Multi-attribute utility theory (very superficial in textbook!) 164 / 401 Multiple objectives – an example Consider the problem of deciding upon treatment for a patient with esophageal cancer. The problem can be (schematically) modelled as follows: radio therapy chemo therapy surgery endo prosthesis L.E. Q.L. X X X X X X X X X X X X X When a decision problem concerns multiple objectives, captured by multiple attributes, consequences are no longer ’simple’; this is apparent from the above consequence matrix. 165 / 401 Multiple objectives – another example The City of Utrecht is considering four different sites (A,B,C and D) for a new electric power generating station. The objectives of the city are to • minimise the cost of building the station; • minimise the acres of land damaged by building it. Factors influencing the objectives include the land type at the different sites, the architect and construction company hired, the cost of material and machines used, the weather, etc. Costs, however, are estimated to fall between e 15 million and e 60 million; between 200 and 600 acres of land will be damaged. The possible consequences of the decision alternatives are captured by two attributes. We therefore need to determine a two-attribute utility function: u(Cost, Acres) (or u(C, A)). 166 / 401 The multiattribute utility function – assessment Let X 1 ,...,X n , n ≥ 2, be a set of attributes associated with the consequences of a decision problem. The utility of a consequence (x 1 ,...,x n ) can be determined from • direct assessment: estimate the combined utility u(x 1 ,...,x n ) over the given values of all n attributes; • decomposed assessment: 1 estimate n conditional utilities u i (x i ) for the given values of the n attributes; 2 compute u(x 1 ,...,x n ) by combining the u i (x i ) of all attributes: u(x 1 ,...,x n )= f [u 1 (x 1 ),...,u n (x n )] 167 / 401
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Multi-attribute utility theory(very superficial in textbook!)
164 / 401
Multiple objectives – an example
Consider the problem of deciding upon treatment for a patientwith esophageal cancer. The problem can be (schematically)modelled as follows:
radiotherapy
chemotherapy
surgery
endoprosthesis
L.E. Q.L.X
X
X
X
X
X
X XX
X
XXX
When a decision problem concerns multiple objectives,captured by multiple attributes, consequences are no longer’simple’; this is apparent from the above consequence matrix.
165 / 401
Multiple objectives – another example
The City of Utrecht is considering four different sites (A,B,C andD) for a new electric power generating station.
The objectives of the city are to• minimise the cost of building the station;• minimise the acres of land damaged by building it.
Factors influencing the objectives include the land type at thedifferent sites, the architect and construction company hired, thecost of material and machines used, the weather, etc.
Costs, however, are estimated to fall between e 15 million ande 60 million; between 200 and 600 acres of land will bedamaged.
The possible consequences of the decision alternatives arecaptured by two attributes. We therefore need to determine atwo-attribute utility function: u(Cost, Acres) (or u(C,A)).
166 / 401
The multiattribute utility function – assessment
Let X1, . . . , Xn, n ≥ 2, be a set of attributes associated with theconsequences of a decision problem.
The utility of a consequence (x1, . . . , xn) can be determined from
• direct assessment: estimate the combined utilityu(x1, . . . , xn) over the given values of all n attributes;
• decomposed assessment:1 estimate n conditional utilities ui(xi) for the given values of
the n attributes;2 compute u(x1, . . . , xn) by combining the ui(xi) of all
attributes:
u(x1, . . . , xn) = f [u1(x1), . . . , un(xn)]
167 / 401
Direct assessment – an example
15
60Cost 200
600
Acres
u
Assign a utility of 0 to the worst consequence (60, 600), and autility of 1 to the best consequence (15, 200).
The utility of, for example, consequence (50, 300) can bedetermined from:
1.0(50, 300) ∼
p
(1− p)
(15, 200)
(60, 600)
To find a good representation through direct assessment,utilities must be assessed for a substantial number of points.
