Let's analyze and solve this problem step-by-step. --- # **Given Data** | System | Cost | Fuel Consumption | Lifetime | Salvage Value | Other info | |---------|-------|------------------|------------|--------------|------------| | Heating/AC | \$8,000 | \$800/year | 15 years | None | | | Heat Pump | \$12,000 | \$400/year | 15 years | \$700 | | ### Interest rate - \( i = 8\% \) per year ### Variables - \( S \): Monthly expense - \( P \): Present value or worth - \( F \): Future worth - \( m \): number of compounding periods per year (monthly system implies \( m=12 \)) - \( n \): total number of periods (months) --- # **Step 1: Understand the formulas** The formulas given are: \[ S = P \times \frac{S/P(i/m, m, n)}{} \] where \[ S/P(i/m, m, n) = \frac{(i/m)(1 + i/m)^{n}}{(1 + i/m)^{n} - 1} \] and \[ S = F \times \frac{S/F(i/m, m, n)}{} \] where \[ S/F(i/m, m, n) = \frac{(i/m)}{(1 + i/m)^n - 1} \] --- # **Step 2: Calculate the monthly interest rate and total periods** \[ i = 8\% = .08 \] \[ m = 12 \] \[ n = 15 \text{ years} \times 12 \text{ months/year} = 180 \text{ months} \] \[ i/m = .08/12 = .0066667 \] --- # **Step 3: Compute the factors \( S/P \) and \( S/F \)** ### For the heating/AC system: \[ \text{Annual cost } = \$800 \] - Since costs are annual, convert to monthly: \[ \text{Monthly cost} = \frac{\$800}{12} \approx \$66.67 \] ### For the heat pump: \[ \text{Annual cost} = \$400 \] - Monthly: \[ \text{Monthly cost} = \frac{\$400}{12} \approx \$33.33 \] --- # **Step 4: Calculate \( S/P \) and \( S/F \) factors** \[ S/P(i/m, m, n) = \frac{(i/m)(1 + i/m)^n}{(1 + i/m)^n - 1} \] \[ S/F(i/m, m, n) = \frac{(i/m)}{(1 + i/m)^n - 1} \] Compute \((1 + i/m)^n\): \[ (1 + .0066667)^{180} \approx e^{180 \times \ln(1.0066667)} \approx e^{180 \times .006644} \approx e^{1.196} \approx 3.305 \] ### For \( S/P \): \[ S/P = \frac{.0066667 \times 3.305}{3.305 - 1} = \frac{.02203}{2.305} \approx .00956 \] ### For \( S/F \): \[ S/F = \frac{.0066667}{3.305 - 1} = \frac{.0066667}{2.305} \approx .00289 \] --- # **Step 5: Compute the present worth \( P \) for each system** ### For the heating/AC system: \[ P = \frac{\text{Monthly cost}}{S/P} = \frac{\$66.67}{.00956} \approx \$6,981 \] ### For the heat pump: \[ P = \frac{\$33.33}{.00956} \approx \$3,491 \] --- # **Step 6: Adjust for salvage value** The heat pump has a salvage value of \$700 at the end of 15 years. Calculate the future value of the salvage: \[ F_{\text{salvage}} = 700 \] The present worth of the salvage: \[ P_{\text{salvage}} = F / (1 + i)^n \] \[ (1 + .08)^{15} \approx 3.172 \] \[ P_{\text{salvage}} = \frac{700}{3.172} \approx \$220.7 \] The adjusted present worth considering salvage: \[ \text{Net Present Worth} = P - P_{\text{salvage}} \approx 3491 - 220.7 \approx \$3,270 \] --- # **# Final Answers** | System | Approximate Present Worth | Notes | |---------|-----------------------------|--------| | Heating/AC | **\$6,981** | No salvage value | | Heat Pump | **\$3,270** | Considering salvage value | --- # **Conclusion** - The **heat pump** has a lower equivalent monthly cost and is more economical over 15 years considering the salvage value. - **Final decision**: The heat pump is more cost-effective based on the present worth calculations. --- # **Part (b): Timeline Sketch** - Heating/AC: - Purchase at Year - Use for 15 years, then replaced or discarded - Heat Pump: - Purchase at Year - Use for 15 years, salvage value realized at Year 15 *Use 3 dots between years for each process, e.g., ... 5 ... 10 ... 15.*

# **Subject: Engineering Economics** This question falls under the discipline of Engineering Economics, which involves the analysis and evaluation of the financial aspects of engineering projects and systems. The focus is on calculating costs, present worth, future worth, and understanding cash flows over time, particularly in the context of heating and cooling systems.