### Investment Analysis for Emission Reduction Equipment**Scenario:**Carlisle Company has been cited and must invest in equipment to reduce stack emissions or face EPA fines of $20,500 per year. An emission reduction filter will cost $75,000 and have an expected life of 5 years. Carlisle's Minimum Acceptable Rate of Return (MARR) is 10% per year.**Problem:****Part A**- *Question:* What is the future worth of this investment?**Instructions:**Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±10.### Steps for Calculation:1. **Initial Data:** - Cost of the emission reduction filter = $75,000 - Life span of the filter = 5 years - Fines avoided per year = $20,500 - Minimum Acceptable Rate of Return (MARR) = 10% per year2. **Formula for Future Worth (F):** The future worth of the savings can be calculated using the Future Worth formula for uniform series savings: \[ F = P \times \left( (1 + i)^n - 1 \right) \div i \] Where: - $ F $ = Future worth - $ P $ = Annual savings ($20,500) - $ i $ = Interest rate (10% or 0.10) - $ n $ = Number of years (5)3. **Calculation:** \[ F = 20,500 \times \left( (1 + 0.10)^5 - 1 \right) \div 0.10 \]4. **Detailed Computation:** - Calculate $ (1 + 0.10)^5 $ - $ (1.10)^5 = 1.61051 $ - Subtract 1 from the result: - $ 1.61051 - 1 = 0.61051 $ - Divide by the interest rate, $ i $: - $ 0.61051 \div 0.10 = 6.1051 $ - Multiply by the annual savings: - $ 20,500 \times 6.1051 = 125,148.55 $5. **R

Future worth(FW) is the value of an asset at a specific date. It measures the nominal future sum of…

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What is the future worth of this investment?

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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### Investment Analysis for Emission Reduction Equipment

**Scenario:**
Carlisle Company has been cited and must invest in equipment to reduce stack emissions or face EPA fines of $20,500 per year. An emission reduction filter will cost $75,000 and have an expected life of 5 years. Carlisle's Minimum Acceptable Rate of Return (MARR) is 10% per year.

**Problem:**
**Part A**
- *Question:* What is the future worth of this investment?

**Instructions:**
Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±10.

### Steps for Calculation:
1. **Initial Data:**
- Cost of the emission reduction filter = $75,000
- Life span of the filter = 5 years
- Fines avoided per year = $20,500
- Minimum Acceptable Rate of Return (MARR) = 10% per year

2. **Formula for Future Worth (F):**
The future worth of the savings can be calculated using the Future Worth formula for uniform series savings:
\[ F = P \times \left( (1 + i)^n - 1 \right) \div i \]

Where:
- $ F $ = Future worth
- $ P $ = Annual savings ($20,500)
- $ i $ = Interest rate (10% or 0.10)
- $ n $ = Number of years (5)

3. **Calculation:**
\[ F = 20,500 \times \left( (1 + 0.10)^5 - 1 \right) \div 0.10 \]

4. **Detailed Computation:**
- Calculate $ (1 + 0.10)^5 $
- $ (1.10)^5 = 1.61051 $
- Subtract 1 from the result:
- $ 1.61051 - 1 = 0.61051 $
- Divide by the interest rate, $ i $:
- $ 0.61051 \div 0.10 = 6.1051 $
- Multiply by the annual savings:
- $ 20,500 \times 6.1051 = 125,148.55 $

5. **R

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