   ### 5.47.1  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 ≤ 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.
Example
Find

max(X+2Y)   where

 (X,Y) ≥ 0 −3X +2Y ≤ 3 X +Y ≤ 4

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 complicated 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+yz+4)      where

 x ≤ 1 y ≥ 2 x+3y−z = 2 2x−y+z ≤ 8 −x+y ≤ 5
(5)

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 :

 X ≥ 0 Y ≥ 0 2(1−X)−(Y+2)+ 5−X+3Y ≤ 8 −(1−X) +(Y+2) ≤ 5

or to find the minimum of :

(−X−2Y+3)     where

 X ≥ 0 Y ≥ 0 −3X+2Y ≤ 3 X +Y ≤ 4

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 Axb) 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+3yz+t with x,y,z,t≥ 0 and :

 −x−y+t = 1 y−z+t = 3
This is equivalent to find the opposite of the maximum of −(2x+3yz+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

 −1/2 0 −1/2 1 1/2 1/2 2 1/2 1 −1/2 0 −1/2 1/2 1 0 0 0 0 1 1 0

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`.   