168 / 401
Notation
Let X be an attribute and Y a set of n− 1, n > 0, attributes.Compare the following:
• if n = 1 then u(X) is a uti-lity function for X in a one-dimensional decision pro-blem
X
Y
u(X, Y )
u(X, yi)
UX(X)
(possibly rescaled)
• if n > 1 then• u(X,Y ) is an n-attribute utility function;• u(X, yi) is a subutility function for X given a fixed
value-assignment yi to attributes Y ;• uX(X) is a conditional utility function for X: a (possibly)
re-scaled function u(X, yk) for some — no longer explicit— yk.
169 / 401
Different functional forms
Let X and Y be two attributes (generalisation to n > 2 attributes isstraightforward). Consider the utility function u(X,Y ) for X and Y .
• u has an additive form if for constants kX and kY
u(X,Y ) = kX · uX(X) + kY · uY (Y )
• u has a multilinear form if for constants kX , kY , and kXY
Reconsider the City of Utrecht’s two attribute utility functionu(C,A) = 3 · uC(C) + 1 · uA(A). Suppose that the City hasexplicitly expressed the indifference (50, 600) ∼ (60, 350), whichindeed holds given all current functions:
Now, however, the City finds out that all alternatives result in atleast a loss of 350 acres of land, and we rescale uC such thatuC(350) = 1. The expressed indifference still holds, implying that
The City decides that it is willing to sacrifice 200 acres of land ifthat would save e 10 million. This implies that, for example,
u(40, 200) = u(30, 400)
that is, a · uC(40) + b · uA(200) = a · uC(30) + b · uA(400)
We conclude thata
b=
uA(400)− uA(200)
uC(40)− uC(30)=
0.5− 1.0
0.4− 0.7=−0.5
−0.3
And thus find that u(C,A) = 0.5 · uC(C) + 0.3 · uA(A)
Is this valid?173 / 401
Assessing scaling constants (I)
Pricing out :Assess the marginal rate of substitution, that is, determine thevalue of one objective in terms of another.
Let X and Y be two attributes with values x1� . . .�xn, n ≥ 2,and y1� . . .�ym, m ≥ 2. Let eX and eY be the units ofmeasurement for the two attributes, respectively.
Suppose you are willing to sacrifice s units of Y for 1 unit of X.Then for all xi, yj: u(xi, yj) = u(xi + eX , yj − s · eY )
If u(X,Y ) is additive and uX(X), uY (Y ) are linear functions,then this implies that
a · uX(xi) + b · uY (yj) = a · uX(xi + eX) + b · uY (yj − s · eY )
As a result,a
b=
uY (yj − s · eY )− uY (yj)
uX(xi)− uX(xi + eX)
174 / 401
Assessing scaling constants (II)
Swing weighting :Consider a set of n ≥ 2 attributes X1, . . . , Xn. Swing weightingassigns weights to attributes based on either a rank-order or arating on attributes and one consequence.
Take the (theoretically) worst possible consequence asbenchmark. Then apply the following procedure:
rate(benchmark)← 0 ;rank(benchmark)← n + 1 ;for i = 1 to n
do Z ← answer to:if you could swing one attribute from worstto best value, which would you swing?;
Swing(Z);rank(Z)← i.if i = 1 then rate(Z)← 100
else rate(Z)← answer ∈ 〈0, 100〉.175 / 401
Swing weighting – cntd
Consider a set of n ≥ 2 attributes X1, . . . , Xn. Let rank(Xi) andrate(Xi) denote a ranking and a rating for attribute Xi,respectively.
The weight w(Xi) for attribute Xi can now be determined usingeither of the following two approaches:
Lottery weights :Consider two attributes X and Y (the following extendsstraightforwardly to n > 2 attributes).
Let (x0, y0) denote the worst possible consequence, and(x+, y+) the best possible consequence.
The weight w(X) for attribute X equals p, where p is theindifference probability that follows from:
1.0(x+, y0) ∼
p
(1− p)
(x+, y+)
(x0, y0)
178 / 401
Lottery weights: an example
Assume once more that the City of Utrecht decides to model theutility function u(Cost, Acres) as an additive function:
u(C,A) = kC · uC(C) + kA · uA(A)
The weight kC is assessed from:
1.0(15, 600) ∼
kC
(1− kC)
(15, 200)
(60, 600)
The weight kA is assessed using:
1.0(60, 200) ∼
kA
(1− kA)
(15, 200)
(60, 600)
179 / 401
MAUT with n = 2 attributes:when can we use additive and multilinear forms?
