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Linear Programming

Solve a linear programming problem with given constraints and maximize the objective function using tableau method.

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Linear Programming

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  1. Linear Programming Graphical

  2. Linear Programming (Lp '((-0.6 -0.7 -0.5 0) (2.4 3 2 1200) (0 2.5 1.5 600) (5 0 2.5 1500))) Max 0.6X1 + 0.7X2 + 0.5x3 s.t. 2.4X1 + 3.0X2 + 2.0X3 <= 1200 0.0X1 + 2.5X2 + 1.5X3 <= 1600 5.0X1 + 0.0X2 + 2.5X3 <= 1500 Initial Tableau Basis X1 X2 X3 X4 X5 X6 Q Z-row -0.6 -0.7 -0.5 0.0 0.00 0.0 0.00 X4 2.4 3.2 2.0 1.0 0.0 0.0 1200 X5 0.0 2.5 1.5 0.0 1.0 0.0 1600 X6 5.0 0 2.5 0.0 0.0 1.0 1500 continued  … rd

  3. Final Tableau LP Optimum = 295.2 0.0 0.0 0 0.22 0.016 0.0144 295.2 1.0 0.0 0 0.5 -0.6 -0.04 180.0 0.0 1.0 0 0.6 -0.32 -0.288 96 0.0 0.0 1 -1.0 1.2 0.48 240 rd

  4. Linear Programming P9-27 (LP '((-7.8 -9.4 -2.6 0)(4.2 11.7 3.5 1800)(0.8 4.3 1.9 2700) (12.7 3.8 2.5 950))) 0 0.0 0.74 0.67 0.0 0.39 1588.8 0 1 0.26 0.096 0.0 -0.032 142.28 0 0.0 0.70 -0.39 1.0 0.066 2062.43 1 0.0 0.12 -0.03 0.0 0.09 32.23 rd

  5. LP Example (LP '((-2 -3 0)(3 5 15)(6 2 12)))  Max 2x1 + 3x2 -1/5 0 3/5 0 9 subject to 3x1 + 5x2 <= 15 3/5 1 1/5 0 3 x2 6x1 + 2x2 <= 12 24/5 0 -2/5 1 6 x1, x2 >= 0 0 0 7/12 1/24 37/4 0 1 1/4 -1/8 9/4 1 0 -1/12 5/24 5/4 LP Optimum 37/4 0 0 7/12 1/24 37/4 0 1 1/4 -1/8 9/4 1 0 -1/12 5/24 5/4 6 (5/4, 9/4) 0 3 x1 Feasible. Evaluate all corner points rd

  6. Max 2x1 + 3x2 Min 15Y1 + 12Y2 subject to 3x1 + 5x2 <= 15 s.t. 3Y1 + 6Y2 >= 2 6x1 + 2x2 <= 12 5Y1 + 2Y2 >= 3 Basis X1 X2 X3 X4 Q 1.5 Y2 Z-row -2 -3 0 0 0 W(0, 1.5) = 18 X3 3 5 1 0 15 W(2/3, 0) = 10 X4 6 2 0 1 12 1/3 W(7/12, 1/24) = 37/4 -1/5 0 3/5 0 9 (7/12, 1/24) 3/5 1 1/5 0 3 3/5 2/3 Y124/5 0 -2/5 1 6 0 0 7/12 1/24 37/4 0 1 1/4 -1/8 9/4 1 0 -1/12 5/24 5/4 rd

  7. LP Example Max 2X1 + 3X2 s.t. 3X1 + 5X2 <= 15 6X1 + 2X2 <= 12 x1 x2 x3 x4 -2 -3 0 0 0X3 3 5 1 0 15X4 6 2 0 1 12 -1/5 0 3/5 0 9 X2 3/5 1 1/5 0 3X4 24/5 0 -2/5 1 6 rd

  8. Final Tableau Max 2X1 + 3X2 0 0 7/12 1/24 37/40 1 1/4 -1/8 9/4 1 0 -1/12 5/24 5/4

  9. Tableaus X2 5 4 Max 3x1 + 4x2s.t. 2x1 + 3x2 <= 12 AX = B 5x1 + 3x2 <= 15 Initial Tableau at corner point (0, 0) -3 -4 0 0 0 2 3 1 0 12 5 3 0 1 15 Tableau at corner point (0, 4) -1/3 0 4/3 0 16 2/3 1 1/3 0 43 0 -1 1 3 Final Tableau Tableau at corner point (3, 0) 0 0 11/9 1/9 49/3 0 -7/5 0 ¾ 9 0 1 5/9 -2/9 10/3 0 7/5 1 -2/5 61 0 -1/3 1/3 1 1 3/5 0 1/5 3 X1 0 3 6 rd

  10. Final Tableau 0 0 11/9 1/9 49/3 Max 3x1 + 4x2 0 1 5/9 -2/9 10/3 s.t. 2x1 + 3x2 <= 12 1 0 -1/3 1/3 1 5x1 + 3x2 <= 15 AX = B A-1AX = A-1B = X = A-1B Objective Function row = (0 0 11/9 1/9 49/3) x1 = 1 x2 = 10/3 rd

  11. (LP '((-5 -3 -2 0)(2 3 2 24)(4 2 1 8)(1 4 6 36))) Max 5x1 + 3x2 + 2x3 How much for another yard of wool? s.t. 2x1 + 3x2 + 2x3 <= 24 (wool) cotton? 4x1 + 2x2 + 1x3 <= 8(cotton) silk? 1x1 + 4x2 + 6x3 <= 36(silk) Look at tableau; solve problem mentally without tableau. Specify the values for x1 to x6 using the final tableau below. 1/8 0 0 0 5/4 1/8 29/2 -25/8 0 0 1 -5/4 -1/8 19/2 23/8 1 0 0 3/4 -1/8 3/2 -7/4 0 1 0 -1/2 1/4 5 rd

  12. Final Tableau Basis X1 X2 X3 X4 X5 X6 Q Z 1/8 0 0 0 5/4 1/8 29/2 X4 -25/8 0 0 1 -5/4 -1/8 19/2 X2 23/8 1 0 0 3/4 -1/8 3/2 X3 -7/4 0 1 0 -1/2 1/4 5 How much more another unit of cotton ? wool? silk? rd

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