Unidad 3. Participación 9. Dos fases

f) Max z= 3x1 + 5x2

4x1 + x2 ≥ 4

-x1 + 2x2 ≥ 2

 x2 ≤ 3

 x1, x2 ≥ 0

	x1	x2	x3	x4	x5	a1	a2	Sol
wj-cj	0	0	0	0	0	-1	-1	0
zj-cj	-3	-5	0	0	0	0	0	0
a1	4	1	-1	0	0	1	0	4
a2	-1	2	0	-1	0	0	1	2
x5	0	1	0	0	1	0	0	3
	x1	x2	x3	x4	x5	a1	a2	Sol
wj-cj	3	3	-1	-1	0	0	0	6
zj-cj	-3	-5	0	0	0	0	0	0
a1	4	1	-1	0	0	1	0	4
a2	-1	2	0	-1	0	0	1	2
x5	0	1	0	0	1	0	0	3
	x1	x2	x3	x4	x5	a1	a2	Sol
wj-cj	0	2   1/4	-   1/4	-1	0	-   3/4	0	3
zj-cj	0	-4   1/4	-   3/4	0	0	3/4	0	3
x1	1	1/4	-   1/4	0	0	1/4	0	1
a2	0	2   1/4	-   1/4	-1	0	1/4	1	3
x5	0	1	0	0	1	0	0	3
	x1	x2	x3	x4	x5	a1	a2	Sol
wj-cj	0	0	0	0	0	-1	-1	0
zj-cj	0	0	-1   2/9	-1   8/9	0	1   2/9	1   8/9	8   2/3
x1	1	0	-   2/9	1/9	0	2/9	-   1/9	2/3
x2	0	1	-   1/9	-   4/9	0	1/9	4/9	1   1/3
x5	0	0	1/9	4/9	1	-   1/9	-   4/9	1   2/3
	x1	x2	x3	x4	x5	Sol
zj-cj	0	0	-   3/4	0	4   1/4	15   3/4	-
x1	1	0	-   1/4	0	-   1/4	1/4	-
x2	0	1	0	0	1	3	-
x5	0	0	1/4	1	2   1/4	3   3/4	15
	x1	x2	x3	x4	x5	Sol
zj-cj	0	0	0	3	11	27
x1	1	0	0	1	2	4
x2	0	1	0	0	1	3
x5	0	0	1	4	9	15

 Solución óptima :

X1 = 4 ,  x2 = 3 , x3 = x4 = 0, x5 = 15  z = 27