I finished the entry about 'lpsolve' for cascmd_en manual. Below is the TeX file containing the updated section "Linear Programmation". I also added the label on line 47 (subsection: simplex_algorithm) to make a reference to an existing example. The new subsection "Solving general linear programming problems" is appended to the section.
Code:
\section{Linear Programmation}\index{simplex\_reduce|textbf}
Linear programming problems are maximization problem of a linear
functional under linear equality or inequality constraints.
The most simple case can be solved directly by the so-called simplex
algorithm. Most cases requires to solve an auxiliary linear
programming problem to find an initial vertex for the simplex
algorithm.
\subsection{Simplex algorithm: {\tt simplex\_reduce}}
{\bf The simple case}\\
The function {\tt simplex\_reduce} makes the reduction
by the simplex algorithm to find :
\[ \mbox{max}(c.x), \quad A.x \leq b,\ x \geq 0,\ b\geq 0 \]
where $c,x$ are vectors of $\mathbb R^n$, $b\geq 0$ is a vector of
$\mathbb R^p$ and $A$ is a matrix of $p$ rows and $n$ columns.\\
{\tt simplex\_reduce} takes as argument {\tt A,b,c} et
returns {\tt max(c.x)}, the augmented solution of {\tt x}
(augmented since the algorithm works by adding rows($A$) auxiliary
variables) and the reduced matrix.\\
{\bf Example}\\
Find \[ \mbox{max}(X+2Y) \mbox{ where }
\left\{
\begin{array}{rcl}
(X,Y) & \geq & 0 \\
-3X +2Y & \leq & 3\\
X +Y & \leq & 4
\end{array}
\right.
\]
Input :
\begin{center}{\tt simplex\_reduce([[-3,2],[1,1]],[3,4],[1,2])}\end{center}
Output :
\begin{center}{\tt 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]]}\end{center}
Which means that the maximum of {\tt X+2Y} under these conditions
is {\tt 7}, it is obtained for {\tt X=1,Y=3}
because {\tt [1,3,0,0]} is the augmented solution and the reduced matrix is :\\
{\tt [[0,1,1/5,3/5,3],[1,0,(-1)/5,2/5,1], [0,0,1/5,8/5,7]]}.
{\bf A more complicate case that reduces to the simple case}\\
With the former call of {\tt simplex\_reduce}, we have to :
\begin{itemize}
\item rewrite constraints to the form $x_k \geq 0$,
\item remove variables without constraints,
\item add variables such that all the constraints have positive components.
\end{itemize}
For example, find :
\begin{equation}\label{eq:lpexample}
\mbox{min}(2x+y-z+4) \quad \mbox{ where }
\left\{
\begin{array}{rcl}
x & \leq & 1 \\
y & \geq & 2 \\
x+3y-z & = & 2 \\
2x-y+z & \leq & 8\\
-x+y & \leq & 5
\end{array}
\right.
\end{equation}
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 :
\[
\left\{
\begin{array}{rcl}
X & \geq & 0 \\
Y & \geq & 0 \\
2(1-X)-(Y+2)+ 5-X+3Y & \leq & 8\\
-(1-X) +(Y+2) & \leq & 5
\end{array}
\right.
\]
or to find the minimum of~:
\[ (-X-2Y+3) \quad \mbox{ where }
\left\{
\begin{array}{rcl}
X & \geq & 0 \\
Y & \geq & 0 \\
-3X+2Y & \leq & 3\\
X +Y & \leq & 4
\end{array}
\right.
\]
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 {\tt 7},
hence, the result here is {\tt 7-3=4}.
{\bf 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,
{\tt 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 {\tt A} has $p$ rows and $n$ columns and if we define :
\begin{center}
{\tt B:=augment(A,idn(p));} {\tt C:=border(B,b);} \\
{\tt d:=append(-c,0\$(p+1));} {\tt D:=augment(C,[d]);}
\end{center}
{\tt simplex\_reduce} may be called with {\tt D} as single argument.\\
For the previous example, input :
\begin{center}{\tt A:=[[-3,2],[1,1]];B:=augment(A,idn(2)); C:=border(B,[3,4]);
D:=augment(C,[[-1,-2,0,0,0]])}\end{center}
Here
{\tt C=[[-3,2,1,0,3],[1,1,0,1,4]]}\\
and {\tt D=[[-3,2,1,0,3],[1,1,0,1,4],[-1,-2,0,0,0]]}\\
Input :
\begin{center}{\tt simplex\_reduce(D)}\end{center}
Output is the same result as before.
