Let's solve the problem step by step. ### **Given Data** - Current insurance coverage: **$700,000** - Current insurance premium: **$1.00 per $100** of coverage - Sprinkler system cost: **$20,000** - Estimated life of sprinkler system: **20 years** - No salvage value - Annual maintenance cost: **$400** - If installed: - Shop value for insurance = $700,000 + $20,000 = **$720,000** - New insurance premium = **$0.40 per $100** of coverage - MARR (Minimum Attractive Rate of Return): **15% per year** --- ## **A. Present Worth Calculation** Let's compute the present worth (PW) of installing the sprinkler system: ### **1. Initial Investment** - Sprinkler system cost (Year 0): **$20,000** (Outflow) ### **2. Annual Maintenance Cost** - **$400** per year for 20 years ### **3. Insurance Premium Savings** - **Current premium per year:** = ($700,000 / $100) × $1.00 = 7000 × $1.00 = **$7,000 per year** - **New premium per year (with sprinkler):** = ($720,000 / $100) × $0.40 = 7200 × $0.40 = **$2,880 per year** - **Annual insurance premium savings:** = $7,000 - $2,880 = **$4,120 per year** --- ### **4. Net Annual Savings** Net annual cash flow (savings): = Insurance savings - Maintenance cost = $4,120 - $400 = **$3,720** (positive cash flow / savings per year) --- ### **5. Present Worth of Annual Savings** Use the **Present Worth of an Annuity** formula (P/A, i%, n): \[ PW = A \times (P/A, i, n) \] Where: - \(A = \$3,720\) - \(i = 15\%\) - \(n = 20\) The Present Worth Factor (P/A, 15%, 20): \[ (P/A, i, n) = \frac{1 - (1 + i)^{-n}}{i} \] \[ = \frac{1 - (1 + 0.15)^{-20}}{0.15} \] First, calculate \( (1 + 0.15)^{-20} \): \[ (1.15)^{-20} = \frac{1}{(1.15)^{20}} \] Calculate \( (1.15)^{20} \): Take natural logs for easier calculation: \[ \ln(1.15) = 0.13976 \] \[ 0.13976 \times 20 = 2.7952 \] \[ e^{2.7952} \approx 16.391 \] So, \[ (1.15)^{20} \approx 16.37 \] \[ (1.15)^{-20} \approx 0.061 \] Now plug into the formula: \[ (P/A, 15\%, 20) = \frac{1 - 0.061}{0.15} = \frac{0.939}{0.15} = 6.26 \] --- Now, calculate PW of the annual savings: \[ PW_{savings} = \$3,720 \times 6.26 = \$23,287.20 \] --- ### **6. Total Present Worth (PW) of Investment** \[ PW_{total} = - \text{Initial Investment} + PW_{savings} \] \[ = -\$20,000 + \$23,287.20 \] \[ = \$3,287.20 \] --- ### **Final Answer for (a):** \[ \boxed{PW = \$3,287.20} \] --- ## **B. Decision Rule Based on Present Worth** **The decision rule:** - If **PW > 0**, the investment is **economically justified** (i.e., it earns at least the MARR). - If **PW < 0**, the investment is **not justified**. --- ## **C. Is the Sprinkler System Economically Justified?** Since the present worth is **positive (\$3,287.20)**, **the sprinkler system is economically justified.** --- ## **Summary Table** | Part | Answer | |------|--------| | a | PW = **\$3,287.20** | | b | If PW > 0, invest; if PW < 0, do not invest | | c | Yes, the sprinkler system is economically justified | --- **If you need to see the calculations for a different interest rate or if you want a cash flow diagram, let me know!**