Download - Multiattribute decision making3 - BME · î ì í ó x í ì x ì ò x ( ( (\ ¦ z ¦,,\ ,,\ ,,\

$: Multiattribute decision making3 - BME · î ì í ó x í ì x ì ò x ( ( (\ ¦ z ¦,,\ ,,\ ,,\$
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


Methods

• Pairwise comparison?• Multicriteria optimization?• Multiattribute Optimization?


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,


Ej

Oi

Ttj

t(i)

O1

Om

E1 En

sjs1 sn

y

Given Find


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)


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


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


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


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


Utility Functions

• Transform every attributes to 1-5, 0-10, 0-100, 1-10, 1-100 scales


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?


2017.10.06.

Dominance method

O *

Tt j

t( i)

d

O i

D O IIO D t IIdj jj

n

11

1

* ( )

y


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)


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)


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


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


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


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


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)(


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


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, ,...,


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


Houses and the problem of the well

• Houses are given, where is the well?


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


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


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


2017.10.06.

Why median? a robust estimation

Least squares


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


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


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


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