Multi-attribute utility theory (very superficial in textbook!)

Transcription

1 Multiple objectives an example Multi-attribute utility theory (very superficial in textbook!) Consider the problem of deciding upon treatment for a patient with esophageal cancer. The problem can be (schematically) modelled as follows: endo prosthesis surgery L.E. Q.L. 164 / 401 radio therapy chemo therapy 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 1,..., 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

2 Direct assessment an example u 15 Cost Acres Assign a utility of 0 to the worst consequence (60, 600), and a utility of 1 to the best consequence (15, 200). The utility of, for example, consequence (50, 300) can be determined from: p (15, 200) 1.0 (50, 300) (1 p) (60, 600) To find a good representation through direct assessment, utilities must be assessed for a substantial number of points. 168 / Notation Let be an attribute and a set of n 1, n > 0, attributes. Compare the following: if n = 1 then u() is a utility function for in a onedimensional decision problem u(, ) U () (possibly rescaled) if n > 1 then u(, y i ) u(, ) is an n-attribute utility function; u(,y i ) is a subutility function for given a fixed value-assignment y i to attributes ; u () is a conditional utility function for : a (possibly) re-scaled function u(,y k ) for some no longer explicit y k. 169 / 401 Different functional forms Let and be two attributes (generalisation to n > 2 attributes is straightforward). Consider the utility function u(, ) for and. u has an additive form if for constants k and k u(, ) = k u () + k u ( ) u has a multilinear form if for constants k, k, and k u(, ) = k u ()+k u ( )+k u () u ( ) u has a multiplicative form if for constants k, k, c, and c u(, ) = (k u () + c ) (k u ( ) + c ) The constants are often called weights or scaling constants. The different functional forms are only valid under certain assumptions; the values of the scaling constants are thereby often constrained! 170 / 401 An example The City of Utrecht decides to model the utility function u(cost, Acres) as an additive function. Using standard assessment techniques, they find for the conditional utility functions u C and u A that u C (15) = 1.0 u C (30) = 0.5 u C (50) = 0.2 u C (60) = 0.0 u A (200) = 1.0 u A (300) = 0.8 u A (400) = 0.5 u A (600) = 0.0 The City finds Cost three times as important as Acres lost. The resulting utilities for a number of Cost-Acres pairs are then: u(50, 300) = 3 u C (50) + 1 u A (300) = = 1.4 u(30, 400) = = 2.0 u(60, 200) = = 1.0 u(15, 600) = = 3.0 u(15, 200) = = / 401

3 Interpreting weights or scaling constants Reconsider the City of Utrecht s two attribute utility function u(c,a) = 3 u C (C) + 1 u A (A). Suppose that the City has explicitly expressed the indifference (50, 600) (60, 350), which indeed holds given all current functions: u(50, 600) = 3 u C (50) + u A (600) = = 0.6 u(60, 350) = 3 u C (60) + u A (350) = = 0.6 Now, however, the City finds out that all alternatives result in at least a loss of 350 acres of land, and we rescale u C such that u C (350) = 1. The expressed indifference still holds, implying that u(60, 350) = k C 0 + k A 1 = u(50, 600) = k C k A 0 From this it follows that k C = 5 k A. Did C just turn from three times as important as A to five times as important? 172 / 401 Pricing out an example Suppose the following utilities are assessed by the City of Utrecht for Cost and Acres lost: u C (15) = 1.0 u C (30) = 0.7 u C (50) = 0.4 u C (60) = 0.0 u A (200) = 1.0 u A (300) = 0.8 u A (400) = 0.5 u A (600) = 0.0 The City decides that it is willing to sacrifice 200 acres of land if that would save e 10 million. This implies that, for example, u(40, 200) = u(30, 400) that is, a u C (40) + b u A (200) = a u C (30) + b u A (400) We conclude that a b = u A(400) u A (200) = u C (40) u C (30) = And thus find that u(c,a) = 0.5 u C (C) u A (A) Is this valid? 173 / 401 Assessing scaling constants (I) Pricing out: Assess the marginal rate of substitution, that is, determine the value of one objective in terms of another. Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. Let e and e be the units of measurement for the two attributes, respectively. Suppose you are willing to sacrifice s units of for 1 unit of. Then for all x i, y j : u(x i,y j ) = u(x i + e,y j s e ) If u(, ) is additive and u (), u ( ) are linear functions, then this implies that a u (x i ) + b u (y j ) = a u (x i + e ) + b u (y j s e ) As a result, a b = u (y j s e ) u (y j ) u (x i ) u (x i + e ) 174 / 401 Assessing scaling constants (II) Swing weighting: Consider a set of n 2 attributes 1,..., n. Swing weighting assigns weights to attributes based on either a rank-order or a rating on attributes and one consequence. Take the (theoretically) worst possible consequence as benchmark. 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 worst to best value, which would you swing?; Swing(Z); rank(z) i. if i = 1 then rate(z) 100 else rate(z) answer 0, / 401

4 Swing weighting cntd Consider a set of n 2 attributes 1,..., n. Let rank( i ) and rate( i ) denote a ranking and a rating for attribute i, respectively. The weight w( i ) for attribute i can now be determined using either of the following two approaches: rate( i ) direct rating: w( i ) = n j=1 rate( j) Swing weighting an example Attribute consequence to swing mill. e acres rank rate (benchmark) Cost Acres total: rank-sum weighing: w( i ) = reversed rank( i) number of ranks = n rank( i) + 1 n + 1 direct rating: w(c) = = 0.77 w(a) = = 0.23 rank-sum weighting: w(c) = w(a) = = 2 3 = / / 401 Assessing scaling constants (III) Lottery weights: Consider two attributes and (the following extends straightforwardly to n > 2 attributes). Let (x 0,y 0 ) denote the worst possible consequence, and (x +,y + ) the best possible consequence. The weight w() for attribute equals p, where p is the indifference probability that follows from: 1.0 (x +,y 0 ) p (1 p) (x +,y + ) (x 0,y 0 ) Lottery weights: an example Assume once more that the City of Utrecht decides to model the utility function u(cost, Acres) as an additive function: u(c,a) = k C u C (C) + k A u A (A) The weight k C is assessed from: 1.0 (15, 600) The weight k A is assessed using: 1.0 (60, 200) k C (15, 200) (1 k C ) (60, 600) k A (15, 200) (1 k A ) (60, 600) 178 / / 401

