Answer to Question #170740 in Operations Research for ERIC KOOMSON

Question

Daily requirements of 70 g of protein, 1 g calcium, 12 mg iron, and 3000 calories are needed for a balanced diet. The following foods are available for consumption with the cost and nutrients per 100 g as shown.

 

Protein

(g)

Calories

Calcium

(g)

Iron

Cost

GH¢

Brown Bread

12

246

0.1

3.2

0.5

Cheese

24.9

423

0.2

0.3

2

Butter

0.1

793

0.03

0

1

Baked Beans

6

93

0.05

2.3

0.25

Spinach

3

26

0.1

2

0.25

 

The objective is to find a balanced diet with minimum cost.

(a) Formulate a linear programming model for this problem.

(b) Use solver to find optimal solution and sensitivity report.     

Answer 1

a) The objective is to find a balanced diet with minimum cost.

Let:

$x_1$ - number of 100 g units of brown bread

$x_2$ - number of 100 g units of cheese

$x_3$ - number of 100 g units of butter

$x_4$ - number of 100 g units of baked beans

$x_5$ - number of 100 g units of baked beans

The linear programming problem is then:

Minimize: $0.5x_1 + 2x_2 + x_3 + 0.25x_4 + 0.25x_5$

Constraints:

$12x_1 + 24.9x_2 + 0.1x_3 + 6.0x_4 + 3.0x_5 \geq 70$

$246x_1 + 423x_2 + 793x_3 + 93x_4 + 26x_5 \geq 3000$

$0.1x_1 + 0.2x_2 + 0.03x_3 + 0.05x_4 + 0.1x_5 \geq 1$

$3.2x_1 + 0.3x_2 + 2.3x_4 + 2.0x_5 \geq 12$

$x_1,x_2,x_3,x_4,x_5\geq0$

b) Using online solver (https://cbom.atozmath.com), we get:

It was used Two-Phase Simplex method.

Phase 1:

after 1st step: maz $z_j-c_j=793.13$

after 2nd step: max $z_j-c_j=25.33$

after 3rd step: max $z_j-c_j=3.06$

after 4th step: max $z_j-c_j=0.075$

after 5th step: all $z_j-c_j\leq0$

optimal solution:

$min\ z=0,x_1=0.56,x_2=1.95,x_3=2.41,x_4=0,x_5=4.81$

Phase 1:

we eliminate the artificial variables and change the objective function for the original,

after 1st step: max $z_j-c_j=0.17$

after 2nd step: max $z_j-c_j=0.001$

after 3rd step: all $z_j-c_j\leq0$

Finally, optimal solution:

$z=\$5.64$

$x_1=9.77, x_3=0.75,x_2=x_4=x_5=0$

Conclusion:

To get minimal cost ($\$5.64$ ) of diet, it's enough to use $977\ g$ of brown bread and $75\ g$ of butter. And we don't need any cheese, baked beans or spinach.