My country: colours. Joanna Jurgielewicz Justyna Sterniczuk.
Justyna Pastwa - Flood vulnerability estimation using FHA geometric mean method
description
Transcript of Justyna Pastwa - Flood vulnerability estimation using FHA geometric mean method
Evaluation of flood vulnerability in Lower SilesianVoivodeship using fuzzy arithmetic operations
Justyna Pastwa
Palacky University
October 15, 2013
Justyna Pastwa (UPOL) Short title October 15, 2013 1 / 18
Overview
1 Vulnerability
2 Data
3 MethodologyTheoryEstimation of weights
Justyna Pastwa (UPOL) Short title October 15, 2013 2 / 18
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)
Justyna Pastwa (UPOL) Short title October 15, 2013 12 / 18
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
Justyna Pastwa (UPOL) Short title October 15, 2013 15 / 18
Real data description
Justyna Pastwa (UPOL) Short title October 15, 2013 16 / 18
Real data description
Justyna Pastwa (UPOL) Short title October 15, 2013 17 / 18
The End
Justyna Pastwa (UPOL) Short title October 15, 2013 18 / 18