Solutions to Chapter 16 Problems - Electrical Engineering Site · 769 . 14-6 (a) Dominance . Attribute Vendor I vs. Vendor II I vs. III I vs. Retain II vs. III III vs. Retain . Reduction

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Solutions to Chapter 14 Problems

14-1 Left to student.

Engineering Economy, Fourteenth Edition, by Sullivan, Wicks, and Koelling. ISBN 0-13-614297-4.© 2009 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction,storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.

For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.

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14-2 Noncompensatory Models: Full dimensional

Advantages:

1) Quick and easy to apply to eliminate one or more of the alternatives.

2) All attributes are considered in the analysis.

3) Simple, easy to understand, requires little computation if any.

Disadvantages:

1) Very often does not lead to a final selection.

2) May not eliminate any of the alternatives.

3) Tends to “satisfice” rather than optimize.

Compensatory Models: Single dimensional

Advantages:

1) Trade offs are taken into account in arriving at the final decision.

2) Will almost always arrive at a final choice, and method may be developed to break a tie

quantitatively.

3) Numerical answers seem to parallel intuitive choices.

4) All “worths” reduced to a single scale, makes complex problem computationally tractable.

Disadvantages:

1) Weighting is still subjective.

2) Compression to numerical values for qualitative subjective data is often difficult and time

consuming, and may not be meaningful;

3) Translation of numerical or subjective values to a single scale may not be plausible for all

individuals.




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14-4 Some of the difficulties of developing nonlinear functions or nondimensional scaling of qualitative

(subjective) data are as follows:

(a) Dimensionless attributes will contain implicit weighting factors

(b) Dimensionless attributes will not follow same trend with respect to desirability

(c) A non-dimensionalizing procedure could inaccurately rate each attribute in terms of its fractional

accomplishment of the highest attainable value.

(d) Higher (lower) values could dominate the solution.




768

14-5 (a) Assume “ideal” means “maximum”

Attribute A for alternatives 2 and 3 is acceptable: 70 A100

Attribute B for all alternatives is acceptable: 6 B 10

Only alternative 2 attribute C is acceptable: Good C2 Excellent

Attribute D for all alternatives is accceptable: 6 D 10

Only alternative 2 is acceptable because it is the only one whose attributes all lie in acceptable

ranges.

(b) No alternative dominates another.

(c) Alternative 3 has the best value of the top ranked attribute “D”.




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14-6 (a) Dominance

Attribute

Vendor I

vs.

Vendor II

I vs. III I vs.

Retain

II vs.

III

III vs.

Retain

Reduction in throughput time Better Worse Better Worse Better

Flexibility Worse Worse Better Equal Better

Reliability Better Better Better Equal Better

Quality Worse Worse Better Equal Better

Cost of System Better Worse Worse Worse Worse

Dominance? No No No Yes No

Vendor II is removed from consideration.

(b) Satisficing

Attribute

“Worst”

Acceptable

Value

Unacceptable

Alternative

Reduction in throughput time 50% Retain

Flexibility Good Retain

Reliability Good Retain

Quality Good Retain

Cost of System $350,000 None

Remove “Retain Existing System” from consideration

(c) Disjunctive Resolution All alternatives still available (“Retain” already eliminated) pass because all options are acceptable in

at least one attribute.

(d) Lexicography Attribute Number of

times “greater”

Alternative

Ranking

Reduction in throughput time 0 III > I > II

Flexibility 2 II = III > I

Reliability 1 I > II = III

Quality 2 II = III > I

Cost of System 4 I > III > II

Select Vendor III




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14-8 (a)

Dominance:

Paired Comparison

Attribute A vs. B B vs. C A vs. C

1 better Worse worse

2 worse better better

3 better * Worse worse



Dominance? no no no

No alternatives can be eliminated based on the dominance method.

*Assume that knowing the safety value is better than any unknown value.

