costs = $57,310.66

e. Order quantity = 16,970.56 units

(44)

12.11 Inventory Item $Value per Case #Ordered per Week Total $ Value/Week (52 Weeks) Total ($*Weeks) Rank Percent of Inventory Cumulative Percent of Inventory Fish Fillets 143 10 $1,430.00 $74,360.00 1 17.54% 34.43% French Fries 43 32 $1,376.00 $71,552.00 2 16.88% 47.31% Chickens 75 14 $1,050.00 $54,600.00 3 12.88% 59.53% Prime Rib 166 6 $996.00 $51,792.00 4 12.22% 69.83% Lettuce (case) 35 24 $840.00 $43,680.00 5 10.31% 78.85% Lobster Tail 245 3 $735.00 $38,220.00 6 9.02% 83.82% Rib Eye Steak 135 3 $405.00 $21,060.00 7 4.97% 87.25% Bacon 56 5 $280.00 $14,560.00 8 3.44% 90.64% Pasta 23 12 $276.00 $14,352.00 9 3.39% 93.74% Tomato Sauce 23 11 $253.00 $13,156.00 10 3.10% 95.71% Table Cloths 32 5 $160.00 $8,320.00 11 1.96% 97.60% Eggs (case) 22 7 $154.00 $8,008.00 12 1.89% 98.28% Oil 28 2 $56.00 $2,912.00 13 0.69% 98.72% Trash Can Liners 12 3 $36.00 $1,872.00 14 0.44% 99.13% Garlic Powder 11 3 $33.00 $1,716.00 15 0.40% 99.42% Napkins 12 2 $24.00 $1,248.00 16 0.29% 99.72% Order Pads 12 2 $24.00 $1,248.00 17 0.29% 99.83% Pepper 3 3 $9.00 $468.00 18 0.11% 99.93% Sugar 4 2 $8.00 $416.00 19 0.10% 99.93% Salt 3 2 $6.00 $312.00 20 0.07% 100.0% $8,151.00 $423,852.00 100.00% a. Fish filets total $74,360

b. C items are items 10 through 20 in the above list (although this can be one or two items more or less)

c. Total annual $ volume = $423,852

12.12 _{Incremental Costs}

Safety Stock Carrying Cost Stockout Cost Total Cost 0 0 70(100 × 0.4 + 200 × 0.2) = 5,600 5,600 100 100 × 15 = 1,500 (100 × 0.2) × (70) = 1,400 2,900 200 200 × 15 = 3,000 0 3,000

The safety stock which minimizes total incremental cost is 100 kilos. The re-order point then becomes 200 kilos + 100 kilos or, 300 kilos.

(45)

12.13 S=$10 order _,_LT=_{4 days}_,_d= _{80 day}_, σ =₂₀_{, H 10%}= _{of $250,}

(

for term

)

80 100 8, 000 D = × =

a. Q 2 DS 800 calzones H

= = _{. Order every}Q _{days 10 days} d

=

b. ROP for calzone, σ

₍

demand during lead time

₎

= LT × σ = 40_,_Z=₁_{(from Table)} for 0.1587, ROP= × d LT + × Z σ _dLT = 360

c. On hand=_{85, days left} =_{1. Since ROP}

dalt d LT Z σ = × + × _, _Z ROP d LT LT σ − × = _. Thus, Z of 0.25 gives 0.5987. So there is a 40.13% chance of stockout.

d. 400 average inventory 2 Q = = _, D _{10 orders term} Q = _.

( )( )

Holding cost term 400 0.25 $100 2

Q H

= = =

( )( )

Order cost term S D 10 10 $100 Q

= = =

12.14 S =$16_,_H=_{$0.40 calzone term}_, _p =₁₆₀_, _d=₈₀_,_D =_8,000_{for 100 days of the term} a.

(

)

2 1,131.37 1 DS Q H d p = = − b. Cycle 1,131.37 14.14 80 Q d = = =

c. Run production for days 1,131 7.07 days 160

Q p

(46)

12.15 Under present price of $6.40 per box:

Economic Order Quantity: 2 2 5000 25 395.3 or 395 boxes 0.25 6.40 DS Q H ∗ × × = = = × where D =_{annual demand, S}=_{set-up or order cost, H}=_{holding cost}

Total cost order cost holding cost purchase cost

2 5, 000 25 395 0.25 6.40 6.4 5, 000 316.46 316.00 32, 000 395 2 $32,632.46 DS QH CD Q = + + = + + × × × = + + × = + + =

Note: Order and carrying costs are not exactly equal due to rounding of the EOQ to a whole number. Under the quantity discount price of $6.00 per box:

Total cost order cost holding cost purchase cost

2 5, 000 25 5, 000 0.25 6.00 5, 000 6.00 41.67 3, 750.00 30, 000 3, 000 2 $33,791.67 DS QH Q = + + = + × × × = + + × = + + =

Therefore, the old supplier with whom they would incur a total cost of $32,632.46, is preferable.

