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## Simplex algorithm: simplex_reduce

The simple case
The function simplex_reduce makes the reduction by the simplex algorithm to find :

max(c.x),    A.x bx 0, b 0

where c, x are vectors of n, b 0 is a vector of p and A is a matrix of p rows and n columns.
simplex_reduce takes as argument A,b,c et returns max(c.x), the augmented solution of x (augmented since the algorithm works by adding rows(A) auxiliary variables) and the reduced matrix.
Example
Find

max(X + 2Y) where  Input :
simplex_reduce([[-3,2],[1,1]],[3,4],[1,2])
Output :
7,[1,3,0,0],[[0,1,1/5,3/5,3],[1,0,(-1)/5,2/5,1], [0,0,1/5,8/5,7]]
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 :
[[0,1,1/5,3/5,3],[1,0,(-1)/5,2/5,1], [0,0,1/5,8/5,7]].

A more complicate case that reduces to the simple case
With the former call of simplex_reduce, we have to :

• rewrite constraints to the form xk 0,
• remove variables without constraints,
• add variables such that all the constraints have positive components.
For example, find :

min(2x + y - z + 4)     where  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 :

(- X - 2Y + 3)     where  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 :

B:=augment(A,idn(p)); C:=border(B,b);
d:=append(-c,0\$(p+1)); D:=augment(C,[d]);
simplex_reduce may be called with D as single argument.
For the previous example, input :
A:=[[-3,2],[1,1]];B:=augment(A,idn(2)); C:=border(B,[3,4]); D:=augment(C,[[-1,-2,0,0,0]])
Here C=[[-3,2,1,0,3],[1,1,0,1,4]]
and D=[[-3,2,1,0,3],[1,1,0,1,4],[-1,-2,0,0,0]]
Input :
simplex_reduce(D)
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).

• The first problem is to find an x in the Ax = b, x 0 domain. Let m be the number of lines of A. Add artificial variables y1,..., ym and maximize - yi under the conditions Ax = b, x 0, y 0 starting with initial value 0 for x variables and y = b (to solve this with Xcas, call simplex_reduce with 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).
• Now we make a second call to simplex_reduce with the original c and the value of x we found in the domain.
• Example : find the minimum of 2x + 3y - z + t with x, y, z, t 0 and :  This is equivalent to find the opposite of the maximum of - (2x + 3y - z + t). Let us add two artificial variables y1 and y2,
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   Columns 2 and 4 are the columns of the identity (in lines 1 and 2). Hence x = (0, 1, 0, 2) is an initial point in the domain. We are reduced to solve the initial problem, after replacing the lines of Ax = b by the two first lines of the answer above, removing the last columns corresponding to the artificial variables. We add c.x as last line
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.

For more details, search google for simplex algorithm.     suivant: Different matrix norm monter: Linear Programmation précédent: Linear Programmation   Table des matières   Index
giac documentation written by Renée De Graeve and Bernard Parisse