1. Duality Theory and Sensitivity Analysis

2. Learning Objectives
Learn how sensitivity analysis can be obtained from the final simplex tableau
Explain the concept of duality
Formulate a dual problem from its primal and vice versa

3. Review
Introduced topic of sensitivity analysis and dual values within context of graphical analysis
Sensitivity ranges and dual values were also provided as part of the computer solution output
Will show how sensitivity analysis and dual values are derived

4. Sensitivity Analysis
Will show how sensitivity ranges are mathematically performed
Provide a more thorough understanding of the topic
Proceed this topic by considering changes in objective function coefficients followed by changes in RHS of constraints

5. Changes in Objective Function Coefficients
Sensitivity analysis for coefficients in the objective function
Max Z = 40X1+50X2
subject to: 1X1+2X2 ≤ 40 hrs
4X1+3X2 ≤ 120 lb
X1,X2 ≥ 0
Coefficients in objective function symbolically cj
Consider a change in one of cj values by 
Change c1=40 by =10
Changing c1 from $40 to $50
Saw the effect of this change graphically in module 3

6. Changing in Initial Tableau
Optimal simplex tableau shown as
By changing from c1= 40 to c1=40+, initial tableau is
Range of c1 can be determined directly from the optimal tableau
Basic Z x1 x2 s1 s2
Solution 1
-40
-50 2 40 4 3
120
Basic Z x1 x2 s1 s2
Solution 1
-40-
-50 2 40 4 3
120

7. Changing in Optimal Tableau
Optimal tableau is:
Performing simplex operations, optimal tableau is:
z-row values are changed by coefficients of x1-row
x1 is a basic solution variable
Solution will remain optimal as long as the z-row values remain positive
Basic Z x1 x2 s1 s2
Solution 1 16 6
1360
4/5
-1/5 8
-3/5
2/5 24
Basic Z x1 x2 s1 s2
Solution
1+0
0+ 1
0+0
16-3/5
6+2/5  1
4/5
-1/5 8
-3/5
2/5 24

8. Determining Optimality Range for c1
Basic Z x1 x2 s1 s2
Solution
1+0
0+ 1
0+0
16-3/5
6+2/5  1
4/5
-1/5 8
-3/5
2/5 24
Solution to remain optimal
16-3/5 ≥ 0,
6+2/5 ≥ 0,  ≥0
Solving these inequality
-15≤≤26.66
1st yields ≤26.66, 2nd yields ≥-15, and 3rd yields ≥-56.66
Thus ≥-15

9. Determining Optimality Range for c1-Cont.
Recall c1= 40+; therefore,  = c1 - 40
Substituting c1-40 for 
≤26.66, c1-40 ≤26.66, c1≤66.66
≥-15, c1-40 ≥-15 c1≥25
Optimality range for c1
25≤c1≤66.66

10. Optimality Range for c2 c2 = 50 +  Effect of this change is:
Basic Z x1 x2 s1 s2
Solution
1+0
0+ 0
0+1
16+4/5
6-/5
1360+8 1
4/5
-1/5 8
c2 = 50 + 
Effect of this change is:
Solution will remain optimal as long as the row-z values remain positive or zero
Thus, we must have:
16+4/5≥0, 6-/5≥0, ≥0,
Solving these inequalities gives: -20≤≤30
 = c2 – 50
-20≤c2-50≤30
30≤ c2 ≤ 80

11. Changes in Constraint Quantity Values
Use our example problem as:
Max Z = 40X1+50X2
subject to: X1+2X2 ≤ 40 hrs
4X1+3X2 ≤ 120 lb; X1,X2 ≥ 0
40 and 120 will be represented as q1 and q2
Consider a  in q1
Purpose is to determine the range for q1 over which the optimal variable mix will remain the same and the shadow price will remain the same
Range for qi can be determined directly from the optimal tableau
Model constraints become
X1+2X2 ≤ 40+1
4X1+3X2 ≤ 120+0

12. Initial Tableau
Augment  to the right side of the labor constraint
Initial tableau the same as the coefficients in the s1 column
s1 column values will duplicate the  changes in the quantity column in the final tableau
Basic Z x1 x2 s1 s2
Solution 1
-40
-50 2
40+ 4 3
120
Basic Z x1 x2 s1 s2
Solution 1 16 6 
4/5
-1/5
8+4/5
-3/5
2/5
24-3/5

13. Simplex Requirement
Observe just the slack (s1) column corresponding to the model constraint quantity being changed
Simplex requirement is that the quantity values not be negative
≥0, 8+4/5≥0, 24-3/5≥0
≥-85, ≥-10, ≤40
40≥≥-10
 = q1 – 40
40≥ q1 – 40 ≥-10; 80≥q1≥30
Present basic solution will remain same and feasible
Quantity values of those basic variables may change
Their values can change
Basic Z x1 x2 s1 s2
Solution 1 16 6 
4/5
-1/5
8+4/5
-3/5
2/5
24-3/5

