EL METODO ALGORITMO DE SIMPLES PARTE 2

SOLUCION DE LOS PROBLEMAS DE APLICACIÓN

1.- Sea los siguientes problemas :

Max Z  =  2X1 + 3X2 + X4 – X3 – 2X5

S.a.

                        X1 – X2            + X4  –   X5   = 100

                                X2 – X3  + 2X4              = 20

                                       X3   –   X4              ≤ 50

                                        X1, X2, X3, X4, X5  ≥  0

Forma Estandarizada:

Max Z  =  2X1 + 3X2 – X3 + X4 – 2X5 – MA1 – MA2 + 0S3

            S.a.

                        X1 – X2          +     X4  – X5 + A1                    =  100 

                                X2 – X3  +  2X4                   + A2            =  20

                                       X3  –     X4                            + S3   =  50

                                   X1 , X2 , X3 , X4 , X5 , A1 , A2 , S3      ≥  0

		Cj	2	3	-1	1	2	-M	-M	0
CK	XK	Bi	X1	X2	X3	X4	X5	A1	A2	S3	θK
-M	A1	100	1	-1	0	1	-1	1	0	0	100
-M	A2	20	0	1	-1	2	0	0	1	0	10
0	S3	50	0	0	1	-1	0	0	0	1	-50
	Zj	-120M	-M	0	M	-3M	M	-M	-M	0
	Zj             +	Cj	-M - 2	-3	M – 1	-3M - 1	M - 2	0	0	0
-M	A1	90	1	-3/2	-1/2	0	-1	1	-1/2	0	90
1	X4	10	0	1/2	-1/2	1	0	0	1/2	0	-
0	S3	60	0	1/2	1/2	0	0	0	1/2	1	-
	Zj	-90M+10	-M	3M/2+1/2	-M/2-1/2	1	M	-M	M/2+1/2	0
	Zj             +	Cj	-M - 2	3M/2-5/2	-M/2+1/2	0	M - 2	0	3M/2+1/2	0
2	X1	90	1	-3/2	1/2	0	-1	1	-1/2	0	-
1	X4	10	0	1/2	-1/2	1	0	0	1/2	0	20
0	S3	60	0	1/2	1/2	0	0	0	1/2	1	120
	Zj	190	2	-5/2	1/2	1	-2	2	-1/2	0
	Zj             +	Cj	0	-11/2	3/2	0	0	M + 2	M – 1/2	0
2	X1	120	1	0	-1	3	-1	1	1	0	-
3	X2	20	0	1	-1	2	0	0	1	0	-
0	S3	50	0	0	1	-1	0	0	0	0	50
	Zj	300	2	3	-5	12	-2	2	5	0
	Zj             +	Cj	0	0	-4	11	0	M + 2	M + 5	0
2	X1	170	1	0	0	2	-1	1	1	1
3	X2	70	0	1	0	1	0	0	1	1
-1	X3	150	0	0	1	-1	0	0	0	1
	Zj	500	2	3	-1	8	-2	2	5	4
	Zj             +	Cj	0	0	0	7	0	M + 2	M + 5	4