180 / 401
Additive independence – the formal definition
Use of an additive utility function is justified given theassumption of additive independence [AI].
Two attributes X and Y are additive independent if preferencesfor lotteries over X × Y can be established by comparing thevalues one attribute at a time. More formally,
DEFINITION
Two attributes X and Y are additive independent if the pairedpreference comparison of any two lotteries, defined by two jointprobability distributions on X × Y , depends only on theirmarginal distributions.
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Interpreting additive independence
The ability to establish preferences for lotteries over X × Y bycomparing the values one attribute at a time entails thefollowing:
• and u(60, A) for fixed cost:u(60, 200) = 0.25 u(60, 400) = 0.15u(60, 300) = 0.20 u(60, 600) = 0.00
These two functions intersect at u(60, 600) = 0.
Assuming additive independence, we have that for all ci and aj:
u(ci, aj) + u(60, 600) = u(ci, 600) + u(60, aj)
All other points can now be computed. For example,
u(15, 200) = u(15, 600) + u(60, 200)− 0 = 1.0
187 / 401
The additive utility function – Ia
THEOREM
Let X and Y be two attributes with values x1� . . .�xn, n ≥ 2,and y1� . . .�ym, m ≥ 2. Attributes X and Y are additiveindependent iff the two-attribute utility function is additive:
u(X,Y ) = kX · uX(X) + kY · uY (Y ),
where
• u(X,Y ) is normalised with u(x1, y1) = 0 and u(xn, ym) = 1;
• uX(X) is a conditional utility function on X, normalised byuX(x1) = 0 and uX(xn) = 1;
• uY (Y ) is a conditional utility function on Y , normalised byuY (y1) = 0 and uY (ym) = 1;
• kX = u(xn, y1) and kY = u(x1, ym) are positive scalingconstants, summing to 1.
188 / 401
The additive utility function – IIa
LEMMA
Let X and Y be two attributes with values x1� . . .�xn, n ≥ 2,and y1� . . .�ym, m ≥ 2.
Attributes X and Y are additive independent iff the two-attributeutility function is additive:
u(X,Y ) = u(X, y1) + u(x1, Y ),
where u(x1, y1) = 0.
189 / 401
The additive utility function – IIb
Proof of Lemma :(⇒) Let xi and yk be arbitrary values of X resp. Y .
Additive independence implies
u(x1, y1) + u(xi, yk) = u(x1, yk) + u(xi, y1)
Setting u(x1, y1) = 0, we get u(xi, yk) = u(x1, yk) + u(xi, y1).
(⇐) Suppose u(X,Y ) = u(X, y1) + u(x1, Y ), and let xj and yl bearbitrary values of X resp. Y .Then for all values xi and yk of X resp. Y , we have that
For scaling constants kA and kC we know that kA = 1− kC andthat kC = u(15, 600). This latter utility is assessed using thefollowing lottery:
1.0(15, 600) ∼
kC
(1− kC)
(15, 200)
(60, 600)
The indifference probability kC is found to be 0.75. We thereforeconclude that
u(C,A) = 0.75 · uC(C) + 0.25 · uA(A)192 / 401
Utility independence – the formal definition
Use of a multilinear or multiplicative utility function is justifiedunder (mutual) utility independence [(M)UI].
Attribute X is utility independent of attribute Y if conditionalpreferences for lotteries over X given a fixed value for Y do notdepend on the particular value for Y . More formally,
DEFINITION
An attribute X is utility independent of an attribute Y iff for anylotteries [(X, yk)]1 and [(X, yk)]2 over X × Y with Y fixed to valueyk, we have
[(X, yk)]1 � [(X, yk)]2 =⇒ [(X, yl)]1 � [(X, yl)]2 ∀ y of Y
NB: [(X, y)] represents a conditional lottery over X × Y
involving consequences over different values of X combinedwith a fixed value for Y .
