Here is the Standard LP form for Problem 1 in HW-1 Maximize Z=15 x1 +15 x2 −1267.2S .T .6 x1+ 4 x2 <= 4 325 x 1 +5 x 2 <= 412.8 4 x1 +6 x2 <= 422.4x1 , x2 >= 0 A. Determine the optimal solution using the Sim
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Given Problem: Maximize \( Z = 15x_1 + 15x_2 - 1267.2 \) Subject to: 1. \( 6x_1 + 4x_2 \leq 4 \) 2. \( 325x_1 + 5x_2 \leq 412.8 \) 3. \( 4x_1 + 6x_2 \leq 422.4 \) 4. \( x_1, x_2 \geq 0 \) Show more…
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6. (15 points) Consider the following standard form linear programming problem: Maximize f(x) = -x1 + 4x2 + 2x3 + 2x5 subject to: Ax = b, x >= 0. (a) How many basic solutions can this problem have? (b) Give the initial tableau for the problem. (c) Give the initial basic feasible solution. (d) Put the initial tableau in canonical form, i.e., price out the initial tableau. What are the reduced cost coefficients for the non-basic columns? (e) Is the initial basic feasible solution optimal? If not, find the optimal solution using the simplex algorithm.
Adi S.
5. There is an LP model: max z = 6x1 + 5x2 + x3 x1 + x2 + 2x3 <= 15 3x1 + 2x2 + 3x3 <= 40 x1, ..., x3 >= 0 (1) What are the optimal solution and value of objective function? (2) Write the optimal solution of dual problem according to the final table. (3) Sensitivity analysis was conducted on the basis of the final table of question (1): (Note: Sensitivity analysis must be used, otherwise no score will be given ) (1) What's the allowable range of a23 that enable the optimal base to remain unchanged? (2) Introduce a new constraint x1+x2+x3 <= 12 , does the optimal basic matrix change? If changes, what are the new optimal solution and the value of objective function?
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Minimize Z = 0.90X1 + 0.60X2 Subject to: X1 + X2 = 1 10X1 + 6X2 >= 9 12X1 + 9X2 >= 10 X1, X2 >= 0 Solve the above problem with the simplex method. Provide the value of the basic variables (BV) at tableau #3 and the optimal solution of the LP. Notes: - Set M = 100. - Tableau #1 is the initial tableau where an inconsistency exists on row z. - Tableau #2 is the tableau where the inconsistency on row z is fixed. - Set s1 as the slack variable for constraint 2. Select one: BV: A1 = 2/6, A2 = 1/3, X1 = 3/6 | Optimal sol: X1 = 3/4, X2 = 1/4, Z = 33/40 BV: A1 = 1/6, A2 = 2/3, X1 = 5/6 | Optimal sol: X1 = 4/5, X2 = 1/5, Z = 21/25 BV: A1 = 1/6, A2 = 2/3, X1 = 5/6 | Optimal sol: X1 = 3/4, X2 = 1/4, Z = 33/40 None of the options provided BV: A1 = 1, A2 = 9, X1 = 10 | Optimal sol: X1 = 1, X2 = 0, Z = 9/10.
Lien L.
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