Dual Simplex Algorithm and Post Optimal Analysis

Group 1-C

Siddharth Aggarwal (2012ABPS435H)Ashish Garg (2012ABPS551H)

Abhishek Goyal (2012A4PS142H)Yash Pratap Singh (2011B2A4632H)

What is a dual problem• This is derived from the concept of solving the linear

programming problems starting with an infeasible solution.• This is a problem defined directly from its primal

counterpart.• The conversion from one of the problems to other is such

that the optimal solution of one is also an optimal for the other.

Conversion of a primal problem into its dual• Add all the slack and surplus variables where ever required in the

primal equations.• If any unconstrained variable is there in any of the original constraint

equations replace it by (+x&-x) there it self before we proceed to calculating its dual.• Graphically, the dual has a point corresponding to each edge of a

primal graph.

Dual primal graphs

primal Problem dual problem

Constraint VariableObjective function Right sides

i i

Relationships between Primal and Dual Problems

Minimization Maximization

Variables

Variables

Constraints

Constraints

0

0

0

0

Unrestricted

Unrestricted

Minimize 4x1 + 2x2 − x3=0 subject to x1 + x2 + 2x3 ≥ 3 2x1 − 2x2 + 4x3 ≤ 5 x1, x2, x3 ≥ 0. In standard form: Minimize 4x1 + 2x2 − x3 + 0s1 + MR + 0s2=0 subject to x1 + x2 + 2x3 − s1 + R = 3 2x1 − 2x2 + 4x3 + s2 = 5 x1, x2, x3, s1, R, s2 ≥ 0

Example

Minimize 4x1 + 2x2 − x3 + 0s1 + MR + 0s2=0 subject to x1 + x2 + 2x3 − s1 + R = 3….y1 2x1 − 2x2 + 4x3 + s2 = 5…..y2 x1, x2, x3, s1, R, s2 ≥ 0

Maximize 3y1 + 5y2=0 subject to y1 + 2y2 ≤ 4 y1 − 2y2 ≤ 2 2y1 + 4y2 ≤ −1 dual form of problem −y1 ≤ 0, i.e. y1 ≥ 0 y2 ≤ 0

Dual Simplex Algorithm• Dual simplex, unlike primal simplex, starts with an infeasible solution

and these solutions are better than the optimal feasible solution.• The feasibility is restored in final step and thus the solution is

obtained.• Thus, we move towards feasibility and optimality.• Why Dual Simplex?When there is in the constraint equations, artificial variables are required. More variables, more iterations, more chances of error.

When can Dual Simplex be applied?• If the objective function is like

• If the objective function is like

• Easy way to remember:Take all terms to one side. Use “Maximization”. If the coefficients of all terms is positive, we can apply dual simplex.

Dual Simplex Procedure1. Ensure that the objective function is maximization. If not, make it

“Maximization”.2. Convert all inequalities to ‘’.3. Add slack variables in the constraint equations to convert them into

equalities.4. Form the matrix. It should be an optimal matrix.5. If there is an ‘’ constraint, replace it by two inequalities.e.g.:

In the usual Simplex method, leaving variable is determined first, then entering variable.In Dual Simplex, entering variable is determined first, then leaving variable.• Leaving variable:Most value on the RHS. It is the pivot row.

• Entering variable:, where is the pivot row element.

Minimize subject to

Objective function for Dual Simplex:Maximize Constraint equations made compatible for Dual Simplex:

The starting solution is infeasible if at least one of the right-hand sides of the inequalities is .

x1 x2 x3 s1 s2 s3 RHS

s1 -3 -1 -1 1 0 0 -3

s2 3 -3 -1 0 1 0 -6

s3 1 1 1 0 0 1 3

z 3 2 1 0 0 0 0

2/3 1

x1 x2 x3 s1 s2 s3 RHS

s1 -4 0 -2/3 1 -1/3 0 -1

x2 -1 1 1/3 0 -1/3 0 2

s3 2 0 2/3 0 1/3 1 1

z 5 2 1/3 0 2/3 0 -4

5/4=1.25 (1/3)/(2/3)=0.5

x1 x2 x3 s1 s2 s3 RHS

x3 6 0 1 -3/2 1/2 0 3/2

x2 -3 1 0 -1/2 -1/2 0 3/2

s3 -2 0 0 1 0 1 0

z 3 0 0 1/2 1/2 0 -9/2

Feasible and optimalAll positive

Post Optimal AnalysisMaking changes in the parameters of the model and finding the new optimal solution.