193 / 401
Interpreting utility independence
Independence of conditional preferences for lotteries over X ofthe value of Y , entails the following:
• if [p, (x1, y1); (1− p), (x2, y1)] � [p, (x3, y1); (1− p), (x4, y1)]then the decision maker should also prefer
• X is utility independent of Y :− [1.0, (x1, y0)] ∼ [0.6, (x0, y0); 0.4, (x2, y0)]− [1.0, (x1, y1)] ∼ [0.6, (x0, y1); 0.4, (x2, y1)]− [1.0, (x1, y2)] ∼ [0.6, (x0, y2); 0.4, (x2, y2)]
• Y is not utility independent of X, as in the context of x1 wehave y1 ≻ y0 ≻ y2, and in the context of x2 we havey1 ≻ y2 ≻ y0!
196 / 401
Utility independence – the implication for u(X,Y )
Consider two attributes X and Y with values x1, . . . , xn, n ≥ 2,and y1, . . . , ym, m ≥ 2.
First we observe that x1 � xj � xn for each value xj of X.
Now, if X is utility independent of Y , then we have by definitionthat for any xj there exists a pj such that for all yl,
u(xj, yl) = pj · u(x1, yl) + (1− pj) · u(xn, yl)
that is, pj =u(xj, yl)− u(xn, yl)
u(x1, yl)− u(xn, yl)is a linear function of u(X, yl).
This means that all subutility functions u(X, yk) are the same, upto (positive) scaling and translation:
u(X, yk) = ckl · u(X, yl) + dkl for some constants ckl > 0 and dkl.
197 / 401
Utility independence – an equivalence
PROPOSITION
Consider two attributes X and Y with values x1, . . . , xn, n ≥ 2,and y1, . . . , ym, m ≥ 2.
X is utility independent of Y , iff for each value yl of Y , thereexist real functions gl > 0 and hl, such that
u(X,Y ) = gl(Y ) · u(X, yl) + hl(Y )
for all values of X and Y .
198 / 401
Proof of the Proposition (sketch) :
(⇒) First observe that for all x, x1 � x � xn. Utilityindependence holds iff for each x a probability p exists such that(x, Y ) ∼ [p, (x1, Y )]; (1− p), (xn, Y ) and therefore (main theorem)
(I) u(X,Y ) = p · u(x1, Y ) + (1− p) · u(xn, Y )
= p ·(
u(x1, Y )− u(xn, Y ))
+ u(xn, Y )
1 solve p from (I) with Y set to yl;2 substitute this result in (I) to get the desired result.
(⇐) Let gl > 0 and hl be such that for arbitrary yl:
(II) u(X,Y ) = gl(Y ) · u(X, yl) + hl(Y )
1 for arbitrary xj and yj, choose xi and xk s.t.(xi, yj) � (xj, yj) � (xk, yj);
2 continuity: ∃p : u(xj, yj) = p · u(xi, yj) + (1− p) · u(xk, yj);3 apply (II) to u(xi, yj), u(xj, yj) and u(xk, yj).4 rewrite to find the desired result. �
199 / 401
Exploiting utility independence
Consider two attributes X and Y with values x1, . . . , xn, n ≥ 2,and y1, . . . , ym, m ≥ 2.
If X is utility independent of Y , then one subutility functionu(X, yk) and two points u(xi, yl), and u(xj, yl) serve forestablishing a second function: u(X, yl)
Y
X
u(X,Y ) can be constructed en-tirely from three subutility functi-ons:
1. u(X, yk) for some value yk of Y
2. u(xi, Y ) for some value xi of X
3. u(xj, Y ) for some value xj of X,xj 6= xi
200 / 401
Three subutility functions
THEOREM
Let X and Y be two attributes with values x1� . . .�xn, n ≥ 2,and y1� . . .�ym, m ≥ 2.
If X is utility independent of Y then
u(X,Y ) = u(x1, Y ) · [1− u(X, y1)] + u(xn, Y ) · u(X, y1)
where u(X,Y ) is normalised by u(x1, y1) = 0 and u(xn, y1) = 1.
