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TRANSCRIPT
2017.10.06.
Multiattribute decisionmakingMályusz Levente
Department of Construction Technology and Management
Problem / raising questions
Given: • Objects and know their properties, Object: house, plot, offer, plan ....
• Features/attributes: cost of construction, running costs, deadline ...
• Let's put the objects in some order of importance
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Methods
• Pairwise comparison?• Multicriteria optimization?• Multiattribute Optimization?
Department of Construction Technology and Management
2017.10.06.
Evaluating steps
• I select objects to compare,• II. defining the evaluation factors and their weights, which can
usually be achieved by organizing the evaluation factors by tree structure,
• III. then the objects must be evaluated according to the evaluation factors, ie the value of each object must be determined according to each evaluation factor,
• IV. selecting the evaluation procedure/method and carrying out the evaluation,
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Ej
Oi
Ttj
t(i)
O1
Om
E1 En
sjs1 sn
y
Given Find
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Scales 1.
• Nominal scale: dwelling house, not quantifiable, only frequency of occurrence can be investigated
• Ordinary Scale: Grades, "Better" or "Multiple" are the catchy (median, mean?)
• Interval Scale: Temperature, Time; the difference can be grasped, the ratio does not make sense Scale: consumption, Kelvin (temperature scale)
• Ratio scale: consumption (Kelvin temperature scale)
Department of Construction Technology and Management
2017.10.06.
Daniel Bernoulli : Risk aversion
• Assumption: feeling is proportional to a relative increase in wealth
AA-1 A+1
F(A+t)-F(A)>F(a)-F(a-t)
FF=ln(x)+c
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physical stimulus and the perceived change
• Earnst Heinrich Weber (1795-1878) quantify human response to a physical stimulus
• Weber–Fechner law: logaritmic function, f=const1*ln(x)+const2• Stanley Smith Stevens, 1906-1973, • Stevens’s power law: power function, f=const*xα
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Divergence functions, „sensation function”
érzet=x3.5
érzet=x0.17
érzet=x0.3
inger, x
érzet
érzet=x0.17
érzet=x0.3
érzet=ln(x)
inger, x
érzet
Department of Construction Technology and Management
2017.10.06.
Stevens’s law…Continuum Exponent ( a {\displaystyle a} ) Stimulus condition
Loudness 0.67 Sound pressure of 3000 Hz tone
Vibration 0.95 Amplitude of 60 Hz on finger
Vibration 0.6 Amplitude of 250 Hz on finger
Brightness 0.33 5° target in dark
Brightness 0.5 Point source
Brightness 0.5 Brief flash
Brightness 1 Point source briefly flashed
Lightness 1.2 Reflectance of gray papers
Visual length 1 Projected line
Visual area 0.7 Projected square
Redness (saturation) 1.7 Red-gray mixture
Taste 1.3 Sucrose
Taste 1.4 Salt
Taste 0.8 Saccharine
Smell 0.6 Heptane
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Utility Functions
• Transform every attributes to 1-5, 0-10, 0-100, 1-10, 1-100 scales
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Scales 3. 1-10 or 1-100…………………
T1. T2.O1. 16 26 42
O2. 24 19 43
O1. 2 3 5
O2. 2 2 4
1-10 vagy 1-100?
Department of Construction Technology and Management
2017.10.06.
Dominance method
O *
Tt j
t( i)
d
O i
D O IIO D t IIdj jj
n
11
1
* ( )
y
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Divergence functions: Euclidean Difference
• Definition: Divergence the vector of a=(a1,...,aj,...,an) from vectorb=(b1,...,b2,...bn) is the value of the following expression:
n
j
bababIIaD1
222
211)(
1. D(aIIb)0,2. D(aIIb)=0 if and only if, ha a=b.3. D(aIIb)=D(bIIa)
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Divergence functions: Kullback-Leibler• Divergence the vector of a=(a1,...,aj,...,an) >0 from vector
b=(b1,...,b2,...bn) >0 is the value of the following expression:
n
jjj
j
jj ba
b
aabIIaD
1
ln)(
1. D(aIIb)0,2. D(aIIb)=0 if and only if a=b.3. D(aIIb)D(bIIa)
Department of Construction Technology and Management
2017.10.06.
