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 $50,000 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 11.00% per year?

To calculate the equivalent annual worth (A) of Nippon Steel's heating and cooling costs from years 1 to 12, we need to analyze the situation in two parts:

1. **Constant costs in year 0 and year 1**: The cost is $550,000 for these two years.
2. **Arithmetic gradient starting in year 2**: The cost increases by $50,000 each year from year 2 through year 12.

### Steps to Solve:

1. **Step 1: Determine the annual worth of the constant cost in years 1-12.**

   The constant cost in year 0 and year 1 is $550,000. Since this amount stays constant, it does not involve any gradient. We will use the formula for converting a present value into an annual worth (uniform series) over 12 years:

   \[
   A_{\text{constant}} = P \times \frac{i(1+i)^n}{(1+i)^n - 1}
   \]
   Where:
   - $P = 550,000$
   - $i = 0.11$ (interest rate of 11%)
   - $n = 12$ (number of years)

2. **Step 2: Determine the annual worth of the arithmetic gradient.**

   The costs increase by $50,000 each year from year 2 to year 12. The gradient starts at $0 in year 2 and increases by $50,000 each subsequent year.

   To find the annual worth of an arithmetic gradient, we use the formula for converting an arithmetic gradient to an equivalent annual worth:

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

   Where:
   - $G = 50,000$ (annual increase)
   - $i = 0.11$ (interest rate)
   - $n = 11$ (years from year 2 to year 12)

3. **Step 3: Combine the results of the two parts to find the total equivalent annual worth**.

Now, let's perform the calculations.The equivalent annual worth of Nippon Steel's energy costs from years 1 to 12 is approximately **$283,655.43** at an interest rate of 11%.

Would you like more details or have any questions?

Here are 5 related questions to consider:
1. How would the equivalent annual worth change with a lower interest rate?
2. What if the gradient starts from year 3 instead of year 2?
3. How would this calculation differ if the costs increased geometrically instead of arithmetically?
4. What is the total amount spent over the 12 years, including the gradient?
5. How can the equivalent annual worth help in budget planning for future years?

**Tip:** When dealing with arithmetic gradients, always distinguish between when the gradient begins and the base value stays constant to avoid calculation errors.

Analyze Nippon Steel's energy costs, with expenses starting at $550,000 and increasing by $50,000 annually from year 2 to 12. Calculate the equivalent annual worth from years 1 to 12 at an interest rate of 11%. Explore the arithmetic gradient and time value of money concepts with step-by-step calculations.

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