This document contains three examples of linear programming problems solved using the simplex method. The first maximizes Z=40X+50Y given constraints on X and Y. It finds the optimal solution is Z=200 with X=4 and Y=0. The second maximizes Z=20X+30Y given different constraints, and finds the optimal solution is Z=60 with X=0 and Y=2. The third maximizes Z=20X+50Y subject to additional constraints, and determines the optimal solution is Z=150 with X=0 and Y=3.

1. UNIVERSIDAD NACIONAL DE CHIMBORAZO
FACULTAD DE CIENCIAS POLÍTICAS Y ADMINISTRATIVAS
CARRERA DE CONTABILIDAD Y AUDITORÍA CPA
Nombre: RAUL LOGROÑO
Curso: 5to Semestre
Paralelo: “A”
Fecha: 3 de Junio del 2015
Tema: Método Simplex |
EJERCICIO 01
MAX.
F.O
Z= 40X+50Y
S.A
4X+6Y≤24
2X+Y≤6
X+Y≤4
X;Y≥0
FORMA ESTANDAR
Z=40X+6Y
4X+6Y=24
2X+Y=6
X+Y=4
Z
-40X -50Y -0H1 -0H2 -0H3 =0
4X 6Y +H1 =24
2X Y +H2 =6
X Y +H3 =4
Z X Y H1 H2 H3 VALOR V. E
Z 1 -40 -50 0 0 0 0 0

2. H1 0 4 6 1 24 4
H2 0 2 1 1 6 6
H3 0 1 1 1 4 4
Z X Y H1 H2 H3 VALOR
Z 1 10 0 0 0 0 200
H1 0 -2 0 1 0 -6 0
H2 0 1 0 0 1 -1 2
H3 0 1 1 0 0 1 4
0 -1 -1 0 0 -1 -4
0 2 1 0 1 0 6
0 -6 -6 0 0 6 -24
0 4 6 1 0 0 24
0 50 50 0 0 50 200
1 -40 -50 0 0 0 0
S.O
Z =200
X =4
Y =0
H1 =0
H2 =2
H3 =0
METODO DUAL
MIN
F.O
Z=24Y1+6Y2+4Y3
S.A
4Y1+2Y 2+2Y3=40
6Y1+Y 2+Y3=50
COMPROBACIÓN
4 Y1 +2 Y 3 =40 4Y1 +2Y2 =40
6 Y1 +Y 3 =50 (-2) -12Y1 -2Y2 =-100
Y1 =
15
2
Y3 =5
Z=24(
15
2
)+4(5)
Z=200
EJERCICIO 02

3. MAX.
F.O
Z= 20X+30Y
S.A
5X+3Y≤15
2X+2Y≤10
X+Y≤2
X;Y≥0
FORMA ESTANDAR
Z=20X+30Y
5X+3Y=15
2X+2Y=10
X+Y=2
Z -20X -30Y -0H1 -0H2 -0H3 =0
5X 3Y +H1 =15
2X 2Y +H2 =10
X Y +H3 =2
Z X Y H1 H2 H3 VALOR V. E
Z 1 -20 -30 0 0 0 0 0
H1 0 5 3 1 15 5
H2 0 2 2 1 10 5
H3 0 1 1 1 2 2
Z X Y H1 H2 H3 VALOR
Z 1 10 0 0 0 30 60
H1 0 2 0 1 0 12 9
H2 0 0 0 0 1 -2 6
H3 0 1 1 0 0 1 2
0 -2 -2 0 0 -2 -4
0 2 2 0 1 0 10
0 -3 -3 0 0 -3 -6
0 5 3 1 0 15 15
0 30 30 0 0 30 60
1 -20 -30 0 0 0 0
S.O

4. Z =60
X =0
Y =2
H1 =9
H2 =6
H3 =0
METODO DUAL
MIN
F.O
Z=15Y1+10Y2+2Y3
S.A
5Y1+2Y 2+Y3=20
3Y1+2Y 2+Y3=30
Y3=30
Z=2(30)
Z=60
EJERCICIO 02
MAX.
F.O
Z= 20X+50Y
S.A
8X+2Y≤16
2X+4Y≤12
X+2Y≤8
X;Y≥0
FORMA ESTANDAR
Z=20X+50Y
8X+4Y=16
2X+2Y=12
X+2Y=8
Z -20X -50Y -0H1 -0H2 -0H3 =0

5. 8X 4Y +H1 =16
2X 2Y +H2 =12
X 2Y +H3 =8
Z X Y H1 H2 H3 VALOR V. E
Z 1 -20 -50 0 0 0 0 0
H1 0 8 2 1 16 8
H2 0 2 4 1 12 3
H3 0 1 1 1 8 8
Z X Y H1 H2 H3 VALOR
Z 1 10 0 0
25
2
0 150
H1 0 2 0 1 −
1
2
2 10
H2 0 0 0 0
1
4
0 3
H3 0 1 1 0 −
1
4
1 5
0 25 50 0
25
2
0 150
0 -20 -50 0 0 0 0
0 -1 -2 0 −
1
2
0 -6
0 8 2 1 0 0 16
0 −
1
2
-1 0 −
1
4
0 -3
0 1 1 0 0 0 8
S.O
Z =150
X =0
Y =3
H1 =10
H2 =0
H3 =5
METODO DUAL
MIN
F.O
Z=16Y1+12Y2+8Y3
S.A
8Y1+2Y 2+Y3=20
2Y1+4Y 2+2Y3=50
Y2=
25
2
Z=12(
25
2
)
Z=150