jit

Chapter 15Multicriteria Decision Problems

Learning Objectives

1. Understand the concept of multicriteria decision making and how it differs from situations and procedures involving a single criterion.

2. Be able to develop a goal programming model of a multiple criteria problem.

3. Know how to use the goal programming graphical solution procedure to solve goal programming problems involving two decision variables.

4. Understand how the relative importance of the goals can be reflected by altering the weights or coefficients for the decision variables in the objective function.

5. Know how to develop a solution to a goal programming model by solving a sequence of linear programming models using a general purpose linear programming package.

6. Know what a scoring model is and how to use it to solve a multicriteria decision problem.

7. Understand how a scoring model uses weights to identify the relative importance of each criterion.

8. Know how to apply the analytic hierarchy process (AHP) to solve a problem involving multiple criteria.

9. Understand how AHP utilizes pairwise comparisons to establish priority measures for both the criteria and decision alternatives.

10. Understand the following terms:

multicriteria decision problem analytic hierarchy process (AHP)goal programming hierarchydeviation variables pairwise comparison matrixpriority levels synthesizationgoal equation consistencypreemptive priorities consistency ratioscoring model

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Chapter 15

Solutions:

1. a.

Raw Material

1

2

3

Amount Needed toAchieve Both P1 Goals

2/5 (30) + 1/2 (15)1/5 (15)

3/5 (30) + 3/10 (15)

=

=

=

12 + 7.5

3

18 + 4.5

= 19.5

= 22.5

Since there are only 21 tons of Material 3 available, it is not possible to achieve both goals.

b. Letx1 = the number of tons of fuel additive producedx2 = the number of tons of solvent base produced

= the amount by which the number of tons of fuel additive produced exceeds the target value of 30 tons

= the amount by which the number of tons of fuel additive produced is less than the target of 30 tons

= the amount by which the number of tons of solvent base produced exceeds the target value of 15 tons

= the amount by which the number of tons of solvent base is less than the target value of 15 tons

Min +

s.t.25

x1 + 12x2 20 Material 1

15x2 5 Material 2

35x1 + 310

x2 21 Material 3

x1 - + = 30 Goal 1x2 - + = 15 Goal 2

x1, x2, , , , 0

c. In the graphical solution, point A minimizes the sum of the deviations from the goals and thus provides the optimal product mix.

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Multicriteria Decision Problems

.

0 10 20 30 40 50

10

20

30

40

50

60

70

Material 2

Material 3

Material 1

Goal 2

Goa

l 1

A (30, 10)B (27.5, 15)

Tons

of S

olve

nt B

ase

Tons of Fuel Additive

x2

x1

d. In the graphical solution shown above, point B minimizes and thus provides the optimal product mix.

2. a. Letx1 = number of shares of AGA Products purchasedx2 = number of shares of Key Oil purchased

To obtain an annual return of exactly 9%

0.06(50)x1 + 0.10(100)x2 = 0.09(50,000) 3x1 + 10x2 = 4500

To have exactly 60% of the total investment in Key Oil

100x2 = 0.60(50,000) x2 = 300

Therefore, we can write the goal programming model as follows:

Min P1( ) + P2( )s.t.

50x1 + 100x2 50,000 Funds Available 3x1 + 10x2 - + = 4,500 P1 Goal

x2 - + = 300 P2 Goal

x1, x2, , , , 0

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Chapter 15

b. In the graphical solution shown below, x1 = 250 and x2 = 375.. .

0 500 1000 1500

500

1000

Goal

Goal

Funds Available

Points that satisfy the funds availaableconstraint and satisfy the priority 1 goal

(250, 375)

x2

x1

P1

P2

3. a. Let

x1 = number of units of product 1 producedx2 = number of units of product 2 produced

Min P1( ) + P1()

+ P1( ) + P1( ) + P2( )

s.t.1x1 + 1x2 - + = 350 Goal 12x1 + 5x2 - + = 1000 Goal 24x1 + 2x2 - + = 1300 Goal 3

x1, x2, , , , , , 0

b. In the graphical solution, point A provides the optimal solution. Note that with x1 = 250 and x2 = 100, this solution achieves goals 1 and 2, but underachieves goal 3 (profit) by $100 since 4(250) + 2(100) = $1200.

