justyna pastwa - flood vulnerability estimation using fha geometric mean method

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Evaluation of flood vulnerability in Lower SilesianVoivodeship using fuzzy arithmetic operations

Justyna Pastwa

Palacky University

pastwa.justyna@gmail.com

October 15, 2013

Justyna Pastwa (UPOL) Short title October 15, 2013 1 / 18

Overview

1 Vulnerability

2 Data

3 MethodologyTheoryEstimation of weights

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Vulnerability

Justyna Pastwa (UPOL) Short title October 15, 2013 3 / 18

Vulnerability

Justyna Pastwa (UPOL) Short title October 15, 2013 4 / 18

Vulnerability

Justyna Pastwa (UPOL) Short title October 15, 2013 5 / 18

Data

which data are suitable?

which variables are meaningful?

are they available?

Used data:

registered events of floods, 2007–2011, Lower Silesian Voivodeship

magnitude, number of injured, fatalities and evacuated, flood damageand frequency of events

Justyna Pastwa (UPOL) Short title October 15, 2013 6 / 18

Data

which data are suitable?

which variables are meaningful?

are they available?

Used data:

registered events of floods, 2007–2011, Lower Silesian Voivodeship

magnitude, number of injured, fatalities and evacuated, flood damageand frequency of events

Justyna Pastwa (UPOL) Short title October 15, 2013 6 / 18

Problems:

imprecise

conflicting attributes

different importance

Justyna Pastwa (UPOL) Short title October 15, 2013 7 / 18

Problems:

imprecise

conflicting attributes

different importance

Justyna Pastwa (UPOL) Short title October 15, 2013 7 / 18

Problems:

imprecise

conflicting attributes

different importance

Justyna Pastwa (UPOL) Short title October 15, 2013 7 / 18

Methodology

Lets consider the problem of ranking m alternatives (Ai ; i = 1, 2, . . . ,m)by a decision maker (DM). DM wish to rank m alternatives with the helpof information supplied by n experts (Ej ; j = 1, 2, . . . , n) on each of Kcriteria (Ck ; k = 1, 2, . . . ,K ). DM wish to find which from alternativessatisfy criteria the best.

Justyna Pastwa (UPOL) Short title October 15, 2013 8 / 18

Methodology – weights estimation

The average of fuzzy numbers across all the experts:

q̃k = (1/n) � ( ˜ck1 ⊕ ˜ck2 ⊕ . . .⊕ ˜ckn) (1)

where � and ⊕ are fuzzy multiplication and addition, respectively

Justyna Pastwa (UPOL) Short title October 15, 2013 9 / 18

Methodology – fuzzy rank score matrix

For each criteria, rank the degree of satisfiability for each system withrespect to each criteria item by integer numbers 1, 2, . . . , etc:

A =

C1 C2 . . . Cn

A1 p̃11 p̃12 . . . p̃1nA2 p̃21 p̃22 . . . p̃2n...

......

. . ....

Am ˜pm1 ˜pm2 . . . ˜pmn

(2)

Justyna Pastwa (UPOL) Short title October 15, 2013 10 / 18

Methodology – ranking alternatives

Perform following operations:

R = A�W T =

p̃11 � w̃1 ⊕ p̃12 � w̃2 ⊕ . . .⊕ p̃1n � w̃n

p̃21 � w̃1 ⊕ p̃22 � w̃2 ⊕ . . .⊕ p̃2n � w̃n...

˜pm1 � w̃1 ⊕ ˜pm2 � w̃2 ⊕ . . .⊕ ˜pmn � w̃n

=

R(1)R(2)

...R(m)

(3)

Deffuzzification.

Justyna Pastwa (UPOL) Short title October 15, 2013 11 / 18

Weights estimation of the attributes

W =

E1 E2 E3 E4 E5 E6 E7 E8 E9 E10

freq 5 7 7 10 0 10 9 1 4 10mag 7 10 5 10 0 10 4 1 2 8inj 0 4 0 0 0 0 0 0 0 0dead 0 4 0 0 0 0 0 0 0 0eva 4 8 1 6 0 0 7 0 2 1loss 5 10 3 8 1 0 9 4 2 5

(4)

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Weights

Justyna Pastwa (UPOL) Short title October 15, 2013 13 / 18

Real data description

Justyna Pastwa (UPOL) Short title October 15, 2013 14 / 18

Real data description

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Real data description

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Real data description

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The End

Justyna Pastwa (UPOL) Short title October 15, 2013 18 / 18

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