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

for babysitting

**at least 3**

hours but no more than 8 hours a week tutoring.

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