11/25/13 An Example of Dantzig-Wolfe Decomposition

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MATP6640/DSES6780 Linear Programming

An Example of Dantzig-Wolfe Decomposition

Consider the linear programming problem:

Thus, cT=(1,-1,-2), A=(1,1,1) and b=3. The Master Problem (MP) is:

where is the set of extreme points of X, and is the set of extreme rays of X.

11/25/13 An Example of Dantzig-Wolfe Decomposition

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Initialization:

Start with the two extreme points x1:=(2,0,0)T and x2:=(2,0,2)T of X. Then when using the revised

simplex method, our basic variables are and . So, in terms of the basic variables, we get the

Revised Master Problem (RMP):

Here, the coefficient of in the objective function is cTx1, the coefficient of 4 for in the first

constraint comes from Ax2, etc. The second constraint is the convexity constraint . We

will denote costs in (MP) by , columns by , etc.

Initial basis:

11/25/13 An Example of Dantzig-Wolfe Decomposition

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Thus, the initial basic feasible solution is:

the corresponding dual solution is:

the corresponding solution to the initial problem is

Thus the subproblem is

By inspection, the optimal solution is x=(0,0,0)T, giving . Therefore, our current

primal solution is not optimal, and we need to introduce a column for x3:=(0,0,0)T into the Revised

Master Problem.

First iteration:

Then enters the basis. Objective function value is ; column of constraint matrix is

In order to determine which variable leaves the basis, we need to

calculate . We then compare this column of ``B-1N'' with the current bfs

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using the minimum ratio test. Thus, leaves the basis. The new basis matrix is:

Thus, the new primal solution to (MP) is

and the new dual solution is

Hence, the new solution for the original LP is:

The subproblem is

This subproblem has an unbounded optimal solution, and the extreme ray is d=(0,1,0)T.

Set d4:=(0,1,0)T; then and . We introduce

into the basis.

Second iteration:

We need to determine which variable leaves the basis. Therefore, we calculate

11/25/13 An Example of Dantzig-Wolfe Decomposition

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. Using the minimum ratio test with shows that

leaves the basis. The new basis matrix is:

Thus, the new primal solution to (MP) is

and the new dual solution is

Hence, the new solution for the original LP is:

The subproblem is

An optimal solution is x=(0,0,2)T, giving . So we do not have the optimal solution to the

master problem (MP). Hence, we set x5:=(0,0,2)T, and we introduce a column for this extreme point

into the Revised Master Problem.

Third iteration:

Then enters the basis. Objective function value is ; column of constraint matrix

11/25/13 An Example of Dantzig-Wolfe Decomposition

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is . In order to determine which variable leaves the basis, we need to

calculate . Using the minimum ratio test with shows that

leaves the basis. The new basis matrix is:

Thus, the new primal solution to (MP) is

and the new dual solution is

Hence, the new solution for the original LP is:

The subproblem is

An optimal solution is x=(0,0,2)T, giving . Hence, our optimality criterion is satisfied,

and is an optimal solution to our original LP.

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About this document ...

John E Mitchell

2000-02-03