Satisficing:

Attribute Feasible Range Unacceptable

Alternatives

1 $80,000 - $100,000 none

2 Fair - Excellent none

3 Good - Excellent Alternative A has unknown value *

4 94 - 99% none

5 Fair - Excellent none

*if the same assumption is used (as in the dominance model) alternative A would be

eliminated using the satisficing model

Lexicography:

Paired comparisons - using given weighting: 5 > 1 > 4 > 3 > 2

Attribute

Ordinal

Ranking **

Ranking

1 3 C > A > B

2 0 B > A > C

3 1 C > B > A

4 2 B > A > C

5 4 B > A = C

** 4 is most important rank.

The selection, based on highest ranked attribute (#5), would be Alternative B.

Because all alternatives meet at least one acceptability range, no alternatives are rejected.




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14-8 (a) continued Non-dimensional scaling Attribute Value Rating

Procedure

Dimensionless

Value

1 $100,000

$140,000

$180,000

($180,000-Cost)/$80,000 1.0

0.5

0.0

2 Excellent

Good

Fair

Relative rank - 1

2

1.0

0.5

0.0

3 Excellent

Good

Not known

Relative rank - 1

2

1.0

0.5

0.0

4 99%

98%

94%

Reliability % - 94

99-94

1.0

0.8

0.0

5 Excellent

Good

Relative rank - 1

2

1.0

0.0

Non-Dimensional Value

Attribute A B C

1 0.5 0.0 1.0

2 0.5 1.0 0.0

3 0.0 0.5 1.0

4 0.8 1.0 0.0

5 0.0 1.0 0.0

Additive Weighting (using given weights) Attribute Weight A B C

1. Initial Cost 0.25 0.5(0.25)=0.125 0.0(0.25)=0 1.0(0.25)=0.25

2. Maintenance 0.10 0.5(0.10)=0.05 1.0(0.10)=0.1 0.0(0.10)=0.0

3. Safety 0.15 0.0(0.15)=0.00 0.5(0.15)=0.075 1.0(0.15)=0.15

4. Reliability 0.20 0.8(0.20)=0.16 1.0(0.20)=0.2 0.0(0.20)=0.0

5. Prod. Quality 0.30 0.0(0.30)=0.00 1.0(0.30)=0.3 0.0(0.30)=0

Using Additive Weighting Alternative B would be selected.

(b) If two (or more) attributes are dependent, including them in the analysis will result in the same ranking of

alternatives for these attributes, essentially double (or triple, or more) counting those decision elements.

This can be a false representation of the decision maker's true attitudes. When attributes are entirely

dependent, they are more appropriately modeled as a single attribute and weighted appropriately (note that

in these cases, the weight of these combined attributes may not necessarily be the sum of the individual

weights of the original attributes). Once this is accomplished, the decision analysis proceeds as in part (a).




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14-9 (a)

Attribute Relative Rank Normalized Rank

Social Climate 1.00 1/2.08 = 0.481

Starting Salary 0.50 0.5/2.08 = 0.240

Career Adv. 0.33 0.33/2.08 = 0.159

Weather/Sports 0.25 0.25/2.08 = 0.120

2.08 1.00

(b) Alternatives

Attribute Apex (N.Y.) Sycon (L.A.) Sigma (GA.) Mc-Graw-Wesley (AZ.)

Starting Salary $35,000 $30,000 $34,500 $31,500

Dimensionless

Equivalent (DE)

1.0

0.0

0.9

0.3

DE = Worst Outcome - Outcome Being Made Dimensionless

Worst Outcome - Best Outcome

(c)

Attribute Normalized

Weight

Apex Sycon Sigma Mc-Graw

Wesley

Social

Climate

0.48 1x0.48 1x0.48 0.5x0.48 0x0.48

Starting

Salary

0.24 1x0.24 1x0.24 0.9x0.24 0.3x0.25

Career

Adv.

0.16 0x0.16 0x0.16 0.6x0.16 1x0.16

Weather/

Sports

0.12

0x0.12

0x0.12

0.33x0.12

0.67x0.12

Sum 0.72 0.63 0.59 0.31

Using lexicography we conclude that social climate is the most important attribute and Apex is selected.

