Lecture 3: Making Decisions with Multiple Objectives Under Certainty
Keywords
Lecture Outline
Preferential independence
EWL 5.1 Value function and preference
Additive value function
EWL 5.2 Methods for determining value
functions
Non-additive value function
Bisection method
Difference standard sequence
technique (DSST)
Direct-rating method
EWL 6.1 Value functions for multiple
attributes
EWL 6.2 The additive model
EWL 6.3 Requirements for the
applicability of the additive model
CC 5 Multi Attribute Utility Theory
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In this lecture, we will talk about how to approach to decision problems that
satisfy the following assumptions:
1. Preferences are rational (complete and transitive)
2. No uncertainty about the outcome of each alternative
3. Multiple (conflicting) objectives
The procedure for solving a decision problem under certainty:
1. Determine
the fundamental objectives
how to measure the achievement of these objectives (attributes)
the set of alternatives that might achieve these goals
2. Apply the Multi-Attribute Value Theory (MAVT)
Assign value scores to each attribute level for all alternatives
Determine the weight (the relative importance) of each attribute
Rank all alternatives according to weighted-average total score
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What is “conflicting objectives”?
Example: Choosing an automobile
Objectives: Minimize price, maximize life span
Alternatives: Model a, b and c
Example: Choosing an automobile with conflicting objectives
Objectives: Minimize price, maximize life span
Alternatives: Model a, b and c
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Which automobile to choose?
The answer depends on how much you are willing to pay to increase the life span.
Because the alternatives with longer expected life span are more expensive, there is no clear
“winner” among the three alternatives.
The tradeoff between the objectives “Price” and “Life span” must be considered to determine
the most preferred alternative.
The three alternatives can be ranked only if some procedure is used to combine the two
attributes into a single index of the overall desirability of an alternative.
Example
Ultimate objective: Best car
Lower-level fundamental objectives: Minimize price and maximize life span
Alternatives: Model a, Model b, Model c
Attributes: Price (attribute 1) and Life span (attribute 2)
Alternatives: a = (a1 , a2 ), b = (b1 , b2 ), c = (c1 , c2 )
Example: a = ($17K , 12 years)
Attribute value functions: v1 (.) and v2 (.) (subjective value scores)
Example: v1 ($17K ) < v1 ($10K )
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Overall value of alternative a: V (a) = f (v1 (a1 ), v2 (a2 ))
Example: V (a) = f (v1 ($17K ), v2 (12 years))
How to aggregate attribute values?
1. Additive value function (also called “weighted average method”)
V (a) = w1 v1 (a1 ) + w2 v2 (a2 ) + w3 v3 (a3 )
Simple
Requires Mutual Preferential Independence
2. Non-additive value function
V (a)
=
w1 v1 (a1 ) + w2 v2 (a2 ) + w3 v3 (a3 )
+
kw1 w2 v1 (a1 )v2 (a2 ) + kw1 w3 v1 (a1 )v3 (a3 ) + kw2 w3 v2 (a2 )v3 (a3 )
+
k 2 w1 w2 w3 v1 (a1 )v2 (a2 )v3 (a3 )
Permits interactions across many attributes
Complex
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When is it safe to assume an additive multi-attribute value function?
If mutual preferential independence is satisfied, then
preferences can be represented by an additive ordinal value function.
If additive difference independence is satisfied, then
preferences can be represented by an additive cardinal value function.
Ordinal versus Cardinal Value Functions (See the Math Handout)
If a value function is ordinal, then
BMW Audi ⇔ v (BMW ) > v (Audi)
What else? Nothing! It only ranks the alternatives, and the value differences are not
meaningful.
