Calculate the value of total production for 2020 valued at the base year. Sugar per kg 1 Shirt Petrol per liter Base year: Price
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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 ≥
3x₁ + x₂ + 3x₃ ≥
x₁ + 0x₂ + 2x₃ ≥
6x₁ + 8x₂ + 4x₃ ≥
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
Answer: $0.51
Kristen Lu purchased a used automobile for $25,950 at the beginning of last year and incurred the following operating costs:
Depreciation ($25,950 ÷ 5 years) $5,190
Insurance $2,700
Garage rent $1,400
Automobile tax and license $710
Variable operating cost $0.11 per mile
The variable operating cost consists of gasoline, oil, tires, maintenance, and repairs. Kristen estimates that, at her current rate of usage, the car will have zero resale value in five years, so the annual straight-line depreciation is $5,190. The car is kept in a garage for a monthly fee.
Explanation:
To calculate the fixed cost per mile, calculate the total fixed cost and divide by the number of miles (25000).
The total fixed cost combines all the cost that do not change depending on the level of usage. These costs do not reduce if Kristen uses her car less or increase if she uses her car more. The variable cost is the opposite of this.
Total fixed cost = $5,190 + $2,700 + $1,400 + $710 = 10,000
Fixed cost per mile = 10000/25000 = $0.40
Variable Operating Cost/Mile = $0.11
Average Operating Cost/Mile = $0.51