5 Additive independence the formal definition Use of an additive utility function is justified given the assumption of additive independence [AI]. MAUT with n = 2 attributes: when can we use additive and multilinear forms? Two attributes and are additive independent if preferences for lotteries over can be established by comparing the values one attribute at a time. More formally, DEFINITION Two attributes and are additive independent if the paired preference comparison of any two lotteries, defined by two joint probability distributions on, depends only on their marginal distributions. 180 / / 401 Interpreting additive independence The ability to establish preferences for lotteries over by comparing the values one attribute at a time entails the following: the decisionmaker should be indifferent between [p, (x 1,y 1 ); (1 p), (x 2,y 2 )] and [p, (x 1,y 2 ); (1 p), (x 2,y 1 )] DEFINITION Additive independence a practical definition Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. and are additive independent iff for an arbitrary pair (x j,y l ), we have for all pairs (x i,y k ) that since both have the same probability of achieving x 1 vs x 2 ; the decisionmaker should also be indifferent between [p, (x 1,y 1 ); (1 p), (x 2,y 2 )] and [p, (x 2,y 1 ); (1 p), (x 1,y 2 )] (x i,y k ) (x j,y l ) (x i,y l ) (x j,y k ) since both have the same probability of achieving y 1 vs y 2 ; this can only hold if p = 1 p = 0.5 Note that additive independence is a symmetric property. 182 / / 401

6 Additive independence an example Let u(, ) be a two-attribute utility function defined as follows: u(x 0,y 0 ) = 1.0 u(x 0,y 1 ) = 0.7 u(x 0,y 2 ) = 0.5 u(x 1,y 0 ) = 0.9 u(x 1,y 1 ) = 0.6 u(x 1,y 2 ) = 0.4 u(x 2,y 0 ) = 0.5 u(x 2,y 1 ) = 0.2 u(x 2,y 2 ) = 0.0 and are additive independent: [0.5, (x 1,y 1 ); 0.5, (x 0,y 0 )] [0.5, (x 1,y 0 ); 0.5, (x 0,y 1 )] [0.5, (x 2,y 1 ); 0.5, (x 0,y 0 )] [0.5, (x 2,y 0 ); 0.5, (x 0,y 1 )] [0.5, (x 1,y 2 ); 0.5, (x 0,y 0 )] [0.5, (x 1,y 0 ); 0.5, (x 0,y 2 )] [0.5, (x 2,y 2 ); 0.5, (x 0,y 0 )] [0.5, (x 2,y 0 ); 0.5, (x 0,y 2 )]... Additive independence the implication for u(, ) Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. If and are additive independent, then for any pair (x j,y l ), and all pairs (x i,y k ), we have by definition that 0.5 u(x i,y k ) u(x j,y l ) = 0.5 u(x i,y l ) u(x j,y k ) that is, the change in utility for values of one attribute is independent of the values of the other attribute: u(x i,y k ) u(x j,y k ) = u(x i,y l ) u(x j,y l ) This means that all subutility functions u(,y k ) are the same, up to translation: u(,y k ) = u(,y l ) + c kl for some constant c kl. 184 / / 401 Exploiting additive independence an example Exploiting Additive independence Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. If and are additive independent, then three points u(x j,y l ), u(x j,y k ), and u(x i,y l ) serve for establishing a forth: u(x i,y k ) u(, ) can be constructed entirely from two intersecting subutility functions: 1. u(,y k ) for some value y k of 2. u(x i, ) for some value x i of To establish u(c, A), the City of Utrecht assesses u(c, 600) for fixed acres: u(15, 600) = 0.75 u(50, 600) = 0.15 u(30, 600) = 0.35 u(60, 600) = 0.00 and u(60,a) for fixed cost: u(60, 200) = 0.25 u(60, 400) = 0.15 u(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 c i and a j : u(c i,a j ) + u(60, 600) = u(c i, 600) + u(60,a j ) All other points can now be computed. For example, u(15, 200) = u(15, 600) + u(60, 200) 0 = / / 401

7 The additive utility function Ia THEOREM Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. Attributes and are additive independent iff the two-attribute utility function is additive: where u(, ) = k u () + k u ( ), u(, ) is normalised with u(x 1,y 1 ) = 0 and u(x n,y m ) = 1; u () is a conditional utility function on, normalised by u (x 1 ) = 0 and u (x n ) = 1; u ( ) is a conditional utility function on, normalised by u (y 1 ) = 0 and u (y m ) = 1; k = u(x n,y 1 ) and k = u(x 1,y m ) are positive scaling constants, summing to 1. LEMMA The additive utility function IIa Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. Attributes and are additive independent iff the two-attribute utility function is additive: u(, ) = u(,y 1 ) + u(x 1, ), where u(x 1,y 1 ) = / / 401 The additive utility function IIb Proof of Lemma: ( ) Let x i and y k be arbitrary values of resp.. Additive independence implies u(x 1,y 1 ) + u(x i,y k ) = u(x 1,y k ) + u(x i,y 1 ) Setting u(x 1,y 1 ) = 0, we get u(x i,y k ) = u(x 1,y k ) + u(x i,y 1 ). ( ) Suppose u(, ) = u(,y 1 ) + u(x 1, ), and let x j and y l be arbitrary values of resp.. Then for all values x i and y k of resp., we have that u(x i,y k ) + u(x j,y l ) = u(x i,y 1 ) + u(x 1,y k ) + u(x j,y 1 ) + u(x 1,y l ) = u(x i,y 1 ) + u(x 1,y l ) + u(x j,y 1 ) + u(x 1,y k ) = u(x i,y l ) + u(x j,y k ) Proof of the Theorem: The additive utility function Ib ( ) From the previous lemma we have that for arbitrary values x i and y k of resp. u(x i,y k ) = u(x 1,y k ) + u(x i,y 1 ) if u(x 1,y 1 ) = 0 Normalising u(x 1, ) gives us u(x 1, ) = k u ( ) with k = u(x 1,y m ); similarly, u(,y 1 ) = k u () with k = u(x n,y 1 ). ( ) Analogous to proof of lemma. We conclude that and are additive independent. 190 / / 401