• Changes affecting feasibility1) The right hand side of the constraints are changed.2) A new constraint is added to the model.

• Changes affecting optimality1) Changes in the original objective coefficients2) Addition of a new economic activity (variable) to the model

Changes affecting feasibilityMaximize z = 3x1+ 2x2 + 5x3

Subjected to x1 + 2x2 + x3 ≤ 430 3x1 + 2x3 ≤ 460 x1 + 4x2 ≤ 420 x1, , x2 , x3 ≥ 0

Basic x1 x2 x3 s1 s2 s3 RHSs1 1 2 1 1 0 0 430s2 3 0 2 0 1 0 460s3 1 4 0 0 0 1 420z -3 -2 -5 0 0 0 0

s1 -1/2 2 0 1 -1/2 0 200x3 1.5 0 1 0 0.5 0 230s3 1 4 0 0 0 1 420z 9/2 -2 0 0 5/2 0 1150

x2 -1/4 1 0 1/2 -1/4 0 100x3 3/2 0 1 0 1/2 0 230s3 2 0 0 -2 1 1 20

z 4 0 0 1 2 0 1350 optimal

The right hand side of the constraints are changed[New right-hand side of tableau in iteration i]

= [Inverse in iteration i] X [New right hand side of constraints]

600

640

590

14032030

Example 1 Solution is feasible

1/2 -1/4 0

0 1/2 0

-2 1 1

The associated optimum value is 1880.

450460400

110

230

-40

Example 2:

Solution is not feasible

1/2 -1/4 0

0 1/2 0

-2 1 1

Basic x1 x2 x3 s1 s2 s3 RHS

x2 -1/4 1 0 1/2 -1/4 0 100

x3 3/2 0 1 0 1/2 0 230

s3 2 0 0 -2 1 1 -40

z 4 0 0 1 2 0 1370

The slack capacity of operation 3 (s3=20) is shifted to the capacity of operation 1

Basic x1 x2 x3 s1 s2 s3 RHS

x2 -1/4 1 0 1/2 -1/4 0 100

x3 3/2 0 1 0 1/2 0 230

s3 2 0 0 -2 1 1 20

z 4 0 0 1 2 0 1350

The optimum solution remains the same as in the original model. This means that the proposed shift in capacity allocation is not advantageous because it simply shifts surplus capacity from op3 to op1.

A new constraint is added to the modelSituation 1Example 3x1+ x2+ x3 ≤ 500 redundant

Situation 2Example 3x1+ 3x2+ x3 ≤ 500

x1 x2 x3 s1 s2 s3 s4 RHS

X2 -1/4 1 0 1/2 -1/4 0 0 100

X3 3/2 0 1 0 1/2 0 0 230

S3 2 0 0 -2 1 1 0 20

s4 3 3 1 0 0 0 1 500 coefficients of basic variables to zero

z 4 0 0 1 2 0 0 1350

A new constraint is added to the model

x1 x2 x3 s1 s2 s3 s4 RHS

x2 -1/4 1 0 1/2 -1/4 0 0 100

x3 3/2 0 1 0 1/2 0 0 230

s3 2 0 0 -2 1 1 0 20

s4 9/4 0 0 -3/2 1/4 0 1 -30

z 4 0 0 1 2 0 0 1350

x1 x2 x3 s1 s2 s3 s4 RHS

x2 1/3 1 0 0 -1/6 0 0 90

x3 3/2 0 1 0 1/2 0 0 230

s3 -1 0 0 0 2/3 1 -4/3 60

s1 -3/2 0 0 1 -1/6 0 -2/3 20

z 11/2 0 0 0 13/6 0 2/3 1330 optimal

Changes Affecting Optimality:Changes in the original objective coefficients

• Example 1: New pricing policy : $2, $3, $4 respectively for train, truck and car toys, respectively.

