function [t,inb,otb,x,v,n] = simplex(f,A,b,tol)
%
%  returns the solutions to the linear programming problem
%
%      max f'*x subject to A*x <= b, and x >= 0.
%
%  on entry:
%    f         coefficients of the linear objective function;
%    A         coefficients of the linear inequalities;
%    b         right hand side vector of the constraints;
%    tol       tolerance to decide whether a number is zero.
%
%  on return:
%    t         final tableau;
%    inb       indices to variables in the basis,
%              inb(i) gives column for i-th unit vector;
%    otb       indices to variables outside the basis;
%    x         extended solution vector, the first size(A,2)
%              entries gives values for the original x variables;
%    v         value of the objective function at the solution;
%    n         number of pivoting steps.
%
%  example:
%    f = [3 5]'
%    A = [1 0; 0 2; 3 2]
%    b = [4 12 18]'
%    tol = 1.0e-8;
%    [t,inb,otb,x,v,n] = simplex(f,A,b,tol)
%    f'*x(1:2)'  % check value of f'*x
%    A*x(1:2)'   % verify if x satisfies A*x <= b
%
%  simplex needs initialize.m, entering.m, leaving.m, diagonalize.m,
%                passing.m, and solution.m
%
[t,inb,otb] = initialize(f,A,b);          % initialize tableau
e = entering(t,otb);                      % compute entering variable
n = 0;
while (e ~= 0)                            % as long as not optimal
  l = leaving(t,e,tol);                   % compute leaving variable
  if (l == 0) break; end;                 % solution may be unbounded
  t = diagonalize(t,e,l,tol);             % diagonalize tableau
  [inb,otb] = passing(inb,otb,e,l);       % update the basis
  n = n+1;                                % count #pivoting steps
  e = entering(t,otb);                    % new entering variable
end;
[x,v] = solution(t,inb);                  % compute solution