{\bf 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 $Ax\leq b$)
under the conditions $x\geq 0$. We may further assume that $b\geq 0$
(if not, one can change the sign of the corresponding line).
\begin{itemize}
\item The first problem is to find an $x$ in the $Ax=b, x\geq 0$ domain.
Let $m$ be the number of lines of $A$. Add artificial variables
$y_1,...,y_m$ and maximize
$-\sum y_i$ under the conditions $Ax=b, x \geq 0, y \geq 0$
starting with initial value $0$ for $x$ variables
and $y=b$
(to solve this with {\tt Xcas}, call \verb|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).
\item Now we make a second call to \verb|simplex_reduce|
with the original $c$ and the value of $x$ we found in the domain.
\item
Example~: find the minimum of $2x+3y-z+t$ with
$x,y,z,t\geq 0$ and~:
\[ \left\{ \begin{array}{rcl}
-x-y+t&=&1\\
y-z+t&=&3
\end{array}
\right. \]
This is equivalent to find the opposite of the maximum of $-(2x+3y-z+t)$.
Let us add two artificial variables $y_1$ and $y_2$,
\begin{verbatim}
simplex_reduce([[-1,-1,0,1,1,0,1],
[0,1,-1,1,0,1,3],
[0,0,0,0,1,1,0]])
\end{verbatim}
Output: optimum=0, artificial variables=0, and the matrix
\[
\left(\begin{array}{ccccccc}
-1/2 & 0 & -1/2 & 1 & 1/2 & 1/2 & 2 \\
1/2 & 1 & -1/2 & 0 & -1/2 & 1/2 & 1 \\
0 & 0 & 0 & 0 & 1 & 1 & 0
\end{array}\right)
\]
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
\begin{verbatim}
simplex_reduce([[-1/2,0,-1/2,1,2],
[1/2,1,-1/2,0,1],[2,3,-1,1,0]])
\end{verbatim}
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 $t=2$.
\end{itemize}
For more details, search google for \verb|simplex algorithm|.
\subsection{Solving general linear programming problems: {\tt lpsolve}}\index{lpsolve|textbf}
Linear programming problems where a multivariate linear function should be maximized or minimized subject to linear equality or inequality constraints, as well as mixed integer problems, can be solved with the function {\tt lpsolve}. It takes (at most) four arguments, in the following order :
\begin{itemize}
\item {\tt obj} : symbolic expression representing the objective function,
\item {\tt constr} (optional) : list of linear constraints which may be equalities or inequalities or bounded expressions entered as {\tt expr=a..b},
\item {\tt bd} (optional) : sequence of expressions of type {\tt var=a..b} specifying that the variable {\tt var} is bounded with {\tt a} below and with {\tt b} above,
\item {\tt opts} (optional) : sequence of {\tt option=value} parameters for the solver, where {\tt option} may be one of {\tt assume}, {\tt lp\_maximize}, {\tt lp\_variables}, {\tt lp\_integervariables}, {\tt lp\_binaryvariables} or {\tt lp\_depthlimit}.
\end{itemize}
The return value is in the form {\tt [optimum,soln]} where {\tt optimum} is the minimum/maximum value of the objective function and {\tt soln} is a list of coordinates corresponding to the point at which the optimal value is attained. If there is no feasible solution, an empty list is returned. When the objective function is unbounded, {\tt optimum} is returned as {\tt +infinity} (for maximization problems) or {\tt -infinity} (for minimization problems), while {\tt soln} is generally meaningless.
If {\tt obj} is given as constant (for example, zero) then only a feasible point is returned as a list, if one exists. If the problem is infeasible, an empty list is returned. This may be used as a test to check whether a set of linear constraints is feasible or not.