12.16 Economic Order Quantity, non-instantaneous delivery:

where: D = period demand, S = set-up or order cost, H = holding cost, d = daily demand rate, p = daily production rate

(

)

(

50

)

200 2 2 10, 000 200 2,309.4 1.00 1 1 d p DS Q H ∗ × × = = = − − or 2,309 units

(47)

13

CHAPTER

Aggregate

Planning

13.1 The total production required over the year is 8,400 units, of 700 per month. Thus, the schedule is to produce 700 per month and have no costs associated with work force

variation. The only costs incurred will be the monthly production cost, the inventory cost, and the shortage cost. The costs are calculated as follows.

Month Beginning Inventory Produc-tion Production Cost Demand Ending Inventory Shortage Inventory Cost Shortage Cost January 0 700 $49,000 500 200 0 $600 $0 February 200 700 49,000 600 300 0 900 0 March 300 700 49,000 600 400 0 1,200 0 April 400 700 49,000 700 400 0 1,200 0 May 400 700 49,000 700 400 0 1,200 0 June 400 700 49,000 800 300 0 900 0 July 300 700 49,000 900 100 0 300 0 August 100 700 49,000 900 0 100 0 1,000 September 0 700 49,000 800 0 200 0 2,000 October 0 700 49,000 700 0 200 0 2,000 November 0 700 49,000 600 0 100 0 1,000 December 0 700 49,000 600 0 0 $ 0 0 8,400 $588,000 8,400 2,100 600 $ 6,300 $6,000

(48)

13.2 a. Total hotel demand for the year =_7,000,000 Total restaurant demand for the year =_2,080,000 Level staffing Quarter Hotel (Room) Demand Personnel Required Restaurant

Demand Req. Total Hire Terminate Winter 800,000 33 160,000 12 45 45 – Spring 2,200,000 33 800,000 12 45 – – Summer 3,300,000 33 960,000 12 45 – – Fall 700,000 33 160,000 12 45 – – Totals 7,000,000 132 2,080,000 48 180 45

Personal cost=_{180 quarters of labor}×_5,000= _$900,000 Hiring cost = 45 hires at $1,000 = 45,000 Termination cost = none = 0 $945,000 b. Total hotel demand for the year = 7,000,000

Total restaurant demand for the year = 2,080,000 Chase staffing (staffing to meet the forecasted demand)

Personnel Personnel Totals

Quarter

Hotel (Room)

Demand Req. Hire

Termi-nate

Restaurant

Demand Req. Hire

Termi-nate Quarter Hires Quarter Termi-nations Winter 800,000 8 8 160,000 2 2 10 Spring 2,200,000 22 14 800,000 10 8 22 Summer 3,300,000 33 11 960,000 12 2 13 Fall 700,000 7 0 26 160,000 2 0 10 36 Totals 7,000,000 70 2,080,000 26 45 36

Personnel cost = 96 quarters of labor ×_{5,000 =} _$480,000 Hiring cost = 45 hires @ $1,000 = 45,000 Termination cost = 36 terminations @ $2,000 = 72,000 $597,000

(49)

13.3 Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000

Hiring from local staffing agency all personnel above base requirements.

Quarter

Hotel (Room)

Demand Req. Hire

Personnel from Agency

Restaurant

Demand Req. Hire

Personnel from Agency Quarter Hires Quarter from Agency Fall 800,000 8 7 1 160,000 2 2 0 2 1 Spring 2,200,000 22 0 15 800,000 10 0 8 0 23 Summer 3,300,000 33 0 26 960,000 12 0 10 0 36 Fall 700,000 7 0 0 160,000 2 0 0 0 0 Totals 7,000,000 28* 42 2,080,000 26 8** 18 35 60

*On Hotel Grand payroll (7 each quarter×_{4 quarters = 28)} ** On Hotel Grand payroll (2 each quarter×_{4 quarters = 8)} Total on Grand Hotel payroll = 28+ 8 = 36

