Question: Question 2. Consider the LP model in Question 1, and answer the following sensitivity analysis related questions based on the graphical representation and optimal solution

Question 2. Consider the LP model in Question 1, and answer the following sensitivity analysis related questions based on the graphical representation and optimal solution you obtained. You can use the solution provided by the instructor. Make sure you perform the sensitivity analyses by hand (do the computations on paper/iPad, not Excel). You need to show all the steps and you are not allowed to use Excel Solver Sensitivity report.

  1. a) Perform sensitivity analysis for the objective coefficients. Find the allowable increase and decrease for both of the objective function coefficients. What do we mean by allowable increase and decrease for an objective coefficient? What happens to the optimal solution and the optimal objective value if we change the coefficients within allowable increase and decrease? What happens if we go beyond the allowable range?

  2. b) Compute the shadow price for all the constraints except the nonnegativity constraints. Explain in words what each of them means.

  3. c) Compute the allowable increase and decrease for the RHS of the nonbinding constraints only. State the values and explain what they mean. Remember, do not use Excel.Question 2. Consider the LP model in Question 1,Question 2. Consider the LP model in Question 1,Question 2. Consider the LP model in Question 1,

3.5 1.5 10 12 14 16 18 20 Oljectar function values Extreme Parts (34) Z -2+34 A 1.43*2K) | 1(143) +3(2.14) - 7-86 B(2.24,3-35) 1(2-24) +3(3-35) = 12-29 13 12 C (4,3) 1(4) +3(3) 0 (6,2) 1 (6) + 3(2) (4) +3(%) - 4.643 Now to answer the questions: The decision variables are xfy .so. Nors of decisions notes = 2 - (b) do we had discussed in the problem The total number of constraints = 5 (c) Number of conner points = 5 (d) Number of feasible solutions = 5 (e) Number of optional solution 6) The optimal solution is x=4&y=3 The optimal objecture value as Max 2 = 13

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