# Answer in Operations Research for Unique #248702

A pharmacy has determined that a healthy person should receive 70 units of proteins, 100 units of carbohydrates and 20 units of fat daily. If the store carries the six types of health food with their ingredients as shown in the table below, what blend of foods satisfies the requirements at minimum cost to the pharmacy? Make a mathematical model for the given problem

Foods Protein units Carbohydrates units Fat units Cost per unit

A 20 50 4 2

B 30 30 9 3

C 40 20 11 5

D 40 25 10 6

E 45 50 9 8

F 30 20 10 8

minimize cost:

“z=2x_1+3x_2+5x_3+6x_4+8x_5+8x_6”

subject to:

“20x_1+30x_2+40x_3+40x_4+45x_5+30x_6ge70” : amount of protein

“50x_1+30x_2+20x_3+25x_4+50x_5+20x_6ge100” : amount of carbohydrate

“4x_1+9x_2+11x_3+10x_4+9x_5+10x_6ge20” : amount of fat

where x1, x2, x3, x4, x5, x6 are units of 6 foods

solution using Simplex method:

After introducing artificial variables:

Min Z=2x1+3x2+5x3+6x4+8x5+8x6+0S1+0S2+0S3+MA1+MA2+MA3

subject to

20x1+30x2+40x3+40x4+45x5+30x6-S1+A1“ge” 70

50x1+30x2+20x3+25x4+50x5+20x6-S2+A2“ge” 100

4x1+9x2+11x3+10x4+9x5+10x6-S3+A3=20

and

x1,x2,x3,x4,x5,x6,A1,A2,A3≥0

Positive maximum ZjCj is 104M-8 and its column index is 5. So, the entering variable is x5.

Minimum ratio is 1.5556 and its row index is 1. So, the leaving basis variable is A1.

The pivot element is 45.

Entering =x5, Departing =A1, Key Element =45

Positive maximum ZjCj is 27.7778M+1.5556 and its column index is 1. So, the entering variable is x1.

Minimum ratio is 0.8 and its row index is 2. So, the leaving basis variable is A2.

The pivot element is 27.7778.

Entering =x1, Departing =A2, Key Element =27.7778

Positive maximum ZjCj is 4M-1.92 and its column index is 6. So, the entering variable is x6.

Minimum ratio is 1.3636 and its row index is 1. So, the leaving basis variable is x5.

The pivot element is 0.88.

Entering =x6, Departing =x5, Key Element =0.88

Positive maximum ZjCj is 0.3818M-0.3273 and its column index is 7. So, the entering variable is S1.

Minimum ratio is 1.4286 and its row index is 3. So, the leaving basis variable is A3.

The pivot element is 0.3818.

Entering =S1, Departing =A3, Key Element =0.3818

Positive maximum ZjCj is 3.8571 and its column index is 3. So, the entering variable is x3.

Minimum ratio is 1.2766 and its row index is 1. So, the leaving basis variable is x6.

The pivot element is 1.119.

Entering =x3, Departing =x6, Key Element =1.119

Positive maximum ZjCj is 1.1489 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 1.8182 and its row index is 1. So, the leaving basis variable is x3.

The pivot element is 0.7021.

Entering =x2, Departing =x3, Key Element =0.7021

Since all ZjCj≤0

Hence, optimal solution is arrived with value of variables as :

x1=0.9091,x2=1.8182,x3=0,x4=0,x5=0,x6=0

Min Z=7.2727

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