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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?
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