Lecture 5: Incomplete Information
Keywords
Lecture Outline
Error minimization
Sensitivity analysis
Dominance test
Irem Demirci
CC 501 Decison Anlaysis
EWL 6.5 Incomplete information about
the weights
CC 6 Value and Risk Management for
Multi-Objective Decisions
Lecture 5: Incomplete Information
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Additive value function
n
V (a) =
∑ wr · vr (ar ) = w1 · v1 (a1 ) + . . . + wn · vn (an )
r =1
wr : Compare different attributes
in terms of importance
∑nr=1 wr = 1 and wr > 0
vr : Compare the value of
different attribute levels
vr (xr− ) = 0 and vr (xr+ ) = 1
Where do vr come from? Value function elicitation methods:
X
Bisection method
DSST
Direct-Rating method
Where do the wr come from? Decision weight elicitation methods:
X
Swing method
Trade-off method
Direct-ratio method
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Incomplete or inconsistent information
So far, we assumed that we can obtain the point estimates for
weights by conducting an interview with the DM.
But what if point estimates are not available
Preferences are less robust (i.e. inconsistent or unsure answer)
DM is unwilling to provide more detailed info
Can we still make a decision?
Error minimization (Inconsistent Information)
Sensitivity analysis (How robust is the decision?/Incomplete
information)
Dominance tests (Incomplete information)
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Example: Inconsistent Information
Tradeoff method: 2 attributes
Statement 1: w1 = 2w2
Statement 2: w2 = 0.45w1
Normalization: w1 + w2 = 1
Example: Incomplete Information
Tradeoff method: 3 attributes
Statement 1: w1 = 2w2
Normalization: w1 + w2 + w3 = 1
Example: Incomplete Information
Swing method: 3 attributes
t1 = 100 points
t2 = 50 points
t3 ∈ [40,60]
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Inconsistent Information
Example: Job choice (trade-off method)
Consider the following indifference statements
1:
(55000 , 40 , ∗) ∼ (45000 , 28 , ∗)
2:
(55000 , 40 , ∗) ∼ (70000 , 52 , ∗)
3:
(55000 , ∗ , good) ∼ (45000 , ∗ , excellent)
4:
(43000 , ∗ , good) ∼ (80000 , ∗ , bad)
and the equations that they imply under the additive model:

0.212w1 − 0.3w2 = 0 



−0.209w1 + 0.3w2 = 0 


0.212w1 − 0.3w3 = 0
Over-determined



−0.563w1 + 0.7w3 = 0 



w1 + w2 + w3 = 1
Question: How can you handle such inconsistencies?
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Inconsistent Information
Example cont.: Job choice (trade-off method)
Are there any redundancies? (Can you express any of these
equations as a linear combination of the others?)

0.212w1 − 0.3w2 = 0 



−0.209w1 + 0.3w2 = 0 


0.212w1 − 0.3w3 = 0


−0.563w1 + 0.7w3 = 0 




w1 + w2 + w3 = 1
For example: 0.212w1 − 0.3w3 = 0 and −0.495w1 + 0.7w3 = 0
would be redundant since they yield the same equation:
multiplying −0.495w1 + 0.7w3 = 0 by 3/7 yields 0.212w1 − 0.3w3 = 0
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Error minimization
Example cont.: Job choice (trade-off method)
Idea: Minimize errors of each trade-off equation
 
0.212w1 − 0.3w2 = ε1
0.212w1 − 0.3w2 = 0 









−0.209w1 + 0.3w2 = 0 
 
 −0.209w1 + 0.3w2 = ε2
0.212w1 − 0.3w3 = ε3
0.212w1 − 0.3w3 = 0
⇒






−0.563w1 + 0.7w3 = 0 
 −0.563w1 + 0.7w3 = ε4
 

 

