A GP-AHP method for solving group decision-making fuzzy AHP ...
A GP-AHP method for solving group decision-making fuzzy AHP ...
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 1995<br />
where log ae 21j − log 3+de 21j ¿ 0, ae 21j ;de 21j ¿ 0, <strong>for</strong> j =2; 3, and e =1; 2; 3;<br />
log ae 312 = − 0:066949999 (log ae 312) − 1:8455416666 de 312 +0:544072297; (6.17)<br />
where log ae 312 − log 3+de 312 ¿ 0, ae 312 ;de 312 ¿ 0, <strong>for</strong> e =1; 2;<br />
log a3 312 = − 0:096910023 (log a3 312) − 1:775652326 d3 312 +0:397939017; (6.18)<br />
where log a3 312 − log 2+d3 312 ¿ 0, a3 312 ;d3 312 ¿ 0;<br />
log ae 313 = − 0:066949999 (log ae 313) − 1:845541666 de 313 +0:544072297; (6.19)<br />
where log ae 313 − log 3+de 313 ¿ 0, ae 313 ;de 313 ¿ 0, <strong>for</strong> e =1; 2; 3;<br />
log ae 323 = − 0:17609013 (log ae 323) − 2 de 323 +0:176091259; (6.20)<br />
where log ae 323 − log 1+de 323 ¿ 0, ae 323 ;de 323 ¿ 0, <strong>for</strong> e =1; 2; 3;<br />
log ae 413 = − 0:096910023 (log ae 413) − 1:775652326 de 413 +0:397939017; (6.21)<br />
where log ae 413 − log 2+de 413 ¿ 0, ae 413 ;de 413 ¿ 0, <strong>for</strong> e =1; 2;<br />
log a3 413 = − 0:124940028 (log a3 413) − 2:289237416 d3 413 − 0:17608998; (6.22)<br />
where log a 3 413<br />
1 − log( 2 )+d3 413 ¿ 0, a3 413 ;d3 413 ¿ 0;<br />
log ae 423 = − 0:124940028 (log ae 423) − 2:289237416 de 423 − 0:17608998; (6.23)<br />
where log ae 423 − log 2+de 423 ¿ 0, ae 423 ;de 423 ¿ 0, <strong>for</strong> e =1; 2; 3.<br />
Step 2: In the <strong>for</strong>m of the proposed model 2, Example 4 can be <strong>for</strong>mulated as follows:<br />
⎧<br />
⎨ 4� 4� 3�<br />
Minimize<br />
(log vq − log vq<br />
⎩<br />
′ − log aeqq ′ +2 e qq ′)<br />
q=1 q ′ ¿q e=1<br />
+<br />
⎧<br />
⎨<br />
−<br />
⎩<br />
4�<br />
3�<br />
3�<br />
q=1 i=1 j¿i e=1<br />
4�<br />
4�<br />
3�<br />
q=1 q ′ ¿q e=1<br />
Subject to (6:1)–(6:23);<br />
3�<br />
(log vqi − log vqj − log ae qij +2 e ⎫<br />
⎬<br />
qij)<br />
⎭<br />
(log a e qq ′)+<br />
4�<br />
3�<br />
3�<br />
3�<br />
q=1 i=1 j¿i e=1<br />
(log ae ⎫<br />
⎬<br />
qij)<br />
⎭<br />
log vq − log vq ′ − log ae qq ′ + e qq ′ ¿ 0; log vqi − log vqj − log a e qij + e qij ¿ 0;<br />
vq;vqi;a e qq ′;ae qij; e qq ′; e qij;d e ij;d e qij ¿ 0;<br />
(q; q ′ ) ∈{(q; q ′ ) | 1 6 q¡q ′ 6 4}; (i; j) ∈{(i; j) | 1 6 i¡j6 3}:<br />
Step 3: Using Theorem 1 to obtain M 1 =2:25871, M 0 =0:9860733, and =47=(M 1 − M 0 )=<br />
37:71697:<br />
Step 4: After running on the LINDO or EXCEL, the acquired solution set is (log v1 =0:26061,<br />
log v2 =0:405622, log v3 =0, log v4 =0:21752, log v11 =0:175915, log v12 =0, log v13 =0:276157,