Let's break down the problem and calculate the **Rate of Return (ROR) on the extra investment** required for the EMT alternative using the **annual worth (AW) method**. ### Given Data | Parameter | Hydraulic Press (HP) | Electric Machine Tech (EMT) | |----------------|---------------------|----------------------------| | First Cost | $680,000 | $800,000 | | Annual Cost | $200,000 | $152,000 | | Salvage Value | $70,000 | $130,000 | | Life | 5 years | 5 years | ### Step 1: Find the "Difference" Project Calculate the **incremental cash flows** (EMT - HP): - **First cost difference:** $800,000 - $680,000 = **$120,000** (extra initial investment for EMT) - **Annual cost difference:** $152,000 - $200,000 = **-$48,000** (EMT saves $48,000 per year) - **Salvage value difference:** $130,000 - $70,000 = **$60,000** (EMT has $60,000 higher salvage) So, the incremental project is: - Initial investment: $120,000 (outflow at time 0) - Annual saving: $48,000 (inflow, years 1-5) - Additional salvage: $60,000 (inflow, year 5) ### Step 2: Set Up the AW Equation We want the interest rate \( i \) that makes the **annual worth (AW) = 0** for the incremental investment. \[ 0 = -120,\!000(A/P, i, 5) + 48,\!000 + 60,\!000(A/F, i, 5) \] #### Where: - \( (A/P, i, 5) \) is the capital recovery factor - \( (A/F, i, 5) \) is the sinking fund factor Recall: \[ (A/P, i, n) = \frac{i(1+i)^n}{(1+i)^n - 1} \] \[ (A/F, i, n) = \frac{i}{(1+i)^n - 1} \] ### Step 3: Substitute Cash Flows Rearrange: \[ -120,\!000(A/P, i, 5) + 48,\!000 + 60,\!000(A/F, i, 5) = 0 \] \[ 48,\!000 = 120,\!000(A/P, i, 5) - 60,\!000(A/F, i, 5) \] \[ 48,\!000 = 120,\!000(A/P, i, 5) - 60,\!000(A/F, i, 5) \] Or, \[ \frac{48,\!000}{120,\!000} = (A/P, i, 5) - 0.5(A/F, i, 5) \] \[ 0.4 = (A/P, i, 5) - 0.5(A/F, i, 5) \] ### Step 4: Solve for \( i \) (Iterative/Numerical Solution) Let’s try values for \( i \): #### Try \( i = 20\% \): - \( (A/P, 20\%, 5) = \frac{0.2(1.2)^5}{(1.2)^5 - 1} = \frac{0.2 \times 2.48832}{2.48832 - 1} = \frac{0.497664}{1.48832} \approx 0.3344 \) - \( (A/F, 20\%, 5) = \frac{0.2}{2.48832 - 1} = \frac{0.2}{1.48832} \approx 0.1344 \) - So \( 0.3344 - 0.5 \times 0.1344 = 0.3344 - 0.0672 = 0.2672 \) (too low) #### Try \( i = 10\% \): - \( (A/P, 10\%, 5) = \frac{0.1(1.1)^5}{(1.1)^5 - 1} = \frac{0.1 \times 1.61051}{0.61051} \approx 0.2639 \) - \( (A/F, 10\%, 5) = \frac{0.1}{0.61051} \approx 0.1637 \) - \( 0.2639 - 0.5 \times 0.1637 = 0.2639 - 0.0818 = 0.1821 \) (still too low) #### Try \( i = 30\% \): - \( (1.3)^5 = 3.71293 \) - \( (A/P, 30\%, 5) = \frac{0.3 \times 3.71293}{2.71293} = \frac{1.113879}{2.71293} \approx 0.4105 \) - \( (A/F, 30\%, 5) = \frac{0.3}{2.71293} = 0.1106 \) - \( 0.4105 - 0.5 \times 0.1106 = 0.4105 - 0.0553 = 0.3552 \) (closer, but still less than 0.4) #### Try \( i = 40\% \): - \( (1.4)^5 = 5.37824 \) - \( (A/P, 40\%, 5) = \frac{0.4 \times 5.37824}{4.37824} = \frac{2.151296}{4.37824} \approx 0.4913 \) - \( (A/F, 40\%, 5) = \frac{0.4}{4.37824} \approx 0.0914 \) - \( 0.4913 - 0.5 \times 0.0914 = 0.4913 - 0.0457 = 0.4456 \) (now too high) ### Now, interpolate between 30% and 40%: At 30%: 0.3552 At 40%: 0.4456 Target: 0.4 Slope: (0.4456 - 0.3552) / (40% - 30%) = 0.0904 / 10% = 0.00904 per percent Difference to target: 0.4 - 0.3552 = 0.0448 So, increment from 30%: 0.0448 / 0.00904 ≈ 4.96 ≈ 5% Therefore, **ROR ≈ 30% + 5% = 35%** --- ## Final Answer **The ROR on the extra investment for the EMT alternative is** \[ \boxed{35\%} \]