Programaç˜ao Linear - Notas de aula - CEUNES

CAPÍTULO 11. ANÁLISE DE SENSIBILIDADE 125Para λ = 1 o QS ficaz x 1 x 2 x 3 x 4 RHSz 1 0 0 −5/3 + 5/3λ −2/3 − 1/3λ −14 + 8λx 1 0 1 0 2/3 −1/3 2x 2 0 0 1 1/3 1/3 4z x 1 x 2 x 3 x 4 RHSz 1 0 0 0 −1 −6x 1 0 1 0 2/3 −1/3 2x 2 0 0 1 1/3 1/3 4e x 3 entra na base. Observe que x 1 sai, e o novo QS éz x 1 x 2 x 3 x 4 RHSz 1 0 0 0 −1 −6x 3 0 3/2 0 0 −1/2 3x 2 0 −1/2 1 1 1/2 3Queremos agora encontrar o intervalo [1, λ 2 ] no qual a base corrente [ ]a 3 a 2 permaneçaótima. Lembrando que o vetor de custos original é c = (−1, −3, 0, 0) e a perturbação na direçãoc ′ = (2, 1, 0, 0), temosz ′ 1 − c ′ 1 = c ′ By 1 − c ′ 1 = [ 0 1 ] [ 3/2−1/2z ′ 4 − c ′ 4 = c ′ By 4 − c ′ 4 = [ 0 1 ] [ −1/21/2z 4 − c 4 = c B y 4 − c 4 = [ 0 −3 ] [ −1/21/2Logo S = {4} e ˆλ = 3. Para λ ∈ [1, 3] temos aindaz(λ) = c B b + λc ′ Bb = [ 0 −3 ] [ ]3+ λ [ 0 1 ] [ 333]− 2 = −5/2]− 0 = 1/2]+ 4 = −3/2.]= −9 + 3λ,(z 1 − c 1 ) + λ(z ′ 1 − c ′ 1) = 5/2 − 5/2λ,(z 4 − c 4 ) + λ(z ′ 4 − c ′ 4) = −3/2 + 1/2λ,e o QS ótimo éz x 1 x 2 x 3 x 4 RHSz 1 5/2 − 5/2λ 0 0 −3/2 + 1/2λ −9 + 3λx 3 0 3/2 0 0 −1/2 3x 2 0 −1/2 1 1 1/2 3Para λ = 3, a VNB x 4 entra na base e temos o QSz x 1 x 2 x 3 x 4 RHSz 1 −5 0 0 0 0x 3 0 1 1 1 0 6x 4 0 −1 2 0 1 6