Matlab linear programming optimization algorithm (1)

In Matlab solution form linear programming of the formula:

wherein the object comprises optimizing f '* x, inequality constraints, equality constraints and bounds constraints variables.
Providing linprog function in Matlab solving linear optimization:
EG:
[X, FVAL, exitflag, Output, the lambda] = linprog (f, A, B, Aeq, BEQ, LB, UB, Options)
enter the f function, optimization of the object that is f, a, b, Aeq, beq, lb, ub are specifically expressed in the above formula. The last options is to optimize process parameters, mainly include:

options = optimoptions(‘linprog’,‘Algorithm’,‘interior-point’,‘Display’,‘iter’,‘MaxIterations’,10)

Selection optimization: linear optimization
selection method: the interior point method (linear optimization algorithms provide further)
: Displays the cycle number
limit cycles

There are other parameters can be set: specific search Optimization Options Matlab reference obtained in Reference.
For example:
we solve the following linear programming problem:

f = [-2;-1;1];

A = [1 4 -1; 2 -2 1];

b = [4;12];

Aeq = [1 1 2];
beq = 6;

lb = zeros(3,1);

To give the corresponding optimization parameters set and solved:

options = optimoptions('linprog','Algorithm','dual-simplex','Display','iter','MaxIterations',10)

[x,fval,exitflag,output,lambda]  = linprog(f,A,b,Aeq,beq,lb,ub,options)

We can obtain the following results:

options = 

  linprog options:

   Options used by current Algorithm ('dual-simplex'):
   (Other available algorithms: 'interior-point', 'interior-point-legacy')

   Set properties:
              Algorithm: 'dual-simplex'
                Display: 'iter'
          MaxIterations: 10

   Default properties:
    ConstraintTolerance: 1.0000e-04
                MaxTime: Inf
    OptimalityTolerance: 1.0000e-07


LP preprocessing removed 0 inequalities, 0 equalities,
0 variables, and added 0 non-zero elements.

 Iter      Time            Fval  Primal Infeas    Dual Infeas  
    0     0.001    0.000000e+00   6.999174e+00   1.113300e+00  
    2     0.002   -1.080000e+01   1.750843e+00   0.000000e+00  
    4     0.002   -8.666667e+00   0.000000e+00   0.000000e+00  

Optimal solution found.


x =

    4.6667
         0
    0.6667


fval =

   -8.6667


exitflag =

     1


output = 

  struct with fields:

         iterations: 4
    constrviolation: 8.8818e-16
            message: 'Optimal solution found.'
          algorithm: 'dual-simplex'
      firstorderopt: 7.4015e-16


lambda = 

  struct with fields:

      lower: [3×1 double]
      upper: [3×1 double]
      eqlin: 0.3333
    ineqlin: [2×1 double]

Wherein the parameter comprises a display, data of each step of the cycle, and the final value of the function value of the most advantageous.