For instance, if v (Audi) = 0.5, v (BMW ) = 0.7, v (Mercedes) = 0.8, then the following
does not necessarily hold:
Audi → BMW BMW → Mercedes
With a cardinal value function, the value differences reflect preferences over
transitions: if v (Audi) = 0.5, v (BMW ) = 0.7, v (Mercedes) = 0.8, then
Audi → BMW BMW → Mercedes
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Simple Preferential Independence (required for mutual independence)
Example: The best meal
Wine (Attribute 1): red or white
Dish (Attribute 2): beef or chicken
Preference Statement 1: “I prefer beef over chicken”
⇒ (white, beef ) (white, chicken) and (red, beef ) (red, chicken)
Preference Statement 2: “I prefer white wine with chicken and red wine with beef”
⇒ (white, chicken) (red, chicken) and (red, beef ) (white, beef )
Is “dish” independent of wine? Yes!
Is “wine” independent of dish? No!
Definition: Simple Preferential Independence
Let a = (a1 , ..., ai , ..., am ) and b = (a1 , ..., bi , ..., am ) be two alternatives that differ in attribute i only. Let
0
0
a0 = (a10 , ..., ai , ..., am
) and b0 = (a10 , ..., bi , ..., am
) be two alternatives that also differ in attribute i only, and
have the same values in attribute i as alternatives a and b respectively.
We call attribute Xi (simple) preferential independent of the remaining attributes if and only if
a (∼)b ⇐⇒ a0 (∼)b0 for any a, a’, b, b’.
Example: Attribute wine is (simple) preferential independent of dish if
(red, beef ) (white, beef ) and (red, chicken) (white, chicken)
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Mutual Preferential Independence
Definition: Mutual Preferential Independence
The attributes X1 , ..., Xm are mutually preferential independent if each possible subset
of attributes is preferential independent of the complementary set.
Example: Car choice with attributes “color”, “price” and “maximum speed”
“color” is preferential independent of “price” and “maximum speed”.
“price” is preferential independent of “color” and “maximum speed”.
“maximum speed” is preferential independent of “color” and “price”.
“color” and “price” are preferential independent of “maximum speed”.
Example: (Black, $30,000, 220km/hr) (White, $20,000, 220km/hr)
⇐⇒ (Black, $30,000, 250km/hr) (White, $20,000, 250km/hr)
“price” and “maximum speed” are preferential independent of “color”.
“maximum speed” and “color” are preferential independent of “price”.
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Example: Mutual Preferential Independence
(White, $X, Ykm/hr) (Black, $X, Ykm/hr) for any X and Y
(Zcolor, $20,000, Ykm/hr) (Zcolor, $30,000, Ykm/hr) for any Z and Y
Simple preferential independence is satisfied BUT
(White, $20,000, 220km/hr) (Black, $30,000, 220km/hr)
(Black, $30,000, 250km/hr) (White, $20,000, 250km/hr)
Mutual preferential independence is NOT
REMARK: Simple preferential independence is about preferences over a single
attribute whereas mutual preferential independence is about preferences over a
subset of attributes.
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Additive Difference Independence
Simple preferential independence: preferences over attribute levels of a particular
attribute should not depend on the level of other attributes
Additive difference independence: preferences over transitions between attribute
levels of a particular attribute should not depend on the level of other attributes
Example: The car choice problem
Example: A=(Black, $30,000, 220km/hr) → B=(Black, $30,000, 250km/hr)
∼ C=(White, $20,000, 220km/hr) → D=(White, $20,000, 250km/hr)
The attribute “maximum speed” is difference independent of the attributes “price” and
“color” if the additional value attached to a particular increase in maximum speed (e.g.
from 200 km/hr to 220 km/hr) is independent of the vehicle being a $30,000 black car
or a $20,000 white car.
Definition: Additive Difference Independence
Let a = (a1 , ..., ai , ..., am ) and b = (a1 , ..., bi , ..., am ) be two alternatives that differ in attribute i
0 ) and b 0 = (a0 , ..., b , ..., a0 ) be two alternatives that also differ in
only. Let a0 = (a10 , ..., ai , ..., am
i
m
1
attribute i only, and have the same values in attribute i as alternatives a and b respectively.
We call attribute Xi additive difference independent of the remaining attributes if and only if
(a → b) ∼ (a0 → b0 ) for any a, a’, b, b’.
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What if preferences are not independent?
Cannot use additive value functions!
Try to redefine attributes in order to eliminate existing dependencies.
