1. Chapter 3
Linear Programming: Sensitivity Analysis and Interpretation of Solution
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2. 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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3. 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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4. Operations With Linear Inequalities
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5. Sherwood – Graph Solution
Line 2 5 4 3
Line 1 1 2
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6. 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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7. 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
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8. Sherwood – Calculate Slope of Line 2
8x1 + 2x2 <= 160
2x2 = -8x
x2 = -4x1 + 80
Slope of Intercept of
Line Line 2 on x2 axis
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9. Sherwood – Optimal Solution
Extreme Point 3 is optimal if:
-4 <= Slope of objective function <= -4/3
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10. 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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11. Sherwood – Optimal Solution
Extreme Point 3 is optimal if:
-4 <= -(Cx1/Cx2) <= -4/3 Or
4/3 <= (Cx1/Cx2) <= 4
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12. 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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13. 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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14. Sherwood – Compute the Range of Optimality
From the right-hand inequality, we have
(Cx1/10) <= 4
Thus,
Cx1 <= 40
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15. Sherwood – Compute the Range of Optimality
Summarizing these limits
40/3 <= Cx1 <= 40
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16. 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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17. 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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18. Sherwood – Compute the Range of Optimality
Summarizing these limits
40/3 <= Cx1 <= 40
5 <= Cx2 <= 15
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19. 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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20. Sherwood – Graph Solution
Line 2 5 4 3
Line 1 1 2
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21. 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 =
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22. 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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23. 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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24. 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 =
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25. 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
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26. 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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27. Sherwood – Range of Feasibility
Constraint 3 RHS
Slack = 12
Shadow Price = 0
Range of Feasibility
20 <= Constraint 3 RHS <= Infinite
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28. 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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29. 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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30. 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
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31. 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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32. 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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33. 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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37. 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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38. Pet Food Co. – Linear Equations
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39. Pet Food Co. – Graph Solution3
Line 3
Line 2 2 1
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40. 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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41. Pet Food Co. – Calculate Slope of Line 2
0P1 + 1P2 >= 200
1P2 >= -0P
Slope of Intercept of
Line Line 2 on P2 axis
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42. 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
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43. Pet Food Co. – Optimal Solution
Extreme Point 2 is optimal if:
-1 <= Slope of objective function <= 0
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44. 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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45. Pet Food Co. – Optimal Solution
Extreme Point 2 is optimal if:
-1 <= -(CP1/CP2) <= 0 Or
0 <= (CP1/CP2) <= 1
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46. 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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47. 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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48. 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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49. Pet Food Co. – Compute the Range of Optimality
Summarizing these limits
0 <= CP1 <= 8
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50. 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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51. 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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52. 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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53. Pet Food Co. – Compute the Range of Optimality
Summarizing these limits
0 <= CP1 <= 8
5 <= CP2 <= Infinite
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54. 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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55. Pet Food Company – Graph Solution
Line 3
Line 2
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56. 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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57. 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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58. 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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59. 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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60. 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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61. 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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62. 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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63. 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
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64. 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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65. 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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66. Pet Food Co. – Constraint Limits
Range of Feasibility
300 <= Constraint 1 <= Infinite
100 <= Constraint 2 <= 500
200 <= Constraint 3 <= 600
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