201 / 401
Three subutility functions
Proof of the Theorem (sketch) :
Utility independence implies the existence of functions g > 0, hsuch that
(I) u(X,Y ) = gl(Y ) · u(X, yl) + hl(Y ) ∀yl
1 set yl to y1;
2 solve (I) for x1 to get h1(Y ), using u(x1, y1) = 0;
3 solve (I) for xn to get g1(Y ), using u(xn, y1) = 1;
4 substitute these results in (I) to get the desired result. �
202 / 401
An Example
Suppose the City of Utrecht assesses the following utilities forCost and Acres lost, given A fixed at 600, and C fixed at 15 and60, respectively:
Given that u(C, 300) and u(C, 600) are known functions, wecan compute an f60(ci) and k60(ci) for each ci.
211 / 401
The multilinear utility function — IIa
THEOREM
Let X and Y be two attributes with values x1� . . .�xn, n ≥ 2,and y1� . . .�ym, m ≥ 2. If X and Y are mutually utilityindependent then the two-attribute utility function is multilinear:
u(X,Y ) = kX · uX(X) + kY · uY (Y ) + kXY · uX(X) · uY (Y )
where
• u(X,Y ) is normalised by u(x1, y1) = 0 and u(xn, ym) = 1;
• uX(X) is a conditional utility function on X, normalised byuX(x1) = 0 and uX(xn) = 1;
• uY (Y ) is a conditional utility function on Y , normalised byuY (y1) = 0 and uY (ym) = 1;
Constants kC = u(15, 600) and kA = u(60, 200) are assessed:
1.0(15, 600) ∼
kC
(1− kC)
(15, 200)
(60, 600)
1.0(60, 200) ∼
kA
(1− kA)
(15, 200)
(60, 600)
The indifference probabilities found: kC = 0.75 and kA = 0.45.So, u(C,A) = 0.75 · uC(C) + 0.45 · uA(A)− 0.2 · uC(C) · uA(A)
216 / 401
Interpreting scaling constants (I)
Let X and Y be two attributes such that Y ranges from 0 to 100and u(X,Y ) = 0.25 · uX(X) + 0.75 · uY (Y ).
Suppose that u(0, 10) = u(100, 0).
0
100
100
0
10
100
X
Y
Y
X
10
u = 1
u′ = 1
u = 0.25 = kX
u = 0.25
u = 0
u′ = 0
u = 0.75 = kY
u′ = a = k′X
u′ = a = k′Y
Suppose we decide it is suffi-cient for Y to range from 0 to 10.
Then rescaling results in:
k′
X = U ′(100, 0) = U ′(0, 10) = k′
Y
Did Y just turn from three timesas important as X to equally im-portant?
217 / 401
Interpreting scaling constants (II)
Consider a multilinear utility function u(X,Y ) over attributes X
and Y with values x1� . . .�xn, n ≥ 2, resp. y1� . . .�ym, m ≥ 2.• kXY : consider two lotteries over values xi � xj and yk � yl:
0.5
0.5
(xi, yk)
(xj , yl)
A:
0.5
0.5
(xi, yl)
(xj , yk)
B:
A ∼ B ⇐⇒ kXY = 0A ≻ B ⇐⇒ kXY > 0 (X, Y are complements)A ≺ B ⇐⇒ kXY < 0 (X, Y are substitutes)
• kX , kY : if you would rather ’swing’ x1 to xn than y1 to ym, thenkX > kY , and vice-versa;
Even a mighty important attribute will have a small scalingconstant if its range is relatively small!
