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Programaç˜ao Linear - Notas de aula - CEUNES

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CAPÍTULO 8. MÉTODO SIMPLEX: INICIALIZAÇÃO E CICLAGEM 85Como b ≥ 0, a base I formada pelas colunas 1, 2 e 3 é viável.z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 0 0 0 3/4 −20 1/2 −6 0x 1 0 1 0 0 1/4 −8 −1 9 0x 2 0 0 1 0 1/2 −12 −1/2 3 0x 3 0 0 0 1 0 0 1 0 1z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 −3 0 0 0 4 7/2 −33 0x 4 0 4 0 0 1 −32 −4 36 0x 2 0 −2 1 0 0 4 3/2 −15 0x 3 0 0 0 1 0 0 1 0 1z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 −1 −1 0 0 0 2 −18 0x 4 0 −12 8 0 1 0 8 −84 0x 5 0 −1/2 1/4 0 0 1 3/8 −15/4 0x 3 0 0 0 1 0 0 1 0 1z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 2 −3 0 −1/4 0 0 3 0x 6 0 −3/2 1 0 1/8 0 1 −21/2 0x 5 0 1/16 −1/8 0 −3/64 1 0 3/16 0x 3 0 3/2 −1 1 −1/8 0 0 21/2 1z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 1 −1 0 1/2 −16 0 0 0x 6 0 2 −6 0 −5/2 56 1 0 0x 7 0 1/3 −2/3 0 −1/4 16/3 0 1 0x 3 0 −2 6 1 5/2 −56 0 0 1z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 0 2 0 7/4 −44 −1/2 0 0x 1 0 1 −3 0 −5/4 28 1/2 0 0x 7 0 0 1/3 0 1/6 −4 −1/6 1 0x 3 0 0 0 1 0 0 1 0 1z x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 0 0 0 3/4 −20 1/2 −6 0x 1 0 1 0 0 1/4 −8 −1 9 0x 2 0 0 1 0 1/2 −12 −1/2 3 0x 3 0 0 0 1 0 0 1 0 1O último QS é idêntico ao primeiro. Note também que todos os QS correspon<strong>de</strong>m aovértice x = (0, 0, 1, 0, 0, 0, 0), ou seja, o processo acima gerou as bases B 1 , B 2 , . . . , B 7 , on<strong>de</strong>B 7 = B 1 = [ ]a 1 a 2 a 3 todas associadas à x. Se a mesma sequência <strong>de</strong> pivoteamentos fosseutilizada, o Simplex ciclaria entre essas bases e não convergiria.No Exemplo acima, logo no primeiro quadro temos um empate na escolha da variável <strong>de</strong>saída: x 1 empata com x 2 pois b 1 /y 14 = b 2 /y 24 = 0. Po<strong>de</strong>mos então, ao invés <strong>de</strong> pivotear sobre

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