If not, use a non-additive value function:
V (a) = w1 v1 (a1 ) + w2 v2 (a2 ) + kw1 w2 v1 (a1 )v2 (a2 )
quality of life
duration
The last term captures the interaction between the attributes.
If k is positive, then the two attributes complement each other.
If k is negative, then the two attributes are substitutes.
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What if preferences are not independent?
Example 1: Medical treatment with attributes “quality of life” and “duration”
A = (good, long) B = (good, short)
(( + w2 v2 (long) > (
(( + w2 v2 (short) = V (B)
V (A) = (
w1(
v1 (good)
w1(
v1 (good)
(
(
C = (bad, short) D = (bad, long)
+ w2 v2 (long) = V (D)
+ w2 v2 (short) > V (C) = w1
v1 (bad)
w1
v1 (bad)
CONTRADICTION! MUTUAL PREFERENTIAL INDEPENDENCE IS VIOLATED
Example 2: Medical treatment with attributes “quality of life” and “duration”
A = (good, 10years) → B = (good, 12years) C = (bad, 10years) → D = (bad, 12years)
V (B) − V (A) = w2 [v2 (12years) − v2 (10years)]
> V (D) − V (C) = w2 [v2 (12years) − v2 (10years)]
CONTRADICTION! ADDITIVE DIFFERENCE INDEPENDENCE IS VIOLATED
REMARK: In general, we observe preferences satisfying mutual preferential
independence more often than additive difference independence.
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Additive value function
n
V (a) =
∑ wr · vr (ar ) = w1 · v1 (a1 ) + . . . + wn · vn (an )
r =1
vr : Compare the value of
different attribute levels
wr : Compare different attributes
in terms of importance
∑nr=1 wr = 1 and wr > 0
vr (xr− ) = 0 and vr (xr+ ) = 1
Is this a sensible approach? Concepts of preference independence:
X
Ordinal: Mutual Preferential Independence
Cardinal: Additive Difference Independence
Where do vr come from? Value function elicitation methods:
?
Bisection method
DSST
Direct-Rating method
Where do the wr come from?
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?
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Common forms of attribute value functions
Attribute value functions convert attribute levels into levels of “desirability”,
“worth”, “utility”
There is no right or wrong value function for an attribute
The shape of a value function depends on the decision maker’s personal
preferences.
While attribute levels are objective, the level of “desirability” is inherently
subjective.
Value functions do not need to be perfect but they should capture the
preferences of the DM well enough to understand and analyze the current
situation.
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General procedure for deriving value functions
1. Choose xmin and xmax (also called x + and x − )
2. Determine some points on the value function curve
3. Use these data points to generate the complete curve
Linear interpolation
Best fitting curve
Note: We will use linear interpolation throughout the course!
4. Normalize the function to the interval [0 , 1]
5. Check for consistency
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Methods for determining attribute value functions
1. Bisection method (mid-value splitting technique): find the attribute level
whose value is halfway between the most and the least preferred attribute
level
2. Difference standard sequence technique (DSST): find the attribute levels
x0 , x1 , ..., xn , such that the increments in the strength of preference from xi
to xi+1 are equal for all i = 0, ..., n − 1
3. Direct rating method: directly assign values to the attribute levels
“The assumptions”:
Preferences are rational
Value functions are cardinal (therefore value differences reflect
preferences over transitions)
Attribute values are continuous (What if they are discrete?)
Monotonically increasing (or decreasing) value functions (What if they are
non-monotonic?)
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1) Bisection Method: Value function for the attribute “Salary”
Normalization condition yields 2 data points on the original value function
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1) Bisection Method: Value function for the attribute “Salary”
We are looking for 3 more points that satisfy the original function
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1) Bisection Method: Value function for the attribute “Salary”
If the original value function is linear, then we know those 3 data points
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1) Bisection Method: Value function for the attribute “Salary”
But the original value function is not necessarily linear
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1) Bisection Method: Value function for the attribute “Salary”
Check if the value function is linear: (30K → 55K ) ? (55K → 80K )
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1) Bisection Method: Value function for the attribute “Salary”
If (30K → 55K ) (55K → 80K ), then adjust the transitions until the DM is indifferent between
the two: (30K → 50K ) ? (50K → 80K )
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1) Bisection Method: Value function for the attribute “Salary”
1. Determine the most-preferred outcome (x + = 80, 000) and the least-preferred
(x − = 30, 000) outcome.