w1 + w2 + w3 = 1
w1 + w2 + w3 = 1
|
{z
}
Under-determined
Solve for w1 , w2 , w3 that minimize the sum of absolute errors:
min ∑ |εi |
w1 ,w2 ,w3
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How to set up the objective function (Optional)
The problem:
min ∑ |εi | subject to w1 , w2 , w3 ≥ 0
w1 ,w2 ,w3
Remark 1: εi can take on negative values.
Remark 2: This is not a linear programming problem because of the
nonlinear absolute value function.
It can be transformed into a linear problem by writing εi as the difference
of two non-negative variables: εi = εi+ − εi−
Furthermore, the absolute value terms can be simplified whenever either
εi+ or εi− equals zero:
|εi+ − εi− | = |εi+ | + |εi− | = εi+ + εi−
Therefore, the problem can be rewritten as:
min ∑ εi+ + εi−
w1 ,w2 ,w3
subject to w1 , w2 , w3 ≥ 0 and εi+ , εi− ≥ 0 for i = 1, 2, 3, 4.
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Shortcut: Excel Solver
⇓ Excel Solver
Minimize the sum of absolute errors by changing w1
and w2 subject to w1 , w2 , w3 ≥ 0
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Sensitivity Analysis
Does the decision change if the weights are slightly altered?
How robust is the decision (i.e. by how much could the parameters
change without changing the result)?
Sensitivity analysis:
- Determine the relationship between the attribute weights (i.e. express each
weight in terms of the weight that you would like to alter)
- Rewrite the value of each alternative as a function of the attribute weight
that you would like to alter
- Vary the objective weight in order to see how sensitive the decision is
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Sensitivity Analysis
Example: Apartment search
Your friend is looking for an apartment close to campus. You obtained the following data
regarding your friend’s problem:
w1 + w2 + w3 = 1 and w2 = w3 and w1 = 2w2
Given w1 = 0.5, w2 = 0.25, and w3 = 0.25, the DM prefers the apartment in
Neckarstadt.
After conducting couple of consistency checks, you realized that the weight of
“distance to campus” is actually 0.6, and size and rent are equally important.
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Sensitivity Analysis
Example cont.: Apartment search
Is Neckarstadt still the optimal alternative if w1 = 0.6?
−−→
−−−−
−−−
−−−
→
w1 + w2 + w3 = 1 and assume that w2 /w3 = 1
Therefore, the optimal alternative changes if w1 increases from 0.5 to 0.6.
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Example cont.: How sensitive are the results to w1 ?
Express w2 and w3 in terms of the weight that you would like to alter (w1 )
w1 + w2 + w3 = 1
w2 /w3 = 1
⇒ w2 = w3 =
1 − w1
2
Rewrite the value of each alternative as a function of the attribute weight
that you would like to alter
v (a) = w1
1
v (b) = 0.5w1 + 0.3w2 + w3 = 0.5w1 + 1.3 1−w
2
1−w1
v (c) = w2 + 0.75w3 = 1.75 2
Vary the objective weight in order to see how sensitive the decision is
1
v(a) > v(b) ⇒ w1 > 0.5 × w1 + 1.3 × 1−w
⇒ w1 > 0.565
2
1
v(a) > v(c) ⇒ w1 > 1.75 × 1−w
⇒ w1 > 0.467
2
1
1
v(b) > v(c) ⇒ 0.5 × w1 + 1.3 × 1−w
> 1.75 × 1−w
⇒ w1 > 0.310
2
2
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Sensitivity Analysis
Example cont.: Apartment search
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Exercise: Sensitivity Analysis
Daniel wants to buy a portable CD player for jogging. He is only interested in the
attributes “price” and “weight”. Daniel made sure that preference independence
between the attributes holds so that he can use the additive model. On the relevant
intervals [100e, 600e] for price and [200g, 500g] for weight, Daniel has a linear and
monotonically decreasing value function. Daniel wants to determine his objective
weights. Therefore, he took the following notes:
(A) (100 e, 500g) (600 e, 350g)
(B) (600 e, 260g) (100 e, 500g)
(C) (100 e, 500g) ∼ (600 e, 300g)
a) Which objective weights for “price” and “weight” does Daniel have?
b) Which one of the alternatives Allstar: (100e, 350g), BestBuy: (500e, 200g) and
CD-Supreme: (300e, 230g) is optimal for Daniel?
c) How sensitive is the result to w1 ? Illustrate the sensitivity graphically and
