Contoh : Kasus Maksimasi

Maks, Z = 60 X1 + 30 X2 + 20 X3 s/t 8 X1 + 6 X2 + X3 ≤ 48 4 X1 + 2 X2 + 1,5 X3 ≤ 20 2 X1 + 1,5 X2 + 0,5 X3 ≤ 8 X2 ≤ 5

X1, X2, X3 ≥ 0

Maks, Z = 60 X1 + 30 X2 + 20 X3 s/t 8 X1 + 6 X2 + X3 + S1 = 48 4 X1 + 2 X2 + 1,5 X3 + S2 = 20 2 X1 + 1,5 X2 + 0,5 X3 + S3 = 8 X2 + S4 = 5

X1, X2, X3, S1, S2, S3, S4 ≥ 0

Iterasi 0

BV Z X1 X2 X3 S1 S2 S3 S4 Solusi RasioZ 1 -60 -30 -20 0 0 0 0 0S1 0 8 6 1 1 0 0 0 48 48/8 = 6S2 0 4 2 1,5 0 1 0 0 20 20/4 = 5S3 0 2 1,5 0,5 0 0 1 0 8 8/2 = 4S4 0 0 1 0 0 0 0 1 5 5/0 = ~

Iterasi 1

BV Z X1 X2 X3 S1 S2 S3 S4 Solusi RasioZ 1 0 15 -5 0 0 30 0 240S1 0 0 0 -1 1 0 -4 0 16 16/-1 = -16S2 0 0 -1 0,5 0 1 -2 0 4 4/0,5 = 8X1 0 1 0,75 0,25 0 0 0,5 0 4 4/0,25 = 16S4 0 0 1 0 0 0 0 1 5 5/0 = ~

Iterasi 2

BV Z X1 X2 X3 S1 S2 S3 S4 Solusi RasioZ 1 0 5 0 0 10 10 0 280S1 0 0 -2 0 1 2 -8 0 24X3 0 0 -2 1 0 2 -4 -0 8X1 0 1 1,25 0 0 -0,5 1,5 0 2S4 0 0 1 0 0 0 0 1 5

Solusi Optimum :

Z = 280X1 = 2; X2 = 0; X3 = 8S1 = 24; S2 = 0; S3 = 0; S4 = 5