2. Normalize the value function by assuming v (30, 000) = 0 and v (80, 000) = 1.
(Why do we normalize the value function? What kind of a transformation is this?)
3. Ask the DM for the outcome x0.5 such that:
(30, 000 → x0.5 ) ∼ (x0.5 → 80, 000) , e.g. x0.5 = 50, 000
Let M = (Best level + Worst level)/2 = $55,000 = midpoint of total range
Ask the DM which change produces a greater value improvement:
- Change 1: Improve from $55,000 to $80,000
- Change 2: Improve from $30,000 to $55,000
If, for example, the answer is that Change 2 has a greater impact, this implies
v(55,000) - v(30,000) > v(80,000) - v(55,000) (How is this related to cardinality?)
Repeat the question with a salary less than $55,000 until the DM is indifferent between
Change 1 and 2 (Let’s assume that the DM is indifferent if we replace $55,000 with
$50,000).
4. Assign the evaluation 0.5 to this outcome
⇒ v (50, 000) − v (30, 000) = v (80, 000) − v (50, 000) ⇒ v (50, 000) = 0.5
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1) Bisection Method: Value function for the attribute “Salary”
5. Determine the outcomes x0.25 and x0.75 in the same way.
0.25
v($30,000)=0
0.25
v (x0.25 )
0.25
0.25
v (x0.75 )
v($50,000)=0.5
v($80,000)=1
(30, 000 → 40, 000) ∼ (40, 000 → 50, 000) and (50, 000 → 65, 000) ∼ (65, 000 → 80, 000)
| {z }
| {z }
| {z }
| {z }
=0
=0.5
v(50,000)-v(40,000) = v(40,000)-v(30,000)
⇒ v(40,000) = 0.25
=0.5
and
=1
v(65,000)-v(50,000) = v(80,000)-v(65,000)
⇒ v(65,000) = 0.75
6. Use linear interpolation to complete the value function.
7. Check consistency
Use a different method
Ask additional questions:
(40, 000 → a) ∼ (a → 65, 000)
Is a close to x0.5 ?
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Exercise: Bisection Method
An entrepreneur has made the following statements concerning his value function
for profits: Given that the profits will lie within the range between $4 million and $8
million, the value of the difference between $4 million and $5 million is the same as
the value of the transition from $5 million to $8 million. Furthermore, the transition
from $4 million to $4.25 million is valued the same as the transition from $4.25 to $5
million. Similarly, the transition from $5 million to $6.25 million is valued the same as
the transition from $6.25 million to $8 million.
a) Draw the value function for the interval [0,1]. Which method is used?
b) How would you check for consistency?
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Exercise: Bisection Method (Cont.)
An entrepreneur has made the following statements concerning his value function for profits:
Given that the profits will lie within the range between $4 million and $8 million,
the value of the difference between $4 million and $5 million is the same as the value of
the transition from $5 million to $8 million. Furthermore, the transition from $4 million to $4.25
million is valued the same as the transition from $4.25 to $5 million. Similarly, the transition
from $5 million to $6.25 million is valued the same as the transition from $6.25 million to $8
million.
Bisection method
0
0.25
0.25
v($4.25M)
v($4M)
0.25
v($5M)
1
0.25
v($6.25M)
v($8M)
($4M → $5M) ∼ ($5M → $8M)
v ($5M) − v ($4M) = v ($8M) − v ($5M) ⇒ v ($5M) = 0.5
($4M → $4.25M) ∼ ($4.25M → $5M)
v ($4.25M) − v ($4M) = v ($5M) − v ($4.25M) ⇒ v ($4.25M) = 0.25
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Exercise: Bisection Method (Cont.)
a) Draw the value function for the interval [0,1]. Which method is used?