The given objective function is minimized by default. To maximize it, include the option {\tt lp\_maximize=true} or simply {\tt lp\_maximize}.
By default, all variables are considered continuous and unrestricted in sign.
The solver combines the two-phase simplex method and the dual simplex method to find the optimal solution.
\subsubsection{Problems with continuous variables}
For example, to solve the problem specified in~\eqref{eq:lpexample}, input :
\begin{center}
{\tt constr:=[x<=1,y>=2,x+3y-z=2,3x-y+z<=8,-x+y<=5];}\\
{\tt lpsolve(2x+y-z+4,constr)}
\end{center}
Output :
\begin{center}
\tt [-4,[x=0,y=5,z=13]]
\end{center}
Therefore, the minimum value of $ f(x,y,z)=2\,x+y-z+4 $ is equal to $ -4 $ under the given constraints. The optimal value is attained at point $ (x,y,z)=(0,5,13) $.
Constraints may also take the form {\tt expr=a..b} for bounded linear expressions.
\noindent Input :
\begin{center}
{\tt lpsolve(x+2y+3z,[x+y=1..5,y+z+1=2..4,x>=0,y>=0])}
\end{center}
Output :
\begin{center}
{\tt [-2,[x=0,y=5,z=-4]]}
\end{center}
Use the {\tt assume=lp\_nonnegative} option to specify that all variables are nonnegative. That is easier than entering the nonnegativity constraints explicitly.
\noindent Input:
\begin{center}
{\tt lpsolve(-x-y,[y<=3x+1/2,y<=-5x+2],}\\
{\tt assume=lp\_nonnegative)}
\end{center}
Output:
\begin{center}
{\tt [-5/4,[x=3/16,y=17/16]]}
\end{center}
Bounds can be added separately for some variables. They are entered after the list of constraints.
\noindent Input :
\begin{center}
{\tt constr:=[5x-10y<=20,2z-3y=6,-x+3y<=3];}\\
{\tt lpsolve(-6x+4y+z,constr,x=1..20,y=0..inf)}
\end{center}
Output :
\begin{center}
{\tt [-133/2,[x=18,y=7,z=27/2]]}
\end{center}
\subsubsection{Integer programming}
Use the {\tt assume=integer} or {\tt assume=lp\_integer} option to solve integer programming problems. The function {\tt lpsolve} uses branch and bound method in such cases. The total number of investigated nodes is printed out before the function returns. To limit branch and bound tree depth, use the option : \begin{center}
{\tt lp\_depthlimit=<positive integer>}
\end{center} In that case optimality of the solution can't be guaranteed.
\noindent Input :
\begin{center}
{\tt lpsolve(-5x-7y,[7x+y<=35,-x+3y<=6],assume=integer)}
\end{center}
Output :
\begin{center}
{\tt [-41,[x=4,y=3]]}
\end{center}
Use the option {\tt assume=lp\_binary} to specify that all variables are binary, i.e.~the only allowed values are 0 and 1. Binary variables usually represent {\tt true} and {\tt false} values, giving them a certain meaning in logical context.
\noindent Input :
\begin{center}
{\tt lpsolve(8x1+11x2+6x3+4x4,[5x1+7x2+4x3+3x4<=14],}\\
{\tt assume=lp\_binary,lp\_maximize)}
\end{center}
Output :
\begin{center}
{\tt [21,[x1=0,x2=1,x3=1,x4=1]]}
\end{center}
Options \begin{center}
{\tt lp\_integervariables=<list of integer variables>}
\end{center} and \begin{center}
{\tt lp\_binaryvariables=<list of binary variables>}
\end{center} are used for specifying mixed integer/binary programming problems. Also, the \begin{center}
{\tt lp\_variables=<list of continuous variables>}
\end{center} option may be used, which overrides integer and binary settings. For example, a linear programming problem with mostly integer variables may be specified using the option {\tt assume=integer} and specifying continuous variables with {\tt lp\_variables}, which overrides the global integer setting.