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0 100 200 300 400 500

100

200

300

400

500

600

700

Goal 1

Goal 2

A (250, 100)

B (281.25, 87.5)

Goal 3

x2

x1

c.Max 4x1 + 2x2

s.t.1x1 + 1x2 350 Dept. A2x1 + 5x2 1000 Dept. B

x1, x2 0

The graphical solution indicates that there are four extreme points. The profit corresponding to each extreme point is as follows:

Extreme Point Profit1 4(0) + 2(0) = 02 4(350) + 2(0) = 14003 4(250) + 2(100) = 12004 4(0) + 2(200) = 400

Thus, the optimal product mix is x1 = 350 and x2 = 0 with a profit of $1400.

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Chapter 15

. x2

x10 100 200 300 400 500

100

200

300

400

1

3

4

2

(250,100)

(0,250)

(0,0)

(350,0)

Department B

Feasible Region

Department A

d. The solution to part (a) achieves both labor goals, whereas the solution to part (b) results in using only 2(350) + 5(0) = 700 hours of labor in department B. Although (c) results in a $100 increase in profit, the problems associated with underachieving the original department labor goal by 300 hours may be more significant in terms of long-term considerations.

e. Refer to the graphical solution in part (b). The solution to the revised problem is point B, with x1 = 281.25 and x2 = 87.5. Although this solution achieves the original department B labor goal and the profit goal, this solution uses 1(281.25) + 1(87.5) = 368.75 hours of labor in department A, which is 18.75 hours more than the original goal.

4. a. Letx1 = number of gallons of IC-100 producedx2 = number of gallons of IC-200 produced

Min P1( ) + P1( ) + P2( ) + P2( ) + P3( )s.t.

20x1 + 30x2 - + = 4800 Goal 120x1 + 30x2 - + = 6000 Goal 2 x1 - + = 100 Goal 3

x2 - + = 120 Goal 4 x1 + x2 - + = 300 Goal 5

x1, x2, all deviation variables 0

b. In the graphical solution, the point x1 = 120 and x2 = 120 is optimal.

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.

x2

x10 100 200 300

100

200

300

Goal 2

Goal 3

Goal 4

Goal 5

Goal 1

Solution points that satisfy goals 1-4

Optimal solution (120, 120)

5. a.May: = 200June: = 600July: = 600August: = 600 (no need for ending inventory)

b.May to June:June to July:July to August:

c. No. For instance, there must be at least 200 pumps in inventory at the end of May to meet the June requirement of shipping 600 pumps.

The inventory variables are constrained to be nonnegative so we only need to be concerned with positive deviations.

June:July:August:

d. Production capacity constraints are needed for each month.

May:June:July:August:

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Chapter 15

e. Min s.t.

3 Goal equations in (b)3 Goal equations in (c)4 Demand constraints in (a)4 Capacity constraints in (d)

Optimal Solution:

f. Yes. Note in part (c) that the inventory deviation variables are equal to the ending inventory variables. So, we could eliminate those goal equations and substitute and for and

in the objective function. In this case the inventory variables themselves represent the deviations from the goal of zero.

6. a. Note that getting at least 10,000 customers from group 1 is equivalent to x1 = 40,000 (25% of 40,000 = 10,000) and getting 5,000 customers is equivalent to x2 = 50,000 (10% of 50,000 = 5,000). Thus, to satisfy both goals, 40,000 + 50,000 = 90,000 letters would have to be mailed at a cost of 90,000($1) = $90,000.

Letx1 = number of letters mailed to group 1 customersx2 = number of letters mailed to group 2 customers

= number of letters mailed to group 1 customers over the desired 40,000= number of letters mailed to group 1 customers under the desired 40,000= number of letters mailed to group 2 customers over the desired 50,000= number of letters mailed to group 2 customers under the desired 50,000= the amount by which the expenses exceeds the target value of $70,000= the amount by which the expenses falls short of the target value of $70,000

Min P1( ) + P1( ) + P2( )s.t.

x1 - + = 40,000 Goal 1 x2 - 1 + 1 = 50,000 Goal 2

1x1 + 1x2 - + = 70,000 Goal 3

x1, x2, , , , , , 0

b. Optimal Solution: x1 = 40,000, x2 = 50,000

c. Objective function becomes

min P1( ) + P1(2 ) + P2( )

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Optimal solution does not change since it is possible to achieve both goals 1 and 2 in the original problem.

