Let's solve the problem step by step. ## Given Data - **System cost (initial investment)**: $120,000 - **Expected life**: 7 years - **Salvage value (end of year 7)**: $7,500 - **O&M expenses**: $2,000 per year (if purchased) - **Alternative (no purchase)**: Pay $40,000 per year to Bay City for water purification - **If purchased**: No need to pay Bay City - **Loan**: ½ of purchase price ($60,000), 3 equal annual payments, 1st payment at end of year 2, interest rate 8% compounded annually - **Company's MARR**: 10% compounded annually We are to find **the annual worth (AW) of this investment**. --- ## Step 1: Calculate Net Cash Flows ### If System is Purchased #### Outflows: - **Year 0**: Pay $60,000 cash. Borrow $60,000 (½ of purchase price). - **Operating and Maintenance (O&M)**: $2,000 per year for 7 years. - **Loan Repayment**: 3 equal annual payments starting from end of year 2 (years 2, 3, 4). - **Salvage Value**: $7,500 in year 7 (inflow). #### Inflows: - **Avoided cost**: No longer pay $40,000 per year to Bay City (so, $40,000/year saving for 7 years). --- ## Step 2: Calculate the Loan Repayment Amount **Loan Amount:** $60,000 **Interest Rate:** 8% **Payment Years:** End of years 2, 3, and 4 (3 payments) **First payment:** End of year 2 This is a **deferred annuity** (payments start after 2 years). We need to find the payment \( A \) such that the present value at year 0 is $60,000. The present value (PV) of 3 payments of amount \( A \), made at the end of years 2, 3, and 4 at 8% is: \[ PV = \frac{A}{(1+0.08)^2} + \frac{A}{(1+0.08)^3} + \frac{A}{(1+0.08)^4} \] Set PV = $60,000, solve for A: Calculate the denominators: - \( (1.08)^2 = 1.1664 \) - \( (1.08)^3 = 1.259712 \) - \( (1.08)^4 = 1.360489 \) \[ \frac{1}{1.1664} = 0.8573 \] \[ \frac{1}{1.259712} = 0.7939 \] \[ \frac{1}{1.360489} = 0.7349 \] Sum: \( 0.8573 + 0.7939 + 0.7349 = 2.3861 \) \[ 60,000 = A \times 2.3861 \] \[ A = \frac{60,000}{2.3861} \approx \$25,146 \] --- ## Step 3: Find Equivalent Cash Flows for Analysis Let’s lay out the cash flows for years 0–7: | Year | Cash Flow Component | Value | Net Cash Flow | |------|--------------------------------------------------------|----------|-----------------| | 0 | Pay ½ price in cash | -60,000 | -60,000 | | 0 | Receive loan | +60,000 | 0 | | 1 | O&M | -2,000 | -2,000 | | 1 | Avoided Bay City cost | +40,000 | +38,000 | | 2 | O&M | -2,000 | -2,000 | | 2 | Avoided Bay City cost | +40,000 | +38,000 | | 2 | Loan payment | -25,146 | +12,854 | | 3 | O&M | -2,000 | -2,000 | | 3 | Avoided Bay City cost | +40,000 | +38,000 | | 3 | Loan payment | -25,146 | +12,854 | | 4 | O&M | -2,000 | -2,000 | | 4 | Avoided Bay City cost | +40,000 | +38,000 | | 4 | Loan payment | -25,146 | +12,854 | | 5 | O&M | -2,000 | -2,000 | | 5 | Avoided Bay City cost | +40,000 | +38,000 | | 6 | O&M | -2,000 | -2,000 | | 6 | Avoided Bay City cost | +40,000 | +38,000 | | 7 | O&M | -2,000 | -2,000 | | 7 | Avoided Bay City cost | +40,000 | +38,000 | | 7 | Salvage value | +7,500 | +45,500 | But, since the $60,000 outflow is offset by the $60,000 inflow (the loan), we can ignore both for equivalence calculations, and just treat the loan repayment as an outflow. So, the net cash flows for each year are: | Year | Cash Flow | |------|------------------| | 0 | -60,000 | | 1 | +38,000 | | 2 | +12,854 | | 3 | +12,854 | | 4 | +12,854 | | 5 | +38,000 | | 6 | +38,000 | | 7 | +45,500 | --- ## Step 4: Find Present Worth (PW) at MARR = 10% We use the formula: \[ PW = \sum_{t=0}^{7} \frac{CF_t}{(1 + 0.10)^t} \] Let’s calculate each term step by step. ### Year 0: -60,000 \[ PW_0 = -60,000 \] ### Year 1: +38,000 \[ PW_1 = \frac{38,000}{1.10^1} = \frac{38,000}{1.10} = 34,545.45 \] ### Year 2: +12,854 \[ PW_2 = \frac{12,854}{1.10^2} = \frac{12,854}{1.21} = 10,629.75 \] ### Year 3: +12,854 \[ PW_3 = \frac{12,854}{1.10^3} = \frac{12,854}{1.331} = 9,658.09 \] ### Year 4: +12,854 \[ PW_4 = \frac{12,854}{1.10^4} = \frac{12,854}{1.4641} = 8,778.26 \] ### Year 5: +38,000 \[ PW_5 = \frac{38,000}{1.10^5} = \frac{38,000}{1.61051} = 23,601.44 \] ### Year 6: +38,000 \[ PW_6 = \frac{38,000}{1.10^6} = \frac{38,000}{1.771561} = 21,453.48 \] ### Year 7: +45,500 \[ PW_7 = \frac{45,500}{1.10^7} = \frac{45,500}{1.9487171} = 23,340.47 \] --- ### Add up present values: \[ PW = -60,000 + 34,545.45 + 10,629.75 + 9,658.09 + 8,778.26 + 23,601.44 + 21,453.48 + 23,340.47 \] \[ PW = -60,000 + 34,545.45 = -25,454.55 \] \[ -25,454.55 + 10,629.75 = -14,824.80 \] \[ -14,824.80 + 9,658.09 = -5,166.71 \] \[ -5,166.71 + 8,778.26 = 3,611.55 \] \[ 3,611.55 + 23,601.44 = 27,212.99 \] \[ 27,212.99 + 21,453.48 = 48,666.47 \] \[ 48,666.47 + 23,340.47 = 72,006.94 \] So, \[ PW = \$72,007 \text{ (rounded to the nearest dollar)} \] --- ## Step 5: Find the Equivalent Annual Worth (AW) \[ AW = PW \times \frac{A/P, i, n} \] Where: - \( PW = 72,007 \) - \( i = 10\% \) - \( n = 7 \) The capital recovery factor \( (A/P, 10\%, 7) \): \[ A/P = \frac{i(1 + i)^n}{(1 + i)^n - 1} \] \[ A/P = \frac{0.10 \times (1.10)^7}{(1.10)^7 - 1} \] \[ (1.10)^7 = 1.9487171 \] \[ A/P = \frac{0.10 \times 1.9487171}{1.9487171 - 1} = \frac{0.19487171}{0.9487171} = 0.2054 \] \[ AW = 72,007 \times 0.2054 = \$14,792 \] --- ## **Final Answer** \[ \boxed{\$14,792 \text{ per year}} \] --- ### **Summary of Steps** 1. **Calculate loan repayments** using deferred annuity formula. 2. **Lay out net cash flows** for each year. 3. **Find present worth (PW)** at MARR 10%. 4. **Find annual worth (AW)** over 7 years at 10%. --- **The annual worth of this investment is \$14,792 per year.**