help linprog;
 LINPROG Linear programming.
    X=LINPROG(f,A,b) attempts to solve the linear programming problem:
         
             min f'*x    subject to:   A*x <= b 
              x
 
    X=LINPROG(f,A,b,Aeq,beq) solves the problem above while additionally
    satisfying the equality constraints Aeq*x = beq.
    
    X=LINPROG(f,A,b,Aeq,beq,LB,UB) defines a set of lower and upper
    bounds on the design variables, X, so that the solution is in
    the range LB <= X <= UB.  Use empty matrices for LB and UB
    if no bounds exist. Set LB(i) = -Inf if X(i) is unbounded below; 
    set UB(i) = Inf if X(i) is unbounded above.
 
    X=LINPROG(f,A,b,Aeq,beq,LB,UB,X0) sets the starting point to X0.  This
    option is only available with the active-set algorithm.  The default
    interior point algorithm will ignore any non-empty starting point.
 
    X=LINPROG(f,A,b,Aeq,beq,LB,UB,X0,OPTIONS) minimizes with the default 
    optimization parameters replaced by values in the structure OPTIONS, an 
    argument created with the OPTIMSET function.  See OPTIMSET for details.  
    Options are Display, Diagnostics, TolFun, LargeScale, MaxIter. 
    Currently, only 'final' and 'off' are valid values for the parameter 
    Display when LargeScale is 'off' ('iter' is valid when LargeScale is 'on').
 
    [X,FVAL]=LINPROG(f,A,b) returns the value of the objective function at X:
    FVAL = f'*X.
 
    [X,FVAL,EXITFLAG] = LINPROG(f,A,b) returns an EXITFLAG that describes the
    exit condition of LINPROG. Possible values of EXITFLAG and the corresponding 
    exit conditions are
 
      1  LINPROG converged to a solution X.
      0  Maximum number of iterations reached.
     -2  No feasible point found.
     -3  Problem is unbounded.
     -4  NaN value encountered during execution of algorithm.
     -5  Both primal and dual problems are infeasible.
     -7  Search direction became too small; no further progress can be made. 
 
    [X,FVAL,EXITFLAG,OUTPUT] = LINPROG(f,A,b) returns a structure
    OUTPUT with the number of iterations taken in OUTPUT.iterations, the type
    of algorithm used in OUTPUT.algorithm, the number of conjugate gradient
    iterations (if used) in OUTPUT.cgiterations, and the exit message in
    OUTPUT.message.
 
    [X,FVAL,EXITFLAG,OUTPUT,LAMBDA]=LINPROG(f,A,b) returns the set of 
    Lagrangian multipliers LAMBDA, at the solution: LAMBDA.ineqlin for the 
    linear inequalities A, LAMBDA.eqlin for the linear equalities Aeq, 
    LAMBDA.lower for LB, and LAMBDA.upper for UB.
    
    NOTE: the LargeScale (the default) version of LINPROG uses a primal-dual
          method. Both the primal problem and the dual problem must be feasible 
          for convergence. Infeasibility messages of either the primal or dual, 
          or both, are given as appropriate.  The primal problem in standard 
          form is 
               min f'*x such that A*x = b, x >= 0.
          The dual problem is
               max b'*y such that A'*y + s = f, s >= 0.

    Reference page in Help browser
       <a href="matlab:doc linprog">doc linprog</a>



% 1. defining an optimization problem
% to maximize profit, we would like to determine
% the values of x and y
x = 0:80;
y1 = max(75 - x,0); % constraint on available land
y2 = max((4000-110*x)/30,0); % storage
y3 = max((15000-120*x)/210,0); % expenses
ytop = min([y1; y2; y3]); 
area(x,ytop) % feasible region
% we can solve the optimization problem
% graphically
hold on
[u v] = meshgrid(0:80 ,0:80);
contour(u,v,143*u + 60*v); % plots the profit
% the contour lines give the lines of equal
% profit, the solution to our optimization
% problem occurs at the corner of the feasible
% region where the line of highest profit meets
% 2. using linprog
f = [-143 -60]; % target
A = [120 210; 110 30; 1 1; -1 0; 0 -1];
% the matrix a contains in its three rows
% the original constraints; the last two
% rows indicate that x and y must be >= 0
b = [15000; 4000; 75; 0; 0]; % right hand side
% now we defined the following problem:
%   min f'*x subject to A*x <= b
linprog(f,A,b)
Optimization terminated.

ans =

   21.8750
   53.1250

% the answer gives us the values for x and y
% for which the profit is maximized:
sol = ans;
-f*sol

ans =

  6.3156e+003

A*sol

ans =

  1.0e+004 *

    1.3781
    0.4000
    0.0075
   -0.0022
   -0.0053

format long
b - A*sol

ans =

  1.0e+003 *

   1.21875000063205
   0.00000000000707
   0.00000000000263
   0.02187500000090
   0.05312499999648

diary off