\noindent Input :
\begin{center}
{\tt lpsolve(x+3y+3z,[x+3y+2z<=7,2x+2y+z<=11],}\\
{\tt assume=lp\_nonnegative,lp\_maximize,}
{\tt lp\_integervariables=[x,z])}
\end{center}
Output:
\begin{center}
{\tt [10,[x=0,y=1/3,z=3]]}
\end{center}
Use the {\tt assume=lp\_nonnegint} or {\tt assume=nonnegint} option to get nonnegative integer values.
\noindent Input :
\begin{center}
{\tt lpsolve(2x+5y,[3x-y=1,x-y<=5],assume=nonnegint)}
\end{center}
Output :
\begin{center}
{\tt [12,[x=1,y=2]]}
\end{center}
\subsubsection{Entering linear programs in matrix form}
The function {\tt lpsolve} supports entering linear programming problems in matrix form, which is convenient for problems with relatively large number of variables and/or constraints.
To enter a problem in matrix form, the parameter {\tt obj} must be a vector of coefficients $ \mathbf{c} $ and {\tt constr}, which is mandatory this case, must be a list $ [\mathbf{A},\mathbf{b},\mathbf{A}_\mathrm{eq},\mathbf{b}_\mathrm{eq}] $, where $ \mathbf{A},\mathbf{A}_\mathrm{eq} $ are matrices and $ \mathbf{b},\mathbf{b}_\mathrm{eq} $ are vectors of real numbers. Without any other parameters, this minimizes $ \mathbf{c}^T\,\mathbf{x} $ under conditions $ \mathbf{A}\,\mathbf{x}\leq\mathbf{b} $ and $ \mathbf{A}_\mathrm{eq}\,\mathbf{x}=\mathbf{b}_\mathrm{eq} $. If a problem does not contain equality constraints, parameters $ \mathbf{A}_\mathrm{eq} $ and $ \mathbf{b}_\mathrm{eq} $ may be omitted. For a problem that does not contain inequality constraints, empty lists must be passed as $ \mathbf{A} $ and $ \mathbf{b} $. Also, {\tt constr} may be an empty list.
The parameter {\tt bd} is entered as a list of two vectors $ \mathbf{b}_l $ and $ \mathbf{b}_u $ of the same length as the vector $ \mathbf{c} $ such that $ \mathbf{b}_l\leq\mathbf{x}\leq\mathbf{b}_u $. For unbounded variables use {\tt +infinity} or {\tt -infinity}.
When specifying mixed problems in matrix form, variables are entered as the corresponding indexes, which are 1-based, i.e.~the first variable has index 1, the second variable has index 2 and so on. Other options for {\tt lpsolve} are passed to a problem in matrix form in the same way as if it was in symbolic form.
\noindent Input :
\begin{center}
{\tt c:=[-2,1];A:=[[-1,1],[1,1],[-1,0],[0,-1]];}\\
{\tt b:=[3,5,0,0];lpsolve(c,[A,b])}
\end{center}
Output :
\begin{center}
{\tt [-10,[5,0]]}
\end{center}
Input :
\begin{center}
{\tt c:=[-2,5,-3];bl:=[2,3,1];bu:=[6,10,3.5];}\\
{\tt lpsolve(c,[],[bl,bu])}
\end{center}
Output :
\begin{center}
{\tt [-7.5,[6.0,3.0,3.5]]}
\end{center}
Input :
\begin{center}
{\tt c:=[4,5];Aeq:=[[-1,1.5],[-3,2]];beq:=[2,3];}\\
{\tt lpsolve(c,[[],[],Aeq,beq])}
\end{center}
Output :
\begin{center}
{\tt [5.2,[-0.2,1.2]]}
\end{center}
Input :
\begin{center}
{\tt c:=[2,-3,-5];A:=[[-5,4,-5],[2,5,7],[2,-3,4]];}\\
{\tt b:=[3,1,-2];lpsolve(c,[A,b],lp\_integervariables=[1,3])}
\end{center}
Output :
\begin{center}
{\tt [19,[1,3/4,-1]]}
\end{center}