A priori (prior to the event), a posteriori (afterthe event)• The vector „a” is after the "a posteriori" event
Vector „b” in a pre-event (prior to the event) role
D aIIb aa
ba bj
j
jj j
j
n
( ) ln
1
Prior to the eventnotion
After event, fact
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Dominance methodE1. E2. E3. E4. E5. E6.
O1. 1 2 4 2 1 4O2. 2 3 3 2 1 1O3. 3 3 2 3 4 1O4. 4 2 1 1 3 2
domináns (d) 4 3 4 3 4 4
D t IIdj jj
n( ) ln ln ln1
1
11
41 4 2
2
32 3 4
4
44 4
22
32 3 1
1
41 4 4
4
44 4 3 605ln ln ln .
y1=D(t(1)IId)=3.605y2=D(t(2)IId)=4.167y3=D(t(3)IId)=2.364y4=D(t(4)IId)=3.454
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Hellinger, Pearson, Fischer divergencefunctions
D aIIb b aH j jj
n
( )
2
1
D aIIb
b a
bPj j
jj
n
( )
2
1
D aIIb bb
aF jj
jj
n
( ) log
1
a jj
n
1
1
b jj
n
1
1
Department of Construction Technology and Management
2017.10.06.
I. elv. Bridgman model (minimize averagedeviation)
Ej
Oi
Ttj
t(i)
O1
Om
E1 En
sj
y
s1 sn
Evaluating vector
Let the sum of the divergences among y and vectors be minimalnj ttt ,...,...1
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Bridgman model 1.
• Our task is therefore to look for a mean vector y, for which the average deviation isminimum.
•
•
• It is minimum if
n
jjj tIIyDs
1
min
miy
tIIyD
i
,...1;0)(
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Bridgman model 1
n
jij
i
iij
i
n
jiji
ij
iij
ty
yys
y
tyty
ys
y
tIIyD
1
11lnln
ln)(
0lnln1
n
jijij tys
;,...,1,)ln(exp11
mittsyn
j
sij
n
jijji
j
Department of Construction Technology and Management
2017.10.06.
Bridgman model 2.
• Our task is therefore to look for a mean vector y, for which the average deviation is
• minimum.
• It is minimum if
n
jjj yIItDs
1
min
y s t i mi j ijj
n
1
1, ,...,
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II. elv. Arimoto- Blahut model (minimizingmaximum deviation)
Ej
Oi
Ttj
t(i)
O1
Om
E1 En
sj
y
s1 sn
evaluating vector
Let the biggest divergences among y and vectors be minimalnj ttt ,...,...1
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Houses and the problem of the well
• Houses are given, where is the well?
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2017.10.06.
III. Robust estimattion (minimizing median deviation)
Ej
Oi
Ttj
t(i)
O1
Om
E1 En
sj
y
s1 sn
Evaluating vector
Let median divergence among y vectors be minimalnj ttt ,...,...1
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Principles for determining the evaluation vector
• Let's examine the deviation from the idealLet minimum the sum of the difference between y andvectors Let the largest difference between y and vectors be minimalLet the median deviation between y and the vectors be minimal
nj ttt ,...,...1
nj ttt ,...,...1
nj ttt ,...,...1
nj ttt ,...,...1
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Mathematical problems:
),,...1(),...,1(,0, njmit ji ),,...1(),...,1(,0, njmit ji
Problem A1.:
n
jjj tIIyDw
1
min Problem A2.:
n
jjj yIItDw
1
min
Problem B1.: jjy
tIIyDmaxmin Problem B2.: yIItD jjy
maxmin
Problem C1.: jjy
tIIyDmedmin Problem C2.: yIItDmed jjy
min
Department of Construction Technology and Management
2017.10.06.
Why median? a robust estimation
Least squares
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example
Y=2x+3
X 2 3 4 5 6Y 7 9 11 13 155
X 2 3 4 5 6Y 7 9 11 13 15
Y=30x-81
min)(1
2
n
jjj xfy
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Example cont.
Y=2x+3
X 2 3 4 5 6Y 7 9 11 13 155
X 2 3 4 5 6Y 7 9 11 13 15
Y=2x+3
min2 baxymed jjj
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2017.10.06.
Example cont.
X 2 3 4 5 6Y 7 9 11 133 155
Y=42X-105 Y=2X+3
min2 baxymed jjj
min)(1
2
n
jjj xfy
Department of Construction Technology and Management