# Present Worth Calculation of Water Supply Project Let's break down the problem and solve step by step. ## 1. **Timeline Overview** - **Today:** End of 2025 (beginning of 2026) - **Construction:** 2030–2034 (5 years, $30M/year) - **Maintenance starts:** 2035 (first year after construction) - **Maintenance period:** 2035–2080 - **Interest rate:** \( i = 9\% \) - **No salvage value** ## 2. **Step 1: Present Worth of Construction Costs** ### Construction Details - **Years:** 2030, 2031, 2032, 2033, 2034 (5 payments) - **Each payment:** $30M - **Timing:** Each payment at **year-end** (i.e., at end of 2030, etc.) ### Find Present Worth (PW) at end of 2025 (beginning of 2026) #### **Step 2.1: Find PW at end of 2029 (before first payment)** This is an **ordinary annuity** (equal payments at regular intervals). \[ N = 5, \quad A = \$30M, \quad i = 9\% \] \[ PW_{2029} = A \cdot \frac{1 - (1+i)^{-N}}{i} \] \[ PW_{2029} = 30 \cdot \frac{1 - (1+0.09)^{-5}}{0.09} \] Calculate \( (1+0.09)^{-5} \): \[ (1.09)^5 = 1.53862 \implies (1.09)^{-5} = \frac{1}{1.53862} = 0.64993 \] So, \[ PW_{2029} = 30 \cdot \frac{1 - 0.64993}{0.09} = 30 \cdot \frac{0.35007}{0.09} = 30 \cdot 3.8897 = \$116.69\,M \] #### **Step 2.2: Bring PW back to end of 2025 (4 years earlier)** Discount for 4 years at 9%: \[ PW_{2025} = PW_{2029} \cdot (1+i)^{-4} \] \[ (1.09)^4 = 1.4116 \implies (1.09)^{-4} = 0.70843 \] \[ PW_{2025} = 116.69 \cdot 0.70843 = \$82.62\,M \] --- ## 3. **Step 2: Present Worth of Maintenance & Repairs** - **First payment:** End of 2035 (\$3M) - **Increases:** 11% per year (geometric gradient) - **Number of payments:** From end of 2035 to end of 2080: \[ N = 2080 - 2035 + 1 = 46 \text{ years} \] - **Gradient rate:** \( g = 11\% \) - **Interest rate:** \( i = 9\% \) - **First payment (A1):** \$3M ### **Step 3.1: Find PW at end of 2034 (before first maintenance payment)** For a geometric gradient: \[ PW_{2034} = A_1 \cdot \frac{1 - \left( \frac{1+g}{1+i} \right)^N }{i-g} \] \[ A_1 = 3, \quad g = 0.11, \quad i = 0.09, \quad N = 46 \] \[ \frac{1+g}{1+i} = \frac{1.11}{1.09} = 1.01835 \] \[ (1.01835)^{46} = e^{46 \ln 1.01835} \approx e^{46 \times 0.01818} = e^{0.836} \approx 2.3078 \] \[ PW_{2034} = 3 \cdot \frac{1 - 2.3078}{0.09 - 0.11} = 3 \cdot \frac{-1.3078}{-0.02} = 3 \cdot 65.39 = \$196.17\,M \] ### **Step 3.2: Discount Maintenance PW to end of 2025** Time between end of 2025 and end of 2034: \( 2034 - 2025 = 9 \) years \[ PW_{2025} = PW_{2034} \cdot (1+i)^{-9} \] \[ (1.09)^9 = 2.172 \implies (1.09)^{-9} = 0.46022 \] \[ PW_{2025} = 196.17 \cdot 0.46022 = \$90.32\,M \] --- ## 4. **Step 3: Total Present Worth** \[ PW_{total} = PW_{construction} + PW_{maintenance} \] \[ PW_{total} = 82.62 + 90.32 = \boxed{\$172.94\,M} \] --- ## 5. **Final Answer** ### **The present worth of the project (as of end of 2025) is:** \[ \boxed{\$172.94\,\text{million}} \] --- ### **Summary Table** | Cost Component | Present Worth (\$M) | |---------------------|:------------------:| | Construction | 82.62 | | Maintenance/Repair | 90.32 | | **Total** | **172.94** | --- **Notes:** - All calculations are rounded to two decimal places. - Cash flows are assumed to occur at year-end as specified.