8 Assessing the additive utility function an example Suppose the City of Utrecht assesses the following conditional utilities for Cost and Acres lost: u C (15) = 1.0 u A (200) = 1.0 u C (30) = 0.5 u A (300) = 0.8 u C (50) = 0.2 u A (400) = 0.5 u C (60) = 0.0 u A (600) = 0.0 For scaling constants k A and k C we know that k A = 1 k C and that k C = u(15, 600). This latter utility is assessed using the following lottery: k C (15, 200) 1.0 (15, 600) (1 k C ) (60, 600) The indifference probability k C is found to be We therefore conclude that u(c,a) = 0.75 u C (C) u A (A) 192 / 401 Utility independence the formal definition Use of a multilinear or multiplicative utility function is justified under (mutual) utility independence [(M)UI]. Attribute is utility independent of attribute if conditional preferences for lotteries over given a fixed value for do not depend on the particular value for. More formally, DEFINITION An attribute is utility independent of an attribute iff for any lotteries [(,y k )] 1 and [(,y k )] 2 over with fixed to value y k, we have [(,y k )] 1 [(,y k )] 2 = [(,y l )] 1 [(,y l )] 2 y of NB: [(, y)] represents a conditional lottery over involving consequences over different values of combined with a fixed value for. 193 / 401 Interpreting utility independence Independence of conditional preferences for lotteries over of the value of, entails the following: if [p, (x 1,y 1 ); (1 p), (x 2,y 1 )] [p, (x 3,y 1 ); (1 p), (x 4,y 1 )] then the decision maker should also prefer [p, (x 1,y 2 ); (1 p), (x 2,y 2 )] over [p, (x 3,y 2 ); (1 p), (x 4,y 2 )] since there is only a change in sure outcome of attribute (y 1 vs y 2 ), which should not affect preferences among lotteries over ; similarly, if (x C,y 1 ) is the certainty equivalent of the first lottery, then (x C,y 2 ) should be the certainty equivalent of the third lottery; Note that the above means that also the preference order on values of should be independent of the value of! 194 / 401 DEFINITION Utility independence a practical definition Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. is utility independent of iff for an arbitrary triple x i,x j,x k with x i x j x k, there exists a probability p such that for all values y l of, we have that p (x i,y l ) 1.0 (x j,y l ) (1 p) (x k,y l ) If is utility independent of and is utility independent of then and are mutually utility independent. Note: the above definition, in terms of a probability equivalent, can also be rephrased in terms of a certainty equivalent for a lottery. 195 / 401

9 An example Let u(, ) be a two-attribute utility function defined as follows: u(x 0,y 0 ) = 1.0 u(x 0,y 1 ) = 1.0 u(x 0,y 2 ) = 0.7 u(x 1,y 0 ) = 0.6 u(x 1,y 1 ) = 0.8 u(x 1,y 2 ) = 0.5 u(x 2,y 0 ) = 0.0 u(x 2,y 1 ) = 0.5 u(x 2,y 2 ) = 0.2 Then we have: is utility independent of : [1.0, (x 1,y 0 )] [0.6, (x 0,y 0 ); 0.4, (x 2,y 0 )] [1.0, (x 1,y 1 )] [0.6, (x 0,y 1 ); 0.4, (x 2,y 1 )] [1.0, (x 1,y 2 )] [0.6, (x 0,y 2 ); 0.4, (x 2,y 2 )] is not utility independent of, as in the context of x 1 we have y 1 y 0 y 2, and in the context of x 2 we have y 1 y 2 y 0! 196 / 401 Utility independence the implication for u(, ) Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. First we observe that x 1 x j x n for each value x j of. Now, if is utility independent of, then we have by definition that for any x j there exists a p j such that for all y l, u(x j,y l ) = p j u(x 1,y l ) + (1 p j ) u(x n,y l ) that is, p j = u(x j,y l ) u(x n,y l ) u(x 1,y l ) u(x n,y l ) is a linear function of u(,y l). This means that all subutility functions u(,y k ) are the same, up to (positive) scaling and translation: u(,y k ) = c kl u(,y l ) + d kl for some constants c kl > 0 and d kl. 197 / 401 PROPOSITION Utility independence an equivalence Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. is utility independent of, iff for each value y l of, there exist real functions g l > 0 and h l, such that u(, ) = g l ( ) u(,y l ) + h l ( ) for all values of and. 198 / 401 Proof of the Proposition (sketch): ( ) First observe that for all x, x 1 x x n. Utility independence holds iff for each x a probability p exists such that (x, ) [p, (x 1, )]; (1 p), (x n, ) and therefore (main theorem) (I) u(, ) = p u(x 1, ) + (1 p) u(x n, ) ( ) = p u(x 1, ) u(x n, ) + u(x n, ) 1 solve p from (I) with set to y l ; 2 substitute this result in (I) to get the desired result. ( ) Let g l > 0 and h l be such that for arbitrary y l : (II) u(, ) = g l ( ) u(,y l ) + h l ( ) 1 for arbitrary x j and y j, choose x i and x k s.t. (x i,y j ) (x j,y j ) (x k,y j ); 2 continuity: p : u(x j,y j ) = p u(x i,y j ) + (1 p) u(x k,y j ); 3 apply (II) to u(x i,y j ), u(x j,y j ) and u(x k,y j ). 4 rewrite to find the desired result. 199 / 401

