Fuzzy subgrad ien t a lgor ithm for solv ing Lagrang ian relaxa tion dua l problem

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1 19 11 V o l. 19 N o. 11 Con trol nd D ecision N ov : (2004) , (, ) :,,,,, ; ;. : ; ; ; : O 232: A Fuzzy subgrd ien t lgor ith for solv ing Lgrng in relx tion du l proble ZH OU W ei, J IN Y i2hu i (D eprt ent of A uto tion, T singhu U niversity, Beijing , Ch in. Co rrespondent: ZHOU W ei, E2 il: zhouw 99@ ils. tsinghu. edu. cn) Abstrct: To the p roble of zigzging hppened in so lving the undifferentil L grngin dul p roble s by subgrdi2 ent lgo rith, subgrdient lgo rith bsed on fuzzy theo ry is p resented. In th is ethod, the resulting subgrdient direction is ttined by co bining ll h isto ry subgrdient directions, w h ich re ch ieved in the itertion p rocess, fo l2 low ing si p le e bersh ip function. T he resulting subgrdient direction uses the h isto ry info r tion suitbly, thereby significntly reduces the so lution zigzgging difficulty w ithout uch dditionl co puttionl require ents. T he convergence of the lgo rith is p roved. T h is ethod is then pp lied in the trveling sles n p roble, nd the results show tht th is ethod leds to significnt i p rove ent over the trditionl subgrdient lgo rith. Key words: L grngin relxtion; subgrdient lgo rith ; fuzzy theo ry; dul 1 (L R ),,.,,., [1, 2 ],., : [ 2 ] ; [ 3 ], ; [ 4 ],,.,., : ; : : ( ). : (1977),,,, ; (1936),,,,.

2 , T SP.,. 2 (P) : in f (x), s. t. g (x) 0, x D. : D, f g D, g (x) r 1., : L (x, u) = f (x) + g T (x) u, q (u) = inf L (x, u), x D (D ) : x q (u), s. t. u 0. : L (x, u) ; q (u) f (x), (D ) (P) ; u 1 r,,, L (x, u) x. L R (D ) (P). q (u), (D ), k : u k = u k- 1 + tkg k. (1) : g k = g (x k) k, tk k.,,,., [15 ]., [ 2 ], u k = u k- 1 + tkd k (2) (1),. d k = k w k j g j, (3) g j j, w k j k j. [1, 5 ],. [ 6 ] Bundle,.,,., ( j = 1,, k ) : j L (u j, x j) = in x D L (u j, x) = g T j u j + cj. (4) : g j = g (x j), cj = f (x j). (D ) L 3 = L (u 3, x 3 ) = x inl (u, x) = g 3 T u 3 + c 3. (5) u0 x D : g 3 = g (x 3 ), c 3 = f (x 3 ). w k j = w { k j k w { k j. (6) w { k j : w { k j = A (x j) = (L (u k, x k) + Ε- L (u k, x j) ) gε, L (u k, x j) < L (u k, x k) + Ε L (u k, x j) L 3 ; 0,. (7) 0 Ε(L 3 - L (u k, x k) ) g. (8) tk 0 tk 2 ( - 1) (L 3 - L (u k, x k) ) d k 2, > 1. (9), (7),. : Step 1: : k = 0, u 0 0; Step 2: u k (D )., Step 5;, k = k + 1; Step 3: (6) (8), (9) ; Step 4: (2) u k, Step 2; Step 5: x k u k ;., O (X ), k, O (X + k). X,, k,, O (X )., k 200,

3 11 : Π j = 1, 2,, k, L (u k, x j) L 3, : 0 L 3 - L (u k, x j) g T j (u 3 - u k). (10), L 3 - L (u k, x j) 0,. (4) : L (u k, x k) = in L (u k, x), x D L (u k, x k) L (u k, x j). (11) L 3 = L (u 3, x 3 ) L (u 3, x j), (12) (4) (12) L 3 - L (u k, x j) L (u 3, x j) - L (u k, x j) = g T j u 3 + cj - (g T j u k + cj) = g T j (u 3 - u k). 1,,. 2 > 1, : 0 L 3 - L (u k - x k) < d T j (u 3 - u k). (13) (13), : L (u k, x j) < L (u k, x k) + = 1, 2,, k (8), L (u k, x j) < L (u k, x k) + Ε L (u k, x k) + (L 3 - L (u k, x k) ) g, L 3 - L (u k, x j) > L 3 - [L (u k, x k) + (L 3 - L (u k, x k) ) g ] = ( - 1) (L 3 - L (u k, x k) ) g, 1, ( - 1) (L 3 - L (u k, x k) ) g < L 3 - L (u k, x j). ( - 1) (L 3 - L (u k, x k) ) L 3 - L (u k, x j) g T j (u 3 - u k). (3) (6), ( - 1) (L 3 - L (u k, x k) ) < < Ε, Π j w k j g T j (u 3 - u k) = d T k (u 3 - u k). 2,,. 1,. u 3 - u k+ 1< u 3 - u k, Π k. (14) u 3 - u k+ 1 2 = u 3 - u k - tkd k 2 u 3 - u k 2-2tkd T k (u 3 - u k) + t 2 kd k 2. 2 u 3 - u k+ 1 2 < u 3 - u k 2 - tk t 2 kd k 2 = u 2 - u k 2 - tk 2 ( - 1) (L 3 - L (u k, x k) ) (9), 2 ( - 1) (L 3 - L (u k, x k) ) - tkd k 2. u 3 - u k+ 1 2 < u 3 - u k 2. 4 (TSP). : cijx ij, i= 1, ji in s. t. x ij + x j i = 2, i = 1,,., ji, ji : i j ; ; X ; cij i j ; i j x ij = 1,, x ij = 0., q (u) = - 2 u i + in{ i= 1 u i + u j) x ij: x X }. (cij + i= 1, ji XPR ESS, 33, , 110. : gp 0. 1, gp = (L 3 - L ) gl (S) (FS), tk = Χk (L 3 - L (u k, x k) ) gg k 2 (15). : Χk = 1, = 2., L 3 +,, [1 ], [ 7 ]..,, XPR ESS, L 3. PC, CPU 850

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