compute the respective intervals for which each alternative is optimal.
d) If Daniel is not sure about his preference statement (C) but has no doubt about
(A) and (B), is it still possible to determine the optimal alternative?
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Exercise: Sensitivity Analysis (Cont.)
Daniel wants to buy a portable CD player for jogging. He is only interested in the
attributes “price” and “weight”. Daniel made sure that preference independence
between the attributes holds so that he can use the additive model. On the relevant
intervals [100e, 600e] for price and [200g, 500g] for weight, Daniel has a linear
and monotonically decreasing value function. Daniel wants to determine his
objective weights. Therefore, he took the following notes:
(A) (100 e, 500g) (600 e, 350g)
(B) (600 e, 260g) (100 e, 500g)
(C) (100 e, 500g) ∼ (600 e, 300g)
a) Which objective weights for “price” and “weight” does Daniel have?
2
w1 × v1 (100e) +w2 × v2 (500g) = w1 × v1 (600e) +w2 × v2 (300g) ⇒ w1 = w2
| {z }
| {z }
3
| {z }
| {z }
1
0
0
2
3
Solving for w1 = w2 × 2/3 and w1 + w2 = 1 together yields w1 = 0.4 and w2 = 0.6.
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Exercise: Sensitivity Analysis (Cont.)
b) Which one of the alternatives Allstar: (100e, 350g), BestBuy: (500e, 200g) and
CD-Supreme: (300e, 230g) is optimal for Daniel?
Irem Demirci
V (Allstar )
=
0.4 × 1 + 0.6 × (5/3 − 350/300) = 0.4 × 1 + 0.6 × 0.5 = 0.7
V (BestBuy )
=
0.4 × (1.2 − 0.002 × 500) + 0.6 × 1 = 0.4 × 0.2 + 0.6 × 1 = 0.68
V (CD − Supreme)
=
0.4 × (1.2 − 0.002 × 300) + 0.6 × (5/3 − 230/300)
=
0.4 × 0.6 + 0.6 × 0.9 = 0.78
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Exercise: Sensitivity Analysis (Cont.)
c) How sensitive is the result to w1 ? w2 = 1 − w1
Rewrite each alternative’s value in terms of w1 noting that w2 = 1 − w1 :
V (Allstar )
V (BestBuy)
V (CD − Supreme)
=
w1 × v1 (100e) + w2 × v2 (350g) = 1 × w1 + 0.5 × (1 − w1 )
=
0.5 + 0.5w1
=
w1 × v1 (500e) + w2 × v2 (200g) = 0.2 × w1 + 1 × (1 − w1 )
=
1 − 0.8w1
=
w1 × v1 (300e) + w2 × v2 (230g) = 0.6 × w1 + 0.9 × (1 − w1 )
=
0.9 − 0.3w1
V (Allstar ) = 0.5 + 0.5w1
V (BestBuy ) = 1 − 0.8w1
V (CD − Supreme) = 0.9 − 0.3w1
V (Allstar ) = v (CD) ⇒ w1 = 0.5
V (BestBuy ) = v (CD) ⇒ w1 = 0.2
Allstar is optimal for w1 ∈ [0.5,1]; BestBuy is optimal for w1 ∈ [0,0.2] and
CD-Supreme is optimal for w1 ∈ [0.2,0.5].
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Exercise: Sensitivity Analysis (Cont.)
d) If Daniel is not sure about his preference statement (C) but has no doubt about (A)
and (B), is it still possible to determine the optimal alternative?
The decision weights from statements (A) and (B) can be expressed as follows:
(A) (100 e, 500g) (600 e, 350g)
w1 × 1 + w2 × 0>w1 × 0 + w2 × 0.5 ⇒ w1 >(1 − w1 ) × 0.5 ⇒ w1 > 1/3
(B) (600 e, 260g) (100 e, 500g)
w1 × 0 + w2 × 0.8>w1 × 1 + w2 × 0 ⇒ (1 − w1 ) × 0.8>w1 ⇒ w1 < 4/9
From these preferences, we already know that w1 has to be in the interval
[0.33,0.44]. Together with the results from part c) we can conclude that the
optimal alternative is CD-Supreme.
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Dominance Test
What if the information regarding the objective weights is incomplete? (i.e. you do not
have the point estimates but you know the range of values)
Can you still make a decision despite the incomplete information?
Dominance checks allow for narrowing down the number of relevant alternatives on the
basis of a low level of information.
Definition
Let V (I) the set of value functions v that are consistent with the preference
information I. Let a, b ∈ A. Then we define a relation V (I) as:
(
a V (I) b ⇔
v (a) ≥ v (b) for all v ∈ V (I) and
v (a) > v (b) for at least one v ∈ V (I)
a V (I) b ↔ a dominates b with respect to V (I)
(Analogously for ≺)
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Dominance Test
Definition of a dominating b
If a dominates b, then for any set of weights that is consistent with the incomplete
information, V (a) ≥ V (b) should hold, and there should be at least one weight
combination under which V (a) > V (b) holds.
Definition of a NOT dominating b
If there exists at least one admissible weight combination for which V (b) > V (a),
then alternative a CANNOT dominate b.
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Example: Is there a dominance relation between alternatives a and b?
Possible outcomes:
a dominates b
b dominates a
there is no dominance between a and b
Objective 1
Objective 2
Objective 3
w1 ∈ [0.2 , 0.4]
w2 ∈ [0.25 , 0.45]
w3 ∈ [0.1 , 0.4]
Alternative a
v1 (a1 ) = 1
v2 (a2 ) = 0
v3 (a3 ) = 0
Alternative b
v1 (b1 ) = 0.5
v2 (b2 ) = 0.2
v3 (b3 ) = 1
Alternative c
...
Assume that you do not know the objective weights exactly but you know that they lie
within the following intervals: w1 ∈ [0.2 , 0.4], w2 ∈ [0.25 , 0.45] and w3 ∈ [0.1 , 0.4].
Examples: w = (0.2, 0.4, 0.4), w = (0.3, 0.3, 0.4), w = (0.3, 0.35, 0.35), ...
Pick any set of weights that satisfy the boundary conditions (and the normalization
condition), e.g.: w1 = 0.3, w2 = 0.3 and w3 = 0.4
V (a) = 0.3 and V (b) = 0.61 ⇒ Alternative a cannot dominate Alternative b!
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Example cont.: Is there a dominance relation between these alternatives?
Objective 1
Objective 2
Objective 3
w1 ∈ [0.2 , 0.4]
w2 ∈ [0.25 , 0.45]
w3 ∈ [0.1 , 0.4]
Alternative a
v1 (a1 ) = 1
v2 (a2 ) = 0
v3 (a3 ) = 0
Alternative b
v1 (b1 ) = 0.5
v2 (b2 ) = 0.2
v3 (b3 ) = 1
Is Alternative a dominated by b?
What is the value difference between Alternatives a and b?
V (a) − V (b) = w1 − [0.5 · w1 + 0.2 · w2 + w3 ]
=
0.5 · w
| {z 1}
to be maximized
−
0.2 · w2
| {z }
to be minimized
−
w3
|{z}
to be minimized
What are the weights that satisfy w1 + w2 + w3 = 1 and maximize this difference?
max [V (a) − V (b)]
=
0.5 · 0.4 − 0.2 · 0.25 − 0.1 → w1 + w2 + w3 = 0.75
max [V (a) − V (b)]
=
0.5 · 0.4 − 0.2 · 0.45 − 0.1 → w1 + w2 + w3 = 0.95
max [V (a) − V (b)]
=
0.5 · 0.4 − 0.2 · 0.45 − 0.15 = −0.04 < 0
If for the weights that generate the maximum difference, b yields a higher value than
a, then for any set of admissible weights, b yields a higher value. Therefore,
Alternative b dominates a.
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REMARKs
If you’ve already shown that a cannot dominate b, then
max [V (a) − V (b)] = 0 also implies that b dominates a. (In the
worst-case scenario, they have equal values, in others b should have a
better value than a, which is consistent with the definition of dominance.)
If max [V (a) − V (b)] > 0, and you’ve already shown that a cannot
dominate b, then there exists at least one admissible set of weights for
which a yields a higher value than b. Therefore, b cannot dominate a.
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Exercise: Dominance Test
You are planning an internship abroad. You have three offers: New York, London, and Toronto.
Your relevant objectives as well as the attribute valuations of the three alternatives are as follows:
salary v1 (x1 )
city appeal v2 (x2 )
working hours v3 (x3 )
days off v4 (x4 )
New York
0.5
1
0
0.75
London
1
0.5
0.2
0
Toronto
0.75
0
0.54
1
a) To determine your objective weights you used the swing method. Using the swing method,
you derived the following scores for the artificial alternatives:
a− (0 points), b1 (100 points), b2 (62.5 points), b3 (50 points), b4 (37.5 points)
where alternatives b1 , b2 , b3 , and b4 are constructed in order to elicit the evaluations for
decision weights w1 , w2 , w3 , and w4 , respectively. Calculate the decision weights.
b) Assume that using a different method, you determined the following upper and lower bounds:
Attribute
salary
city appeal
working hours
days off
Weight
w1
w2
w3
w4
Lower
bound
0.3
0.35
0.1
0.1
Upper
bound
0.4
0.5
0.2
0.15
Conduct a dominance check between alternatives London and Toronto assuming that the
information in the table is correct.
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Exercise: Dominance Test (Cont.)
You are planning an internship abroad. You have three offers: New York, London, and Toronto. To
