Shifted Gradients
Q5. 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 are estimated to increase by $40,000 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
10% per year?

To find the **equivalent annual worth (EAW)** of Nippon Steel’s heating and cooling costs, we need to account for the **shifted arithmetic gradient**. Here's the breakdown:

- **Initial cost** at year 0 and year 1 is $ C_0 = 550,000 $ (which remains constant in year 0 and 1).
- From **year 2 to year 12**, the cost increases by a gradient of $ G = 40,000 $ per year.
- The **interest rate** is 10%, or $ i = 0.10 $.

### Steps to solve:

1. **Separate constant and gradient costs**:
   - From year 0 to year 12, there is a constant cost of **$550,000** per year.
   - Starting from year 2, the cost increases by **$40,000** per year.

2. **Equivalent Annual Worth of the constant cost**:
   The constant cost is $550,000 from year 1 to year 12. We can use the **capital recovery factor (CRF)** to compute its equivalent annual worth over 12 years. The formula is:

   \[
   CRF = \frac{i(1+i)^n}{(1+i)^n - 1}
   \]
   where $ i = 0.10 $ and $ n = 12 $. This gives us the equivalent annual worth of the constant $550,000 over 12 years.

3. **Equivalent Annual Worth of the shifted arithmetic gradient**:
   Since the arithmetic gradient starts in **year 2**, we can shift it back by 1 year using the appropriate factor. The formula for the **EAW of a shifted arithmetic gradient** is:

   \[
   EAW_G = G \cdot \frac{(1 + i)^n - i \cdot n - 1}{i^2 \cdot (1 + i)^n}
   \]

   Where $ G = 40,000 $, $ i = 0.10 $, and $ n = 11 $ (since the gradient runs from year 2 to year 12).

4. **Sum the equivalent annual worths**:
   The total equivalent annual worth is the sum of the EAW for the constant cost and the EAW for the shifted arithmetic gradient.

Let's compute these steps to get the total EAW.The equivalent annual worth (EAW) of Nippon Steel's energy costs over the 12 years, considering the constant cost and the shifted arithmetic gradient, is approximately **$1,136,571.07** per year.

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

Here are 5 related questions for further exploration:
1. How does a change in the interest rate affect the equivalent annual worth in this scenario?
2. What would be the impact on EAW if the gradient increase starts in year 3 instead of year 2?
3. How is the capital recovery factor used in financial calculations, and why is it important?
4. What if the gradient increases by a different amount every year? How would that change the calculation?
5. How would the calculation change if the costs decreased by $40,000 every year instead?

**Tip**: Always check whether a gradient starts in year 1 or is shifted to a later year, as it changes the formula for calculating equivalent annual worth.

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 are estimated to increase by $40,000 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 10% per year?

Find out how to calculate the equivalent annual worth of Nippon Steel's energy costs, starting with a base of $550,000 and increasing by $40,000 annually from year 2 to year 12, using engineering economic principles at a 10% interest rate.

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