218 / 401
The multiplicative utility function
THEOREM
Let X and Y be two attributes with values x1� . . .�xn, n ≥ 2,and y1� . . .�ym, m ≥ 2. If X and Y are truly mutually utilityindependent then the two-attribute utility function ismultiplicative:
• X PI Y since ∀yi: x0 � x1 � x2;• we have no X UI Y :
(x1, y) ∼ [p, (x0, y); (1− p), (x2, y)]
holds for p ≈ 0.71 if y ≡ y0, and for p = 0, if y ≡ y1.229 / 401
Validating utility independence
If X PI Y , then X is also utility independent of Y if in thefollowing procedure xC is equivalent for all values of Y :
1 choose a lottery [0.5, P ; 0.5, Q] where P = (x1, y) andQ = (xn, y) for some value y of Y and values x1 and xn of X;
2 ask the decision maker whether or not he prefers the lottery[0.5, P ; 0.5, Q] to a consequence (xi, y) for some xi,x1 � xi � xn;
3 repeat step 2 until youconverge to the certaintyequivalent (xC , y) of thelottery;
4 repeat steps 1 – 3 fordifferent values of Y .
Repeat the procedure for dif-ferent pairs of values for X.
x1 xn
X
Y
ym
y1
Q
Q′′
Q′
P ′
P ′′
P xC
xC
xC
230 / 401
AI implies (M)UI
PROPOSITION
Consider two attributes X and Y with values x1, . . . , xn, n ≥ 2,and y1, . . . , ym, m ≥ 2.
If X and Y are additive independent then X and Y are mutuallyutility independent
Proof (sketch) : We prove X UI Y ; Y UI X is analogous.
AI implies u(X, yk) = u(X, y1) + u(x1, yk) for arbitrary yk, whereu(x1, yk) is constant w.r.t the value of X.
Continuity implies, for arbitrary xi, xj and xk with(xi, y1) � (xj, y1) � (xk, y1), that ∃p such that(I) u(xj, y1) = p · u(xi, y1) + (1− p) · u(xk, y1).
Substitution of each u(X, y1) in (I) with u(X, yk)− u(x1, yk) givesu(xj, yk) = p · u(xi, yk) + (1− p) · u(xk, yk) for arbitrary yk. �
231 / 401
MUI implies AI ?
If X and Y are mutually utility independent then X and Y arenot necessarily additive independent.
Counter argument :We have seen that an additional assumption is necessary toconclude additive independence given mutual utilityindependence (see Additive utility function IVa). �
The mentioned corollary can be used to validate AI given MUI;another option is to assume that AI holds. . .
232 / 401
Validating additive independence – example
Suppose the City of Utrecht assesses the following conditionalutilities for Cost and Acres lost:
resulting in an isopreference curve~ıC(A) with u(~ıC(A), A) = 0.40and e.g. ~ıC(600) = 35 (does not follow from above).If Cost is utility independent of Acres lost, then
u(C,A) = u(60, A) +
[
u(35, 600)− u(60, A)
u(~ıC(A), 600)
]
· u(C, 600)
244 / 401
Substitution of u(X,~ıY (X)) for u(X, y1) (a)
COROLLARY
Let X and Y be two attributes with values x1� . . .�xn, n ≥ 1,and y1� . . .�ym, m ≥ 1.
If X is utility independent of Y then
u(X,Y ) =
u(x1, Y ) · u(xn,~ıY (X))− u(xn, Y ) · u(x1,~ı
Y (X))
u(xn,~ıY (X))− u(x1,~ıY (X))
where
• ~ıY (X) is defined such that (X,~ıY (X)) ∼ (x1, yk) for an yk with:
• u(x1, yk) = 0, (x1 6= xn)
245 / 401
Substitution of u(X,~ıY (X)) for u(X, y1) (b)
Proof (sketch) :
Utility independence implies the existence of functions g > 0, hsuch that
(I) u(X,Y ) = gl(Y ) · u(X, yl) + hl(Y ) ∀yl
Now let yk be the point where~ıY (X) intersects the line (x1, Y ),then u(X,~ıY (X)) = u(x1, yk).
1 set yl to yk;2 solve (I) for x1 to get hk(Y ), using u(x1, yk) = 0
3 solve (I) for xn to get gk(Y );4 solve (I) for~ıY (X) to get u(X, yk).5 substitute these results in (I) to get the desired result. �
246 / 401
An example
Suppose the City of Utrecht assesses the following utilities forCost and Acres lost: