10/28/2022
ISE 562; Dr. Decision Models II Decision Theory
10/28/2022
ISE 562; Dr.
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• MAUT computation example from A-Z
• Utility function assessment revisited
• Calculating K
• Uncertainty
• Sensitivity analysis
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ISE 562; Dr. Smith
• Multiattribute utility functions
Let xi be the level of attribute i; ui(xi) be the attribute utility function for attribute i; ki be the attribute tradeoff scaling constant for attribute i. If the xi are mutually utility independent then the expression for the multiattribute utility function takes one of the following forms depending on the sum of the ki:
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If k 1.0, then:U(x) k u (x ) Eq1
k1 n1
If k 1.0, then:U(x)
1Kk u (x )1 Eq2
where the master scaling constant, K, is solved from the equation :
1K1Kk Eq3
ISE 562; Dr. is the relationship between this : 1N
U ( x ) K 1 K k u ( x ) 1 nnn
n1 and this :
N U(x)k u (x )
It’s not obvious… 10/28/2022
nnn n1
ISE 562; Dr. ‘ s easier to see if we expand the multiplicative form and rearrange :
U (x) kiui (xi ) (interaction terms)
For example, if N 2
U(x)K1Kk u(x)1K[1Kku(x)][1Kku(x)]1
nnn 111 222 n1
1 1 Kk u (x ) Kk u (x ) K 2k k u (x )u (x ) 1
111 222 121122 1 Kk1u1(x1)Kk2u2(x2)K2k1k2u1(x1)u2(x2)
One’s cancel Rearranging
2 kiui(xi)Kk1k2u1(x1)u2(x2) i1
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Interaction terms. Now solve for K
ISE 562; Dr. the master scaling constant equation for K with N=2 attributes:
1 K N 1 K k n n1
for N 2 :
1K [1Kk1][1Kk2]
1K1Kk Kk K2kk 1212
K 1(k1 k2) k1k2
(Note that if k1+k2 =1, K=0)
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ISE 562; Dr. to the multiplicative model : 2
Notethatifk 1,
ku(x)Kkku(x)u (x ) iii 121122
thenK1k1k2 0 k1k2
the multiplicative reduces to the additive model!
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ku(x)Kkku(x)u (x ) iii 121122
ku(x)0 iii
ISE 562; Dr.
System Data ($ and mpg)
Alternative
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After collection of system data we extract the ranges:
Cost: 20 to 40 k$ (less is better) Mpg: 16-32 mpg (more is better)
ISE 562; Dr. need to assess the utility functions and scaling constants for these attributes
Cost: 20 to 40 k$ (less is better) Mpg: 16-32 mpg (more is better)
To do this we construct an interview package to meet with the DM
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ISE 562; Dr. Smith
Interview package outline for assessing decision maker value model
– Introduction
– Utility function assessment (2)
– Independence check
– Attribute ranking
– Attribute tradeoff scaling constants
– Check for preference reversals
– Reassessment as required 10/28/2022
ISE 562; Dr.
• Why are we here
• Purpose of study • Process
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ISE 562; Dr. AMBLE
ATTRIBUTE: Cost, k$
20 20 SURETHING
For which value of the sure thing are you indifferent between the sure thing and the gamble?
24 (the CE is 24, U(24)=0.50) Indifference point _______
If you knew that all the other attributes were at their worst states?
Indifference point _______
If you knew that all the other attributes were at their best states?
– 10/28/2022
Indifference point _______
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ISE 562; Dr. AMBLE
ATTRIBUTE: Mpg
32 32 SURETHING
For which value of the sure thing are you indifferent between the sure thing and the gamble?
28 (the CE is 28, U(28)=0.50) Indifference point _______
If you knew that all the other attributes were at their worst states?
Indifference point _______
If you knew that all the other attributes were at their best states?
– 10/28/2022
Indifference point _______
ISE 562; Dr. assess additional points ( .25, .75, etc.), use the following template
SURE THING
ATTRIBUTE: Mpg
For which value of the sure thing are you indifferent between the sure thing and the gamble?
– 10/28/2022
Indifference point _______
1-P= 1⁄2 ____
ISE 562; Dr. have the following:
For cost, u(20)=1.0; u(40)=0.0; u(24)=0.50 For mpg, u(32)=1.0; u(16)=0.0; u(28)=0.50
Now we can generate the utility functions.
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ISE 562; Dr. Utility Function:
For cost, u(20)=1.0; u(40)=0.0; u(24)=0.50 •2 piecewise functions
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– Interval 20-24
– Interval 24-40
ISE 562; Dr. :
M1=(1-.5)/(20-24)= –.125 Y1= –.125(X1-20)+1
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Utility function for cost is: u(cost)= -.125(cost-20) +1
if 20cost<24 if 24cost40
M2=(.5-0)/(24-40)= –.03125 Y2= –.03125 (X1-40)
-.03125(cost-40)
ISE 562; Dr. Utility Function:
For mpg, u(32)=1.0; u(16)=0.0; u(28)=0.50 •2 piecewise functions
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Interval 16-28 Interval 28-32
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ISE 562; Dr. :
M1=(.5-0)/(28-16)= .0417 Y1= .0417(X-16)
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Utility function for mpg is: u(mpg)= .0417(mpg-16)
if 16mpg<28 if 28mpg32
M2=(1-.5)/(32-28)= .125 Y2= .125 (X-32) + 1
.125(mpg-32) +1
ISE 562; Dr. of importance of attributes.
If you could only change one attribute from its worst state to its best state, which would you choose?
Best state
Worst state
Order of importance
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Another example...
ISE 562; Dr. Smith
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ISE 562; Dr. of: Cost
For what value of p are you indifferent between the sure
SURE THING
Indifference
thing and the gamble?
WORST SYSTEM
REFERENCE SYSTEM
Cost tradeoff scaling constant k1 = 0.8
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ISE 562; Dr. of: Mpg
For what value of p are you indifferent between the sure
SURE THING
Indifference
thing and the gamble?
REFERENCE SYSTEM
WORST SYSTEM
Cost tradeoff scaling constant k2 = 0.6
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ISE 562; Dr. completes the collection of value data:
Utility function for cost is:
u(X1)= -.125(cost-20) +1 if 20cost<24
-.03125(cost-40) if 24cost40
Utility function for mpg is:
u(X2)= .0417(mpg-16) if 16mpg<28
.125(mpg-32) +1 if 28mpg32
Alternative
Attribute tradeoff scaling constants k1= .8 (cost)
k2= .6 (mpg)
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ISE 562; Dr. check which MAU model applies:
ki = 1.4 1, so use multiplicative model
First, we need the master scaling constant, K
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1N U ( x ) K 1 K k u ( x ) 1
nnn n1
where the master scaling constant, K,
issolvedfromtheequation:1K1Kk n
ISE 562; Dr. for K, the master scaling constant:
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or, for this example, n 2
1K [1Kk1][1Kk2]
The equation to solve for K :
1K [1.8K][1.6K]
1K 11.4K.48K2
.48K2 .4K0
K(.48K .4) 0
K .8333 26
wehave1K1Kk n
ISE 562; Dr. calculate the multiplicative multiattribute utilities:
1N U(CarA) U(20,16) (.8333) 1(.8333)k u (x )1
(1.2)1.8333(.8)(1.0)1.8333(.6)(0.0)1 .80
nnn n1
1N U(CarB)U(30,25) (.8333)1(.8333)k u (x )1
u1 (30) .03125(30 40) 0.3125; u2 (25) .0417(25 16) 0.3753 (1.2)1.8333(.8)(.3125)1.8333(.6)(.3753)1
1N U(CarC)U(40,32) (.8333)1(.8333)k u (x )1
nnn n1
(1.2)1.8333(.8)(0.0)1.8333(.6)(1.0)1
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ISE 562; Dr. conclusion for this DM:
U (CarA) U (20,16) .80
U (CarB) U (30,25) 0.4282 U (CarC) U (40,32) .60
so CarA CarC CarB
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Alternative
ISE 562; Dr. Smith
•What if the sum of the scaling constants were = 1.0?
– Use the additive multiattribute utility function
– For example, suppose we had obtained k1=.7 and k2=.3...
– Note: no master scaling constant
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1K [1.7K][1.3K] 1 K 1 K .21K 2
ISE 562; Dr. the additive multiattribute utilities:
U(CarA)U(20,16)k u (x ).7u(x).3u (x )
.7u1 (20) .3u2 (16) .7(1) .3(0) .70
nnn 11 22 n1
U(CarB)U(30,25)k u(x).7u(x).3u(x)
U(CarC)U(40,32)
.7u1 (40) .3u2 (32) .7(0) .3(1)
nnn 11 22 n1
u1 (30) .03125(30 40) 0.3125; u2 (25) .0417(25 16) 0.3753 .7u1 (30) .3u2 (25) .7(.3125) .3(.3753)
k u(x).7u(x).3u(x) nnn 11 22
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ISE 562; Dr. conclusion in the additive case would be:
U (CarA) U (20,16) .70
U (CarB) U (30,25) 0.3313 U (CarC) U (40,32) .30
so CarA CarB CarC Multiplicative
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Alternative
Alternative
ISE 562; Dr. ?
• In multiplicative case cost accounted for .8/1.4 = 57% of attribute importance
• In additive case cost increased to 70% so alternative with lower costs was amplified
•Cost of B was 30k, cost of C was 40k so B moved up in the rankings
Multiplicative Additive
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Alternative
Alternative
ISE 562; Dr. ’s try some utility assessments
Suppose you are going to rent an apartment with the following choices
1. Assess a three-point utility function for Cost
2. Assess the attribute tradeoff scaling constant for cost (and distance if enough time).
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Alternative
Cost/month /roomate
Distance to campus, miles
ISE 562; Dr. AMBLE
ATTRIBUTE: __Cost_____________
_800___ __800__
SURE THING
For which value of the sure thing are you indifferent between the sure thing and the gamble?
Indifference point _______
If you knew that all the other attributes were at their worst states?
Indifference point _______
If you knew that all the other attributes were at their best states?
Indifference point _______
ISE 562; Dr. of: __Cost__________
For what value of p are you indifferent between the sure thing and the gamble?
SURE THING
100% P= 50%
REFERENCE SYSTEM
BEST SYSTEM
WORST SYSTEM
Reference $400
ISE 562; Dr. Smith
•By the way, we have a problem when n>4
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Can you see it?
Hint: we need ’s help
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ISE 562; Dr. earlier example, k1=.8, k2=.6; check which MAU model applies: ki = 1.4 1, use multiplicative model
First, we need the master scaling constant, K
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1N U ( x ) K 1 K k u ( x ) 1
nnn n1
where the master scaling constant, K,
issolvedfromtheequation:1K1Kk n
ISE 562; Dr. for K, the master scaling constant:
or, for this example, n 2
1K [1Kk1][1Kk2] The equation to solve for K : .48K2 .4K0
K .8333
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Wehave1K1Kk n
ISE 562; Dr. ’s examine the function: f(K)=.48K2 + .4K=0
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ISE 562; Dr. if k1=0.5 and k2=0.3? ki = 0.8 1
N Wehave1K1Kk
or, for this example, n 2
1K [1Kk1][1Kk2] The equation to solve for K :
1K [1.5K][1.3K]or .15K2 .2K0
K 1.3333
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ISE 562; Dr. ’s examine the function: f(K)=.15K2 – .2K=0
0.5 1 1.5 2 2.5 3
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ISE 562; Dr. ’s compare the functions:
So if ki >1, root on left side of origin
0.5 1 1.5 2 2.5 3
So if ki <1, root on right side of origin
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ISE 562; Dr. to solve if n large?
Newton’s method for finding root of function numerically
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ISE 562; Dr. Smith
Newton singlehandedly contributed more to the
development of science than any other individual in
history. He surpassed all the gains brought about by the
great scientific minds of antiquity, producing a scheme of
the universe which was more consistent, elegant, and
intuitive than any proposed before. Newton stated explicit
principles of scientific methods which applied universally
to all branches of science. This was in sharp
contradistinction to the earlier methodologies of Aristotle
and Aquinas, which had outlined separate methods for
different disciplines.
Source: http://scienceworld.wolfram.com/biography/Newton.html
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ISE 562; Dr. Smith
English physicist and mathematician who was born into a poor farming family. Luckily for humanity, Newton was not a good farmer, and was sent to Cambridge to study to become a preacher. At Cambridge, Newton studied mathematics, being especially strongly influenced by Euclid, although he was also influenced by Baconian and Cartesian philosophies. Newton was forced to leave Cambridge when it was closed because of the plague, and it was during this period that he made some of his most significant discoveries. With the reticence he was to show later in life, Newton did not, however, publish his results. In response to a letter from Hooke, he suggested that a particle, if released, would spiral in to the center of the Earth. Hooke wrote back, claiming that the path would not be a spiral, but an ellipse. Newton, who hated being bested, then proceeded to work out the mathematics of orbits. Again, he did not publish his calculations. Newton devoted the period from August 1684 to spring 1686 to this task, and the result became one of the most important and influential works on physics of all times, (Mathematical Principles of Natural Philosophy) (1687), often shortened to or simply "the Principia." In Book I of Principia, Newton opened with definitions and the three laws of motion now known as Newton's laws (laws of inertia, action and reaction, and acceleration proportional to force). Book II presented Newton's new scientific philosophy which came to replace Cartesianism. Finally, Book III consisted of applications of his dynamics, including an explanation for tides and a theory of lunar motion. To test his hypothesis of universal gravitation, Newton wrote Flamsteed to ask if Saturn had been observed to slow down upon passing Jupiter. The surprised Flamsteed replied that an effect had indeed been observed, and it was closely predicted by the calculations Newton had provided. Newton's equations were further confirmed by observing the shape of the Earth to be oblate spheroidal, as Newton claimed it should be. Newton's equations also described the motion of Moon by successive approximations, and correctly predicted the return of Halley's Comet. Newton also correctly formulated and solved the first ever problem in the calculus of variations. Newton invented a scientific method which was truly universal in its scope. These four concise and universal rules for investigation were truly revolutionary. By their application, Newton formulated the universal laws of nature with which he was able to unravel virtually all the unsolved problems of his day. Newton refined Galileo's experimental method, creating the compositional method of experimentation still practiced today. This analysis consists of making experiments and observations, and in drawing general conclusions from them by induction...by this way of analysis we may proceed from compounds to ingredients, and from motions to the forces producing them; and in general from effects to their causes, and from particular causes to more general ones till the argument end in the most general. This is the method of analysis: and the synthesis consists in assuming the causes discovered and established as principles, and by them explaining the phenomena preceding from them, and proving the explanations."