Personnel cost = 36 quarters of labor× _{5,000 =} _$180,000

Hiring cost = 9

₍

= 7 2+

₎

_{hires @ $1,000 =} _9,000 Termination cost = 0 terminations @ $2,000 = 0

Staffing agency costs = 60 @ 6,500 = 390,000 $579,000

13.4 Plan A:

Month Demand Production Hire Fire Extra Cost Mar 1,000 900 700 56,000 Apr 1,200 1,200 300 12,000 May 1,400 1,400 200 8,000 June 1,200 1,200 200 8,000 July 1,500 1,500 300 12,000 Aug 1,300 1,300 200 16,000 Total extra cost: $112,000

Plan B:

Month Demand Production Inventory Sub-Contracting Extra Cost Mar 1,000 1,100 200 2,000 Apr 1,200 1,100 100 1,000 May 1,400 1,100 200 8,000 June 1,200 1,100 100 4,000 July 1,500 1,100 400 16,000 Aug 1,300 1,100 200 8,000 Total extra cost: $39,000

(50)

13.5 Plan Number 1:

Month Demand Production Inventory Sub-Contracting Extra Cost 1 1,000 1,200 300 3,000 2 1,200 1,200 300 3,000 3 1,400 1,200 100 1,000 4 1,200 1,200 100 1,000 5 1,500 1,200 200 8,000 6 1,300 1,200 100 4,000 Total extra cost: $112,000

Plan Number 2:

Month Demand Production Inventory Overtime Extra Cost 1 1,000 1,200 300 3,000 2 1,200 1,200 300 3,000 3 1,400 1,200 100 1,000 4 1,200 1,200 100 1,000 5 1,500 1,200 200 2,000 6 1,300 1,200 100 1,000 Total extra cost: $39,000

Therefore, Plan Number 2 would be preferred.

13.6 Plan Y:

Month Demand Production Hire Fire Extra Cost 1 1,100 1,100 400 32,000 2 1,600 1,600 500 20,000 3 2,200 2,200 600 24,000 4 2,100 2,100 100 8,000 5 1,800 1,800 300 24,000 6 1,900 1,900 100 4,000 Total extra cost: $112,000

Plan Z:

Month Demand Production Inventory Sub-Contracting Extra Cost 1 1,100 1,600 600 6,000 2 1,600 1,600 600 6,000 3 2,200 1,600 4 2,100 1,600 500 20,000 5 1,800 1,600 200 8,000 6 1,900 1,600 300 12,000 Total extra cost: $52,000

(51)

14

CHAPTER

Materials Requirements

Planning (MRP) & ERP

14.1 Requirement for 3,500 “Get Well” bud vases. 3,500 Vases

3,500×_{8" white ribbons (2,333 ft.)} 3,500×_{8" red ribbons (2,333 ft.)} 3,500 signature cards

7,000 sprigs of baby’s breath 7,000 pink roses

14.2 _Lot Period (week)

Size Lead Time On Hand Safety Stock Allo-cated Low-Level Code Item ID CD Case ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈ Gross Requirements 650 300 550 400 500 Scheduled Receipts Projected On Hand 1,000 1,000 1,000 1,000 350 50 Net Requirements 500 400 500

Planned Order Receipts 500 400 500

Lot for Lot 1 1,000 — — 0 CD Case

Planned Order Releases 500 400 500

Gross Requirements 500 400 500

Scheduled Receipts Projected On Hand

Net Requirements 500 400 500

Lot for Lot 1 — — — 1 CD Top

Lot for Lot 1 — — — 1 CD Bottom

Lot for Lot 1 — — — 1 CD Insert

Scheduled Receipts

Projected On Hand 100 100 100 100

Lot for Lot

2 — — — 2 Black

Dye

(52)

14.3 _Lot Period (day) Size Lead Time On Hand Safety Stock Allo-cated Low-Level Code Item

ID Ball Point Pens ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈

Gross Requirements 10,000

Scheduled Receipts —

Projected On Hand —

Net Requirements 10,000

Planned Order Receipts 10,000

Lot for Lot

1 — — — 0

Planned Order Releases 10,000 10,000

Scheduled Receipts Projected On Hand Net Requirements

Lot for Lot

1 — — — 1 Cap

Planned Order Releases 10,000

Scheduled Receipts —

Projected On Hand —

Lot for Lot

3 — — — 1 Ink CC

Lot for Lot

1 — — — Body

Lot for Lot 1 — — — Clip Fine Pt

Scheduled Receipts

Projected On Hand 3,000 3,000

Lot for Lot 1 3,000 — — Std. Clip

Scheduled Receipts

Projected On Hand 3,000 3,000

Lot for Lot 1 3,000 — — Std. Ball Point

Lot for Lot 1 — — — Fine Point Ball Point

(53)

