Let's solve the problem step-by-step: ### **Given:** | Parameter | Hydraulic Press (HP) | Electric Machine (EMT) | |------------------------ |:--------------------:|:----------------------:| | First Cost | $680,000 | $800,000 | | Annual Cost | $200,000 | $152,000 | | Salvage Value (5 years) | $70,000 | $130,000 | | Life | 5 years | 5 years | **We are to:** Use an **Annual Worth (AW) equation** to determine the **Rate of Return (ROR)** on the **extra investment** required for the EMT alternative. --- ## **Step 1: Find the Incremental Cash Flows** We compare **EMT - HP** (since EMT is the more expensive alternative): - **Incremental First Cost**: $800,000 - $680,000 = **$120,000** - **Incremental Annual Cost**: $152,000 - $200,000 = **- $48,000** (annual *savings*) - **Incremental Salvage Value**: $130,000 - $70,000 = **$60,000** (extra at end) --- ## **Step 2: Set up the Annual Worth (AW) Equation** We want the ROR (**i**) that makes the **AW of the incremental investment zero**: \[ AW = - (Incremental\ First\ Cost) (A/P, i, 5) + (Incremental\ Annual\ Cost) + (Incremental\ Salvage) (A/F, i, 5) = 0 \] Plug in values: - First cost: **$120,000** (negative, as an outflow) - Annual cost: **- $48,000** (annual *savings*, so negative) - Salvage: **$60,000** (inflow at end) \[ -120,000(A/P, i, 5) - 48,000 + 60,000(A/F, i, 5) = 0 \] --- ## **Step 3: Express the P/A and F/A factors** - \((A/P, i, 5) = \dfrac{i(1+i)^5}{(1+i)^5 - 1}\) - \((A/F, i, 5) = \dfrac{i}{(1+i)^5 - 1}\) But for practical purposes, let's use trial and error (or Excel/financial calculator) to find **i**. Let's rearrange: \[ -120,000(A/P, i, 5) - 48,000 + 60,000(A/F, i, 5) = 0 \] \[ -120,000(A/P, i, 5) + 60,000(A/F, i, 5) = 48,000 \] --- ## **Step 4: Guess and Check ROR (Trial and Error)** Let's try **i = 20%** first: ### Compute Factors #### \((A/P, 20\%, 5)\): \[ (A/P, 20\%, 5) = \frac{0.20 \times (1.20)^5}{(1.20)^5 - 1} \] \[ (1.20)^5 = 2.48832 \] \[ = \frac{0.20 \times 2.48832}{2.48832 - 1} = \frac{0.497664}{1.48832} = 0.3345 \] #### \((A/F, 20\%, 5)\): \[ (A/F, 20\%, 5) = \frac{0.20}{2.48832 - 1} = \frac{0.20}{1.48832} = 0.1344 \] #### Plug into equation: \[ -120,000(0.3345) + 60,000(0.1344) = 48,000 \] \[ -40,140 + 8,064 = -32,076 \] Add annual savings: \[ -32,076 - 48,000 = -80,076 \neq 0 \] So, at 20%, the AW is negative. We need a much smaller rate. --- Try **i = 10%**: \[ (1.10)^5 = 1.61051 \] \[ (A/P, 10\%, 5) = \frac{0.10 \times 1.61051}{1.61051 - 1} = \frac{0.161051}{0.61051} = 0.2638 \] \[ (A/F, 10\%, 5) = \frac{0.10}{0.61051} = 0.1637 \] Plug in: \[ -120,000(0.2638) + 60,000(0.1637) = 48,000 \] \[ -31,656 + 9,822 = -21,834 \] \[ -21,834 - 48,000 = -69,834 \] Still negative, but less negative. --- Try **i = 0%**: - \((A/P, 0\%, 5) = 1/5 = 0.2\) - \((A/F, 0\%, 5) = 1/5 = 0.2\) \[ -120,000(0.2) + 60,000(0.2) = 48,000 \] \[ -24,000 + 12,000 = -12,000 \] \[ -12,000 - 48,000 = -60,000 \] --- Try **i = 50%**: \[ (1.5)^5 = 7.59375 \] \[ (A/P, 50\%, 5) = \frac{0.5 \times 7.59375}{7.59375 - 1} = \frac{3.79688}{6.59375} = 0.576 \] \[ (A/F, 50\%, 5) = \frac{0.5}{6.59375} = 0.0758 \] \[ -120,000(0.576) + 60,000(0.0758) = 48,000 \] \[ -69,120 + 4,548 = -64,572 \] \[ -64,572 - 48,000 = -112,572 \] Wait, as we increase i, the AW is getting more negative. Let's try **i = 5%**: \[ (1.05)^5 = 1.27628 \] \[ (A/P, 5\%, 5) = \frac{0.05 \times 1.27628}{1.27628 - 1} = \frac{0.063814}{0.27628} = 0.2310 \] \[ (A/F, 5\%, 5) = \frac{0.05}{0.27628} = 0.1809 \] \[ -120,000(0.2310) + 60,000(0.1809) = 48,000 \] \[ -27,720 + 10,854 = -16,866 \] \[ -16,866 - 48,000 = -64,866 \] --- Try **i = -10%**: \[ (0.9)^5 = 0.59049 \] \[ (A/P, -10\%, 5) = \frac{-0.1 \times 0.59049}{0.59049 - 1} = \frac{-0.059049}{-0.40951} = 0.1442 \] \[ (A/F, -10\%, 5) = \frac{-0.1}{-0.40951} = 0.2441 \] \[ -120,000(0.1442) + 60,000(0.2441) = 48,000 \] \[ -17,304 + 14,646 = -2,658 \] \[ -2,658 - 48,000 = -50,658 \] --- Let’s try **i = -50%**: \[ (0.5)^5 = 0.03125 \] \[ (A/P, -50\%, 5) = \frac{-0.5 \times 0.03125}{0.03125 - 1} = \frac{-0.015625}{-0.96875} = 0.01613 \] \[ (A/F, -50\%, 5) = \frac{-0.5}{-0.96875} = 0.516 \] \[ -120,000(0.01613) + 60,000(0.516) = 48,000 \] \[ -1,935.6 + 30,960 = 29,024.4 \] \[ 29,024.4 - 48,000 = -18,975.6 \] --- ### Let's try **i = -90%**: \[ (0.1)^5 = 0.00001 \] \[ (A/P, -90\%, 5) = \frac{-0.9 \times 0.00001}{0.00001 - 1} = \frac{-0.000009}{-0.99999} = 0.000009 \] \[ (A/F, -90\%, 5) = \frac{-0.9}{-0.99999} = 0.900009 \] \[ -120,000(0.000009) + 60,000(0.900009) = 48,000 \] \[ -1.08 + 54,000.54 = 53,999.46 \] \[ 53,999.46 - 48,000 = 5,999.46 \] So at i = -90%, AW is positive. --- ### Now, let's interpolate between i = -50% (AW = -18,976) and i = -90% (AW = +5,999). \[ i^* = -50\% + \left(\frac{0 - (-18,976)}{5,999 - (-18,976)}\right) \times (-90\% - (-50\%)) \] \[ = -50\% + \left(\frac{18,976}{24,975}\right) \times (-40\%) \] \[ = -50\% + (0.760) \times (-40\%) \] \[ = -50\% - 30.4\% = -80.4\% \] --- ## **Step 5: Final Answer** \[ \boxed{-80\%} \] --- ### **Conclusion** **The ROR on the extra investment for the EMT alternative is approximately \(\boxed{-80\%}\).** This large negative ROR means the extra investment in EMT is not justified by the annual savings and higher salvage value compared to HP. In fact, the extra cost cannot be recovered at any reasonable positive interest rate. --- ### **Summary Table** | | HP | EMT | Difference (EMT-HP) | |---|------------|------------|---------------------| | First Cost | $680,000 | $800,000 | $120,000 | | Annual Cost | $200,000 | $152,000 | -$48,000 | | Salvage Value | $70,000 | $130,000 | $60,000 | | Life (years) | 5 | 5 | | **AW Equation:** \[ -120,000(A/P, i, 5) - 48,000 + 60,000(A/F, i, 5) = 0 \] **Calculated ROR:** \(\boxed{-80\%}\)