Question 1: 1A. Write the general dual problem associated with the given LP.(Do not transform or rewrite the primal problem before writing the general dual) Maximize –4x1 + 2x2Subject To4x1 + x2 + x3 = 202x1 – x2 ≥ 6x1 – x2 + 5x3 ≥ –5–3x1 + 2x2 + x3 ≤ 4x1 ≤ 0, x2 ≥ 0, x3 unrestricted 1B. Given the following information for a product-mix problem with three products and three resources.Primal Decision Variables: x1 = number of unit 1 produced; x2 = # of unit 2 produced; x3 = # of unit 3 producedPrimal Formulation: Dual Formulation:Max Z (Rev.) = 25x1 + 30x2 + 20x3 Min W = 50π1 + 20π2 +25π3Subject To 8x1 + 6x2 + x3 ≤ 50 (Res. 1 constraint) Subject To 8π1 + 4π2 +2π3≥ 25 4x1 + 2x2 + 3x3 ≤ 20 (Res. 2 constraint) 6π1 + 2π2 +π3 ≥ 30 2x1 + x2 + 2x3 ≤ 25 (Res. 3 constraint) π1 + 3π2 +2π3≥ 20 x1, x2, x3 ≥ 0 (Nonnegativity) π1, π2, π3 ≥ 0 Optimal Solution:Optimal Z = Revenue = $268.75x1 = 0 (Number of unit 1) Dual Var. Optimal Value = 22.5 (Surplus variable in 1st dual constraint)x2 = 8.125 (Number of unit 2) Dual Var. Optimal Value = 0 (Surplus variable in 2nd dual constraint)x3 = 1.25 (Number of unit 3) Dual Var. Optimal Value = 0 (Surplus variable in 3rd dual constraint)Resource Constraints:Resource 1 = 0 leftover units Dual Var. Optimal Value = 3.125 = π1Resource 2 = 0 leftover units Dual Var. Optimal Value = 5.625 = π2Resource 3 = 14.375 leftover units Dual Var. Optimal Value = 0 = π3 1Bi. What is the fair-market price for one unit of Resource 3? 1Bii. What is the meaning of the surplus variable value of 22.5 in the 1st dual constraint with respect to the primal problem?

Question

Question 1: 1A.  Write the general dual problem associated with the given LP.(Do not transform or rewrite the primal problem before writing the general dual) Maximize –4x1 + 2x2Subject To4x1 + x2 + x3 = 202x1 – x2 ≥ 6x1 – x2 + 5x3 ≥ –5–3x1 + 2x2 + x3 ≤ 4x1 ≤ 0, x2 ≥ 0, x3 unrestricted  1B.  Given the following information for a product-mix problem with three products and three resources.Primal Decision Variables:  x1 = number of unit 1 produced; x2 = # of unit 2 produced; x3 = # of unit 3 producedPrimal Formulation:                                                                           Dual Formulation:Max Z (Rev.) = 25x1     + 30x2 + 20x3                                                  Min W =           50π1     + 20π2  +25π3Subject To        8x1       + 6x2    + x3      ≤ 50     (Res. 1 constraint)        Subject To        8π1       + 4π2    +2π3≥ 25                        4x1       + 2x2    + 3x3    ≤ 20     (Res. 2 constraint)                                6π1       + 2π2    +π3  ≥ 30                        2x1       + x2      + 2x3    ≤ 25     (Res. 3 constraint)                                π1         + 3π2    +2π3≥ 20                                    x1, x2, x3           ≥ 0       (Nonnegativity)                                    π1, π2, π3                   ≥ 0 Optimal Solution:Optimal Z = Revenue = $268.75x1 = 0 (Number of unit 1)                      Dual Var. Optimal Value = 22.5 (Surplus variable in 1st dual constraint)x2 = 8.125 (Number of unit 2)               Dual Var. Optimal Value = 0 (Surplus variable in 2nd dual constraint)x3 = 1.25 (Number of unit 3)                 Dual Var. Optimal Value = 0 (Surplus variable in 3rd dual constraint)Resource Constraints:Resource 1 = 0 leftover units                 Dual Var. Optimal Value = 3.125 = π1Resource 2 = 0 leftover units                 Dual Var. Optimal Value = 5.625 = π2Resource 3 = 14.375 leftover units        Dual Var. Optimal Value = 0 = π3 1Bi.  What is the fair-market price for one unit of Resource 3? 1Bii.  What is the meaning of the surplus variable value of 22.5 in the 1st dual constraint with respect to the primal problem?

Accepted Answer

1 A)

Write the general dual problem associated with the given LP. (Do not transform or rewrite the primal problem before writing the general dual) Maximize –4x1 + 2x2 Subject To 4x1 + x2 + x3 = 20 2x1 – x2 ≥ 6 x1 – x2 + 5x3 ≥ –5 –3x1 + 2x2 + x3 ≤ 4 x1 ≤ 0, x2 ≥ 0, x3 unrestricted