Answer to Question #169968 in Operations Research for mara

Question

The decision variables represent the amounts of ingredients 1, 2, and 3 to put into a blend. The objective function represents profit. The first three constraints measure the usage and availability of resources A, B, and C. The fourth constraint is a minimum requirement for ingredient 3.

 

LP Problem

 

Maximize 4x₁ + 6x₂ + 7x₃

s.t.

3x₁+2x₂+5x₃ ≤ 120

  x₁+3x₂+3x₃ ≤  80

5x₁+5x₂+8x₃ ≤ 160

    x₃ ≥  10

         x₁, x₂ , x₃ ≥ 0

 

Answer the following questions.

a. How much of ingredient 1 will be put into the blend?

b. How much of ingredient 2 will be put into the blend?

c. How much of ingredient 3 will be put into the blend?

d. How much resource A is used?

e. How much resource B will be left unused?

f. What will the profit be?

g. What will happen to the solution if the profit from ingredient 2 drops to 4?

h. What will happen to the solution if the profit from ingredient 3 increases by 1?

i. What will happen to the solution if the amount of resource C increases by 2?

j. What will happen to the solution if the minimum requirement for ingredient 3 increases to 15?

Answer 1

Using Simplex method online calculator (https://cbom.atozmath.com/CBOM/Simplex.aspx?q=sm), we get:

a) $x_1=0$

b) $x_2=16$

c) $x_3=10$

d) $2\cdot16+3\cdot10=62$

e) $80-(3\cdot16+3\cdot10)=2$

f) Profit: $6\cdot16+7\cdot10=166$

g) There is change in the objective function:

$4x_1+4x_2+7x_3$

Then:

$x_1=x_2=0,x_3=20,z=140$

h) Change in the objective function:

$4x_1+4x_2+8x_3$

Result:

$x_1=0,x_2=16,x_3=10,z=176$

i) Change in the 3rd constraint:

$5x_1+5x_2+8x_3 ≤ 162$

Result:

$x_1=0,x_2=16.4,x_3=10,z=168.4$

j) Change in the 4rd constraint:

$x_3 ≥ 15$

Then:

$x_1=0,x_2=8,x_3=15,z=153$