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?

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

Linear Programming

Sherwood – Linear Equations

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

,

Linear Programming

Sherwood – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

Line 2

Line 1

1 2

3

45

Linear Programming

Sherwood – Optimal Solution Extreme Point 3 is optimal if:

Slope of Line 1 <= Slope of objective function <= Slope of Line 2

Linear Programming

Sherwood – Calculate Slope of Line 1

8x1 + 2x2 <= 160

2x2 = -8x1 + 160

x2 = -4x1 + 80

 

Slope of Intercept of

Line 1 Line 1 on x2 axis

Linear Programming

Sherwood – Calculate Slope of Line 2

4x1 + 3 x2 <= 120

3x2 = -4x1 + 120

x2 = -4/3x1 + 40

 

Slope of Intercept of

Line 2 Line 2 on x2 axis

Linear Programming

Sherwood – Optimal Solution Extreme Point 3 is optimal if:

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

Linear Programming

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

Linear Programming

Sherwood – Optimal Solution Extreme Point 3 is optimal if:

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

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

Linear Programming

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

Linear Programming

Sherwood – Compute the Range of Optimality

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

Thus, 40/3 <= Cx1

Linear Programming

Sherwood – Compute the Range of Optimality

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

Thus, Cx1 <= 40

Linear Programming

Sherwood – Compute the Range of Optimality

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

Linear Programming

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

Linear Programming

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

Linear Programming

Sherwood – Compute the Range of Optimality

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

5 <= Cx2 <= 15

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?

Linear Programming

Sherwood – Graph Solution

0

10

20

30

40

50

60

70

80

90

0 5 10 15 20 25 30 35

x1

x2

Line 2

Line 1

1 2

3

45

Linear Programming

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

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

Linear Programming

Sherwood – Range of Feasibility Constraint 1 RHS = 120

Allowable Increase = 24 Allowable Decrease = 40

Range of Feasibility 80 <= Constraint 1 RHS <= 144

Linear Programming

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

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

Linear Programming

Sherwood – Range of Feasibility Constraint 2 RHS = 160

Allowable Increase = 80 Allowable Decrease = 48

Range of Feasibility 112 <= Constraint 2 RHS <= 240

Linear Programming

Sherwood – Range of Feasibility Constraint 3 RHS

Slack = 12 Shadow Price = 0

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

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

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.

Linear Programming

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

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; S1 = 0 hrs.

Constraint 2; S2 = 0 hrs.

Constraint 3; S3 = 12 Custom

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

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