PROBLEMA DUAL
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This document presents the formulation and solution of two linear programming problems using the simplex method. The first problem involves maximizing an objective function subject to three inequality constraints. The optimal solution is found to be Z=12000 with A=30, B=0, H1=0, H2=10, H3=0. The dual problem is also formulated and the solutions are verified to match. The second problem involves minimizing an objective function subject to three constraints. The optimal solution is found to be Z=57 with X1=2, X2=7, H1=4, H2=0. Again, the dual problem is formulated and the solutions are verified to match.
PROBLEMA DUAL
- 1. EL PROBLEMA DUAL MAXIMIZAR Z= 400A + 300B 2A + B 60 A + 3B 40 A + B 30 A, B 0 FORMA ESTNDAR Z= 400A + 300B + 0H1 + 0H2 + 0H3 2A + B + H1 60 A + 3B +H2 40 A + B + H3 30 A, B, H1, H2, H3 0 FORMA CANNICA O DE ECUACIONES Z - 400A - 300B - 0H1 - 0H2 - 0H3 = 0 2A + B + H1 = 60 A + 3B +H2 = 40 A + B + H3 = 30 A, B, H1, H2, H3 0 TABLA SIMPLEX VB Z A B H1 H2 H3 VALOR Z 1 -400 -300 0 0 0 0 H1 0 2 1 1 0 0 60 H2 0 1 3 0 1 0 40 H3 0 1 1 0 0 1 30 Z 0 0 100 0 0 400 12000 H1 0 0 -1 1 0 -2 0 H2 0 0 2 0 1 -1 10 A 0 1 1 0 0 1 30 VE= A VS= H3 PIVOTE=1
- 2. RESPUESTAS DEL PROBLEMA PRIMAL: Soluci坦n ptima Z= 12000 Valores ptimos A=30 H1=0 B= 0 H2=10 H3=0 FORMULACIN DEL PROBLEMA DUAL MINIMIZAR Z= 60Y1 + 40Y2 + 30Y3 Y1=100 2Y1 + Y2 + Y3 400 Y2= 0 Y1 + 3Y2 + Y3 300 Y3= 200 Yi 0 COMPROBACIN Z= 60(100)+40(0)+30(200) Z= 6000 + 6000 Z= 12000 2Y1 + Y3 = 400 -Y1 - Y3 = 300 Y1 = 100 2(100) + Y3 = 400 Y3 = 400-200 Y3 = 200 MINIMIZAR Z= 4X1 + 7X2 X1 6 2X2 = 14 3X1 + 2X2 20 Xi 0 FORMA ESTNDAR Z= 4X1 + 7X2 +MA1 +MA2 + 0H1 + 0H2 (-1) -Z= -4X1 - 7X2 -MA1 -MA2 - 0H1 - 0H2 X1 + H1 6 2X2 + A1 =14 (-M) 3X1 + 2X2 + A2 H2 20 (-M) X1, X2, A1, A2, H1, H2 0 FORMA CANNICA O DE ECUACIONES
- 3. -Z + 4X1 + 7X2 +MA1 + MA2 +0H1+0H2 = 0 -2MX2 -MA1 = -14M - 3MX1 -2MX2 -MA2 +M = -20M -Z+ (-3M+4) X1+ (-4M+7) X2 +0H1+MH2= -34M X1 + H1 = 6 2X2 + A1 = 14 3X1 + 2X2 + A2 -H2 = 20 X1, X2, A1, A2, H1, H2 0 TABLA SIMPLEX VB Z X1 X2 A1 A2 H1 H2 VALOR Z -1 (-3M+4) (-4M+7) 0 0 0 M (-34M) H1 0 1 0 0 0 1 0 6 A1 0 0 2 1 0 0 0 14 A2 0 3 2 0 1 0 -1 20 Z -1 (-3M+4) 0 (2M-7/2) 0 0 M (-6M-49) H1 0 1 0 0 0 1 0 6 X2 0 0 1 1/2 0 0 0 7 A2 0 3 0 -1 1 0 -1 6 Z -1 0 0 (M-13/6) (M-4/3) 0 1 1/3 -57 H1 0 0 0 1/3 - 1/3 1 1/3 4 X2 0 0 1 1/2 0 0 0 7 X1 0 1 0 - 1/3 1/3 0 - 1/3 2 VE= X2 VS= A1 PIVOTE= 2 VE= X1 VS= A2 PIVOTE= 3 RESPUESTAS: Soluci坦n ptima -Z= -57 Z=57 Valores ptimos X1=2 H1=4 X2=7 H2=0
- 4. FORMULACIN DEL PROBLEMA DUAL MINIMIZAR Z= 6Y1 + 14Y2 + 20Y3 Y1=0 Y1 + 3Y3 4 Y2= 13/6 2Y2 + 2Y3 < > 7 Y3= 4/3 Yi 0 COMPROBACIN Z= 6(0)+14(13/6)+20(4/3) Z= 57