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Chapter 3 Linear Programming: Sensitivity Analysis and Interpretation of Solution MT 235

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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? All of our sensitivity analysis will involve only one parameter at a time MT 235

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**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 MT 235

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**Operations With Linear Inequalities**

MT 235

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**Sherwood – Graph Solution**

Line 2 5 4 3 Line 1 1 2 MT 235

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**Sherwood – Optimal Solution**

Extreme Point 3 is optimal if: Slope of Line 2 <= Slope of objective function <= Slope of Line 1 MT 235

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**Sherwood – Calculate Slope of Line 1**

4x1 + 3 x2 <= 120 3x2 = -4x x2 = -4/3x1 + 40 Slope of Intercept of Line Line 1 on x2 axis MT 235

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**Sherwood – Calculate Slope of Line 2**

8x1 + 2x2 <= 160 2x2 = -8x x2 = -4x1 + 80 Slope of Intercept of Line Line 2 on x2 axis MT 235

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**Sherwood – Optimal Solution**

Extreme Point 3 is optimal if: -4 <= Slope of objective function <= -4/3 MT 235

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**Calculating Slope-Intercept**

General form of objective function P = Cx1x1 + Cx2x2 Slope-intercept for objective function x2 = -(Cx1/Cx2) x1 + P/Cx2 Slope of Intercept of Obj. Function Obj. Function on x2 axis MT 235

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**Sherwood – Optimal Solution**

Extreme Point 3 is optimal if: -4 <= -(Cx1/Cx2) <= -4/3 Or 4/3 <= (Cx1/Cx2) <= 4 MT 235

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**Sherwood – Compute the Range of Optimality**

Extreme Point 3 is optimal if: 4/3 <= (Cx1/Cx2) <= 4 Compute range for Cx1, hold Cx2 constant 4/3 <= (Cx1/10) <= 4 MT 235

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**Sherwood – Compute the Range of Optimality**

From the left-hand inequality, we have 4/3 <= (Cx1/10) Thus, 40/3 <= Cx1 MT 235

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**Sherwood – Compute the Range of Optimality**

From the right-hand inequality, we have (Cx1/10) <= 4 Thus, Cx1 <= 40 MT 235

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**Sherwood – Compute the Range of Optimality**

Summarizing these limits 40/3 <= Cx1 <= 40 MT 235

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**Sherwood – Compute the Range of Optimality**

Extreme Point 3 is optimal if: 4/3 <= (Cx1/Cx2) <= 4 Compute range for Cx2, hold Cx1 constant 4/3 <= (20/Cx2) <= 4 MT 235

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**Sherwood – Compute the Range of Optimality**

From the inequality, we have 4/3 <= (20/Cx2) <= 4 Thus, 4/60 <= (1/Cx2) <= 4/20 5 <= Cx2 <= 15 MT 235

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**Sherwood – Compute the Range of Optimality**

Summarizing these limits 40/3 <= Cx1 <= 40 5 <= Cx2 <= 15 MT 235

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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? MT 235

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**Sherwood – Graph Solution**

Line 2 5 4 3 Line 1 1 2 MT 235

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**Sherwood – Change in the Right-hand Side**

Constraint 1 – add 1 to right-hand side 4x1 + 3x2 <= 121 8x1 + 2x2 <= 160 Solve for x2 2(4x1 + 3x2 = 121) -1(8x1 + 2x2 = 160) 4x2 = 82 x2 = 20.5 Solve for x1 8x1 + 2(20.5) = 160 x1 = MT 235

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**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 MT 235

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**Sherwood – Range of Feasibility**

Constraint 1 RHS = 120 Allowable Increase = 24 Allowable Decrease = 40 Range of Feasibility 80 <= Constraint 1 RHS <= 144 MT 235

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**Sherwood – Change in the Right-hand Side**

Constraint 2 – add 1 to right-hand side 4x1 + 3x2 <= 120 8x1 + 2x2 <= 161 Solve for x2 2(4x1 + 3x2 = 120) -1(8x1 + 2x2 = 161) 4x2 = 79 x2 = 19.75 Solve for x1 4x1 + 3(19.75) = 120 x1 = MT 235

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**Sherwood – Change in the Right-hand Side**

Solve objective function z = 20( ) + 10(19.75) z = Shadow Price – 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 MT 235

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**Sherwood – Range of Feasibility**

Constraint 2 RHS = 160 Allowable Increase = 80 Allowable Decrease = 48 Range of Feasibility 112 <= Constraint 2 RHS <= 240 MT 235

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**Sherwood – Range of Feasibility**

Constraint 3 RHS Slack = 12 Shadow Price = 0 Range of Feasibility 20 <= Constraint 3 RHS <= Infinite MT 235

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**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 MT 235

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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. MT 235

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**Sherwood - Slack Variables**

Max 20x1 + 10x2 + 0S1 + 0S2 + 0S3 s.t. 4x x2 + 1S = 120 8x x S = 160 x S3 = 32 x1, x2, S1 ,S2 ,S3 >= 0 MT 235

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**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; S1 = 0 hrs. Constraint 2; S2 = 0 hrs. Constraint 3; S3 = 12 Custom MT 235

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**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 MT 235

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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 MT 235

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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? MT 235

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**Pet Food Co. – Linear Equations**

MT 235

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**Pet Food Co. – Graph Solution**

3 Line 3 Line 2 2 1 MT 235

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**Pet Food Co. – Optimal Solution**

Extreme Point 2 is optimal if: Slope of Line 3 <= Slope of objective function <= Slope of Line 2 MT 235

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**Pet Food Co. – Calculate Slope of Line 2**