7. a. Letx1 = number of TV advertisementsx2 = number of radio advertisementsx3 = number of newspaper advertisements

Min P1( ) + P2( ) + P3( ) + P4( )s.t.

x1 10 TV x2 15 Radio

x3 20 Newspaper 20x1 + 5x2 + 10x3 - d 1

+ + = 400 Goal 1 0.7x1 - 0.3x2 - 0.3x3 - + = 0 Goal 2-0.2x1 + 0.8x2 - 0.2x3 - + = 0 Goal 3 25x1 + 4x2 + 5x3 - + = 200 Goal 4

x1, x2, x3, , , , , , , , 0

b. Optimal Solution: x1 = 9.474, x2 = 2.105, x3 = 20

Rounding down leads to a recommendation of 9 TV advertisements, 2 radio advertisements, and 20 newspaper advertisements. Note, however, that rounding down results in not achieving goals 1 and 2.

8. Let x1 = first coordinate of the new machine locationx2 = second coordinate of the new machine location

= amount by which x1 coordinate of new machine exceeds x1 coordinate of machine i

= amount by which x1 coordinate of machine i exceeds x1 coordinate of new machine

= amount by which x2 coordinate of new machine exceeds x2 coordinate of machine i

= amount by which x2 coordinate of machine i exceeds x2 coordinate of new machine

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Chapter 15

The goal programming model is given below.

b. The optimal solution is given by

x1 = 5x2 = 7

= 4= 2= 1= 5

The value of the solution is 12.

9. 1

A B C D EScoring Calculations

Analyst Accountant AuditorCriteria Chicago Denver HoustonCareer Advancement 35 20 20Location 10 12 8Management 30 25 35Salary 28 32 16Prestige 32 20 24Job Security 8 10 16Enjoy the Work 28 20 20

Score 171 139 139

The analyst position in Chicago is recommended. The overall scores for the accountant position in Denver and the auditor position in Houston are the same. There is no clear second choice between the two positions.

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10.12

A B C D EKenyon Manufacturing Plant Location

RatingsGeorgetown Marysville Clarksville

Criteria Weight Kentucky Ohio TennesseeLand Cost 4 7 4 5Labor Cost 3 6 5 8Labor Availability 5 7 8 6Construction Cost 4 6 7 5Transportation 3 5 7 4Access to Customers 5 6 8 5Long Range Goals 4 7 6 5

Scoring CalculationsGeorgetown Marysville Clarksville

Criteria Kentucky Ohio TennesseeLand Cost 28 16 20Labor Cost 18 15 24Labor Availability 35 40 30Construction Cost 24 28 20Transportation 15 21 12Access to Customers 30 40 25Long Range Goals 28 24 20

Score 178 184 151

Marysville, Ohio (184) is the leading candidate. However, Georgetown, Kentucky is a close second choice (178). Kenyon Management may want to review the relative advantages and disadvantages of these two locations one more time before making a final decision.

11.A B C D E

Myrtle Beach Smokey BransonCriteria South Carolina Mountains MissouriTravel Distance 10 14 6Vacation Cost 25 30 20Entertainment Available 21 12 24Outdoor Activities 18 12 10Unique Experience 24 28 32Family Fun 40 35 35

Score 138 131 127

Myrtle Beach is the recommended choice.

12. A B C D E FMidwestern State College Handover Techmseh

Criteria University at Newport College StateSchool Prestige 24 18 21 15Number of Students 12 20 32 28Average Class Size 20 25 40 35Cost 25 40 15 30Distance From Home 14 16 14 12Sports Program 36 20 16 24Housing Desirability 24 20 28 24Beauty of Campus 15 9 24 15

Score 170 168 190 183

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Chapter 15

Handover College is recommended. However Tecumseh State is the second choice and is less expensive than Handover. If cost becomes a constraint, Tecumseh State may be the most viable alternative.