b) How would you check for consistency?
v($4M)=0
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v($5M)=0.5
v($8M)=1
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Exercise: Bisection Method (Cont.)
a) Draw the value function for the interval [0,1]. Which method is used?
b) How would you check for consistency?
v($4.25M)=0.25
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?
v($6.25M)=0.75
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Exercise: Bisection Method (Cont.)
a) Draw the value function for the interval [0,1]. Which method is used?
b) How would you check for consistency?
v($4.25M)=0.25
?
v($6.25M)=0.75
($4.25M → ?) ∼ (? → $6.25M)
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Exercise: Bisection Method (Cont.)
a) Draw the value function for the interval [0,1]. Which method is used?
b) How would you check for consistency?
v($4.25M)=0.25
?
v($6.25M)=0.75
($4.25M → ?) ∼ (? → $6.25M)
? roughly = $5M
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2) DSST Method: Value function for the attribute “Salary”
1. Determine the most-preferred outcome (x + = 80, 000) and the least-preferred (x − = 30, 000)
outcome.
2. Define (as the researcher) a unit 4 that is approximately 1/5 of the length of the interval
[30, 000, 80, 000] (10,000). Define x 1 = 30, 000 + 10, 000 = 40, 000.
4
4
4
4
4
$30,000 $40,000 $50,000 $60,000 $70,000 $80,000
4 = $10, 000
3. Ask the DM for the outcome x 2 that satisfies the following indifference statement:
(30, 000 → 40, 000) ∼ (40, 000 → x 2 )
x
v($30,000)=0
x
v($40,000)
v (x 2 )
(30, 000 → 40, 000) ∼ (40, 000 → 50, 000)
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2) DSST Method: Value function for the attribute “Salary”
4. Ask for the outcome x 3 with: (40, 000 → 50, 000) ∼ (50, 000 → x 3 )
x
x
v($30,000)=0 v($40,000)
x
v($50,000)
v (x 3 )
(40, 000 → 50, 000) ∼ (50, 000 → 65, 000)
5. Proceed with the same procedure until x + is reached or exceeded.
x
x
x
v($30,000)=0 v($40,000) v($50,000)
x
v($65,000)
v (x 4 )
(50, 000 → 65, 000) ∼ (65, 000 → 80, 000)
6. v ($80, 000) = 1 ⇒ 4x = 1 ⇒ x = 0.25
0.25
0.25
0.25
v($30,000)=0 v($40,000) v($50,000)
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0.25
v($65,000) v($80,000)=1
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Exercise: DSST Method
An entrepreneur has made the following statements concerning his value function
for profits ranging between -$10 million and $10 million: the value of the difference
between -$10 million and -$7 million is the same as the values of the transitions
from -$7 million to -$5 million, from -$5 million to -$3 million, from -$3 million to $0
million, from $0 million to $1 million, from $1 million to $5 million, and from $5
million to $10 million.
a) Draw the value function for the interval [0,1]. Which method is used?
b) How would you check for consistency?
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Exercise: DSST Method (Cont.)
An entrepreneur has made the following statements concerning his value function for profits
ranging between -$10 million and $10 million:
the value of the difference between -$10 million and -$7 million is the same as the
values of the transitions from -$7 million to -$5 million, from -$5 million to -$3 million, from -$3
million to $0 million, from $0 million to $1 million, from $1 million to $5 million, and from $5
million to $10 million.
The difference standard sequence technique (DSST) has been used.
0
v(-10)
x
x
v(-7)
x
v(-5)
x
v(-3)
⇒x =
x
x
v(0)
v(1)
x
v(5)
1
v(10)
1
7
Each transition has a value of 1/7.
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Exercise: DSST Method (Cont.)
a) Draw the value function for the interval [0,1].
Which method is used?
b) How would you check for
consistency?
Change the size of the first transition:
e.g. instead of 4 =3 (-10 to -7), set
4 =4
What is the attribute level x that
satisfies the following?
(−10 → −6) ∼ (−6 → x).
REMARK: The number of transitions (and the number of data points) depends on the DM’s
preferences, therefore is not known until the interview is completed.