Additive weighting also selects Apex.




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14-10 (a) Wright dominates Alott – Alott is removed from further consideration.

(b) Only Wright meets the minimum performance levels for all attributes.

(c) All candidates would be retained under disjunctive resolution.

(d) Lexicography – Based on project management skills (most important attribute), Busy is eliminated.

Looking next at general attitude, Surley is eliminated. Lastly, looking at years manufacturing

experience, Wright would be selected.




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14-12 (a) Left to student – no unique answer.

(b) Select a mathematical model similar to additive weighting. Let each judge set his/her own

weightings and develop a score for each contestant. Then, sum the three scores for each contestant.

The contestant with the highest total score is the winner.

This method will allow each judge to be as subjective about each attribute as he/she desires while

making the final selection objective.




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14-13 Assume all attributes are of equal importance.

Attribute Alott Surley Busy Wright

Total years 0.33 0.00 1.00 0.67

Manufacturing years 0.33 1.00 0.00 0.67

Project management skills 1.00 1.00 0.00 1.00

Management years 0.00 0.50 0.50 1.00

General attitude 1.00 0.00 0.50 1.00

Total Score 2.66 2.50 2.00 4.34

Wright would be selected.




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14-14 i Wi Rank

1 1.0 1

2 ? 4

3 0.8 2

4 0.7 ? By inspection, Ranki=4=3

10

Vj WiXiji1

n

V2 = 2.3 = (1.0) (0.7) + W2 (1.0) + (0.8) (0.5) + (0.7) (1.0)

2.3 = W2 + 0.7 + 0.4 + 0.7

W2 = 0.50

V1 = 2.69 = (1.0) (1.0) + (0.5) (0.8) + (0.8) X3,1 + (0.7) (0.7)

2.69 = 1.0 + 0.4 + 0.8 X3,1 + 0.49

X3,1 = 2.69 - 1.0 - 0.4 - 0.49 = 1.0

0.8

Vj normalized = V1 = 2.69/2.69 = 1.00

V2 = 2.30/2.69 = 0.86

By inspection Rankj=1 = 2.0

Rankj=2 = 1.0

Filling in blanks,

i Wi Rank

1 1.0 1

2 0.5 4

3 0.8 2

4 0.7 3




779

14-4 continued

Keep Existing

Tool

Purchase

new machine

Tool

Rank 2.0 1.0

X1j 1.0 0.7

Rank 2.0 1.0

X2j 0.8 1.0

Rank 1.0 2.0

X3j 1.0 0.5

Rank 2.0 1.0

X4j 0.7 1.0

Vj 2.69 2.30

Vj norm 1.0 0.86




780





781





782





783

Solutions to Spreadsheet Exercises

14-18

1234567891011121314151617181920

2122232425

A B C D E F GAttribute Dr. Molar Dr. Feelgood Dr. Whoops Dr. Pepper

Cost 50$ 80$ 20$ 40$ Anesthesia Novocaine Acupuncture Hypnosis Laughing GasDistance 15 20 5 30Office Hours 40 25 40 40Quality Excellent Fair Poor Good

Excellent 4 Acupuncture 1Fair 2 Hypnosis 4Good 3 Laughing Gas 2Poor 1 Novocaine 3

Attribute Dr. Molar Dr. Feelgood Dr. Whoops Dr. Pepper

Cost 0.50 0.00 1.00 0.67Anesthesia 0.67 0.00 1.00 0.33Distance 0.60 0.40 1.00 0.00Hours 1.00 0.00 1.00 1.00Quality 1.00 0.33 0.00 0.67Sum = 3.77 0.73 4.00 2.67

^ best choice

With novocaine rated as the preferred method of anesthesia, Dr. Whoops becomes the dentist of choice.

Quality Anesthesia




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Solutions to Chapter 16 Problems - Electrical Engineering Site · 769 . 14-6 (a) Dominance . Attribute Vendor I vs. Vendor II I vs. III I vs. Retain II vs. III III vs. Retain . Reduction