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P ö ly - Chem Plastics is considering two types of injection molding machines: hydraulic and electric. The hydraulic press ( HP ) will have a first cost of $ 6 8 0 , 0 0 0 , annual costs of $ 2 0 0 , 0 0 0 , and a salvage value of $ 7 0 , 0 0 0 after 5 years. Electric machine technology ( EMT ) will have a first cost of $ 8 0 0 , 0 0 0 , annual costs of $ 1 5 2 , 0 0 0 , and a salvage value of $ 1 3 0 , 0 0 0 after 5 years. NOTE: This is a multi - part question. Once an answer is submitted, you will be unable to return to this part. Use an AW - based equation to determine the ROR on the extra investment required for the EMT alternative. The ROR on the extra investment for the EMT alternative is ◻ % .SO a ga JF ~~ A ad oy a % ade ee oe Si RY 52 Aes = ~3 (NS Fo SAD Or Fw Bon TRE fn ei SR 3 Cpr ia SE RK = Fo Si RSSOE gen A, a 5 > XH oa) GD on 2 SESLEELE Ras MTA — a LE 5 ee Lv Ee la __ A - fomework Seved r Required information Pély-Chem Plastics is considering two types of injection molding machines: hydraulic and electric. The hydraulic press (HP) will have a first cost of $680,000, annual costs of $200,000, and a salvage value of $70,000 after 5 years. Electric machine technology (EMT) will have a first cost of $800,000, annual costs of $152,000, and a salvage value of $130,000 after 5 years NOTE: This is a multi-part question. Once an answer is fs EX ” — X( ( (( { TE CM) (Ge (C (

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P ö ly - Chem Plastics is considering two types of injection molding machines: hydraulic and electric. The hydraulic press ( HP ) will have a first cost of $ 6 8 0 , 0 0 0 , annual costs of $ 2 0 0 , 0 0 0 , and a salvage value of $ 7 0 , 0 0 0 after 5 years. Electric machine technology ( EMT ) will have a first cost of $ 8 0 0 , 0 0 0 , annual costs of $ 1 5 2 , 0 0 0 , and a salvage value of $ 1 3 0 , 0 0 0 after 5 years. NOTE: This is a multi - part question. Once an answer is submitted, you will be unable to return to this part. Use an AW - based equation to determine the ROR on the extra investment required for the EMT alternative. The ROR on the extra investment for the EMT alternative is ◻ % .Uploaded ImageSO a ga JF ~~ A ad oy a % ade ee oe Si RY 52 Aes = ~3 (NS Fo SAD Or Fw Bon TRE fn ei SR 3 Cpr ia SE RK = Fo Si RSSOE gen A, a 5 > XH oa) GD on 2 SESLEELE Ras MTA — a LE 5 ee Lv Ee la __ A - fomework Seved r Required information Pély-Chem Plastics is considering two types of injection molding machines: hydraulic and electric. The hydraulic press (HP) will have a first cost of $680,000, annual costs of $200,000, and a salvage value of $70,000 after 5 years. Electric machine technology (EMT) will have a first cost of $800,000, annual costs of $152,000, and a salvage value of $130,000 after 5 years NOTE: This is a multi-part question. Once an answer is fs EX ” — X( ( (( { TE CM) (Ge (C (

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Asked by: Veera manikanta
Created at: 2025-07-29 10:34:21

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Let's break down the problem and solve for the **Rate of Return (ROR)** on the extra investment for the EMT alternative using the **Annual Worth (AW)** approach. ### **Given Data** | | HP | 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 (EMT - HP)** - **ΔFirst Cost**: $800,000 - $680,000 = **$120,000** - **ΔAnnual Cost**: $152,000 - $200,000 = **-$48,000** - **ΔSalvage Value**: $130,000 - $70,000 = **$60,000** --- ### **Step 2: Set Up the AW Equation** The incremental cash flow is: - Initial Cost: **-$120,000** - Annual Cost: **+$48,000** (since it's an annual savings) - Salvage Value: **+$60,000** (received at year 5) The AW equation for ROR (\(i\)) is: \[ 0 = -\$120,000 + \$48,000 \cdot (P/A, i, 5) + \$60,000 \cdot (P/F, i, 5) \] Where: - \( (P/A, i, 5) \) is the Present Worth factor for an annuity (5 years) - \( (P/F, i, 5) \) is the Present Worth factor for a single future amount --- ### **Step 3: Plug in the Factors** \[ 0 = -120,000 + 48,000(P/A, i, 5) + 60,000(P/F, i, 5) \] Where: - \( (P/A, i, 5) = \frac{1 - (1+i)^{-5}}{i} \) - \( (P/F, i, 5) = (1+i)^{-5} \) --- ### **Step 4: Solve for ROR (Iteratively or Using Financial Calculator)** Let's try an approximate solution by trial and error or spreadsheet: #### Let’s try **i = 30%** (0.30): \[ (P/A, 0.30, 5) = \frac{1 - (1.3)^{-5}}{0.3} = \frac{1 - 0.237}{0.3} = \frac{0.763}{0.3} \approx 2.543 \] \[ (P/F, 0.30, 5) = (1.3)^{-5} \approx 0.237 \] Plug into the equation: \[ 0 = -120,000 + 48,000 \times 2.543 + 60,000 \times 0.237 \] \[ 0 = -120,000 + 122,064 + 14,220 \] \[ 0 = 16,284 \] This is positive, so try a higher rate. #### Try **i = 40%** (0.40): \[ (P/A, 0.40, 5) = \frac{1 - (1.4)^{-5}}{0.4} = \frac{1 - 0.186}{0.4} = \frac{0.814}{0.4} = 2.035 \] \[ (P/F, 0.40, 5) = (1.4)^{-5} = 0.186 \] Plug in: \[ 0 = -120,000 + 48,000 \times 2.035 + 60,000 \times 0.186 \] \[ 0 = -120,000 + 97,680 + 11,160 \] \[ 0 = -120,000 + 108,840 = -11,160 \] Now it's negative. So the actual ROR is between **30%** and **40%**. --- ### **Step 5: Interpolate** Linear interpolation: \[ \text{ROR} \approx 30\% + \frac{16,284}{16,284 + 11,160} \times (40\% - 30\%) \] \[ = 30\% + \frac{16,284}{27,444} \times 10\% \] \[ = 30\% + 0.5936 \times 10\% \] \[ = 30\% + 5.94\% \] \[ \approx \boxed{36\%} \] --- ## **Final Answer** **The ROR on the extra investment for the EMT alternative is approximately:** \[ \boxed{36\%} \] --- If you want a more exact number, use a financial calculator or Excel's IRR function for more precise calculation. But **36%** is a well-interpolated answer.

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