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DSST Method: Attribute range
Local versus global attribute range
It is often preferable to use a (global) range that is wider than the minimum and
maximum values of the alternatives (local range).
If the last question in the interview results in an x value that is greater than x +
(maximum attribute level):
Expand the attribute range or
Ask for the value of the transition from the last x elicited to x + , relative to the previous
transitions
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Exercise: Attribute Range
Now, assume that all the transitions mentioned in the previous question hold, except
the last one such that the transition from $5 million to $10 million creates a smaller
value than the other transitions. Can you still generate the value function for the
profits from -$10 million to $10 million if the entrepreneur makes one of the following
statements?
(i) The transition from $1 million to $5 million generates the same value as
the transition from $5 million to $12 million.
0
x
x
v(-10)
x
v(-7)
x
v(-5)
x
x
v(-3)
v(0)
x
v(1)
v(5)
1
v(12)
(ii) The transition from $5 million to $10 million generates half as much value as the
transition from $1 million to $5 million.
0
v(-10)
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x
x
v(-7)
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x
v(-5)
x
v(-3)
x
v(0)
x
v(1)
1
x/2
v(5) v(10)
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Exercise: Attribute Range (Cont.)
Now, assume that all the transitions mentioned in the question hold, except the last one such that the
transition from $5 million to $10 million creates a smaller value than the other transitions. Can you
still generate the value function for the profits from -$10 million to $10 million if the entrepreneur
makes one of the following statements?
(i) The transition from $1 million to $5 million generates the same value as
the transition from $5 million to $12 million.
0
v(-10)
x
x
v(-7)
x
v(-5)
x
v(-3)
v (12) = 1, v (5) = 1 −
x
x
v(0)
v(1)
1
x
v(5)
v(12)
1
10 − 5 1
= 0.857 ⇒ v (10) = 0.857 +
× = 0.96
7
12 − 5 7
(ii) The transition from $5 million to $10 million generates half as much value as the transition from
$1 million to $5 million.
0
x
v(-10)
x
v(-7)
x
v(-5)
v (−10) = 0, v (−7) =
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x
v(-3)
x
v(0)
x
v(1)
x/2
1
v(5) v(10)
1
1
, ..., v (5) = 1 −
= 0.923, v (10) = 1
6.5
13
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3) Direct Rating Method: Value function for the attribute “Salary”
1. Determine the most-preferred outcome (x + = 80, 000) and the least-preferred
(x − = 30, 000) outcome.
2. Order the outcomes of all alternatives (a:30,000, b:50,000, c:65,000, d:40,000,
e:80,000) from the most preferred to the least preferred. (e c b d a)
3. Assign 100 and 0 points to the best and worst outcomes.
4. Assign points to the intermediate outcomes, s.t. the point differences truly reflect
the strength of preference.
e:100, c:75, b:50, d:25, a:0
5. Normalize: Divide points by 100 to receive evaluations.
e:1, c:0.75, b:0.5, d:0.25, a:0
6. Check consistency
Use a different method
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What if “the assumptions” do not hold?
Attribute values are continuous (What if they are discrete?)
→ Bisection method and DSST cannot be used
→ Use the direct method
Example: Car choice problem
Can you use DSST in order to elicit the attribute value function for “Color”?
Monotonically increasing (or decreasing) value functions (What if they are
non-monotonic?)
→ Split the objective into monotonic lower-level objectives
→ Or split the interval into subintervals on which the value function is
monotonically increasing or decreasing. Then, apply the value
function elicitation methods separately on both intervals.
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Exercise: Generating non-monotonic value functions
You are thinking about buying a house in a nice village. This village, as most of the
villages in this area, consists of only one street that goes alongside a canal for about
10 km. However, in the middle of the village there’s a new highway connecting the
village with the bigger cities around. The distance of a house to this highway is an
important objective that determines your choice of a house. Assume that your
valuation of “distance to highway” is non-monotonic such that you find a distance of
500 m optimal and better than the extremes 20 m and 5 km. Why is it impossible to
represent your preferences using a monotone value function? How could
non-monotonicity of the value function be resolved?