0P1 + 1P2 >= 200 1P2 >= -0P Slope of Intercept of Line Line 2 on P2 axis MT 235

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**Pet Food Co. – Calculate Slope of Line 3**

1P1 + 1P2 >= 500 1P2 >= -1P P2 >= -P Slope of Intercept of Line Line 3 on P2 axis MT 235

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**Pet Food Co. – Optimal Solution**

Extreme Point 2 is optimal if: -1 <= Slope of objective function <= 0 MT 235

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**Calculating Slope-Intercept**

General form of objective function Z = CP1P1 + CP2P2 Slope-intercept for objective function P2 = -(CP1/CP2) P1 + Z/CP2 Slope of Intercept of Obj. Function Obj. Function on x2 axis MT 235

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**Pet Food Co. – Optimal Solution**

Extreme Point 2 is optimal if: -1 <= -(CP1/CP2) <= 0 Or 0 <= (CP1/CP2) <= 1 MT 235

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**Pet Food Co. – Compute the Range of Optimality**

Extreme Point 2 is optimal if: 0 <= (CP1/CP2) <= 1 Compute range for CP1, hold CP2 constant 0 <= (CP1/8) <= 1 MT 235

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**Pet Food Co. – Compute the Range of Optimality**

From the left-hand inequality, we have 0 <= (CP1/8) Thus, 0 <= CP1 MT 235

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**Pet Food Co. – Compute the Range of Optimality**

From the right-hand inequality, we have (CP1/8) <= 1 Thus, CP1 <= 8 MT 235

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**Pet Food Co. – Compute the Range of Optimality**

Summarizing these limits 0 <= CP1 <= 8 MT 235

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**Pet Food Co. – Compute the Range of Optimality**

Extreme Point 2 is optimal if: 0 <= (CP1/CP2) <= 1 Compute range for CP2, hold CP1 constant 0 <= (5/CP2) <= 1 MT 235

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**Pet Food Co. – Compute the Range of Optimality**

From the left-hand inequality, we have 0 <= (5/CP2) Thus, (1/5) * 0 <= (1/CP2) Invert 5/0 >= CP2 Division by zero is undefined (infinite). This means the cost of P2 can increase to infinity without changing the optimal solution MT 235

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**Pet Food Co. – Compute the Range of Optimality**

From the right-hand inequality, we have (5/CP2) <= 1 Thus, CP2 >= 5 MT 235

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**Pet Food Co. – Compute the Range of Optimality**

Summarizing these limits 0 <= CP1 <= 8 5 <= CP2 <= Infinite MT 235

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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? MT 235

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**Pet Food Company – Graph Solution**

Line 3 Line 2 MT 235

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**Pet Food Co. – Range of Feasibility**

Constraint 1 – is not binding Therefore, the shadow price is zero Slack is 100 Range of Feasibility 300 <= Constraint 1 RHS <= Infinite MT 235

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**Pet Food Co. – Change in the Right-hand Side**

Constraint 2 – add 1 to right-hand side 0P1 + 1P2 >= 201 1P1 + 1P2 >= 500 Solve for P1 -1(0P1 + 1P2 = 201) 1P1 + 1P2 = 500 P1 = 299 Solve for P2 1(299) + 1P2 >= 500 P2 = 201 MT 235

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**Pet Food Co. – Change in the Right-hand Side**

Solve objective function z = 5(299) + 8(201) z = 3103 Shadow Price 3103 – 3100 = 3 Thus cost increases at $3.00 per lb. added of P2 per batch Conversely, if we decrease lbs. of P2 per batch by 1 the objective function will decrease by $3.00 MT 235

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**Pet Food Co. – Range of Feasibility**

Constraint 2 RHS = 200 Allowable Increase = 300 Allowable Decrease = 100 Range of Feasibility 100 <= Constraint 2 RHS <= 500 MT 235

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**Pet Food Co. – Change in the Right-hand Side**

Constraint 3 – add 1 to right-hand side 0P1 + 1P2 >= 200 1P1 + 1P2 >= 501 Solve for P1 -1(0P1 + 1P2 = 200) 1P1 + 1P2 = 501 P1 = 301 Solve for P2 1(301) + 1P2 >= 501 P2 = 200 MT 235

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**Pet Food Co. – Change in the Right-hand Side**

Solve objective function z = 5(301) + 8(200) z = 3105 Shadow Price 3105 – 3100 = 5 Thus cost increases at $5.00 per lb. added of P2 per batch Conversely, if we decrease lbs. of P2 per batch by 1 the objective function will decrease by $5.00 MT 235

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**Pet Food Co. – Range of Feasibility**

Constraint 3 RHS = 500 Allowable Increase = 100 Allowable Decrease = 300 Range of Feasibility 200 <= Constraint 3 RHS <= 600 MT 235

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**Pet Food Co. – Linear Equations Slack/ Surplus Variables**

Min 5P1 + 8P2 + 0S1 + 0S2 + 0S3 s.t. 1P S = 400 1P S = 200 1P1 + 1P S3 = 500 P1, P2, S1 ,S2 ,S3 >= 0 MT 235

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**Pet Food Co. – Slack Variables**

For each ≤ constraint the difference between the RHS and LHS (RHS-LHS). It is the amount of resource left over. Constraint 1; S1 = 100 lbs. MT 235

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**Pet Food Co. – Surplus Variables**

For each ≥ constraint the difference between the LHS and RHS (LHS-RHS). It is the amount by which a minimum requirement is exceeded. Constraint 2; S2 = 0 lbs. Constraint 3; S3 = 0 lbs. MT 235

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**Pet Food Co. – Constraint Limits**

Range of Feasibility 300 <= Constraint 1 <= Infinite 100 <= Constraint 2 <= 500 200 <= Constraint 3 <= 600 MT 235

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