Question

The Beef-up ranch feeds cattle for midwestern farmers and delivers them to processing plants in Topeka,Kansas and Tulsa, Oklahoma. The ranch must determine the amounts of catlle feed to buy so that variouis nutritional requirements are met while minimizing total feed costs. The mixture fed to the cows must contain different levels of four key nutrients and can be made by blending three different feeds. The amount of each nutrient (in ounces) found in each pound of feed is summarized as follows:
Nutrient a- feed 1 (3) feed 2 (2) feed 3 (4)

b- feed 1 (3) feed 2 (1) feed 3 (3)

c- feed 1 ( 1) feed 2 (0) feed 3 (2)

d- feed 1 (6) feed 2 (8) feed 3 (4)

The cost per pound of feeds 1,2, and 3 are $2.00, $2.50, and $3.00, respectively. The minimum requirement per cow each month is 4 pounds of nutrient A, 5 pounds of nutrient B, 1 pound of nutrient C, and 8 pounds of nutrient D. However, cows should not be fed more than twice the minimum requirement for any nutrient each month. Additionally, the ranch can only obtain 1,500 pounds of each type of feed each month. Because there aree usually 100 cows at the beef-up ranch at any given time, this means that no more than 15 pounds of each type of feed can be used per cow each month.

a. Formulate a linear programming problem to determine how much of each type of feed a cow should be fed each month.

b. Create a spreadsheet model for this problem, and solve it using Solver

c. what is the optimal solution?

Answer 1

Answer:

a. Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

3x₁ + 2x₂ + 4x₃ ≥ 64            ..........(5)

3x₁ + x₂ + 3x₃ ≥ 80              ..........(6)

x₁ + 0x₂ + 2x₃ ≥ 16              ...........(7)

6x₁ + 8x₂ + 4x₃ ≥ 128         ............(8)

x₁ ≤ 15                                 ...........(9)

x₂≤ 15                                  ...........(10)

x₃ ≤ 15                                 ...........(11)

x₁ , x₂, x₃ ≥ 0

b. x₁ = 15 ,  x₂ = 9.5 , x₃ = 8.5

c. The optimal solution is Z = 79.25

Explanation:

Given - The table is as follows :

• Nutrient           Feed 1                    Feed 2                      Feed 3
•    A                      3                              2                                4
•    B                      3                               1                                 3
•    C                      1                                0                                2
•    D                      6                               8                                4

The minimum requirement per cow each month is 4 pounds of nutrient A, 5 pounds of nutrient B, 1 pound of nutrient C, and 8 pounds of nutrient D. However, cows should not be fed more than twice the minimum requirement for any nutrient each month. Additionally, the ranch can only obtain 1,500 pounds of each type of feed each month. Because there are usually 100 cows at the beef-up ranch at any given time, this means that no more than 15 pounds of each type of feed can be used per cow each month.

To find - a. Formulate a linear programming problem to determine how  

                  much of each type of feed a cow should be fed each month.

              b. Create a spreadsheet model for this problem, and solve it using

                   Solver.

              c. What is the optimal solution?

Proof -

a.

Let feed 1 per cow per month = x₁

     feed 2 per cow per month = x₂

     feed 3 per cow per month = x₃

Now,

As given, The cost per pound of feeds 1,2, and 3 are $2.00, $2.50, and $3.00, respectively.

So, we have to minimize the cost , Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 4(32)

3x₁ + x₂ + 3x₃ ≤ 5(32)

x₁ + 0x₂ + 2x₃ ≤ 1(32)

6x₁ + 8x₂ + 4x₃ ≤ 8(32)

∴ we get

Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

Now, as given

However, cows should not be fed more than twice the minimum requirement for any nutrient each month.

∴ we have

3x₁ + 2x₂ + 4x₃ ≥ $\frac{128}{2}$

3x₁ + x₂ + 3x₃ ≥ $\frac{160}{2}$

x₁ + 0x₂ + 2x₃ ≥ $\frac{32}{2}$

6x₁ + 8x₂ + 4x₃ ≥ $\frac{256}{2}$

and also

No more than 15 pounds of each type of feed can be used per cow each month.

⇒x₁ , x₂, x₃ ≤ 15

So,

The LPP model becomes

Min Z = 2x₁ + 2.50x₂ + 3x₃

Subject to constraints :

3x₁ + 2x₂ + 4x₃ ≤ 128           .......(1)

3x₁ + x₂ + 3x₃ ≤ 160             ........(2)

x₁ + 0x₂ + 2x₃ ≤ 32              .........(3)

6x₁ + 8x₂ + 4x₃ ≤ 256          .........(4)

3x₁ + 2x₂ + 4x₃ ≥ 64            ..........(5)

3x₁ + x₂ + 3x₃ ≥ 80              ..........(6)

x₁ + 0x₂ + 2x₃ ≥ 16              ...........(7)

6x₁ + 8x₂ + 4x₃ ≥ 128         ............(8)

x₁ ≤ 15                                 ...........(9)

x₂≤ 15                                  ...........(10)

x₃ ≤ 15                                 ...........(11)

x₁ , x₂, x₃ ≥ 0

b.)

We use simplex method calculator  to solve this LPP Problem

we get

x₁ = 15 ,  x₂ = 9.5 , x₃ = 8.5

c.)

The optimal solution is Z = 79.25