13. A B C D ECriteria Park Shore The Terrace Gulf ViewCost 25 30 25Location 28 16 36Appearance 35 20 35Parking 10 16 10Floor Plan 32 28 20Swimming Pool 7 2 3View 15 12 27Kitchen 32 28 24Closet Space 18 24 12

Score 202 176 192

Park Shore is the preferred condominium.

14. a. A B C D ECriteria 220 Bowrider 230 Overnighter 240 SundancerCost 40 25 15Overnight Capability 6 18 27Kitchen/Bath Facilities 2 8 14Appearance 35 35 30Engine/Speed 30 40 20Towing/Handling 32 20 8Maintenance 28 20 12Resale Value 21 15 18

Score 194 181 144

Clark Anderson prefers the 220 Bowrider.

b. A B C D ECriteria 220 Bowrider 230 Overnighter 240 SundancerCost 21 18 15Overnight Capability 5 30 40Kitchen/Bath Facilities 5 15 35Appearance 20 28 28Engine/Speed 8 10 6Towing/Handling 16 12 4Maintenance 6 5 4Resale Value 10 12 12

Score 91 130 144

Julie Anderson prefers the 240 Sundancer.

15. Synthesization

Step 1: Column totals are 8, 10/3, and 7/4

Step 2:Price Accord Saturn Cavalier

Accord 1/8 1/10 1/7

15 - 12


Saturn 3/8 3/10 2/7Cavalier 4/8 6/10 4/7

Step 3:Price Accord Saturn Cavalier Row Average

Accord 0.125 0.100 0.143 0.123Saturn 0.375 0.300 0.286 0.320

Cavalier 0.500 0.600 0.571 0.557

Consistency Ratio

Step 1:

Step 2: 0.369/0.123 = 3.0060.967/0.320 = 3.0191.688/0.557 = 3.030

Step 3: max = (3.006 + 3.019 + 3.030)/3 = 3.02

Step 4: CI = (3.02 - 3)/2 = 0.010

Step 5: CR = 0.010/0.58 = 0.016

Since CR = 0.016 is less than 0.10, the degree of consistency exhibited in the pairwise comparison matrix for price is acceptable.

16. Synthesization

Step 1: Column totals are 17/4, 31/21, and 12

Step 2:Style Accord Saturn Cavalier

Accord 4/17 7/31 4/12Saturn 12/17 21/31 7/12Cavalier 1/17 3/31 1/12

Step 3:Style Accord Saturn Cavalier Row Average

Accord 0.235 0.226 0.333 0.265Saturn 0.706 0.677 0.583 0.656

Cavalier 0.059 0.097 0.083 0.080

Consistency Ratio

15 - 13

Chapter 15

Step 1:

0.265

1

3

1/4

+ 0.656

1/3

1

1/7

+ 0.080

4

7

1

0.265

0.795

0.066

+

0.219

0.656

0.094

+

0.320

0.560

0.080

=

0.802

2.007

0.239

Step 2: 0.802/0.265 = 3.0282.007/0.656 = 3.0620.239/0.080 = 3.007

Step 3: max = (3.028 + 3.062 + 3.007)/3 = 3.032

Step 4: CI = (3.032 - 3)/2 = 0.016

Step 5: CR = 0.016/0.58 = 0.028

Since CR = 0.028 is less than 0.10, the degree of consistency exhibited in the pairwise comparison matrix for style is acceptable.

17. a.Reputation School A School BSchool A 1 6School B 1/6 1

b. Step 1: Column totals are 7/6 and 7

Step 2:Reputation School A School BSchool A 6/7 6/7School B 1/7 1/7

Step 3:Reputation School A School B Row AverageSchool A 0.857 0.857 0.857School B 0.143 0.143 0.143

18. a. Step 1: Column totals are 47/35, 19/3, 11

Step 2:Desirability City 1 City 2 City 3

15 - 14


City 1 35/47 15/19 7/11City 2 7/47 3/19 3/11City 3 5/47 1/19 1/11

Step 3:Desirability City 1 City 2 City 3 Row Average

City 1 0.745 0.789 0.636 0.724City 2 0.149 0.158 0.273 0.193City 3 0.106 0.053 0.091 0.083

b. Step 1:

