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Linear Programming Sensitivity Analysis How will a change in a coefficient of the objective function affect the optimal solution? How will a change in the right-hand side value for a constraint affect the optimal solution?

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Linear Programming Solving Linear Equations All operations that apply to linear equations also apply to linear inequalities with the following exceptions: If you multiply or divide by a negative number it will switch the direction of the inequality. If you invert an inequality it will also switch the direction of the inequality

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Linear Programming Sherwood – Linear Equations

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Linear Programming Sherwood – Graph Solution Line 2 Line 1 12 3 45

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Linear Programming Sherwood – Optimal Solution Extreme Point 3 is optimal if: Slope of Line 1 <= Slope of objective function <= Slope of Line 2

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Linear Programming Sherwood – Calculate Slope of Line 1 8x 1 + 2x 2 <= 160 2x 2 = -8x 1 + 160 x 2 = -4x 1 + 80 Slope of Intercept of Line 1 Line 1 on x 2 axis

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Linear Programming Sherwood – Calculate Slope of Line 2 4x 1 + 3 x 2 <= 120 3x 2 = -4x 1 + 120 x 2 = -4/3x 1 + 40 Slope of Intercept of Line 2 Line 2 on x 2 axis

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Linear Programming Sherwood – Optimal Solution Extreme Point 3 is optimal if: -4 <= Slope of objective function <= -4/3

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Linear Programming Calculating Slope-Intercept General form of objective function P = C x1 x 1 + C x2 x 2 Slope-intercept for objective function x 2 = -(C x1 /C x2 ) x 1 + P/C x2 Slope of Intercept of Obj. Function Obj. Function on x 2 axis

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Linear Programming Sherwood – Optimal Solution Extreme Point 3 is optimal if: -4 <= -(C x1 /C x2 ) <= -4/3 Or 4/3 <= (C x1 /C x2 ) <= 4

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Linear Programming Sherwood – Compute the Range of Optimality Extreme Point 3 is optimal if: 4/3 <= (C x1 /C x2 ) <= 4 Compute range for C x1, hold C x2 constant 4/3 <= (C x1 /10) <= 4

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Linear Programming Sherwood – Compute the Range of Optimality From the left-hand inequality, we have 4/3 <= (C x1 /10) Thus, 40/3 <= C x1

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Linear Programming Sherwood – Compute the Range of Optimality From the right-hand inequality, we have (C x1 /10) <= 4 Thus, C x1 <= 40

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Linear Programming Sherwood – Compute the Range of Optimality Summarizing these limits 40/3 <= C x1 <= 40

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Linear Programming Sherwood – Compute the Range of Optimality Extreme Point 3 is optimal if: 4/3 <= (C x1 /C x2 ) <= 4 Compute range for C x2, hold C x1 constant 4/3 <= (20/C x2 ) <= 4

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Linear Programming Sherwood – Compute the Range of Optimality From the inequality, we have 4/3 <= (20/C x2 ) <= 4 Thus, 4/60 <= (1/C x2 ) <= 4/20 5 <= C x2 <= 15

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Linear Programming Sherwood – Compute the Range of Optimality Summarizing these limits 40/3 <= C x1 <= 40 5 <= C x2 <= 15

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Linear Programming Sensitivity Analysis How will a change in a coefficient of the objective function affect the optimal solution? How will a change in the right-hand side value for a constraint affect the optimal solution?

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Linear Programming Sherwood – Graph Solution Line 2 Line 1 12 3 45

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Linear Programming Sherwood – Change in the Right-hand Side Constraint 1 – add 1 to right-hand side 4x 1 + 3x 2 <= 121 8x 1 + 2x 2 <= 160 Solve for x 2 2(4x 1 + 3x 2 = 121) -1(8x 1 + 2x 2 = 160) 4x 2 = 82 x 2 = 20.5 Solve for x 1 8x 1 + 2(20.5) = 160 x 1 = 14.875

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Linear Programming Sherwood – Change in the Right-hand Side Solve objective function z = 20(14.875) + 10(20.5) z = 502.5 Shadow Price 502.5 – 500 = 2.5 Thus profit increases at $2.50 per hour of labor added to assembly Conversely, if we decrease labor for assembly by 1 hour the objective function will decrease by $2.50

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Linear Programming Sherwood – Range of Feasibility Constraint 1 RHS = 120 Allowable Increase = 24 Allowable Decrease = 40 Range of Feasibility 80 <= Constraint 1 RHS <= 144

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Linear Programming Sherwood – Change in the Right-hand Side Constraint 2 – add 1 to right-hand side 4x 1 + 3x 2 <= 120 8x 1 + 2x 2 <= 161 Solve for x 2 2(4x 1 + 3x 2 = 120) -1(8x 1 + 2x 2 = 161) 4x 2 = 79 x 2 = 19.75 Solve for x 1 4x 1 + 3(19.75) = 120 x 1 = 15.1875

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Linear Programming Sherwood – Change in the Right-hand Side Solve objective function z = 20(15.1875) + 10(19.75) z = 501.25 Shadow Price 501.25 – 500 = 1.25 Thus profit increases at $1.25 per hour of labor added to finishing Conversely, if we decrease labor for finishing by 1 hour the objective function will decrease by $1.25

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Linear Programming Sherwood – Range of Feasibility Constraint 2 RHS = 160 Allowable Increase = 80 Allowable Decrease = 48 Range of Feasibility 112 <= Constraint 2 RHS <= 240

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Linear Programming Sherwood – Range of Feasibility Constraint 3 RHS Slack = 12 Shadow Price = 0 Range of Feasibility 20 <= Constraint 3 RHS <= Infinite

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Linear Programming Non-Binding Constraints There is more resource then needed (i.e. there is slack). When you have a non-binding constraint the shadow price is zero Also, the allowable increase will be 1E+30 (infinite) represents that no upper limit exists for the range of feasibility The lower limit allowable decrease equals the amount of slack

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Linear Programming Reduced Costs For each decision variable, the absolute value of the reduced costs indicates how much the objective coefficient would have to improve before that variable could assume a positive value in the optimal solution. If the decision variable is already positive in the optimal solution, its reduced costs variable is zero.

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Linear Programming Sherwood - Slack Variables Max 20x 1 + 10x 2 + 0S 1 + 0S 2 + 0S 3 s.t. 4x 1 + 3x 2 + 1S 1 = 120 8x 1 + 2x 2 + 1S 2 = 160 x 2 + 1S 3 = 32 x 1, x 2, S 1,S 2,S 3 >= 0

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Linear Programming Sherwood – Slack Variables For each ≤ constraint the difference between the RHS and LHS (RHS-LHS). It is the amount of resource left over. Constraint 1; S 1 = 0 hrs. Constraint 2; S 2 = 0 hrs. Constraint 3; S 3 = 12 Custom

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Linear Programming Binding vs. Non-Binding Constraints Constraints that have zero slack are considered binding constraints Constraints that have slack or unused capacity available are non-binding. They have a shadow price of zero. This shows that additional units of this resource will not increase the value of the objective function

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Linear Programming Summary In summary, the right-hand-side ranges provide limits within which the shadow prices are applicable. For changes outsides the range, the problem must be resolved to find the new optimal solution and the new shadow price. The ranges of feasibility for the Sherwood problem are: 80 <= Constraint 1 <= 144 112 <= Constraint 2 <= 240 20 <= Constraint 3 <= Infinite

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Section 2.6 Solving Linear Inequalities and Absolute Value Inequalities.

Section 2.6 Solving Linear Inequalities and Absolute Value Inequalities.

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