10 Exploiting utility independence Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. If is utility independent of, then one subutility function u(,y k ) and two points u(x i,y l ), and u(x j,y l ) serve for establishing a second function: u(,y l ) THEOREM Three subutility functions Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. u(, ) can be constructed entirely from three subutility functions: 1. u(,y k ) for some value y k of 2. u(x i, ) for some value x i of 3. u(x j, ) for some value x j of, x j x i If is utility independent of then u(, ) = u(x 1, ) [1 u(,y 1 )] + u(x n, ) u(,y 1 ) where u(, ) is normalised by u(x 1,y 1 ) = 0 and u(x n,y 1 ) = / / 401 An Example Three subutility functions Proof of the Theorem (sketch): Utility independence implies the existence of functions g > 0, h such that (I) 1 set y l to y 1 ; u(, ) = g l ( ) u(,y l ) + h l ( ) y l 2 solve (I) for x 1 to get h 1 ( ), using u(x 1,y 1 ) = 0; 3 solve (I) for x n to get g 1 ( ), using u(x n,y 1 ) = 1; 4 substitute these results in (I) to get the desired result. 202 / 401 Suppose the City of Utrecht assesses the following utilities for Cost and Acres lost, given A fixed at 600, and C fixed at 15 and 60, respectively: u(15, 600) = 0.75 u(15, 200) = 1.00 u(60, 200) = 0.20 u(30, 600) = 0.50 u(15, 300) = 0.90 u(60, 300) = 0.15 u(50, 600) = 0.10 u(15, 400) = 0.80 u(60, 400) = 0.10 u(60, 600) = 0.00 u(15, 600) = 0.75 u(60, 600) = 0.00 Cost is utility independent of Acres lost. We normalise u(c, A) such that u(60, 600) = 0 and u(15, 600) = 1: u(15, 600) = 1.00 u(15, 200) = 1.33 u(60, 200) = 0.27 u(30, 600) = 0.67 u(15, 300) = 1.20 u(60, 300) = 0.20 u(50, 600) = 0.13 u(15, 400) = 1.07 u(60, 400) = 0.13 u(60, 600) = 0.00 u(15, 600) = 1.00 u(60, 600) = 0.00 Now, u(c,a) = u(60,a) [1 u(c, 600)] + u(15,a) u(c, 600) 203 / 401

11 The additive utility function IIIa The additive utility function IIIb COROLLAR Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. If is utility independent of, then the two-attribute utility function is additive iff [0.5, (x 1,y 1 ); 0.5, (x n, )] [0.5, (x 1, ); 0.5, (x n,y 1 )] Under the above conditions, we have that u(, ) = u(x 1, ) + u(,y 1 ) where u(x 1,y 1 ) = 0 and u(x n,y 1 ) = / 401 Proof of the Corollary: Utility independence implies that u(, ) = u(x 1, ) [1 u(,y 1 )] + u(x n, ) u(,y 1 ) with u(x 1,y 1 ) = 0 and u(x n,y 1 ) = 1. The lottery equivalence translates into 0 + u(x n, ) = 1 + u(x 1, ) Substitute u(x n, ) with 1 + u(x 1, ) in the above results in u(, ) = u(x 1, ) + u(,y 1 ) with u(x 1,y 1 ) = 0 and u(x n,y 1 ) = / 401 The multilinear utility function Ia COROLLAR Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. If is utility independent of, then the two-attribute utility function is multilinear iff The multilinear utility function Ia Proof of the Corollary: Utility independence implies that u(, ) = u(x 1, ) [1 u(,y 1 )] + u(x n, ) u(,y 1 ) u(x n, ) = 1 + b u(x 1, ) for some constant b > 0. Under the above conditions, we have that u(, ) = u(x 1, ) + u(,y 1 ) + (b 1) u(x 1, ) u(,y 1 ) with u(x 1,y 1 ) = 0 and u(x n,y 1 ) = 1. Substitute u(x n, ) with 1 + b u(x 1, ) in the above results in u(, ) = u(x 1, ) + u(,y 1 ) + (b 1) u(x 1, ) u(,y 1 ) with u(x 1,y 1 ) = 0 and u(x n,y 1 ) = 1. where u(x 1,y 1 ) = 0 and u(x n,y 1 ) = / / 401

12 Exploiting mutual utility independence Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. If and are mutually utility independent, then u(, ) can be constructed from two subutility functions and one point: 1. u(,y k ) for some value y k of 2. u(x i, ) for some value x i of 3. u(x j,y l ) for some values x j x i of, and y l y k of if is utility independent of, but is utility dependent of : if and are mutually utility independent: 208 / 401 Exploiting mutual utility independence example Suppose the utilities, given fixed A = 60 resp. C = 600, assessed by the City of Utrecht are: u(15, 600) = 0.75 u(30, 600) = 0.35 u(50, 600) = 0.15 u(60, 600) u(60, 200) = 0.25 u(60, 300) = 0.20 u(60, 400) = 0.15 = 0.00 In addition, the utility u(30, 300) = 0.30 is assessed. Mutual utility independence implies strategical equivalence of all horizontal functions with u(c, 600) and all vertical functions with u(60, A): 600 A C we first determine the function u(c, 300); we then use u(c, 300) to derive all vertical functions from u(60,a). 209 / 401 Example continued (I) The utilities assessed by the City of Utrecht: u(15, 600) = 0.75 u(30, 600) = 0.35 u(50, 600) = 0.15 u(60, 600) u(60, 200) = 0.25 u(60, 300) = 0.20 u(60, 400) = 0.15 = 0.00 In addition, the utility u(30, 300) = 0.30 is assessed. we first determine the function u(c, 300): since C is UI of A, there exist functions g and h such that e.g.: u(c, 300) = g 600 (300) u(c, 600) + h 600 (300) We require two equations to find g 600 (300) and h 600 (300): u(30, 300) = 0.30 = g 600 (300) u(30, 600) + h 600 (300) = g 600 (300) h 600 (300) u(60, 300) = 0.20 = g 600 (300) h 600 (300) resulting in u(c, 300) = 0.29 u(c, 600) / 401 Example continued (II) The utilities assessed by the City of Utrecht: u(15, 600) = 0.75 u(30, 600) = 0.35 u(50, 600) = 0.15 u(60, 600) u(60, 200) = 0.25 u(60, 300) = 0.20 u(60, 400) = 0.15 = 0.00 In addition, the utility u(30, 300) = 0.30 is assessed. we then use u(c, 300) to derive all vertical functions from u(60,a): since A is UI of C, there exist functions f and k such that e.g.: u(c,a) = f 60 (C) u(60,a) + k 60 (C) We require two equations to find f 60 (C) and k 60 (C): u(c, 300) = f 60 (C) u(60, 300) + k 60 (C) u(c, 600) = f 60 (C) u(60, 600) + k 60 (C) Given that u(c, 300) and u(c, 600) are known functions, we can compute an f 60 (c i ) and k 60 (c i ) for each c i. 211 / 401