0.724

1

1/5

1/7

+ 0.193

5

1

1/3

+ 0.083

7

3

1

0.723

0.145

0.103

+

0.965

0.193

0.064

+

0.581

0.249

0.083

=

2.273

0.588

0.251

Step 2: 2.273/0.724 = 3.1410.588/0.193 = 3.0430.251/0.083 = 3.014

Step 3: max = (3.141 + 3.043 + 3.014)/3 = 3.066

Step 4: CI = (3.066 - 3)/2 = 0.033

Step 5: CR = 0.033/0.58 = 0.057

Since CR = 0.057 is less than 0.10, the degree of consistency exhibited in the pairwise comparison matrix is acceptable.

19. a. Step 1: Column totals are 4/3 and 4

Step 2:A B

A 3/4 3/4B 1/4 1/4

Step 3:A B Row Average

A 0.75 0.75 0.75B 0.25 0.25 0.25

b. The individual's judgements could not be inconsistent since there are only two programs being compared.

20. a.Flavor A B C

A 1 3 2B 1/3 1 5C 1/2 1/5 1

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Chapter 15

b. Step 1: Column totals are 11/6, 21/5, and 8

Step 2:Flavor A B C

A 6/11 15/21 2/8B 2/11 5/21 5/8C 3/11 1/21 1/8

Step 3:Flavor A B C Row Average

A 0.545 0.714 0.250 0.503B 0.182 0.238 0.625 0.348C 0.273 0.048 0.125 0.148

c. Step 1:

0.503

1

1/3

1/2

+ 0.348

3

1

1/5

+ 0.148

2

5

1

Weighted Sum:

0.503

0.168

0.252

+

1.044

0.348

0.070

+

0.296

0.740

0.148

=

1.845

1.258

0.470

Step 2: 1.845/0.503 = 3.6681.258/0.348 = 3.6150.470/0.148 = 3.123

Step 3: max = (3.668 + 3.615 + 3.123)/3 = 3.469

Step 4: CI = (3.469 - 3)/2 = 0.235

Step 5: CR = 0.235/0.58 = 0.415

Since CR = 0.415 is greater than 0.10, the individual's judgements are not consistent.

21. a.Flavor A B C

A 1 1/2 5B 2 1 5C 1/5 1/5 1

b. Step 1: Column totals are 16/5, 17/10, and 11

Step 2:Flavor A B C

15 - 16


A 5/16 5/17 5/11B 10/16 10/17 5/11C 1/16 2/17 1/11

Step 3:Flavor A B C Row Average

A 0.313 0.294 0.455 0.354B 0.625 0.588 0.455 0.556C 0.063 0.118 0.091 0.090

c. Step 1:

0.354

1

2

1/5

+ 0.556

1/2

1

1/5

+ 0.090

5

5

1

0.354

0.708

0.071

+

0.278

0.556

0.111

+

0.450

0.450

0.090

=

1.083

1.715

0.272

Step 2: 1.083/0.354 = 3.0631.715/0.556 = 3.0850.272/0.090 = 3.014

Step 3: max = (3.063 + 3.085 + 3.014)/3 = 3.054

Step 4: CI = (3.054 - 3)/2 = 0.027

Step 5: CR = 0.027/0.58 = 0.046

Since CR = 0.046 is less than 0.10, the individual's judgements are consistent.

22. a. Let D = DallasS = San FranciscoN = New York

Location D S ND 1 1/4 1/7S 4 1 1/3N 7 3 1

b. Step 1: Column totals are 12, 17/4, and 31/21

Step 2:Location D S N

D 1/12 1/17 3/31S 4/12 4/17 7/31N 7/12 12/17 21/31

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Chapter 15

Step 3:Location D S N Row Average

D 0.083 0.059 0.097 0.080S 0.333 0.235 0.226 0.265N 0.583 0.706 0.677 0.656

c. Step 1:

0.080

1

4

7

+ 0.265

1/4

1

3

+ 0.656

1/7

1/3

1

0.080

0.320

0.560

+

0.066

0.265

0.795

+

0.094

0.219

0.656

=

0.239

0.802

2.007

Step 2: 0.239/0.080 = 3.0070.802/0.265 = 3.0282.007/0.656 = 3.062

Step 3: max = (3.007 + 3.028 + 3.062)/3 = 3.035

Step 4: CI = (3.035 - 3)/2 = 0.017

Step 5: CR = 0.017/0.58 = 0.028

Since CR = 0.028 is less than 0.10, the manager's judgements are consistent.