determine your objective weights you used the swing method. You derived the following scores for
the artificial alternatives:
a− (0 points), b1 (100 points), b2 (62.5 points), b3 (50 points), b4 (37.5 points)
What are these alternatives?
a− : (x1− , x2− , x3− , x4− ), b1 : (x1+ , x2− , x3− , x4− ), b2 : (x1− , x2+ , x3− , x4− ),
b3 : (x1− , x2− , x3+ , x4− ), b4 : (x1− , x2− , x3− , x4+ )
T
w1
w2
w3
w4
Irem Demirci
CC 501 Decison Anlaysis
= t1 + t2 + t3 + t4 = 250
t1
T
= 100
250
t2
= T = 62.5
250
t
50
= T2 = 250
t4
37.5
= T = 250
=
= 0.4
= 0.25
= 0.2
= 0.15
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Exercise: Dominance Test (Cont.)
salary v1 (x1 )
city appeal v2 (x2 )
working hours v3 (x3 )
days off v4 (x4 )
Attribute
salary
city appeal
working hours
days off
London
1
0.5
0.2
0
Lower
bound
0.3
0.35
0.1
0.1
Weight
w1
w2
w3
w4
Toronto
0.75
0
0.54
1
Conduct a dominance test between
London and Toronto.
Upper
bound
0.4
0.5
0.2
0.15
Pick an arbitrary weight combination that satisfies the incomplete information: e.g.
w1 = 0.4, w2 = 0.4, w3 = 0.1 and w4 = 0.1:
V (L)
=
0.4 + 0.4 × 0.5 + 0.1 × 0.2 = 0.62
V (T )
=
0.4 × 0.75 + 0.1 × 0.54 + 0.1 × 1 = 0.454
V (L) > V (T ) ⇒ T cannot dominate L!
WHY? There exists at least one weight combination for which V (L) > V (T )!
DOMINANCE: If T dominates L, then there cannot be any weight combination for which V (L) > V (T )!
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Exercise: Dominance Test (Cont.)
salary v1 (x1 )
city appeal v2 (x2 )
working hours v3 (x3 )
days off v4 (x4 )
Attribute
salary
city appeal
working hours
days off
Weight
w1
w2
w3
w4
London
1
0.5
0.2
0
Toronto
0.75
0
0.54
1
Lower
bound
0.3
0.35
0.1
0.1
Upper
bound
0.4
0.5
0.2
0.15
Does L dominate T? (i.e. Max[V (T ) − V (L)] < 0?)
V (T ) − V (L)
=
− w1 ×0.25 − w2 ×0.5 + w3 ×0.34 + w4
|{z}
|{z}
|{z}
|{z}
Max[V (T ) − V (L)]
=
−0.3 × 0.25 − 0.35 × 0.5 + 0.2 × 0.34 + 0.15
=
−0.032 < 0 ⇒ L dominates T!
minimize
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CC 501 Decison Anlaysis
minimize
Lecture 5: Incomplete Information
maximize
maximize
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Dealing with incomplete information under certainty
Sensitivity Analysis
Dominance Test
Does the decision change if the weights are
slightly altered? How robust is the decision?
What if you are given a range of values for
the weights (i.e. incomplete information)?
What if you are given 4 attributes but only 2
relationships between the weights (+
normalization condition)?
and you do not know how one weight is
related to another?
Can you tell which alternative is optimal
given the interval for one of the weights?
Procedure:
Pick an arbitrary set of weights that satisfy
the incomplete information
Procedure:
Express each weight in terms of the weight
that you would like to alter
Rewrite the value of each alternative as a
function of the attribute weight that you
would like to alter
Vary the objective weight in order to see how
sensitive the decision is.
Irem Demirci
CC 501 Decison Anlaysis
Calculate the values of the alternatives with
respect to these arbitrary weights
Eliminate one of the possible dominance
relationships (i.e. a cannot dominate b)
Check for the opposite direction by
investigating the sign of Max[V (a) − V (b)]
Conclude that b dominates a if the sign is
negative
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Answer the following questions:
1. When would you use error minimization?
2. How do you conduct error minimization?
3. How can you check for the sensitivity of the decision to attribute weights?
4. Can you still make decisions if your information about weights is
incomplete?
Irem Demirci
CC 501 Decison Anlaysis
Lecture 5: Incomplete Information
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Reading
EWL Chapter 6, CC Chapter 6
Next Lecture
Lecture 6: Introduction to decision making under uncertainty
Reading: EWL Chapter 2, MC Chapter 2 and 3, CC Chapter 2 and 10
Irem Demirci
CC 501 Decison Anlaysis
Lecture 5: Incomplete Information
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31
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