Let's solve the problem step by step using **incremental rate of return analysis**: ### **Given Data** | | Type Fe | Type Al | |----------------|--------------|--------------| | First Cost | $145,000 | $306,000 | | Annual O&M Cost| $52,000 | $27,000 | | Salvage Value | $20,000 | $70,000 | | Life (years) | 5 | 10 | | MARR | 20% | 20% | --- ### **Step 1: Find the Present Worth (PW) of Each Option** #### **Type Fe:** - \( \text{First Cost} = -\$145,000 \) - \( \text{Annual O\&M} = -\$52,000 \) for 5 years - \( \text{Salvage Value} = +\$20,000 \) at year 5 **PW calculation formula:** \[ PW = -FC - \text{O\&M} \times (P/A, i, n) + S \times (P/F, i, n) \] Where: - \( (P/A, i, n) \) is the Present Worth factor for an annuity - \( (P/F, i, n) \) is the Present Worth factor for a future sum Let's calculate these factors for 20% and 5 years: - \( (P/A, 20\%, 5) = \frac{1 - (1 + 0.2)^{-5}}{0.2} = \frac{1 - (1.2)^{-5}}{0.2} \) - \( (1.2)^{-5} = 0.4013 \) - \( (P/A) = \frac{1 - 0.4013}{0.2} = \frac{0.5987}{0.2} = 2.9935 \) - \( (P/F, 20\%, 5) = (1.2)^{-5} = 0.4013 \) So, \[ PW_{Fe} = -145,000 - 52,000 \times 2.9935 + 20,000 \times 0.4013 \] \[ PW_{Fe} = -145,000 - 155,662 + 8,026 = -145,000 - 155,662 + 8,026 \] \[ PW_{Fe} = -292,636 \] --- #### **Type Al:** - \( \text{First Cost} = -\$306,000 \) - \( \text{Annual O\&M} = -\$27,000 \) for 10 years - \( \text{Salvage Value} = +\$70,000 \) at year 10 For 10 years: - \( (1.2)^{-10} = 0.1615 \) - \( (P/A, 20\%, 10) = \frac{1 - 0.1615}{0.2} = \frac{0.8385}{0.2} = 4.1925 \) - \( (P/F, 20\%, 10) = 0.1615 \) \[ PW_{Al} = -306,000 - 27,000 \times 4.1925 + 70,000 \times 0.1615 \] \[ PW_{Al} = -306,000 - 113,197.5 + 11,305 \] \[ PW_{Al} = -306,000 - 113,197.5 + 11,305 = -407,892.5 \] --- ### **Step 2: Find the IRR of the Incremental Investment** Incremental = Al - Fe - Incremental Cost = $306,000 - $145,000 = $161,000 (at year 0) - Incremental O&M = $27,000 - $52,000 = -$25,000 per year (for 10 years) - Incremental Salvage = $70,000 - $20,000 = $50,000 (at year 10) The cash flow for the incremental investment: - Year 0: -$161,000 - Year 1-10: +$25,000 per year - Year 10: +$50,000 Set up the equation for IRR (\(i^*\)): \[ -161,000 + 25,000 \times (P/A, i^*, 10) + 50,000 \times (P/F, i^*, 10) = 0 \] Let's estimate IRR: Try \(i^* = 20\%\): - \( (P/A, 20\%, 10) = 4.1925 \) - \( (P/F, 20\%, 10) = 0.1615 \) \[ -161,000 + 25,000 \times 4.1925 + 50,000 \times 0.1615 = -161,000 + 104,812.5 + 8,075 = -48,112.5 \] Still negative, try 10%: - \( (P/A, 10\%, 10) = 6.1446 \) - \( (P/F, 10\%, 10) = 0.3855 \) \[ -161,000 + 25,000 \times 6.1446 + 50,000 \times 0.3855 = -161,000 + 153,615 + 19,275 = 11,890 \] So, at 10% → +$11,890, at 20% → -$48,112.5. IRR is between 10% and 20%. Slope: \[ \frac{11,890 - (-48,112.5)}{10\% - 20\%} = \frac{60,002.5}{-10\%} = -6,000.25 \text{ per percent} \] Let’s interpolate: \[ i^* = 10\% + \left(\frac{11,890}{60,002.5}\right) \times 10\% = 10\% + 0.198 \times 10\% = 10\% + 1.98\% = 12\% \] #### **Estimated IRR ≈ 12%** --- ### **Step 3: Compare IRR to MARR** - MARR = 20% - Incremental IRR = 12% < 20% ### **Conclusion** **Select Type Fe nanoparticle**. Because the incremental IRR is less than the company’s MARR, Type Fe (the cheaper option) should be selected. --- #### **Summary Table** | Option | Present Worth (\$) | |----------|-------------------| | Type Fe | -292,636 | | Type Al | -407,892.5 | **Type Fe is preferred.** **Incremental IRR (Al vs. Fe): 12%** (which is less than 20% MARR). --- If you need the step-by-step in Excel or another format, let me know!