23. a. Step 1: Column totals are 94/21, 33/4, 18, and 21/12

Step 2:Performance 1 2 3 4

1 21/94 12/33 7/18 4/212 7/94 4/33 4/18 3/213 3/94 1/33 1/18 2/214 63/94 16/33 6/18 12/21

Step 3:Performance 1 2 3 4 Row Average

1 0.223 0.364 0.389 0.190 0.2922 0.074 0.121 0.222 0.143 0.1403 0.032 0.030 0.056 0.095 0.0534 0.670 0.485 0.333 0.571 0.515

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b. Step 1:

0.292

1

1/3

1/7

3

+ 0.140

3

1

1/4

4

+ 0.053

7

4

1

6

+ 0.515

1/3

1/4

1/6

1

0.292

0.097

0.042

0.876

+

0.420

0.140

0.035

0.560

+

0.371

0.212

0.053

0.318

+

0.172

0.129

0.086

0.515

=

1.257

0.579

0.216

2.270

Step 2: 1.257/0.292 = 4.3050.579/0.140 = 4.1360.216/0.053 = 4.0752.270/0.515 = 4.408

Step 3: max = (4.305 + 4.136 + 4.075 + 4.408)/4 = 4.231

Step 4: CI = (4.231 - 4)/3 = 0.077

Step 5: CR = 0.077/0.90 = 0.083

Since CR = 0.083 is less than 0.10, the judgements are consistent.

24. a. Criteria: Yield and Risk

Step 1: Column totals are 1.5 and 3

Step 2:Criterion Yield Risk PriorityYield 0.667 0.667 0.667Risk 0.333 0.333 0.333

With only two criteria, CR = 0 and no computation of CR is made.

The same calculations for the Yield and the Risk pairwise comparison matrices provide the following:

Stocks Yield Priority Risk Priority

CCC 0.750 0.333SRI 0.250 0.667

b. Overall Priorities:

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Chapter 15

CCC 0.667(0.750) + 0.333(0.333) = 0.611SRI 0.667(0.250) + 0.333(0.667) = 0.389

CCC is preferred.

25. a. Criteria: Leadership, Personal, Administrative

Step 1: Column Totals are 8, 11/6 and 13/4

Step 2:

Criterion Leader Personal Administrative PriorityLeadership 0.125 0.182 0.077 0.128

Personal 0.375 0.545 0.615 0.512Administrative 0.500 0.273 0.308 0.360

CR = 0.094 if computed.

The same calculations for the leadership, personal and administrative pairwise comparison matrices provide the following.

Candidate Leadership Priority

Personal Priority Administrative Priority

Jacobs 0.800 0.250 0.667Martin 0.200 0.750 0.333


Jacobs 0.128(0.800) + 0.512(0.250) + 0.360(0.667) = 0.470Martin 0.128(0.200) + 0.512(0.250) + 0.360(0.333) = 0.530

Martin is preferred.

26. a. Criteria: Price, Sound and Reception

Step 1: Column totals are 19/12, 13/3 and 8

Step 2:

Criterion Price Sound Reception PriorityPrice 0.632 0.692 0.500 0.608Sound 0.211 0.231 0.375 0.272

Reception 0.158 0.077 0.125 0.120

CR = 0.064

The same calculations for the price, sound and reception pairwise comparison matrices provide the following:

System Price Priority Sound Priority Reception PrioritySystem A 0.557 0.137 0.579System B 0.123 0.239 0.187System C 0.320 0.623 0.234

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CR 0.016 0.016 0.046


System A 0.608(0.557) + 0.272(0.137) + 0.120(0.579) = 0.446System B 0.608(0.123) + 0.272(0.239) + 0.120(0.187) = 0.162System C 0.608(0.320) + 0.272(0.623) + 0.120(0.046) = 0.392

System A is preferred.

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