Metodo Simplex
Resultados: I. Maximizar X = 500 Y1 + 300 Y2 Sujeto a 15 Y1 + 5 Y2 ≤ 300 10 Y1 + 6 Y2 ≤ 240 8 Y1 + 12 Y2 ≤ 450 Y1, Y2
Views 263 Downloads 2 File size 82KB
Recommend stories
- Author / Uploaded
- Markoz Cortés Barrios
Citation preview
Resultados: I. Maximizar X = 500 Y1 + 300 Y2 Sujeto a
15 Y1 + 5 Y2 ≤ 300
10 Y1 + 6 Y2 ≤ 240 8 Y1 + 12 Y2 ≤ 450 Y1, Y2 ≥0
X = -500 Y1 - 300 Y₂ = 0 15 Y₁ + 5Y₂ + S₁ = 300 10Y₁ + 6Y₂ +S₂ = 240 8Y₁ + 12Y₂ +S₃ = 450 Tabla
X
Y1
Y2
S1
R
S2
R1
1
-500
-300
0
0
0
0
R2
0
15
5
1
0
300
20
R3
0
10
6
0
1
240
24
R4
0
8
12
0
0
450
56.25
X
Y1
Y2
S1
S2
R
R1
1
-500
-300
0
0
0
0
R2
0
1
3
15
0
20
1.33
R3
0
10
6
0
1
240
24
R4
0
8
12
0
0
450
56.25
1
3
15
0
0
20
500 R₂ + R₁
R2
R₃
0
R3
0
10
6
0
1
0
240
8
12
0
0
1
450
10R₂ + R₃
R4
0
8 R₂ + R₄ 500 R₂ + R₁
R2
0
1
3
15
0
0
20
R1
1
-500
300
0
0
0
0
1
0
1200
7500
0
0
10,000
+
=
R2 + R1
-10 R₂ + R₃
R2
0
1
3
15
0
0
20
R3
0
10
6
0
1
0
240
R2 + R3
0
0
24
150
1
0
40
+
=
-8R₂ + R₄
R2
0
1
3
15
0
0
20
R4
0
8
12
0
0
1
450
+
=
R4
0
0
X
12
Y1
0
12
Y2
S1
1
290
R
S2
R1
1
0
1200
7500
0
0
10,000
R2
0
0
12
120
0
1
290
R3
0
0
1
6.25
24
0
1.6
RESPUESTA: X = 10,000 Y₁ = 290 Y₂ = 1.6
Algebraicamente:
m = 15 n = 300 n-m =285 Cantidad máxima de soluciones básicas o puntos de esquina: C ⁿm = n! m!(n-m)! C ⁿm = 300!
15! (300-15)! C ⁿm = 300! 15! (285)!
II. Maximizar X = 10 Y1 + 20 Y2 Sujeto a
- Y1 + 2 Y2 ≤ 15
Y1 + Y2 ≤ 12 5 Y1 + 3 Y2 ≤ 45 Y1, Y2 ≥0
X - 10 Y₁ - 20 Y₂= 0 -Y₁ + 2Y₂ + S₁ = 15 Y₁ + Y₂ + S₂ = 12 5Y₁ + 3Y₂ + S₃ = 45
X
Y1
Y2
S1
S2
R
S3
R1
1
-10
-20
0
0
0
0
-½
R2
0
-1
2
1
0
0
15
7.5
R3
0
1
1
0
1
0
12
12
R4
0
5
3
0
0
1
45
15
R1
1
-10
-20
0
0
0
0
10
R2
0
½
1
½
0
0
7.5
R3
0
1
1
0
1
0
12
1
R4
0
5
3
0
0
1
45
3
10R₂ + R₁
R2
20
½
1
½
0
0
7.5
+
R1
1
-10
-20
0
0
0
0
-½
R2 + R1
1
0
0
10
0
0
150
-½
R2
-1
0
½
1
½
0
0
7.5
R3
0
1
1
0
1
0
12
R3
0
½
0
-½
1
0
4.5
-3
0
½
1
½
0
0
=
R₂ + R₃
+
=
R₂ + R₄
R2
+
7.5
R4
0
5
3
0
0
1
45
R₂ + R₄
0
3.5
0
-½
0
1
22.5
=
X
Y1
Y2
S1
S2
S3
R
R1
1
0
0
10
0
0
150
R2
0
½
1
½
0
0
7.5
R3
0
½
0
-½
1
0
4.5
R4
0
1
0
-1.75
0
3.5
6.4
RESPUESTA X = 150 Y₁ = 6.4 Y₂ = 7.5 Algebraicamente: m1 n 15 n-m =14 Cantidad máxima de soluciones básicas o puntos de esquina: C ⁿm = n!
m! (n-m)! C ⁿm = 15! 1! (15-1)! C ⁿm = 15! 1! (14)!
III. Minimizar X = 40 Y1 + 50 Y2 Sujeto a
2 Y1 + 3Y2 ≥ 30
Y1 + Y2 ≥ 12 2 Y1 + Y2 ≥ 20 Y1, Y2 ≥0
X + 40 Y₁ + 50 Y₂ + = 0 2Y₁ + 3Y₂ + S₁ =30 Y₁ + Y₂ + S₂ = 12 2 Y₁ + Y₂ + S₃ = 20
TABLA
X
Y1
Y2
S1
S2
S3
R
R1
1
40
50
0
0
0
0
R2
0
2
3
1
0
0
30
10
R3
0
1
1
0
1
0
12
12
R4
0
2
1
0
0
1
20
20
X
Y1
Y2
S1
S2
R
S3
R1
1
-260
0
0
0
0
-4500
R2
0
6
1
⅓
0
0
90
15
R3
0
7
0
⅓
1
0
-101
14.42
R4
0
8
0
⅓
0
1
-69
8.62
R1 + -50 R₂
R1
1
40
50
0
0
0
0
R2
0
-300
-50
0
0
0
-4500
R1+R 2
1
-260
0
0
0
0
-4500
+
=
R3 - 1 R₂
R3
0
1
1
0
1
0
12
R2
0
6
-1
⅓
0
0
-89
+
= R3 - 1 R₂
0
7
0
⅓
1
0
-101
R4 -1 R₂
R4
0
2
1
0
0
1
20
R2
0
6
-1
⅓
0
0
-89
R4 -1 R₂
0
8
0
⅓
0
1
-69
+
=
X
Y1
Y2
S1
S2
S3
R
R1
1
0
0
-260
0
-260
32,440
R2
0
1
0
⅓
0
1
-69
R3
0
0
1
⅙
0
6
504
R4
0
63
0
12/₃
0
7
584
RESPUESTA X = 32,440 Y₁ = -69 Y₂ = 504