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geniusboy [140]
4 years ago
12

For an economy starting at potential output, a decrease in planned investment in the short run results in a(n):

Business
1 answer:
Kruka [31]4 years ago
6 0

Question:

For an economy starting at potential output, a decrease in autonomous expenditure in the short-run results in a(n):

A. increase in potential output

B. recessionary output gap

C. decrease in potential output

D.  expansionary output gap

Answer:

The correct answer is B

Explanation:

A decrease in autonomous expenditure shifts the Planned Aggregate Expenditure curve downward thus creating a lower equilibrium output.

PAE = C + Ip + G + NX

where

PAE  = Planned Aggregate Expenditure

C = consumption

Ip = Investment Spending

G = Government Spending

NX =  Net Export

If an economy has its output equal to its potential, this will create a reduction in short-run equilibrium output leading to a recessionary output gap.

Cheers!

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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

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