Typical Linear Programming Problem

This example solves the typical linear programming problem

$\min_{x} f^{T} x s u c h t h a t {\begin{array}{cccccccccccccccccccc} A \cdot x \leq b, \\ A e q \cdot x = b e q, \\ x \geq 0 . \end{array}$

Load the sc50b.mat file, which is available when you run this example and contains the matrices and vectors A, Aeq, b, beq, f, and the lower bounds lb.

load sc50b

The problem has 48 variables, 30 inequalities, and 20 equalities.

disp(size(A))

    30    48

disp(size(Aeq))

    20    48

Set options to use the dual-simplex algorithm and the iterative display.

options = optimoptions(@linprog,Algorithm="dual-simplex",Display="iter");

The problem has no upper bound, so set ub to [].

ub = [];

Solve the problem by calling linprog.

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

Running HiGHS 1.11.0: Copyright (c) 2025 HiGHS under MIT licence terms
LP   has 50 rows; 48 cols; 118 nonzeros
Coefficient ranges:
  Matrix [3e-01, 3e+00]
  Cost   [1e+00, 1e+00]
  Bound  [0e+00, 0e+00]
  RHS    [3e+02, 3e+02]
Presolving model
37 rows, 37 cols, 93 nonzeros  0s
21 rows, 21 cols, 63 nonzeros  0s
16 rows, 16 cols, 58 nonzeros  0s
Dependent equations search running on 2 equations with time limit of 1000.00s
Dependent equations search removed 0 rows and 0 nonzeros in 0.00s (limit = 1000.00s)
14 rows, 14 cols, 63 nonzeros  0s
Presolve : Reductions: rows 14(-36); columns 14(-34); elements 63(-55)
Solving the presolved LP
Using EKK dual simplex solver - serial
  Iteration        Objective     Infeasibilities num(sum)
          0    -8.6188182450e-01 Ph1: 10(13.6343); Du: 1(0.861882) 0s
         17    -7.0000000000e+01 Pr: 0(0) 0s
Solving the original LP from the solution after postsolve
Model status        : Optimal
Simplex   iterations: 17
Objective value     : -7.0000000000e+01
P-D objective error :  0.0000000000e+00
HiGHS run time      :          0.00

Optimal solution found.

Examine the exit flag, objective function value at the solution, and number of iterations used by linprog to solve the problem.

exitflag,fval,output.iterations

exitflag = 
1

fval = 
-70.0000

ans = 
17

You can also find the objective function value and number of iterations in the iterative display.