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Programaç˜ao Linear - Notas de aula - CEUNES

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CAPÍTULO 8. MÉTODO SIMPLEX: INICIALIZAÇÃO E CICLAGEM 83O pivoteamento na Fase 1 é feito utilizando a linha <strong>de</strong> x 0 , atualizando também a linha <strong>de</strong>z pelo procedimento usual. Após o término da Fase 1, caso x 0 = 0 e as variáveis artificiaisestejam todas fora da base, eliminamos a linha <strong>de</strong> x 0 e as colunas das artificiais, e proce<strong>de</strong>mosnormalmente com a Fase 2.Exemplo 8.1.2. Consi<strong>de</strong>re o PL do Exemplo anterior após inserção <strong>de</strong> variáveis <strong>de</strong> folga eartificiais,min x 1 −2x 2s.a. x 1 +x 2 −x 3 +x 6 = 2−x 1 +x 2 −x 4 +x 7 = 1x 2 +x 5 = 3x 1 , x 2 , x 3 , x 4 , x 5 , x 6 , x 7 ≥ 0A FO original é x 1 − 2x 2 , e logo c = (1, −2, 0, 0, 0) (as variáveis legítimas são x 1 , . . . , x 5 ). OQS inicial éVar. legítimas Artif.z x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 0 −1 2 0 0 0 0 0 0x 0 0 1 0 2 −1 −1 0 0 0 3x 6 0 0 1 1 −1 0 0 1 0 2x 7 0 0 −1 1 0 −1 0 0 1 1x 5 0 0 0 1 0 0 1 0 0 3QS 2:QS 3:Var. legítimas Artif.z x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 0 1 0 0 2 0 0 −2 −2x 0 0 1 2 0 −1 1 0 0 −2 1x 6 0 0 2 0 −1 1 0 1 −1 1x 2 0 0 −1 1 0 −1 0 0 1 1x 5 0 0 1 0 0 1 1 0 −1 2Var. legítimas Artif.z x 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7 RHSz 1 0 0 0 1/2 3/2 0 −1/2 −3/2 −5/2x 0 0 1 0 0 0 0 0 −1 −1 0x 1 0 0 1 0 −1/2 1/2 0 1/2 −1/2 1/2x 2 0 0 0 1 −1/2 −1/2 0 1/2 1/2 3/2x 5 0 0 0 0 1/2 1/2 1 −1/2 −1/2 3/2FASE IIQS 4:z x 1 x 2 x 3 x 4 x 5 RHSz 1 0 0 1/2 3/2 0 −5/2x 1 0 1 0 −1/2 1/2 0 1/2x 2 0 0 1 −1/2 −1/2 0 3/2x 5 0 0 0 1/2 1/2 1 3/2

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