For a given linear programming problem (LP1)
The correspondence between the variables LP1 and LP2 is as follows, so we can see that this transformation can be done twice to get back to the original Variable support
| LP1 | x1 | x2 | c1 | c2 | b1 | b2 | A11 | A12 | A21 | A22 | |
|---|---|---|---|---|---|---|---|---|---|---|---|
| LP2 | y1 | y2 | -b1 | -b2 | -c1 | -c2 | -A11^T | -A21^T | -A12^T | -A22^T |
For simpler problems
minimize
subject to
minimize
subject to
| x | c | b | A | ||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| LP1 | x1 | x2 | c1 | c2 | b1 | b2 | A11 | A12 | A21 | A22 | |
| way of thinking |
Since x is positively constrained, this corresponds to x1.
The coefficient of x1 in the objective function is c1
The constraints are equations, and if we focus on x1, A21 and b2
All other values are zero.
Substituting for a dual problem
A21 hangs on y2, which we will call y
Note that y2 is not positively constrained.
minimize
subject to
Consider $y_1 \times eq1 + y_2 \times eq2$$(y_1 \ge 0, y_2 \ge 0)$
If the left-hand side is less than the objective function, then the objective function is greater than the right-hand side
So, if we choose y so that the right-hand side is as large as possible, we get the following linear programming problem
maximize
subject to
study text Linear Programming (2) Dual Problem and Dual Theorem
Frame Covers integer programming relaxed linear programming Application of Primal Dual Method http://www.akita-pu.ac.jp/system/elect/ins/kusakari/japanese/teaching/InfoMath/2008/note/14.pdf
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