Multiattribute Utility Functions

7/23/2019 Multiattribute Utility Functions

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xi = level of attribute i

u(x1,x2 ) = utility associated with level x1 of attribute 1 and level x2 of attribute 2

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Attribute 1 is utility independent (ui) of attribute 2 if preferences for lotteries

involving different levels of attribute 1 do not depend on the level of attribute 2.

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If attribute 1 is ui of attribute 2, and attribute 2 is ui of attribute 1, then attributes

1 and 2 are mutually utility independent (mui).

u(x1,x2 ) = k1u1(x1)+ k2u2 (x2 )+ k3u1(x1)u2 (x2 )

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12[k1u1(10) k2u2(5) k3u1(10)u2(5)]

12[k1u1(10) 12[ k1u1(30) k2u2(5) k3u1(30)u2(5)]

k1u1(16) k2u2(5) k3u1(16)u2(5)

Simplifying this equation yields (if k1 0)

12[u1(10) u1(30)] u1(16)

Using (11), we find

E(U for L2) 12[k1u1(10) k2u2(20) k3u1(10)u2(20)]

12[k1u1(30) k2u2(20) k3u1(30)u2(20)]

k1u1(16) k2u2(20) k3u1(16)u2(20)

E(U for L2)

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T H EOR EM 2

Attributes 1 and 2 are mui if and only if the decision makers utility function

u(x1, x2) is a multilinear function of the form

u(x1, x2) k1u1(x1) k2u2(x2) k3u1(x1)u2(x2) (10)


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A decision makers utility function exhibits additive independence if the decision

maker is indifferent between

12

12

x1(best), x2(best) x1(best), x2(worst)

and

12

12

x1(worst), x2(worst) x1(worst), x2(best)

(11)

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Then additive independence implies that

12(k1 k2 k3)

12(0) 1

2(k1)

12(k2)

k3 0

Thus, if attributes 1 and 2 are mui and the decision makers utility function exhibits ad-

ditive independence, his or her utility function is of the following additive form:

u(x1, x2) k1u1(x1) k2u2(x2) (12)


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W1 T/< =J[ =\%/< =`U

u(x1(best), x2(best)) 1, u(x1(worst), x2(worst)) 0,

u1(x1(best)) 1, u1(x1(worst)) 0, u2(x2(best)) 1, u2(x2(worst)) 0


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Thus, k1 can be determined from the fact that the decision maker is indifferent between

k1u(x1(best), x2(best))

1 u(x1(best), x2(worst)) and

1 k1u(x1(worst), x2(worst))

u(x1(best), x2 (worst))= k1(1) + k2 (0)+ k3(0) = k1


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To determine k2

k2u(x1(best), x2(best))

1 u(x1(worst), x2(best)) and

1 k2u(x1(worst), x2(worst))

u(x1(worst), x2(best))= k1(0)+ k2 (1)+ k3(0) = k2


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To determine k3, observe that from (10) and

u(x1(best), x2(best)) u1(x1(best)) u2(x2(best)) 1

we find that

1 u(x1(best), x2(best)) k1(1) k2(1) k3(1) k1 k2 k3

k1 + k2 + k3 = 1

k3 = 1! k

1! k

2

If utility function exhibits additive independence, k3 = 0


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Profit

(millions of $)

Market share (%)10 20 30 40 50

30

25

20

15

10

5

0

F I G U R E 14

Possible Levels of Each

Attribute for Fruit

Computer Company


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12

10%, $15

12

50%, $15

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12

10%, $20

12

50%, $20

12

10%, x2

12

50%, x2

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12

12

50%, $30 50%, $5

and

12

12

10%, $5 10%, $30

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k150%, $30

1 50%, $5 and

1 k

1 10%, $5

W1


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k2

50%, $30

1 10%, $30 and

1 k210%, $5

W1


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k3 = 1!

k1!

k2 =!0.1

u(x1, x

2) =0.6u

1(x

1)+0.5u

2(x

2)!0.1u1(x1)u2 (x2 )

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!u(x1, x

2)

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u1(x1)

x1(%)2010 30 40

(a)Market share

50

1.00

.75

.50

.25

0

u2(x2)

x2(millions of $)5 10 15

(b)Profit

20 25 30

1.00

.75

.50

.25

0F I G U R E 15

u1(x1) and u2(x2) forFruit Computer


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Effect of Advertising on Market Share and Profit

CSL Chooses

Fruit Chooses Small Ad Campaign Large Ad Campaign

Small ad campaign 25%, $16 15%, $12

Large ad campaign 35%, $8 25%, $10


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u1(15) =0.125,u1(25) =0.375,u1(35) =0.625

u2 (8) =0.45,u2 (10) =0.53,u2 (12) =0.58,u2 (16) =0.70.

u(25%, $16) 0.6(.375) 0.5(.7) 0.1(.375)(.7) .549u(15%, $12) 0.6(.125) 0.5(.58) 0.1(.125)(.58) .358

u(35%, $8) 0.6(.625) 0.5(.45) 0.1(.625)(.45) .572

u(25%, $10) 0.6(.375) 0.5(.53) 0.1(.375)(.53) .470

Then

E(U for small ad campaign) (12)(.549) (1

2)(.358) .454

E(U for large ad campaign) (12)(.572) (1

2)(.470) .521

Thus, during the current year, Fruit should mount a large ad campaign.

Documents

Multiattribute Utility Functions