Question: 1. In the lecture a so-called RMC problem was discussed which had decision variables 21 tons of fuel additive produced tons of solvent base produced

1. In the lecture a so-called RMC problem was1. In the lecture a so-called RMC problem was

1. In the lecture a so-called RMC problem was discussed which had decision variables 21 tons of fuel additive produced tons of solvent base produced 22 = leading to the following formulation of the RMC problem: maximize profit subject to Material 1 z = 40:21 + 30:22. 2/5x1 +1/2x2 = 20 1/5225 5 3/5x1 +3/10x2 = 21 Material 2 Material 3 21,22 > 0. The final simplex tableau is shown here: 22 11 0 0 1 22 1 0 0 S2 S1 10/3 -2/3 -5/3 -100/3 S2 0 1 0 0 83 -20/9 20 4/9 1 25/9 25 -400/9 -1600 21 0 0 (a) In the final tableau: Are there basic decision variables, if yes, which? Are there non-basic decision variables, if yes, which? Are there basic slack variables, if yes, which? Are there non-basic slack variables, if yes, which? (b) Explain in one sentence what is understood as 'Reduced Cost' and give the Reduced Cost for each of both decision variables. (c) Which of the constraints are binding and which are non-binding? (d) Explain in one sentence what is understood as 'Dual Price and give the Dual Price for each of the constraints. (e) Which of the materials would increase the profit most if one more ton of it would be available? How much should the company be willing to pay for this material? () Compute the right-hand-side ranges for b, b2 and 63. How much of the material de- termined in the previous subquestion should maximally be bought by the company, given that other additional materials can not be bought? (g) Compute the ranges of optimality for c and c2. (h) Formulate the Dual of the above RMC problem. (i) Solve the Dual with the Management Scientist Software. (j) Use the basic solution of the final tableau of the primal problem above, the solution computed with the computer program for the Dual problem and the Complementary Slackness Theorem to prove that both solutions are optimal

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