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Chapter 3

Linear Programming: Sensitivity Analysis and Interpretation of Solution

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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

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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

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Operations With Linear Inequalities

Maximize

Subject To

20 10

4 3 120

8 2 160

32

0

1 2

1 2

1 2

2

1 2

x x

x x

x x

x

x x

,

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Sherwood – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

Line 1

Line 2

1 2

3

45

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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

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Sherwood – Calculate Slope of Line 1

4x1 + 3 x2 <= 120

3x2 = -4x1 + 120

x2 = -4/3x1 + 40

 

Slope of Intercept of

Line 1 Line 1 on x2 axis

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Sherwood – Calculate Slope of Line 2

8x1 + 2x2 <= 160

2x2 = -8x1 + 160

x2 = -4x1 + 80

 

Slope of Intercept of

Line 2 Line 2 on x2 axis

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Sherwood – Optimal Solution Extreme Point 3 is optimal if:

-4 <= Slope of objective function <= -4/3

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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

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Sherwood – Optimal Solution Extreme Point 3 is optimal if:

-4 <= -(Cx1/Cx2) <= -4/3

Or 4/3 <= (Cx1/Cx2) <= 4

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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

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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

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Sherwood – Compute the Range of Optimality

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

Thus, Cx1 <= 40

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Sherwood – Compute the Range of Optimality

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

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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

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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

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Sherwood – Compute the Range of Optimality

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

5 <= Cx2 <= 15

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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?

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Sherwood – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

Line 1

Line 2

1 2

3

45

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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 = 14.875

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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

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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

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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 = 15.1875

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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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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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Sherwood – Range of Feasibility Constraint 3 RHS

Slack = 12 Shadow Price = 0

Range of Feasibility 20 <= Constraint 3 RHS <= Infinite

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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

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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.

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Sherwood - Slack Variables

Max

20x1 + 10x2 + 0S1 + 0S2 + 0S3

s.t.

4x1 + 3x2 + 1S1 = 120

8x1 + 2x2 + 1S2 = 160

x2 + 1S3 = 32

x1, x2, S1 ,S2 ,S3 >= 0

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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

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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

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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

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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?

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Pet Food Co. – Linear Equations

0,

50011

20010

40001

To,Subject

85

Minimize,

21

21

21

21

21

PP

PP

PP

PP

PP

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Pet Food Co. – Graph Solution

0

100

200

300

400

500

600

0 100 200 300 400 500 600 700

P1

P2 Line 2

Line 3

12

3

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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

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Pet Food Co. – Calculate Slope of Line 2

0P1 + 1P2 >= 200

1P2 >= -0P1 + 200

 

Slope of Intercept of

Line 2 Line 2 on P2 axis

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Pet Food Co. – Calculate Slope of Line 3

1P1 + 1P2 >= 500

1P2 >= -1P1 + 500

P2 >= -P1 + 500

 

Slope of Intercept of

Line 3 Line 3 on P2 axis

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Pet Food Co. – Optimal Solution Extreme Point 2 is optimal if:

-1 <= Slope of objective function <= 0

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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

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Pet Food Co. – Optimal Solution Extreme Point 2 is optimal if:

-1 <= -(CP1/CP2) <= 0

Or 0 <= (CP1/CP2) <= 1

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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

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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

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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

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Pet Food Co. – Compute the Range of Optimality

Summarizing these limits 0 <= CP1 <= 8

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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

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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

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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

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Pet Food Co. – Compute the Range of Optimality

Summarizing these limits 0 <= CP1 <= 8

5 <= CP2 <= Infinite

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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?

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Pet Food Company – Graph Solution

0

100

200

300

400

500

600

0 100 200 300 400 500 600 700

P1

P2 Line 2

Line 3

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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

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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

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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

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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

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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

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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

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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

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Pet Food Co. – Linear Equations Slack/ Surplus Variables

Min

5P1 + 8P2 + 0S1 + 0S2 + 0S3

s.t.

1P1 + 1S1 = 400

1P2 - 1S2 = 200

1P1 + 1P2 - 1S3 = 500

P1, P2, S1 ,S2 ,S3 >= 0

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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.

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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.

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Pet Food Co. – Constraint Limits Range of Feasibility

300 <= Constraint 1 <= Infinite 100 <= Constraint 2 <= 500 200 <= Constraint 3 <= 600

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