Question #170740

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.     


1
Expert's answer
2021-03-15T19:31:16-0400

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

Let:

x1x_1 - number of 100 g units of brown bread

x2x_2 - number of 100 g units of cheese

x3x_3 - number of 100 g units of butter

x4x_4 - number of 100 g units of baked beans

x5x_5 - number of 100 g units of baked beans

The linear programming problem is then:

Minimize: 0.5x1+2x2+x3+0.25x4+0.25x50.5x_1 + 2x_2 + x_3 + 0.25x_4 + 0.25x_5

Constraints:

12x1+24.9x2+0.1x3+6.0x4+3.0x57012x_1 + 24.9x_2 + 0.1x_3 + 6.0x_4 + 3.0x_5 \geq 70

246x1+423x2+793x3+93x4+26x53000246x_1 + 423x_2 + 793x_3 + 93x_4 + 26x_5 \geq 3000

0.1x1+0.2x2+0.03x3+0.05x4+0.1x510.1x_1 + 0.2x_2 + 0.03x_3 + 0.05x_4 + 0.1x_5 \geq 1

3.2x1+0.3x2+2.3x4+2.0x5123.2x_1 + 0.3x_2 + 2.3x_4 + 2.0x_5 \geq 12

x1,x2,x3,x4,x50x_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 zjcj=793.13z_j-c_j=793.13

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

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

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

after 5th step: all zjcj0z_j-c_j\leq0

optimal solution:

min z=0,x1=0.56,x2=1.95,x3=2.41,x4=0,x5=4.81min\ 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 zjcj=0.17z_j-c_j=0.17

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

after 3rd step: all zjcj0z_j-c_j\leq0


Finally, optimal solution:

z=$5.64z=\$5.64

x1=9.77,x3=0.75,x2=x4=x5=0x_1=9.77, x_3=0.75,x_2=x_4=x_5=0


Conclusion:

To get minimal cost ($5.64\$5.64 ) of diet, it's enough to use 977 g977\ g of brown bread and 75 g75\ g of butter. And we don't need any cheese, baked beans or spinach.


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