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A GP-AHP method for solving group decision-making fuzzy AHP ...

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C.-S. Yu / Computers & Operations Research 29 (2002) 1969–2001 1983<br />

By using LINDO [35] or EXCEL [36], the obtained solution set is ( (log a 1 24 )=1, log a1 24 =0,<br />

a 1 24 =1, d1 24;1 =0, d1 24;2 = 0, and d1 24;3 =0:17609).<br />

Similar to Corollary 1, Corollary 2 is described as follows.<br />

Corollary 2. A general concave <strong>fuzzy</strong> value log aij can be interpreted as<br />

�<br />

m−2 �<br />

m−1 �<br />

log aij = (log aij) − (sij;m−1 − sij;k)dij;k +<br />

k=1<br />

where log aij + � m−2<br />

‘=1 dij;‘ ¿ log aij;m−2; aij ¿ 0; and sij;0 =0.<br />

k=1<br />

(sij;k − sij;k−1)log aij;k−1<br />

��<br />

(sij;m−1);<br />

Proof. Regarding Proposition 3, a piecewise linear membership function (log aij) can be represented<br />

by<br />

m−1 �<br />

sij;1(log aij − log aij;0)+<br />

k=2<br />

sij;k − sij;k−1<br />

(|log aij − log aij;k−1| + log aij − log aij;k−1):<br />

2<br />

After utilizing Proposition 4 to linearize the absolute terms with negative coe cients, (log aij)<br />

can then be interpreted as<br />

�<br />

�<br />

m−1 �<br />

�k−1<br />

(log aij)=sij;1(log aij − log aij;0)+ (sij;k − sij;k−1) log aij − log aij;k−1 + ;<br />

where log aij + � m−2<br />

‘=1 dij;‘ ¿ log aij;m−2 and aij ¿ 0.<br />

Accordingly, after expanding (2:7), the following occurs:<br />

k=2<br />

(log aij)=sij;1 × log aij − sij;1 × log aij;0 +(sij;2 − sij;1)(log aij − log aij;1 + dij;1)<br />

+(sij;3 − sij;2)(log aij − log aij;2 + dij;1 + dij;2)<br />

‘=1<br />

dij;‘<br />

(4.7)<br />

+(sij;4 − sij;3)(log aij − log aij;3 + dij;1 + dij;2 + dij;3)+···+(sij;m−1 − sij;m−2)<br />

�<br />

m−2 �<br />

log aij − log aij;m−2 +<br />

�<br />

: (4.8)<br />

‘=1<br />

dij;‘<br />

As a result, after reorganizing (4.8), we obtain<br />

m−2 �<br />

m−1 �<br />

(log aij)= (sij;m−1 − sij;k)dij;k − (sij;k − sij;k−1)log aij;k−1 + sij;m−1 × log aij: (4.9)<br />

k=1<br />

k=1<br />

After reversing (log aij) and log aij in (4.9),<br />

log aij =<br />

�<br />

m−2 �<br />

(log aij) −<br />

k=1<br />

�<br />

m−1<br />

(sij;m−1 − sij;k)dij;k +<br />

where log aij + � m−2<br />

‘=1 dij;‘ ¿ log aij;m−2;aij ¿ 0, and sij;0 =0.<br />

There<strong>for</strong>e, Corollary 2 is completed.<br />

k=1<br />

(sij;k − sij;k−1)log aij;k−1<br />

��<br />

(sij;m−1);

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