Linear Programming -- Objective Sensitivity

# Question 2 if x 1 were at least 1 how would the

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Question 2 If X 1 were at least 1, how would the profit be affected?

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Answer to Question 1 Answer to Question 1 When slope of objective function line equals slope of time constraint line, optimal solutions exist with X 1 >0: Objective function: Objective function: C C 1 X 1 + 5X 2 Time Constraint: 3X 1 + 4X 2 ≤ 2400 Thus, C1 C1 /5 = 3/4 or C1 C1 = 3.75 Per unit profit of product 1 has to be increased to \$3.75 to make production of 1 profitable; profit has to be increased Increased Profit = 3.75 – 1 = 2.75 Increased Profit = 3.75 – 1 = 2.75 Reduced Cost Reduced Cost = 1 – 3.75 = -2.75 -2.75
Answer to Question 2 Answer to Question 2 If X 1 ≥ 1, how is the profit affected? Max. 1 1 X 1 + 5X 2 ST 2X 1 + 1X 2 ≤ 1000 (Plastic) 3X 1 + 4X 2 ≤ 2400 (Time) 1X 1 + 1X 2 700 (Limit) 1X 1 - 1X 2 350 (Product mix) X 1 1 (Requirement of X 1 ) Solution is X 1 = 1, 1, X 2 = 599.25, 599.25, New Optimal Profit = 2997.25 Reduced Cost Reduced Cost = 2997.25 – 3000 = -2.75 -2.75

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Reduced Cost on Excel Reduced Cost on Excel Here is the printout out of the sensitivity analysis dealing with the objective function coefficients for the problem. Reduced Costs Final Reduced Objective Allowable Allowable Cell Name Value Cost Coefficient Increase Decrease \$B\$4 Dozen Space Rays - (2.75) 1 2.75 1E+30 \$C\$4 Dozen Zappers 600 - 5 1E+30 3.666666666
Complementary Slackness MAX 1X MAX 1X 1 1 + 5X + 5X 2 2 Sensitivity report for the problem For X 1 , Final Value = 0, but Reduced Cost ≠ 0.

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