Maximize z = 2x1+ 3x2 + 4x3

(New objective coefficients of basic x2, x3 and s3) = (3, 4, 0)

1. Compute dual values using(Optimal dual values) = (Row vector of objective coefficients of optimal basic variables) x (optimal primal

inverse)

New Dual variables:

(y1, y2, y3) = (3,4,0) = (3/2, 5/4, 0)

2. Substitute the new dual values in (z equation coefficient of variable x) = (Left-hand side of dual constraint) – (Right-hand side of dual constraint)to determine the new reduced costs (z-row coefficients)

Minimize z = 2x1+ 3x2 + 4x3 Subjected to y1 + 3y2 + y3 ≥ 2

2y1 + 2y3 ≥ 3 y1 + 4y2 ≥ 4 y1, , y2 , y3 ≥ 0

Reduced cost of x1 = y1 +3y2 + y3 -2 = 3/2 + 3(5/4) + 0 -2 = 13/4Reduced cost of s1 = y1 - 0 = 3/2Reduced cost of s2 = y2 - 0 = 5/4new revenue generated is $1220

Changes in the original objective coefficients

• Example 2: Maximize z = 6x1+ 3x2 + 4x3

(y1, y2, y3) = (3,4,0) = (3/2, 5/4, 0)

Reduced cost of x1 = y1 +3y2 + y3 -6 = 3/2 + 3*5/4 + 0 -6 = -3/4Reduced cost of s1 = y1 - 0 = 3/2Reduced cost of s2 = y2 - 0 = 5/4

1/2 -1/4 0

0 1/2 0

-2 1 1

Basic x1 x2 x3 s1 s2 s3 RHS

x2 -1/4 1 0 1/2 -1/4 0 100

x3 3/2 0 1 0 1/2 0 230

s3 2 0 0 -2 1 1 20

z -3/4 0 0 3/2 5/4 0 1220

Basic x1 x2 x3 s1 s2 s3 RHS

x2 -1/4 1 0 1/2 -1/4 0 100

x3 3/2 0 1 0 1/2 0 230

s3 2 0 0 -2 1 1 20

z -3/4 0 0 3/2 5/4 0 1220

Basic x1 x2 x3 s1 s2 s3 RHS

x2 0 1 0 1/4 -1/8 1/8 102.5

x3 0 0 1 3/2 -1/4 -3/4 215

x1 1 0 0 -1 1/2 1/2 10

z 0 0 0 3/4 13/8 3/8 1227.5

Addition of a new economic activity (variable) to the model• Given that (y1, y2, y3)=(1,2,0)• Toy- remote control car (x4): $4, assembly times (1, 1, 2) on (op1, op2,

op3)• Reduced cost of x4= y1 +y2 + 2y3 – 4 = -1

• x4 – constraint column=

112

1/41/21

=

1/2 -1/4 0

0 1/2 0

-2 1 1

Basic x1 x2 x3 x4 s1 s2 s3 RHS

x2 -1/4 1 0 1/4 1/2 -1/4 0 100

x3 3/2 0 1 1/2 0 1/2 0 230

s3 2 0 0 1 -2 1 1 20

z 4 0 0 -1 1 2 0 1350

Addition of new activity – example continued

Basic x1 x2 x3 x4 s1 s2 s3 RHS

x2 -3/4 1 0 0 1 -1/2 -1/4 95

x3 1/2 0 1 0 1 0 -1/2 220

x4 2 0 0 1 -2 1 1 20

z 6 0 0 0 -1 3 1 1370

Basic x1 x2 x3 x4 s1 s2 s3 RHS

s1 -3/4 1 0 0 1 -1/2 -1/4 95

x3 5/4 -1 1 0 0 1/2 -1/4 125

x4 7/2 2 0 1 0 0 1/2 210

z 21/4 1 0 0 0 5/2 3/4 1465

The new solution is z=$1465, which improves the original revenues by $115. Therefore, introduction of new product is advantageous.

1. When do we use Dual Simplex algorithm?

2. How does Dual Simplex method proceed?a) Better than optimal and Infeasible to optimal and feasibleb) Feasible and non-optimal to optimal and feasible

3. Difference between Sensitivity and Post-optimal analysis.

4. When does feasibility change?a) Changes in the objective function coefficientsb) Changes in the right hand sidec) Addition of new activityd) All of the above

5. If new constraint is added, what will change?a) Feasibilityb) Optimalityc) No changed) Both

Thank You!