The simple case
The function simplex_reduce makes the reduction by the simplex algorithm to find :
|max(c.x), A.x ≤ b, x ≥ 0, b≥ 0|
where c,x are vectors of ℝn, b≥ 0 is a vector in
ℝp and A is a matrix of p rows and n columns.
simplex_reduce takes as argument A,b,c and returns max(c.x), the augmented solution of x (augmented since the algorithm works by adding rows(A) auxiliary variables) and the reduced matrix.
Which means that the maximum of X+2Y under these conditions
is 7, it is obtained for X=1,Y=3
because [1,3,0,0] is the augmented solution and the reduced matrix is :
A more complicated case that reduces to the simple case
With the former call of simplex_reduce, we have to :
For example, find :
Let x=1−X, y=Y+2, z=5−X+3Y the problem is equivalent to finding the minimum of (−2X+Y−(5−X+3Y)+8) where :
or to find the minimum of :
i.e. to find the maximum of −(−X−2Y+3)=X+2Y−3 under the same conditions, hence it is the same problem as to find the maximum of X+2Y seen before. We found 7, hence, the result here is 7-3=4.
The general case
A linear programming problem may not in general be directly reduced like above to the simple case. The reason is that a starting vertex must be found before applying the simplex algorithm. Therefore, simplex_reduce may be called by specifying this starting vertex, in that case, all the arguments including the starting vertex are grouped in a single matrix.
We first illustrate this kind of call in the simple case where the starting point does not require solving an auxiliary problem. If A has p rows and n columns and if we define :
simplex_reduce may be called with D as single argument.
For the previous example, input :
Output is the same result as before.
Back to the general case.
The standard form of a linear programming problem is similar to the simplest case above, but with Ax=b (instead of Ax≤ b) under the conditions x≥ 0. We may further assume that b≥ 0 (if not, one can change the sign of the corresponding line).
simplex_reducewith a single matrix argument obtained by augmenting A by the identity, b unchanged and an artificial c with 0 under A and 1 under the identity). If the maximum exists and is 0, the identity submatrix above the last column corresponds to an x solution, we may forget the artificial variables (they are 0 if the maximum is 0).
simplex_reducewith the original c and the value of x we found in the domain.
simplex_reduce([[-1,-1,0,1,1,0,1], [0,1,-1,1,0,1,3], [0,0,0,0,1,1,0]])Output: optimum=0, artificial variables=0, and the matrix
simplex_reduce([[-1/2,0,-1/2,1,2], [1/2,1,-1/2,0,1],[2,3,-1,1,0]])Output: maximum=-5, hence the minimum of the opposite is 5, obtained for (0,1,0,2), after replacement x=0, y=1, z=0 and t=2.
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