14. Changes in Second Constraint Quantity Value
s2 column values are used to develop the  inequalities
1360+6≥0, 8-1/5≥0, 24+2/5≥0; -60≤≤40
 =q2-120
60≤q2≤160
Present basic solution variables will remain the same and feasible
Basic Z x1 x2 s1 s2
Solution 1 16 6
1360+6
4/5
-1/5
8-1/5
-3/5
2/5
24+2/5

15. Primal and Dual
Every LP model posses two forms: the primal and the dual
Original form called the primal
Examples so far were primal
Dual is derived from primal model
Provides an alternative way of looking at a problem

16. Model Solutions
Primal gives solutions in max. profit, dual provides solutions in min. total cost
How the dual model is derived from the primal and what it means
Consider another constraint, storage space
Limited to 240 square feet to storing tables and chairs
24 x x2 ≤ 240
Complete model
Max Z = 40 x x2
Subject to
x1 + 2 x2 ≤ 40
4 x1 + 3 x2 ≤ 120
x1, x2 ≥ 0
Represents primal form (a maximization problem)
Dual form is a minimization problem
Min Z = 40 y y y3
y1+ 4 y2+ 24 y3 ≥ 40
2 y1+ 3 y2+ 12 y3 ≥ 50
y1, y2, and y3 ≥ 0

17. Relationships between Primal and Dual
Relationships can be demonstrated as follows
Every constraint in the primal represents a variable in the dual
Primal model has three constraints; therefore, the dual has three decision variables y1, y2, and y3
RHS value of the primal constraints represent objective function coefficients in the dual
40, 120, and 240 form dual objective function: Z= 40 y y y3.
Model constraints coefficients in the primal represent decision variable coefficients in the dual
Coefficients in labor constraint, 1 and 4, represent coefficient of the first decision variable y1 of the dual
Objective function coefficients, 40 and 50, represent RHS values of the constraints in the dual
Maximization primal model has the “≤” constraints whereas the minimization dual model has “≥” constraints

19. What Dual Form Means?
Determine what the dual means
What do y1, y2, and y3 mean, what do the “≥” model constraints mean, and what is being minimized?
Solution from final tableau, x1= 0, x2= 20, s2= 60, and z= 1000
Given the availability of each resource, we want to determine the per unit value such that the total resources used is minimized
Basic x1 x2 s1 s2 s3
Solution Z 20
0.83
1000 1
0.67
-0.03
-0.67
-0.14 60
-0.33
0.06

20. What Dual Form Means? Cont.
Focus on primal objective function
Z=40x1+50x2
($ value per unit of Chair) (No. of units of Chair)+($ value per unit of Table) (No. of units of Table)=Total $ value of production
Given the availability of each resource, we want to determine the per unit value such that the total resources used is minimized
Focus on dual objective function
Z=40y1+120y2+240y3
(Units of Resource 1) ($ value of per hour of labor) +(Units of Resource 2) ($ value of per pound of wood) +(Units of Resource 3) ($ value of per sq ft of storage space) =Total $ value of production

21. What Dual Form Means?-Cont.
In z-row, positive values of 20 and 0.83 under s1 and s3, tell us if one unit of either s1 or s3 were entered into solution, profit would decrease by $20 or $0.83, respectively
Positive values of $20 and $0.83 are marginal values or dual values of s1 and s3
Often referred to as shadow prices
Reflect the maximum "price" that we are willing to pay

22. Unused Resources What happened to second resource, amount of wood?
Explained by present solution of primal
Notice z-row value for s2 is zero
Means that amount of wood has a marginal value of zero
Would not be willing to pay anything for an extra pound of wood
Reason is because the solution of primal shows that 60 pound of wood were left unused (i.e., s2= 60)
60 pound of wood left over means that an extra pound of wood would have no additional value

23. More Clarification
Clarify one additional aspect of shadow prices of s1 and s3
Marginal value of $20 for one hour of labor is not necessarily meant what the company would pay for an hour of labor
Are assuming that all of the resources available are already paid for
Are sunk costs
Value of labor or storage space resource is determined by their contribution toward the $1,000 profit
Wanted to assign a value to resources, it could not assign an amount greater than the profit earned by resources
Value of all resources must exactly equal profit

24. Use of the Dual Why do we have to be concerned about the dual form?
Importance of the dual form lies in the information provided by the model resources
More control over use of resources than over accumulation of profits
A dual problem may be easier to solve because of fewer model constraints
Suppose a primal problem that has 1000 constraints and 100 decision variables
Solving would be much easier in dual form which has 100 constraints and 1000 variables
Notice that solution provided by dual problem can also be used for the primal problem

25. Duality and Sensitivity Analysis
Provides the value of the resources, which is important in deciding whether or not to secure more resources
Next question is "How does this affect the original solution?"
Feasible solution area is determined by the values forming the model constraints
Effect on the solution is the subject of sensitivity analysis

26. Summary
Showed how sensitivity analysis can be performed using the information in the final simplex tableau
Saw how the original LP problem, called the primal, can be converted into its associated dual LP problem
Learned that the value of the dual variables identify the economic contribution or value of additional resources in the primal problem