13 THEOREM The multilinear utility function IIa Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. If and are mutually utility independent then the two-attribute utility function is multilinear: u(, ) = k u () + k u ( ) + k u () u ( ) where u(, ) is normalised by u(x 1,y 1 ) = 0 and u(x n,y m ) = 1; u () is a conditional utility function on, normalised by u (x 1 ) = 0 and u (x n ) = 1; u ( ) is a conditional utility function on, normalised by u (y 1 ) = 0 and u (y m ) = 1; k = u(x n,y 1 ) > 0, k = u(x 1,y m ) > 0, k = 1 k k are scaling constants. 212 / 401 LEMMA The multilinear utility function IIIa Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. If and are mutually utility independent then the two-attribute utility function is multilinear: u(, ) = u(,y 1 ) + u(x 1, ) + k u(,y 1 ) u(x 1, ) where u(x 1,y 1 ) = 0, u(x n,y 1 ) > 0 and u(x 1,y m ) > 0; k = u(x n,y m ) u(x n,y 1 ) u(x 1,y m ) is a scaling constant. u(x n,y 1 ) u(x 1,y m ) 213 / 401 The multilinear utility function IIIb Proof of Lemma: (sketch): Mutual utility independence implies the existence of functions f > 0,g > 0,h, and k such that for arbitrary x l and y l (I) u(, ) = g l ( ) u(,y l ) + h l ( ) (II) u(, ) = f l () u(x l, ) + k l () and The multilinear utility function IIb Proof of the Theorem: From the previous lemma we have that for arbitrary values x i and y j of resp., we have u(x i,y j ) = u(x i,y 1 ) + u(x 1,y j ) + k u(x i,y 1 ) u(x 1,y j ) if u(x 1,y 1 ) = 0, u(x n,y 1 ) > 0 and u(x 1,y m ) > 0; 1 set y l to y 1 and x l to x 1 2 solve (I) for x 1 to get h 1 ( ) and for x n to get g 1 ( ) 3 solve (II) for y 1 to get k 1 () and for y m to get f 1 () 4 substitute these results in (I) and (II) (I*) and (II*) 5 now solve (II*) for x n and fill in this result in (I*) to get the desired result. Normalising u(x 1, ) gives us u(x 1, ) = k u ( ) with k = u(x 1,y m ). Similarly, u(,y 1 ) = k u () with k = u(x n,y 1 ). Finally, define k = k k k. 214 / / 401

14 Assessing a multilinear utility function: example Suppose the City of Utrecht assesses the following conditional utilities for Cost and Acres lost: u C (15) = 1.0 u C (30) = 0.5 u C (50) = 0.2 u C (60) = 0.0 u A (200) = 1.0 u A (300) = 0.8 u A (400) = 0.5 u A (600) = 0.0 Constants k C = u(15, 600) and k A = u(60, 200) are assessed: (15, 600) (60, 200) k C (15, 200) (1 k C ) (60, 600) k A (15, 200) (1 k A ) (60, 600) The indifference probabilities found: k C = 0.75 and k A = So, u(c,a) = 0.75 u C (C) u A (A) 0.2 u C (C) u A (A) 216 / 401 Interpreting scaling constants (I) Let and be two attributes such that ranges from 0 to 100 and u(, ) = 0.25 u () u ( ). Suppose that u(0, 10) = u(100, 0). 100 u = 0.75 = k u = 1 u = 0.25 u = 0 u = 0.25 = k 100 u = a = k u = u = 1 u = a = k Suppose we decide it is sufficient for to range from 0 to 10. Then rescaling results in: k = U (100, 0) = U (0, 10) = k Did just turn from three times as important as to equally important? 217 / 401 Interpreting scaling constants (II) Consider a multilinear utility function u(, ) over attributes and with values x 1... x n, n 2, resp. y 1... y m, m 2. k : consider two lotteries over values x i x j and y k y l : A: (x i,y k ) (x j,y l ) B: (x i,y l ) (x j,y k ) A B k = 0 A B k > 0 (, are complements) A B k < 0 (, are substitutes) k, k : if you would rather swing x 1 to x n than y 1 to y m, then k > k, and vice-versa; Even a mighty important attribute will have a small scaling constant if its range is relatively small! 218 / 401 THEOREM The multiplicative utility function Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. If and are truly mutually utility independent then the two-attribute utility function is multiplicative: u(, ) = (k u(,y 1 ) + 1) (k u(x 1, ) + 1) where u(x 1,y 1 ) = 1, (x n,y 1 ) > 1 and u(x 1,y m ) > 1; k = u(x n,y m ) u(x n,y 1 ) u(x 1,y m ) > 0 is a scaling u(x n,y 1 ) u(x 1,y m ) constant. 219 / 401

