This example was found on linkedIn SlideShare **Page 9**:

As part of your weight training regimen, you want to

consume lean sources of protein. You want to consume

at least **300 Calories a day from at least 48**

grams of protein. One ounce of **chicken provides 35 Calories**

and 8.5 g of protein. One ounce of **tofu provides 20 Calories**

and 2.5 g protein. Your local supermarket charges $5 a pound

for chicken and $2.5 a pound for tofu. How much of each food

should you eat each day if you want to **meet your requierments**

with the lowest cost? What is this **daily cost?**

c = number of lbs of chicken

t = number of lbs of tofu

**optimisation equation: P(Price Paid)**

P = 5 * c + 2.5 * t

**subject to: (16 ounces = 1 lb)**

c >= 0

t >= 0

560 * c + 320 * t >= 300 //1 lb chicken = 16 * 35 = 560 calories

//1 lb tofu = 16 * 20 = 320 calories

136 * c + 40 * t >= 48 //1 lb chicken = 16 * 8.5 = 136 g of protein

//1 lb chicken = 16 * 2.5 = 40 g of protein

**Also written as:**

p = 560 * c + 320 * t

q = 136 * c + 40 * t

300 <= p <= +inf

0 <= x1 < +inf

48 <= q <= +inf

0 <= x2 < +inf

// Philipp Siedler
// Linear Programming
// Real Life Example 4
#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_MIN); //calls the routine glp_set_obj_dir to set the
//omptimization direction flag,
//where GLP_MAX means maximization
//ROWS
glp_add_rows(lp, 2); //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_LO, 300.0, INFINITY); //sets the type and bounds of the first row,
//where GLP_LO means that the row has an lower bound.
//300 <= p <= +inf
//row 2
glp_set_row_name(lp, 2, "q"); //assigns name q to second row
glp_set_row_bnds(lp, 2, GLP_LO, 48.0, INFINITY);//48 <= q <= +inf
//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, 5.0); //sets the objective coefficient for thr first column
//P = 5 * c + 2.5 * t
//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, 2.5); //sets the objective coefficient for thr second column
/*
p = 560 * c + 320 * t
q = 136 * c + 40 * t
*/
row_index[1] = 1, col_index[1] = 1, value[1] = 560.0; // a[1,1] = 560.0
row_index[2] = 1, col_index[2] = 2, value[2] = 320.0; // a[1,2] = 320.0
row_index[3] = 2, col_index[3] = 1, value[3] = 136.0; // a[2,1] = 136.0
row_index[4] = 2, col_index[4] = 2, value[4] = 40.0; // a[2,2] = 40.0
for (int i = 1; i < 5; i++) {
cout << value[i];
cout << ((i % 2 == 0) ? "\n" : "\t");
}
glp_load_matrix(lp, 4, 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("\nPrize(z) = %g; chicken(x1) = %g; tofu(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:**
560 320
136 40
GLPK Simplex Optimizer, v4.63
2 rows, 2 columns, 4 non-zeros
0: obj = 0.000000000e+00 inf = 3.480e+02 (2)
2: obj = 2.678571429e+00 inf = 0.000e+00 (0)
* 3: obj = 2.443181818e+00 inf = 0.000e+00 (0)
OPTIMAL LP SOLUTION FOUND
Prize(z) = 2.44318; chicken(x1) = 0.159091; tofu(x2) = 0.659091;
Please enter a character to exit