Solve the linear programming problem using the simple form method (C ++ implementation code)

Solve the linear programming problem using the simple form method (C ++ implementation code)
1 simple form Method

(1) The simple shape method is an important method to solve the linear programming problem.
The basic framework is as follows:
Step 1: Change LP linear planning to a standard type to determine an Initial Feasible Solution (vertex ).
Step 2: determine the optimum of the initial feasible solution. If the optimal solution is used, stop the process. Otherwise, go to the next step.
Step 3: Change the initial base feasible solution to the adjacent base feasible solution (vertex), and improve the target value-increase the value of the target function, repeat steps 2 and 3 until the optimal solution is found.
(2) Before using a program for computation, convert the target function and the constraint equation into a standard form.

The following transformations must be performed in non-standard forms:
A) when the target function is min z = CX, it is converted to the form of max z =-CX;
B) when there is a constraint equation "≤", add a relaxation variable;
C) when the constraint equation contains "≥", subtract a relaxation variable and add an artificial variable;
D) add an artificial variable when the constraint equation contains "=;
E) for all the manual variables, the target function coefficient of the relaxed variable is set to 0.
(3) Linear Programming in the standard form. The block diagram of the calculation step is calculated using the simple form method.

2 program testing and results:

The linear programming problems are as follows:
Max z = 2 * x1-3 * x2 + 3x3;
X1 + x2-x3 <= 7;
X1-x2 + x3 <=-7;
X1-2 * x2 + 2 * x3 <= 4;
X1, x2, x3> = 0;

3 C ++ implementation code

// Simplex. cpp: defines the entry point of the console application. //// ********************************* --------------------------------- Simple method for Solving Linear Programming (C ++ implementation code) ----------------------------------- Author: Muzhi, Date: 2014 Email: bzhou84@163.com **********************************/# include stdafx. h # include
  
   
# Include
   
    
Using namespace std; # define M 10000 // large global variable M float juzhen [11] [31]; // core matrix table int m = 0, n = 0, t = 0; // m: Number of structure vectors // n: Number of constrained inequalities // t: Target Function Type:-1 indicates minimum value, 1 represents the maximum value void input () // input interface function {int I, j; cout <---------- input the parameter NUMBER OF THE SIMPLE METHOD --------- <
    
     
> M; cout <
     
      
> N; for (I = 0; I <= n + 1; I ++) for (j = 0; j <= m + n; j ++) juzhen [I] [j] = 0; // initialization matrix, all elements are 0 // read constraints cout <
      
        =): <
       
         > Juzhen [I] [j];} for (I = 1; I <= n; I ++) {juzhen [I] [0] = juzhen [I] [m + 2]; juzhen [I] [m + 2] = 0;} // read target condition cout <
        
          > Juzhen [0] [I]; cin> t; // adjust the matrix if (t =-1) for (I = 1; I <= m; I ++) juzhen [0] [I] = (-1) * juzhen [0] [I]; for (I = 1; I <= n; I ++) {juzhen [I] [m + I] = juzhen [I] [m + 1]; if (I! = 1) juzhen [I] [m + 1] = 0 ;}// algorithm function void comput () {int I, j, flag, temp1, temp2, h, k = 0, temp3 [10]; float a, B [11], temp, temp4 [11], temp5 [11], f = 0, aa, d, c; // initialize for (I = 1; I <= n; I ++) temp3 [I] = 0; for (I = 0; I <11; I ++) {temp4 [I] = 0; temp5 [I] = 0;} for (I = 1; I <= n; I ++) {if (juzhen [I] [m + I] =-1) {juzhen [I] [m + n + I] = 1; juzhen [0] [m + n + I] = M; temp3 [I] = m + n + I;} else temp3 [I] = m + I ;} for (I = 1; I <= n; I ++) temp4 [I] = juzhen [0] [temp3 [I]; // cyclically solve do {for (I = 1; I <= m + n; I ++) {a = 0; for (j = 1; j <= n; j ++) a + = juzhen [j] [I] * temp4 [j]; juzhen [n + 1] [I] = juzhen [0] [I]-a;} for (I = 1; I <= m + n; I ++) {if (juzhen [n + 1] [I]> = 0) flag = 1; else {flag =-1; break ;}} if (flag = 1) {for (I = 1; I <= n; I ++) {if (temp3 [I] <= m + n) temp1 = 1; else {temp1 =-1; break ;}/// output result cout <