15 The multiplicative utility function Proof of the Theorem: Mutual utility independence implies that u(, ) is multilinear: u(, ) = u(,y 1 ) + u(x 1, ) + k u(,y 1 ) u(x 1, ) where u(x 1,y 1 ) = 0, u(x n,y 1 ) > 0 and u(x 1,y m ) > 0; k = u(x n,y m ) u(x n,y 1 ) u(x 1,y m ) is a scaling constant. u(x n,y 1 ) u(x 1,y m ) Now, u(, ) u (, ) = k u(, ) + 1, where k = k > 0. u (, ) is multiplicative: u (, ) = k (u(,y 1 ) + u(x 1, ) + k u(,y 1 ) u(x 1, )) + 1 = (k u(,y 1 ) + 1) (k u(x 1, ) + 1) = u (,y 1 ) u (x 1, ) where u (x 1,y 1 ) = 1, u (x n,y 1 ) > 1 and u (x 1,y m ) > / 401 COROLLAR The additive utility function IVa Let and be two attributes with values x 1,...,x n, n 2, and y 1,...,y m, m 2. If and are mutually utility independent then the two-attribute utility function is additive if [0.5, (x 1,y k ); 0.5, (x i,y 1 )] [0.5, (x 1,y 1 ); 0.5, (x i,y k )] for some values x i,y k for which (x 1,y 1 ) (x 1,y k ) and (x 1,y 1 ) (x i,y 1 ) Under the above conditions, we have that u(, ) = u(x 1, ) + u(,y 1 ) where u(x 1,y 1 ) = 0, u(x n,y 1 ) > 0 and u(x 1,y m ) > / 401 The additive utility function IVb Proof of the Corollary (sketch): Given mutual utility independence, we have that u(, ) = u(,y 1 ) + u(x 1, ) + k u(,y 1 ) u(x 1, ) where u(x 1,y 1 ) = 0, u(x n,y 1 ) > 0, u(x 1,y m ) > 0. Let x i,y k be values such that u(x 1,y k ) + u(x i,y 1 ) = u(x 1,y 1 ) + u(x i,y k ) that is, u(x 1,y k ) + u(x i,y 1 ) = u(x i,y k ). Under this constraint, we find from the multilinear form that k u(x i,y 1 ) u(x 1,y k ) = u(x 1,y k ) + u(x i,y 1 ) u(x i,y 1 ) u(x 1,y k ) = 0 Since u(x i,y 1 ) u(x 1,y 1 ) and u(x 1,y k ) u(x 1,y 1 ), k must be zero. 222 / 401 An example Let and be two attributes with values x 0,x 1,x 2 and y 0,y 1,y 2. Let u(, ) be a two-attribute utility function defined as follows: u(x 0,y 0 ) = 1.0 u(x 0,y 1 ) = 0.9 u(x 0,y 2 ) = 0.5 u(x 1,y 0 ) = 0.8 u(x 1,y 1 ) = 0.7 u(x 1,y 2 ) = 0.3 u(x 2,y 0 ) = 0.5 u(x 2,y 1 ) = 0.4 u(x 2,y 2 ) = 0.0 and are mutually utility independent and for x 0 and y 0 we have and u(x 0,y 0 ) + u(x 2,y 2 ) = u(x 0,y 2 ) + u(x 2,y 0 ) u(x 0,y 0 ) { u(x0,y 2 ) u(x 2,y 0 ) u(, ) is therefore additive. 223 / 401

16 Preferential independence the formal definition Attribute is preferentially independent of attribute [PI] if conditional preferences for values of given a fixed value for do not depend on the particular value for. More formally, Verifying independences DEFINITION An attribute is preferentially independent of an attribute iff for any consequences (x i,y k ) and (x j,y k ) over with fixed to y k, we have (x i,y k ) (x j,y k ) = (x i,y l ) (x j,y l ) y l of If is preferentially independent of and is preferentially independent of then and are mutually preferential independent [(M)PI]. 224 / / 401 Preferential independence example Let and have values x 0,x 1,x 2 and y 0,y 1,y 2. Assume the two-attribute utility function is defined as: u(x 0,y 0 ) = 1.0 u(x 0,y 1 ) = 1.0 u(x 0,y 2 ) = 0.7 u(x 1,y 0 ) = 0.6 u(x 1,y 1 ) = 0.8 u(x 1,y 2 ) = 0.5 u(x 2,y 0 ) = 0.0 u(x 2,y 1 ) = 0.5 u(x 2,y 2 ) = 0.2 Then we have: is preferentially independent of : x 0 x 1 x 2 for each value of ; is not preferentially independent of : y 1 y 0 y 2 for = x 1 y 1 y 2 y 0 for = x 2! Validating preferential independence Let and be two attributes with values x 1,...,x n, n 2, and y 1,...,y m, m 2. Use the following procedure to verify whether is preferentially independent of : 1 choose some consequence (x j,y 1 ); 2 ask the decision maker for a value x i for which (x i,y 1 ) (x j,y 1 ) for some value x j of and y 1 of ; 3 for a number of values y y 1 of, ask the decision maker whether (x i,y) (x j,y) still holds; 4 repeat steps 1, 2 and 3 for different values of. 5 check orientation (no preference reversal?!) 226 / / 401

17 PROPOSITION UI implies PI Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. If is utility independent of then is preferentially independent of. PI implies UI? If is preferentially independent of then not necessarily utility independent of. Counter example: Consider the following utility function: Proof: Consider x i, x j and y k such that (x i,y k ) u(x j,y k ), that is, u(x i,y k ) u(x j,y k ) 0. UI now implies the existence of functions g > 0 and h such that for arbitrary y l : u(x i,y k ) = g l (y k ) u(x i,y l ) + h l (y k ) and u(x j,y k ) = g l (y k ) u(x j,y l ) + h l (y k ) From g l (y k ) > 0, we now have u(x i,y l ) u(x j,y l ) 0 for arbitrary y l. 228 / 401 u(x 0,y 0 ) = 1.0 u(x 1,y 0 ) = 0.8 u(x 2,y 0 ) = 0.3 u(x 0,y 1 ) = 0.5 u(x 1,y 1 ) = 0.1 u(x 2,y 1 ) = 0.1 PI since y i : x 0 x 1 x 2 ; we have no UI : (x 1,y) [p, (x 0,y); (1 p), (x 2,y)] holds for p 0.71 if y y 0, and for p = 0, if y y / 401 Validating utility independence If PI, then is also utility independent of if in the following procedure x C is equivalent for all values of : 1 choose a lottery [0.5,P; 0.5,Q] where P = (x 1,y) and Q = (x n,y) for some value y of and values x 1 and x n of ; 2 ask the decision maker whether or not he prefers the lottery [0.5,P; 0.5,Q] to a consequence (x i,y) for some x i, x 1 x i x n ; 3 repeat step 2 until you converge to the certainty equivalent (x C,y) of the lottery; 4 repeat steps 1 3 for different values of. Repeat the procedure for different pairs of values for. y m P P P y 1 x 1 x C x C x C x n Q Q Q 230 / 401 PROPOSITION AI implies (M)UI Consider two attributes and with values x 1,...,x n, n 2, and y 1,...,y m, m 2. If and are additive independent then and are mutually utility independent Proof (sketch): We prove UI ; UI is analogous. AI implies u(,y k ) = u(,y 1 ) + u(x 1,y k ) for arbitrary y k, where u(x 1,y k ) is constant w.r.t the value of. Continuity implies, for arbitrary x i, x j and x k with (x i,y 1 ) (x j,y 1 ) (x k,y 1 ), that p such that (I) u(x j,y 1 ) = p u(x i,y 1 ) + (1 p) u(x k,y 1 ). Substitution of each u(,y 1 ) in (I) with u(,y k ) u(x 1,y k ) gives u(x j,y k ) = p u(x i,y k ) + (1 p) u(x k,y k ) for arbitrary y k. 231 / 401

