This example was found on linkedIn SlideShare Page 7:
Kayla works no more than 20 hours per week
during the school year. She is paid $10 an hour
for tutoring Geometry students and $7 an hour
for babysitting. She wants to spend at least 3
hours but no more than 8 hours a week tutoring.
Find Kayla’s maximum weekly earnings.
t = number of hours spent tutoring
b = number of hours spent babysitting
Optimisation equation: E(Earnings)
E = 10 * t + 7 * b
Subject to:
t >= 3
t <= 8 b >= 0
t + b <= 20
Also written as:
p = t
q = b
r = t + b
3 <= p <= 8
0 <= x1 < +inf
0 < q <= +inf
0 <= x2 < +inf
-inf < r <= 20
0 <= x2 < +inf
// Philipp Siedler
// Linear Programming
// Real Life Example 3
#include <stdio.h>
#include <stdlib.h>
#include <glpk.h>
#include <std_lib_facilities.h>
int main(void)
{
glp_prob *lp;
int row_index[1 + 1000]; //Row indices of each element
int col_index[1 + 1000]; //column indices of each element
double value[1 + 1000]; //numerical values of corresponding elements
double z, x1, x2, x3;
lp = glp_create_prob(); //creates a problem object
glp_set_prob_name(lp, "sample"); //assigns a symbolic name to the problem object
glp_set_obj_dir(lp, GLP_MAX); //calls the routine glp_set_obj_dir to set the
//omptimization direction flag,
//where GLP_MAX means maximization
//ROWS
glp_add_rows(lp, 3); //adds three rows to the problem object
//row 1
glp_set_row_name(lp, 1, "p"); //assigns name p to first row
glp_set_row_bnds(lp, 1, GLP_UP, 3.0, 8.0); //sets the type and bounds of the first row,
//where GLP_UP means that the row has an upper bound.
//3 <= p <= 8
//row 2
glp_set_row_name(lp, 2, "q"); //assigns name q to second row
glp_set_row_bnds(lp, 2, GLP_UP, 0.0, INFINITY);//0 < q <= +inf
//row 3
glp_set_row_name(lp, 3, "r"); //assigns name r to second row
glp_set_row_bnds(lp, 3, GLP_UP, 0.0, 20.0);//-inf < r <= 20
//COLUMNS
glp_add_cols(lp, 2); //adds three columns to the problem object
//column 1
glp_set_col_name(lp, 1, "x1"); //assigns name x1 to first column
glp_set_col_bnds(lp, 1, GLP_LO, 0.0, 0.0); //sets the type and bounds to the first column,
//where GLP_LO means that the column has an lower bound
glp_set_obj_coef(lp, 1, 10.0); //sets the objective coefficient for thr first column
//E = 10 * t + 7 * b
//column 2
glp_set_col_name(lp, 2, "x2"); //assigns name x2 to first column
glp_set_col_bnds(lp, 2, GLP_LO, 0.0, 0.0); //sets the type and bounds to the second column
glp_set_obj_coef(lp, 2, 7.0); //sets the objective coefficient for thr second column
/*
E = 10 * t + 7 * b
p = t
q = b
r = t + b
*/
row_index[1] = 1, col_index[1] = 1, value[1] = 1.0; // a[1,1] = 1.0
row_index[2] = 1, col_index[2] = 2, value[2] = 0.0; // a[1,2] = 0.0
row_index[3] = 2, col_index[3] = 1, value[3] = 0.0; // a[2,1] = 0.0
row_index[4] = 2, col_index[4] = 2, value[4] = 1.0; // a[2,2] = 1.0
row_index[5] = 3, col_index[5] = 1, value[5] = 1.0; // a[3,1] = 1.0
row_index[6] = 3, col_index[6] = 2, value[6] = 1.0; // a[3,2] = 1.0
for (int i = 1; i < 7; i++) {
cout << value[i];
cout << ((i % 2 == 0) ? "\n" : "\t");
}
glp_load_matrix(lp, 6, row_index, col_index, value); //calls the routine glp_load_matrix
//loads information from three arrays
//into the problem object
glp_simplex(lp, NULL); //calls the routine glp_simplex
//to solve LP problem
z = glp_get_obj_val(lp); //obtains a computed value of the objective function
x1 = glp_get_col_prim(lp, 1); //obtain computed values of structural variables (columns)
x2 = glp_get_col_prim(lp, 2); //obtain computed values of structural variables (columns)
printf("\nEarnings(z) = %g; tutoring(x1) = %g; babysitting(x2) = %g;\n", z, x1, x2); //writes out the optimal solution
glp_delete_prob(lp); //calls the routine glp_delete_prob, which frees all the memory
keep_window_open(); //leave this line out and run program with ctrl + F5 to keep window open
//that means you won't need std_lib_facilities.h
}
Output: 1 0 0 1 1 1 GLPK Simplex Optimizer, v4.63 3 rows, 2 columns, 4 non-zeros * 0: obj = -nan(ind) inf = 0.000e+00 (2) * 2: obj = -nan(ind) inf = 0.000e+00 (0) OPTIMAL LP SOLUTION FOUND Earnings(z) = 164; tutoring(x1) = 8; babysitting(x2) = 12; Please enter a character to exit
