EL METODO ALGORITMO DE SIMPLES PARTE 4

10.-      Max Z  =  2X1 +  4X2 + X3

S.a.

              X1 + 2X2 –   X3  ≤  5

                        2X1 –   X2 + 2X3  =  2

                        -X1 + 2X2 + 2X3  ≥  1

                           X1 , X2 , X3       ≥  0

Forma Estandarizada :

            Max Z  =  2X1 + 4X2 + X3 + 0S1 + 0S3 – MA2 – MA3

S.a.

                          X1 + 2X2 –   X3 + S1                          =  5

                        2X1 –   X2 + 2X3        + A2                  =  2

                        -X1 + 2X2 + 2X3                - S3 + A3    =  1

    

		CJ	2	4	1	0	0	-M	-M
CB	XB	bI	X1	X2	X3	S3	S1	A1	A2	θ  K
0	S1	5	1	2	-1	0	1	0	0	-
-M	A1	2	2	-1	2	0	0	1	0	1
-M	A2	1	-1	2	2	-1	0	0	1	1/2
	ZJ	-3M	-M	-M	-4M	M	0	-M	-M
	ZJ     +	CJ	-M - 2	-M - 4	-4M - 1	-M	0	0	0
0	S1	11/2	1/2	3	0	-1/2	1	0	1/2	11
-M	A1	1	3	-3	0	1	0	1	-1	1/3
1	X3	1/2	-1/2	1	1	-1/2	0	0	1/2	-
	ZJ	-M +1/2	-3M-1/2	3M+1	1	-M-1/2	0	-M	M+1/2
	ZJ     +	CJ	-3M-5/2	3M-3	0	-M-1/2	0	0	2M+1/2
0	S1	16/3	0	7/2	0	-2/3	1	-1/6	2/3	32/21
2	X1	1/3	1	-1	0	1/3	0	1/3	-1/3	-
1	X3	2/3	0	1/2	1	-1/3	0	1/6	1/3	4/3
	ZJ	4/3	2	-3/2	1	1/3	0	5/6	-1/3
	ZJ     +	CJ	0	-11/2	0	1/3	0	M + 5/6	M – 1/3
0	S1	2/3	0	0	-7	5/3	1	-4/3	-5/3	2/5
2	X1	5/3	1	0	2	-1/3	0	2/3	1/3	-
4	X2	4/3	0	1	2	-2/3	0	1/3	2/3	-
	ZJ	26/3	2	4	12	-10/3	0	8/3	10/3
	ZJ     +	CJ	0	0	11	-10/3	0	M + 8/3	M + 10/3
0	S3	2/5	0	0	-21/5	1	-3/5	-4/5	-1	-
2	X1	9/5	1	0	3/5	0	1/5	2/5	0	3
4	X2	8/5	0	1	-4/5	0	2/5	-1/5	0	-
	ZJ	10	2	4	-2	0	2	0	0
	ZJ     +	CJ	0	0	-3	0	2	M	M
0	S3	13	7	0	0	8	2	2	-8
1	X3	3	5/3	0	1	0	1/3	2/3	0
4	X2	4	4/3	1	0	0	2/3	1/3	0
	ZJ	19	7	4	1	0	3	2	0
	ZJ     +	CJ	5	0	0	0	3	M + 2	M

X2 = 4        X3 = 3          S3 = 13       Z  =  19

Max Z  =  2(0) + 4(4) + 1(3) + 0(S1) + 0(13) – M(0) – M(0)  =  19

Max Z  =  19

11.-Max X        =  X1 + 2X2

            S.a.

                            X1 – 4X2  ≤  1

                        -3X1 + 2X2  ≤  6

                               X1 , X2  ≥  0

Forma Estandarizada:

            Max  Z  =  X1 + 2X2 + 0S1 + 0S2

S.a.

               X1 – 4X2 +0S1             =  1

            -3X1 +2X2           +0S2   =  6

                        X1 , X2 , S1 , S2   ≥  0

		CJ	1	2	0	0
CB	XB	bi	X1	X2	S1	S2	θ K
0	S1	1	1	-4	1	0	-
0	S2	6	-3	2	0	1	3
	ZJ	0	0	0	0	0
	ZJ    +	CJ	-1	-2	0	0
0	S1	13	-5	0	1	2	-
2	X2	3	-3/2	1	0	1/2	-
	ZJ	6	-3	2	0	1
	ZJ   +	CJ	-4	0	0	1