18 MUI implies AI? If and are mutually utility independent then and are not necessarily additive independent. Counter argument: We have seen that an additional assumption is necessary to conclude additive independence given mutual utility independence (see Additive utility function IVa). The mentioned corollary can be used to validate AI given MUI; another option is to assume that AI holds / 401 Validating additive independence example Suppose the City of Utrecht assesses the following conditional utilities for Cost and Acres lost: u C (15) = 1.0 u C (30) = 0.5 u C (50) = 0.2 u C (60) = 0.0 u A (200) = 1.0 u A (300) = 0.8 u A (400) = 0.5 u A (600) = 0.0 Suppose k C is assessed using the following lottery: 1.0 (15, 600) k C (15, 200) (1 k C ) (60, 600) resulting in k C = As a consistency check, k A is assessed as well: k A (15, 200) 1.0 (60, 200) (1 k A ) (60, 600) resulting in k A = We now have that k C + k A = ! 233 / 401 Summary The different functional forms for u(, ) are valid under the following conditions: additive: iff AI (I, II) if UI and a lottery assumption for (III) if MUI and a lottery assumption for (x i,y k ) (IV) multi-lin.: if UI and a specific s.e. assumption (I) if MUI (II, III) if AI (k = 0) (III) MAUT with n = 2 attributes: isopreference curves multiplic.: if multi-linear and k / / 401

19 Isopreference curves definition Utility functions with one utility independent attribute Let and be two attributes. Suppose that is utility independent of, but the reverse does not hold. Recall that the two-attribute utility function can be specified using three subutility functions: u(, ) = u(x 1, ) [1 u(,y 1 )] + u(x n, ) u(,y 1 ) where u(x 1,y 1 ) = 0 and u(x n,y 1 ) = 1. Other options are: replacing one or two subutility functions with one or two, respectively, isopreference curves. An isopreference curve describes a set of consequences that are equally desirable to the decision maker. DEFINITION Consider two (sets of) attributes and and a partial function i : such that for arbitrary values x i x k of and y j,y l of, we have that i(x i ) = y j and i(x k ) = y l (x i,y j ) (x k,y l ) An isopreference curve is an interpolant ı () of i(). Note that for any two points (x i,y j ) and (x k,y l ) on the isopreference curve, we thus have that u(x i,y j ) = u(x k,y l ) = c for some constant c. 236 / 401 As a result, u(, ı ()) = c is a constant subutility function, defined on all values of. 237 / 401 y m Isopreference curves details (I) ı 2 () Isopreference curves details (II) y m (x i, ı 2 ()) ı 2 () y j (x i, ı 2 (x i )) y j ı 1 (x j ) ı 1 () ı 1 (x j ) y i ı 1 () y i y 1 x 1 x j Here we have for all values x of that (x, ı 1 (x)) (x 1,y i ), that is: u(x, ı 1 (x)) = u(x 1,y i ) (x, ı 2 (x)) (x 1,y j ), that is: u(x, ı 2 (x)) = u(x 1,y j ) x n 238 / 401 y 1 x 1 x i x j u(x i, ı 2 (x i )) is the utility of a (and any) point on the isopreference curve ı 2 (); (x i, ı 2 ()) is a line in an -subspace, intersecting ı 2 (); u(x i, ı 2 ()) is a subsub utility function, defined on a single value of and a subset of. 239 / 401 x n

20 An example Jenny wants to treat her friends to some cookies and candy. Let x i denote i cookies and y j indicate j pieces of candy. Jenny is indifferent between (x 2,y 10 ), (x 6,y 6 ), and (x 8,y 4 ). Also, Jenny is indifferent between (x 1,y 9 ), (x 3,y 5 ), and (x 7,y 3 ). y 10 y 2 ı 2 () ı 1 () x 1 x 8 Assessment of one utility for a single point on the ı 1 () curve gives the utility of all points (x, ı 1 (x)). We have for example that ı 1 (x 2 ) = y 10 and ı 2 (x 3 ) = y 5, but also ı 1 (x 4 ) = y 8 and ı 2 (x 4 ) = y / 401 One isopreference curve for one subutility function Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. Recall that if is utility independent of then u(, ) = u(x 1, ) [1 u(,y 1 )] + u(x n, ) u(,y 1 ) Replace subutility function u(x n, ) over all y i by subutility function u( ı ( ), ) over all y i : y m y 1 x 1 x k x n Replace subutility function u(,y 1 ) over all x i by subutility function u(, ı ()) over all x i : y m y k y 1 x 1 x k x n 241 / 401 COROLLAR Substitution of u( ı ( ), ) for u(x n, ) (a) Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. If is utility independent of then [ ] u(xk,y 1 ) u(x 1, ) u(, ) = u(x 1, ) + u(,y u( ı 1 ) ( ),y 1 ) where u(x 1,y 1 ) = 0 ı ( ) is defined such that ( ı ( ), ) (x k,y 1 ) for an arbitrary x k x 1. Proof (sketch): Substitution of u( ı ( ), ) for u(x n, ) (b) Utility independence implies the existence of functions g > 0, h such that (I) 1 set y l to y 1 ; u(, ) = g l ( ) u(,y l ) + h l ( ) y l 2 solve (I) for x 1 to get h l ( ), using u(x 1,y 1 ) = 0; 3 let x k be the point where ı ( ) intersects the line (,y 1 ), then u( ı ( ), ) = u(x k,y 1 ); use this in solving (I) for ı ( ) to get g l ( ); 4 substitute these results in (I) to get the desired result. 242 / / 401