Canal
5km
5km
Highway
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Exercise: Generating non-monotonic value functions (Cont.)
Canal
5km
5km
Highway
Splitting up the objective “distance to
highway” could resolve
non-monotonicity. There are two
underlying fundamental objectives,
“noise” and “time to reach high-way”. A
higher discomfort due to noise would
always lead to lower valuations.
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Exercise: The Additive Model
New York
Paris
Dubai
Flight time
to home
9 hours
3 hours
5 hours
Salary
$5,000
$2,000
$6,000
Paid
vacation days
0 days
30 days
22 days
For the value function of the attribute “Paid vacation days”, you collected the following
indifference statements regarding the transitions between different attribute levels:
(0 days → 12 days) ∼ (12 days → 30 days)
(0 days → 5 days) ∼ (5 days → 12 days)
(12 days → 20 days) ∼ (20 days → 30 days)
a) Plot the attribute value function by using linear interpolation. Which method did you
use? Find v3 (22 days).
b) While you were running several consistency checks, you obtained the following
preference statements. Can you still use the additive model? Why or why not?
(i) (∗, $6, 000, 30 days) (∗, $6, 000, 22 days) and
(∗, $2, 000, 30 days) ∼ (∗, $2, 000, 22 days)
(ii) (5 hours, $6, 000, ∗) → (3 hours, $6, 000, ∗) (5 hours, $2, 000, ∗) → (3 hours, $2, 000, ∗)
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Exercise: The Additive Model (Cont.)
New York
Paris
Dubai
Flight time
to home
9 hours
3 hours
5 hours
Salary
$5,000
$2,000
$6,000
Paid
vacation days
0 days
30 days
22 days
a) For the value function of the attribute “Paid vacation days”, you collected the following
indifference statements regarding the transitions between different attribute levels:
(0 days → 12 days) ∼ (12 days → 30 days) ⇒ x0.5 = 12
(0 days → 5 days) ∼ (5 days → 12 days) ⇒ x0.25 = 5
(12 days → 20 days) ∼ (20 days → 30 days) ⇒ x0.75 = 20
Bisection method has been used.
v3 (22 days) = 0.75 +
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0.25
× (22 − 20) = 0.8
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Exercise: The Additive Model (Cont.)
b) While you were running several consistency checks, you obtained the following preference
statements. Can you still use the additive model? Why or why not? Explain.
(i) (∗, $6, 000, 30 days)(∗, $6, 000, 22 days) and (∗, $2, 000, 30 days)∼(∗, $2, 000, 22 days)
⇒ w3 v3 (30) > w3 v3 (22), w3 v3 (30) = w3 v3 (22) ⇒ contradiction!
Violates simple (and mutual) preferential independence
(ii) (5 hours, $6, 000, ∗) → (3 hours, $6, 000, ∗)(5 hours, $2, 000, ∗) → (3 hours, $2, 000, ∗)
⇒ w1 v1 (3) − w1 v1 (5) > w1 v1 (3) − w1 v1 (5) ⇒ contradiction!
Violates additive difference independence
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Answer the following questions:
1. What does “conflicting objective” mean?
2. How does a non-additive value function differ from an additive value function?
3. What are the requirements for representing preferences with an ordinal or a cardinal
attribute value function?
4. What are the methods for determining value functions? Compare and contrast each
method.
5. What is the problem with eliciting the value function for a discrete attribute?
6. How can you elicit non-monotonic attribute value functions?
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Reading
EWL Chapter 5 and Chapter 6 and CC Chapter 5
Supplementary Reading (Optional) (ILIAS)
Ulvila, Snider, 1980: Negotiation of International Oil Tanker Standards: An
Application of Multiattribute Value Theory. Operations Research, Vol. 28, pp. 81-96.
Treadwell, 1998: Tests of Preferential Independence in the QALY Model. Medical
Decision Making, Vol. 18, pp. 418-428.
Next Lecture
Lecture 4: Determination of the weights
Reading: EWL Chapter 6 and CC Chapter 5
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