Newton formulated the classical theories of mechanics and optics and invented calculus years before Leibniz. However, he did not publish his work on calculus until afterward Leibniz had published his. This led to a bitter priority dispute between English and continental mathematicians which persisted for decades, to the detriment of all concerned. Newton discovered that the binomial theorem was valid for fractional powers, but left it for Wallis to publish (which he did, with appropriate credit to Newton). Newton formulated a theory of sound, but derived a speed which did not agree with his experiments. The reason for the discrepancy was that the concept of adiabatic propagation did not yet exist, so Newton's answer was too low by a factor of , where is the ratio of heat capacities of air. Newton therefore fudged his theory until agreement was achieved (Engineering and Science, pp. 15-16). Newton also formulated a system of chemistry. In this theory, "elements" consisted of different arrangements of atoms, and atoms consisted ofsmall,hard,billiardball-likeparticles. Source: http://scienceworld.wolfram.com/biography/Newton.html
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ISE 562; Dr. to solve if n large?
Newton’s method:
xn1 xn f(xn) f '(xn )
ki1thenx01; if ki 1 thenx0 2; ki 1 thenx0 0;
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ISE 562; Dr. to solve:
ki 1 thenx0 1; if
0.5 1 1.5 2 2.5 3
if k1thenx2;
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ISE 562; Dr. Smith
Master scaling constant solution
Wehave1 K N 1Kkn n1
f (K ) .48K 2 .4K iteration, n f (K ) .96K .4
x x48x2.4x nn
0 1 2 3 4 5 6 7
0.08 0.009796 0.000258 1.98E-07 1.18E-13 5.55E-17 0
-0.56 -0.42286 -0.40062 -0.4 -0.4 -0.4 -0.4
x(n+1) -1.00000 -0.85714 -0.83398 -0.83333 -0.83333 -0.83333 -0.83333 -0.83333
n1 n .96xn .4 Excel solution
Higher order
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ISE 562; Dr. uncertainty:
If the attribute states are represented by probability distributions, we perform a transformation of random variables from the pdf’s of the xn’s to the resulting U(X) using Monte Carlo simulation.
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1N U ( x ) K 1 K k u ( x ) 1
nnn n1
ISE 562; Dr. uncertainty
• The result would be not only the expected utility, but the variance of the expected utility which allows the following comparison:
Which one would you pick?
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Alternative 1
1 Alternative 2
ISE 562; Dr. example (from last lecture)
Utility function for cost is:
u(X1)= -.125(X1-20) +1 if 20X1<24
-.03125(X1-40) if 24X140
Utility function for mpg is:
u(X2)= .0417(X2-16) if 16X2<28
.125(X2-32) +1 if 28X232
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Alternative
Attribute tradeoff scaling constants k1= .7 (cost)
k2= .3 (mpg)
ISE 562; Dr. consider uncertainty
Utility function for cost is:
u(X1)= -.125(X1-20) +1 if 20X1<24 these were
Utility function for mpg is: u(X2)= .0417(X2-16)
if 16X2<28 if 28X232
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-.03125(X1-40) if 24X140
.125(X2-32) +1
Alternative
Attribute tradeoff scaling constants k1= .8 (cost)
k2= .6 (mpg)
ISE 562; Dr. example, for Car B:
Suppose cost is a uniform pdf that
f(c)14 28c32 Suppose the pdf of mpg is triangular:28
Alternative
ranges from 28 to 32
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1 f(m) f(c)192m 20m28
ISE 562; Dr. simulate cost with Monte Carlo simulation Solve CDF to obtain realization function
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f(c)14 28c32 F(c)c 14d14(c28)
ri 14(ci 28)
ci4ri28 0ri1
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