21 An example Suppose the City of Utrecht assesses the following utilities for Cost and Acres lost: u(15, 600) = 0.75 u(60, 100) = 0.25 u(30, 600) = 0.50 u(60, 200) = 0.20 u(45, 600) = 0.20 u(60, 300) = 0.15 u(50, 600) = 0.10 u(60, 400) = 0.10 u(60, 600) = 0.00 u(60, 600) = 0.00 In addition, the City indicates the indifferences (50, 200) (45, 300) (40, 400) resulting in an isopreference curve ı C (A) with u( ı C (A),A) = 0.40 and e.g. ı C (600) = 35 (does not follow from above). If Cost is utility independent of Acres lost, then [ ] u(35, 600) u(60,a) u(c,a) = u(60,a) + u(c, 600) u( ı C (A), 600) 244 / 401 COROLLAR Substitution of u(, ı ()) for u(,y 1 ) (a) Let and be two attributes with values x 1... x n, n 1, and y 1... y m, m 1. If is utility independent of then where u(, ) = u(x 1, ) u(x n, ı ()) u(x n, ) u(x 1, ı ()) u(x n, ı ()) u(x 1, ı ()) ı () is defined such that (, ı ()) (x 1,y k ) for an y k with: u(x 1,y k ) = 0, (x 1 x n ) 245 / 401 Proof (sketch): Substitution of u(, ı ()) for u(,y 1 ) (b) Utility independence implies the existence of functions g > 0, h such that (I) u(, ) = g l ( ) u(,y l ) + h l ( ) y l Now let y k be the point where ı () intersects the line (x 1, ), then u(, ı ()) = u(x 1,y k ). 1 set y l to y k ; 2 solve (I) for x 1 to get h k ( ), using u(x 1,y k ) = 0 3 solve (I) for x n to get g k ( ); 4 solve (I) for ı () to get u(,y k ). 5 substitute these results in (I) to get the desired result. 246 / 401 An example Suppose the City of Utrecht assesses the following utilities for Cost and Acres lost: u(15, 200) = 1.00 u(60, 200) = 0.20 u(15, 300) = 0.90 u(60, 300) = 0.15 u(15, 400) = 0.80 u(60, 400) = 0.10 u(15, 600) = 0.75 u(60, 600) = 0.00 u(15, 900) = 0.50 u(60, 900) = 0.15 In addition, the City indicates the indifferences (15, 1500) (50, 700) (60, 600) resulting in an isopreference curve ı A (C) with u(c, ı A (C)) = 0.00 and e.g. ı A (30) = 900. If Cost is utility independent of Acres lost, then u(c,a) = u(60,a) u(15, ı A (C)) u(15,a) u(60, ı A (C)) u(15, ı A (C)) u(60, ı A (C)) 247 / 401

22 Use of two isopreference curves Recall that if UI then u(, ) = u(x 1, ) [1 u(,y 1 )] + u(x n, ) u(,y 1 ) Replacing u(x 1, ) by u( ı 1 ( ), ), and u(x n, ) by u( ı 2 ( ), ): y m COROLLAR Substitution of u( ı 1 ( ), ) and u( ı 2 ( ), ) for u(x 1, ) and u(x n, ) (a) Let and be two attributes with values x 1... x n, n 2, and y 1... y m, m 2. If is utility independent of then y 1 x 1 x k Is there another possibility with two isopreference curves? x n 248 / 401 u(, ) = u(,y 1) u( ı 1 ( ),y 1 ) u( ı 2 ( ),y 1 ) u( ı 1 ( ),y 1 ) where u(, ) is normalised by u(x k,y 1 ) = 0 and u(x l,y 1 ) = 1; ı 1 ( ) is defined such that ( ı 1 ( ), ) (x k,y 1 ); ı 2 ( ) is defined such that ( ı 2 ( ), ) (x l,y 1 ) 249 / 401 Proof (sketch): Substitution of u( ı 1 ( ), ) and u( ı 2 ( ), ) for u(x 1, ) and u(x n, ) (b) Utility independence implies g > 0,h such that (I) u(, ) = g( ) u(,y l ) + h( ) y l Let x k, x l be the intersection points for ı 1 ( ), ı 2 ( ), resp., with (,y 1 ). 1 set y l to y 1 ; 2 solve (I) for x k and ı 1 ( ) to get h( ) and g( ), using u( ı 1 ( ), ) = u(x k,y 1 ) = 0; 3 solve (I) for ı 2 ( ) to get u(x k, ), using u( ı 2 ( ), ) = u(x l,y 1 ) = 1; 4 substitute these results in (I). 250 / 401 An example Suppose the City of Utrecht assesses the following utilities for Cost and Acres lost: u(5, 600) = 1.00 u(50, 600) = 0.10 u(12, 600) = 0.70 u(60, 600) = 0.00 u(15, 600) = 0.75 u(75, 600) = 0.10 u(30, 600) = 0.50 In addition, the City indicates the indifferences (80, 200) (75, 300) (70, 400) (60, 600) resulting in isopreference curve ı C 1 (A) with u( ı C 1 (A),A) = 0.00, and the indifferences (15, 200) (12, 300) (10, 400) (5, 600) resulting in isopreference curve ı C 2 (A) with u( ı C 2 (A),A) = If Cost is utility independent of Acres lost, then u(c,a) = u(c, 600) u( ı C 1 (A), 600) u( ı C 2 (A), 600) u( ı C 1 (A), 600) 251 / 401