Nippon Steel’s expenses for heating and cooling a large manufacturing facility are expected to increase according to an arithmetic gradient beginning in year 2. If the cost is $550,000 this year (year 0) and will be $550,000 again in year 1, but then it is estimated to increase by $58,000 each year through year 12, what is the equivalent annual worth in years 1 to 12 of these energy costs at an interest rate of 14.00% per year?

To solve this problem, we need to determine the **equivalent annual worth** of Nippon Steel’s energy costs from year 1 to year 12. The costs follow an arithmetic gradient starting in year 2, and we are given the costs for year 0 and year 1. We will break the problem down into steps.

### Step 1: Define key parameters
- Cost in year 0: $ C_0 = 550,000 $
- Cost in year 1: $ C_1 = 550,000 $
- Gradient increase from year 2 onwards: $ G = 58,000 $
- Interest rate: $ i = 14\% = 0.14 $
- Number of years for analysis: $ n = 12 $

### Step 2: Break down the cost structure
- For year 1, the cost is constant: $ 550,000 $.
- From year 2 to year 12, there is a uniform increase of $ 58,000 $ per year, which forms an arithmetic gradient.

Thus, the cash flow consists of:
- A constant value of $ 550,000 $ in years 1 through 12.
- An arithmetic gradient beginning in year 2.

### Step 3: Equivalent annual worth formula

The total equivalent annual worth (A) is the sum of two components:
1. **The equivalent annual worth of the constant amount** ($ C_1 = 550,000 $) over 12 years.
2. **The equivalent annual worth of the arithmetic gradient** ($ G = 58,000 $) over 11 years (since the gradient starts in year 2).

#### 3.1: Equivalent annual worth of the constant cost $ C_1 $
The formula for the equivalent annual worth of a constant cash flow $ C $ over $ n $ years at an interest rate $ i $ is:

\[
A_{\text{constant}} = C \cdot \frac{i(1+i)^n}{(1+i)^n - 1}
\]

#### 3.2: Equivalent annual worth of the arithmetic gradient
The formula for the equivalent annual worth of an arithmetic gradient $ G $ over $ n $ years is:

\[
A_{\text{gradient}} = G \cdot \frac{1}{i} \cdot \left[ \frac{(1+i)^n - i n - 1}{(1+i)^n - 1} \right]
\]

Now, let's calculate these values.The equivalent annual worth of Nippon Steel's energy costs from year 1 to year 12, considering both the constant costs and the arithmetic gradient, is approximately **$313,700** per year at an interest rate of 14%.

Would you like further details on any part of the calculation?

Here are 5 follow-up questions you might consider:
1. How is the arithmetic gradient used in other financial models?
2. How would the result change with a lower interest rate?
3. What if the gradient started in a different year?
4. How do interest rates affect long-term financial planning in manufacturing?
5. What would be the total cost over the 12 years?

**Tip:** When calculating equivalent annual worth, it's important to distinguish between constant cash flows and those following an arithmetic gradient, as each requires a different approach.

Calculate the equivalent annual worth of Nippon Steel’s energy costs for heating and cooling a facility with an arithmetic gradient increase of $58,000 annually. Learn the step-by-step process of solving this problem using interest rate formulas at 14% over 12 years.

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