Let's break down the problem: ### **Given Data** | Machine | First Cost (\(P\)) | Annual Cost (\(A\)) | Salvage Value (\(S\)) | Life (\(n\)) | |------------------|-------------------|---------------------|-----------------------|--------------| | Hydraulic (HP) | $680,000 | $200,000 | $70,000 | 5 years | | Electric (EMT) | $800,000 | $152,000 | $130,000 | 5 years | ### **Step 1: Calculate the incremental cash flows (EMT vs. HP)** - **Extra initial investment**: \( \$800,000 - \$680,000 = \$120,000 \) (outflow at year 0) - **Annual Cost Difference**: \( \$152,000 - \$200,000 = -\$48,000 \) (annual savings, so inflow) - **Salvage Value Difference**: \( \$130,000 - \$70,000 = \$60,000 \) (inflow at year 5) So, the incremental cash flows for "EMT minus HP" are: | Year | Cash Flow | |------|-----------| | 0 | -$120,000 | | 1-5 | +$48,000 | | 5 | +$60,000 | ### **Step 2: Find the ROR on the extra investment** We need to solve for the interest rate \(i\) that makes the net present worth (NPW) of these incremental cash flows zero: \[ -120,000 + 48,000(P/A, i, 5) + 60,000(P/F, i, 5) = 0 \] #### Expand the factors: - \( (P/A, i, 5) = \frac{1 - (1+i)^{-5}}{i} \) - \( (P/F, i, 5) = (1+i)^{-5} \) So, \[ -120,000 + 48,000 \frac{1 - (1+i)^{-5}}{i} + 60,000 (1+i)^{-5} = 0 \] #### Rearranged: \[ -120,000 + \frac{48,000}{i} [1 - (1+i)^{-5}] + 60,000 (1+i)^{-5} = 0 \] \[ -120,000 + \frac{48,000}{i} - \frac{48,000}{i}(1+i)^{-5} + 60,000 (1+i)^{-5} = 0 \] \[ -120,000 + \frac{48,000}{i} + \left[ -\frac{48,000}{i} + 60,000 \right] (1+i)^{-5} = 0 \] Let \( x = (1+i)^{-5} \): \[ -120,000 + \frac{48,000}{i} + \left[ -\frac{48,000}{i} + 60,000 \right] x = 0 \] \[ -120,000 + \frac{48,000}{i} + \left[ -\frac{48,000}{i} x + 60,000 x \right] = 0 \] \[ -120,000 + \frac{48,000}{i} - \frac{48,000}{i} x + 60,000 x = 0 \] \[ -120,000 + \frac{48,000}{i} + x \left(60,000 - \frac{48,000}{i}\right) = 0 \] But it's easiest to **solve numerically**: #### Let's try some values for \(i\): ##### Try \(i = 20\%\) (\(0.20\)): - \( (P/A, 20\%, 5) = \frac{1 - (1+0.20)^{-5}}{0.20} = \frac{1 - (1/2.48832)}{0.20} = \frac{1 - 0.4013}{0.20} = \frac{0.5987}{0.20} = 2.993 \) - \( (P/F, 20\%, 5) = (1+0.20)^{-5} = 0.4013 \) Plug in: \[ -120,000 + 48,000 \cdot 2.993 + 60,000 \cdot 0.4013 \] \[ -120,000 + 143,664 + 24,078 = 47,742 \] Positive, so try a higher rate. ##### Try \(i = 30\%\): - \( (1.3)^5 = 3.71293 \Rightarrow (P/F) = 1/3.71293 = 0.2693 \) - \( (P/A) = \frac{1 - 0.2693}{0.30} = \frac{0.7307}{0.30} = 2.436 \) \[ -120,000 + 48,000 \cdot 2.436 + 60,000 \cdot 0.2693 \] \[ -120,000 + 116,928 + 16,158 = 13,086 \] Still positive, but closer to zero. ##### Try \(i = 35\%\): - \( (1.35)^5 = 4.482 \Rightarrow (P/F) = 1/4.482 = 0.2231 \) - \( (P/A) = \frac{1 - 0.2231}{0.35} = \frac{0.7769}{0.35} = 2.219 \) \[ -120,000 + 48,000 \cdot 2.219 + 60,000 \cdot 0.2231 \] \[ -120,000 + 106,512 + 13,386 = -101 \] Very close to zero! Let's try between 34% and 35%. ##### Try \(i = 34\%\): - \( (1.34)^5 = 4.323 \Rightarrow (P/F) = 0.2313 \) - \( (P/A) = \frac{1 - 0.2313}{0.34} = 0.7687/0.34 = 2.261 \) \[ -120,000 + 48,000 \cdot 2.261 + 60,000 \cdot 0.2313 \] \[ -120,000 + 108,528 + 13,878 = 2,406 \] ##### Try \(i = 36\%\): - \( (1.36)^5 = 4.648 \Rightarrow (P/F) = 0.2151 \) - \( (P/A) = (1 - 0.2151)/0.36 = 0.7849/0.36 = 2.180 \) \[ -120,000 + 48,000 \cdot 2.180 + 60,000 \cdot 0.2151 \] \[ -120,000 + 104,640 + 12,906 = -2,454 \] Now interpolate between 34% (\$2,406) and 36% (\$-2,454): \[ i^* = 34\% + \frac{2,406}{2,406 + 2,454} \cdot (36\% - 34\%) \] \[ i^* = 34\% + \frac{2,406}{4,860} \cdot 2\% \] \[ i^* = 34\% + 0.495 \cdot 2\% = 34\% + 0.99\% = 34.99\% \] **Rounded to the nearest percent:** \[ \boxed{35\%} \] --- ## **Final Answer** **The ROR on the extra investment for the EMT alternative is:** \[ \boxed{35\%} \] --- **Let me know if you need the step-by-step for the other parts!**