14.4 _Lot Period (week, day) Size Lead Time On Hand Safety Stock Allo-cated Low-Level Code Item

ID Ball Point Pens ₁ ₂ ₃ ₄ ₅ ₆ ₇ ₈

Gross Requirements 640 640 128 128

Net Requirements 640 640 128 128

Planned Order Receipts 640 640 128 128

Lot for Lot 1 — — — 0 Coffee Table

Planned Order Releases 640 640 128 128

Lot for Lot

1 — — — 1 Top

Lot for Lot 1 — — — 1 Stain gal

Scheduled Receipts

Projected On Hand 100 100 100 100 100 60 20 12 4 Net Requirements

Planned Order Receipts Lot for Lot 1 — — — 1 Glue gal

Planned Order Releases

Lot for Lot

1 — — — 1 Base

Lot for Lot 1 — — — 2 Long Braces

Lot for Lot 1 — — — 2 Short Braces

Lot for Lot

1 — — — 2 Leg

Scheduled Receipts

Projected On Hand 880 880 880 880 880 880

Lot for Lot 1 — — — 3 Brass Caps

(54)

14.5 Coffee Table Hours Lead

Master Schedule Required Time Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Day 7 Day 8

640 640 128 128 Table Assembly 2 1 1,280 1,280 256 256 Top Preparation 2 1 1,280 1,280 256 256 Assemble Base 1 1 640 640 128 128 Long Braces (2) 0.25 1 320 320 64 64 Short Braces (2) 0.25 1 320 320 64 64 Legs (4) 0.25 1 640 640 128 128 Total Hours 0 1280 3,200 3,456 1,920 640 256

Employees needed @ 8 hrs. each 0 160 400 432 240 80 32

14.6 The following table lists the components used in assembling FG-A. Also included for each component are the following information: the on-hand supply, lead time, and direct

components.

Item On-Hand LT (weeks) Components

PG-A 0 1 SA-B, SA-C(2), SA-D(2) SA-B 0 1 SA-D(2)

SA-C 0 2 E, F(2) SA-D 0 2 E (3) E 10 1 — F 5 3 —

While not required as part of the question, we recommend you make a product structure tree to help you answer the questions on the Bill-of-Materials and lead time.

1 FGA [LT =1] [LT =1] SA-B [LT =2] SA-D(2) E(3) [LT = 1] [LT =2] SA-C(2) E [LT = 1] F(2) [LT = 3] [LT =2] SA-D(2) E(3) [LT = 1]

a. Bill-of-Material associated with 1 unit of FG-A

1 SA-B Note: this is just the master recipe. Not subtracting off on on-ha nd quantities YET

2 SA-C

4 SA-D = 2 1 2× +

14 E = × + × + × = + +2 3 2 1 2 3 6 2 6 4 F = 2 2×

(55)

b. Total lead time (in weeks) associated with making an item of FG-A, assuming we had no starting on-hand for any part? 6 weeks

Look at all branches of the product structure tree (or an assembly time chart, if we had one). The most common mistake people tend to make is to forget the LT associated with final assembly.

Longest branch is 6 weeks = 1wk FGA 2wks SA-C 3wks F+ +

c. Yes, if we wanted to make one FG-A, we need to order more of either E or F. We have enough F on-hand, but need 4 more E.

Now we look at the on-hand quantities and see we have 10 E and 5 F. Compare that to the BOM calculated above. No subassemblies like SA-B, SA-C or SA-D, so we are going to need all the component parts. Our on-hand records show we have 14 E and 4 F, so we are short 4 E.

14.7 a. 17×_{2 (level 2)}=₃₄

b. [17×_{3 (B’s)] – 15} =_{51 – 15} =₃₆ 36×_{5 (D’s)}=₁₈₀

14.8 a. 17×_{2 (level 2) + 17} ×₂×_{3 (level 3)}=₁₃₆ b. 170 – 12×_{2 (level 2 on hand)}=_{170 – 24}=₁₄₆

231437624-Heizer-Operation-Management-Solution-PDF.pdf

Instructor’s Solutions Manual for

Instructor’s Solutions Manual for

Additional Problems

Additional Problems

Operations

Operations

Management

Management

Principles of

Principles of

Operations

Operations

Management

Management

JAY HEIZER

JAY HEIZER

BARRY RENDER

BARRY RENDER

Contents

Contents

Homework Problem Answers

Homework Problem Answers

1

Operations and

Productivity

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