Chapter 4: �Duality and Post Optimal Analysis
A lecture by:
Dr. Moayad Tanash
Introduction
Definition of the dual problem
The Dual problem
The following is a summary of how the dual is constructed from the (equation form) primal:
1. A dual variable is assigned to each primal (equation) constraint and a dual constraint is assigned to each primal variable.
2. The right-hand sides of the primal constraints provide the coefficients of the dual objective function.
3. The dual constraint corresponding to a primal variable is constructed by transposing the primal variable column into a row with (i) the primal objective coefficient becoming the dual right-hand side and (ii) the remaining constraint coefficients comprising the dual left-hand side coefficients.
4. The sense of optimization, direction of inequalities, and the signs of the variables in the dual are governed by the rules in Table 4.1
The Dual problem
Example 1
Example 1
Example 2
Example 2
Rules for constructing the dual
Optimal Dual Solution
The primal and dual solutions are so closely related that the optimal solution of either problem directly yield ( with little additional computation) the optimal solution to the other problem.
The dual of the dual is itself the primal, which means that the dual solution can also used to yield the optimal primal solution automatically.
Optimal Dual Solution
Two methods for determining the dual values
Method 1
Method 2
Example
Example
Example
| | | | | | |
| x1 | x2 | x3 | x4 | R | RHS |
z | 5 | 12 | 4 | 0 | -M | 0 |
y1 | 1 | 2 | 1 | 1 | 0 | 10 |
y2 | 2 | -1 | 3 | 0 | 1 | 8 |
| | | | | | |
Optimal Dual Solution
Using M-Method, the optimal primal tableau is
Method 1
| x1 | x2 | x3 | x4 | R | RHS | ratio | |
Z | 0.00 | 0.00 | 0.60 | 5.80 | 99.60 | 54.80 | | |
x2 | 0.00 | 1.00 | -0.20 | 0.40 | -0.20 | 2.40 | optimal inverse | |
x1 | 1.00 | 0.00 | 1.40 | 0.20 | 0.40 | 5.20 | ||
Optimal Dual Solution
Method 2
Optimal invers =
and
(original objective coefficient ) = (x2,x1) = (12,5)
Thus, the optimal dual values are computed as:
0.40 | -0.20 |
0.20 | 0.40 |
Primal-Dual objective value
The relation between the primal and dual is given by :
Primal-Dual objective value
In Example 4.2-1 ( x1 = 0, x2=0, x3=2.67) and ( y1=6, y2=0) are feasible primal and optimal solution. The associated value of the objective function are
z= 5x1+12x2+4x3 = 10.667
w= 10y1+8y2 = 60
thus, z= (10.67) for the maximization problem (primal) is less than w = (60) for the minimization problem ( dual). The optimal value of z = (54.8) falls within the range (10.67 60)
Simplex Tableau Computations
We can generate any iteration of the entire simplex tableau using the original data of the problem, the invers associated with the problem and the dual problem.
Example 4.2-3
Economical Interpretation of Duality
per unit of activity j
Economical Interpretation of Duality
Economical Interpretation of Duality
Economical Interpretation of Duality
Example
Example
Additional Simplex Algorithms
Dual Simplex method
Example (Dual Simplex method)
In the present example, the first two inequalities are multiplied by -1 to convert them to ≤ constraints. The starting tableau is thus given as follows:
| x1 | x2 | x3 | x4 | x5 | x6 | RHS |
z | -3 | -2 | -1 | 0 | 0 | 0 | 0 |
x4 | -3 | -1 | -1 | 1 | 0 | 0 | -3 |
x5 | 3 | -3 | -1 | 0 | 1 | 0 | -6 |
x6 | 1 | 1 | 1 | 0 | 0 | 1 | 3 |
| | | | | | | |
Example (Dual Simplex method)
#1 | | | | | | | |
| x1 | x2 | x3 | x4 | x5 | x6 | RHS |
z | -3 | -2 | -1 | 0 | 0 | 0 | 0 |
x4 | -3 | -1 | -1 | 1 | 0 | 0 | -3 |
x5 | 3 | -3 | -1 | 0 | 1 | 0 | -6 |
x6 | 1 | 1 | 1 | 0 | 0 | 1 | 3 |
ratio | 1 | 0.666667 | 1 | | | | |
Example (Dual Simplex method)
Notice how the dual simplex works. In all the iterations, optimality is maintained (all reduced costs are ≤ 0) as each new iteration moves the solution toward feasibility. At iteration 3, feasibility is restored for the first time, and the process ends with the optimal feasible solution given as
#2 | | | | | | | |
| x1 | x2 | x3 | x4 | x5 | x6 | RHS |
z | -5 | 0 | -0.33333 | 0 | -0.66667 | 0 | 4 |
x4 | -4 | 0 | -0.66667 | 1 | -0.33333 | 0 | -1 |
x2 | -1 | 1 | 0.333333 | 0 | -0.33333 | 0 | 2 |
x6 | 2 | 0 | 0.666667 | 0 | 0.333333 | 1 | 1 |
ratio | 1.25 | | 0.5 | | 2 | | |
| | | | | | | |
