EL METODO ALGORITMO DE SIMPLES PARTE 3

5.-Max Z  =  4X1 – 2X2 + 3X3

            S.a.                 X1 – X2 – X3  ≤  8

                                X2 – X3  ≤  4

                        X1         + X3  ≤  12

                           X1 , X2 , X3  ≥  0

Forma Estandarizada :

            Max Z  =  4X1 – 2X2 + 3X3 + 0S1 + 0S2 + 0S3

X1 – X2  – X3 + S1                =  8

        X2 -  X3           + S2             =  4

           X1          + X3              + S3    =  12

                                X1, X2, X3, S1, S2, S3     ≥  0

		CJ	4	-2	3	0	0	0
CB	XB	bi	X1	X2	X3	S1	S2	S3	θK
0	S1	8	1	-1	-1	1	0	0	8
0	S2	4	0	1	-1	0	1	0	-
0	S3	12	1	0	1	0	0	1	12
	ZJ	0	0	0	0	0	0	0
	ZJ         +	CJ	-4	2	-3	0	0	0
4	X1	8	1	-1	-1	1	0	0	-
0	S2	4	0	1	-1	0	1	0	-
0	S3	4	0	1	2	-1	0	1	2
	ZJ	32	4	-4	-4	4	0	0
	ZJ         +	CJ	0	-2	-7	4	0	0
4	X1	10	1	-1/2	0	1/2	0	1/2
0	S2	6	0	3/2	0	-1/2	1	1/2
3	X3	2	0	1/2	1	-1/2	0	1/2
	ZJ	46	4	-1/2	3	1/2	0	7/2
	ZJ         +	CJ	0	3/2	0	1/2	0	7/2

X1 = 10           X3 = 2            S2 = 6           Z  =  46

Max Z  =  4(10) - 2(0) + 3(2) + 0(S1) + 0(6) + 0(S3)  =   46

Max Z  =  46

6.-

Max Z  =  3X1 – 2X2 + X3

            S.a.

                        X1 – X2 - X3         ≤  2

                        X1        – X3         ≤  4

                        X1        + X3         ≤  6

                                 X1 , X2 , X3   ≥  0

Forma Estandarizada :

            Max Z  =  3X1 – 2X2 + X3 + 0S1 + 0S2 + 0S3

X1 – X2  – X3 + S1                =  2

X1         -  X3           + S2             =  4

           X1          + X3              + S3    =  6

                                X1, X2, X3, S1, S2, S3     ≥  0

		CJ	3	-2	1	0	0	0
CB	XB	bi	X1	X2	X3	S1	S2	S3	θK
0	S1	2	1	-1	-1	1	0	0	2
0	S2	4	1	0	-1	0	1	0	4
0	S3	6	1	0	1	0	0	1	6
	ZJ	0	0	0	0	0	0	0
	ZJ         +	CJ	-3	2	-1	0	0	0
3	X1	2	1	-1	-1	1	0	0	-2
0	S2	2	0	1	0	-1	1	0	-
0	S3	4	0	1	2	-1	0	1	2
	ZJ	6	3	-3	-3	3	0	0
	ZJ         +	CJ	0	-1	-4	3	0	0
3	X1	4	1	-1/2	0	1/2	0	1/2
0	S2	2	0	1	0	-1	1	0
1	X3	2	0	1/2	1	-1/2	0	1/2
	ZJ	14	3	-1	1	1	0	2
	ZJ         +	CJ	0	1	0	1	0	2

X1 = 4           X3 = 1           S2 = 2           Z  =  14

Max Z  =  3(4) - 2(0) + 1(2) + 0(S1) + 0(2) + 0(S3)  =   14

Max Z  =  14

7.-

Max Z  =  X1 – X2 + X3

            S.a.

                        2X1 – X2            ≤  4

                                 X2  – X3    ≤  6

                                 X2 +  X3    ≤  8

                           X1 , X2 , X3      ≥  0

Forma Estandarizada :

            Max Z  =  X1 – X2 + X3 + 0S1 + 0S2 + 0S3

2X1 – X2          + S1                =  4

          X2 -  X3           + S2             =  6

                      X2 + X3             + S3    =  8