#3 | | | | | | | |
| x1 | x2 | x3 | x4 | x5 | x6 | RHS |
z | -3 | 0 | 0 | -0.5 | -0.5 | 0 | 4.5 |
x3 | 6 | 0 | 1 | -1.5 | 0.5 | 0 | 1.5 |
x2 | -3 | 1 | 0 | 0.5 | -0.5 | 0 | 1.5 |
x6 | -2 | 0 | 0 | 1 | 0 | 1 | 0 |
ratio | | | | | | | |
Generalized Simplex method
Generalized Simplex method
The primal simplex algorithm starts feasible but nonoptimal
The dual simplex starts optimal but infeasible
Generalized Simplex method
The new solution is now feasible but nonoptimal, and we can use the primal simplex to determine the optimal solution. In general, had we not restored feasibility in the preceding tableau, we would repeat the procedure as necessary until feasibility is satisfied or until there is evidence that the problem has no feasible solution.
Post-Optimal Analysis
We deal with making changes in the parameter of the model and finding the new optimal solution.
Changes effecting feasibility
The feasibility of the current optimal solution maybe effected only if:
in both cases, infeasibility occurs when at least one element of the right hand side of the optimal tableau becomes negative, that is, one or more of the current basic variable become negative
Example Changes effecting feasibility
Example Changes effecting feasibility
Situation1: Supposed that TOYCO wants to expand its assembly lines by increasing the daily capacity of operations 1,2 and 3 by 40% to 602,644 and 588 minutes, respectively. How would this change effect the total revenue ?
Ans :
Example Changes effecting feasibility
Situation2: although the new solution is appealing from the standpoint of increasing revenue, TOYCO recognizes that this implementation may take time. Another proposal was thus made to shift the slacks capacity of operation 3 (x6=20 mins) to the capacity of operation 1. How would this changes impact the optimal solution.
The resulting solution is infeasible which required applying the dual simplex method
Addition of new constraints
The addition of a new constraint to an existing model can lead to:
Example: suppose that TOYCO is changing the design for its toys and that required the addition of a fourth operation in the assembly line. The daily capacity of the new operation is 500 mins and the time per unit for the three products on this operation are 3,1 and 1 min respectively. Study the effect of the new operation on the optimal solution
The constraint of operation 4 is
3x1+x2+x3<=500
The constraint is redundant because it is satisfied by the current optimal solution ( x1=0,x2=100 and x3=230)
Addition of new constraints
Now, suppose that TOYCO unit times in the fourth operation are changed to 3,3,1 mins, respectively. All remaining data of the model remain the same. Will the optimal solution change ?
Ans: the constraint for operation 4 is
3x1+3x2+x3<=500
3x1+3x2+x3+x7=500
The new constraint is not satisfied by the current optimal solution. Thus, the new constraint must be added to the current optimal tableau
Addition of new constraints
Addition of new constraints
Changes effecting optimality
Changes objective coefficient
These changes effect only the optimality of the solution. Such changes require recomputing the z-row coefficient as follows:
1- Compute the dual values using method 2
2- use the new dual values to determine the new reduced cost ( z_row)
Changes effecting optimality
Two cases will result:
Example:
In the TOYCO model, suppose that the company is instituting a revised pricing policy to meet the competition. The new unit revenues are $2, $3, and $4 for train, truck, and car toys, respectively. The new objective function is
Changes effecting optimality
Example:
In the TOYCO model, suppose that the company is instituting a revised pricing policy to meet the competition. The new unit revenues are $2, $3, and $4 for train, truck, and car toys, respectively. The new objective function is
Maximize z=2x1+3x2+4x3
Changes effecting optimality
Note that the right-hand side of the first dual constraint is 2, the new coefficient in the modified objective function. The computations show that the current solution, x1 = 0 train, x2 = 100 trucks, and x3 = 230 cars, remains optimal. The corresponding new revenue is computed as 2 * 0 + 3 * 100 + 4 * 230 = $1220. The new pricing policy is not recommended because it lowers revenue