Let's solve the problem step by step. ## 1. **Restate and Organize the Data** ### Used Machine (remaining life: 4 years) - **Original Cost:** $16,050 (sunk cost, ignore) - **Current Salvage Value:** $11,048 (if sold now) - **Future Scrap Value:** $5,804 (after 4 years) - **O&M Costs:** Year 1: $1,722; Years 2-4: $1,983 per year ### New Machine (if bought now, 4 years) - **Initial Cost:** $12,799 - **Future Scrap Value:** $7,060 (after 4 years) - **O&M Costs:** Year 1: $1,360; Years 2-4: $1,592 per year ### **MARR:** 11% (use as discount rate) ## 2. **Calculate EUAC for Both Alternatives** ### Step 1: **Used Machine** #### **A. Find Present Worth (PW) of Costs if Kept** **Relevant Costs:** - Opportunity cost: If kept, forgo $11,048 (could have by selling now) - O&M: Year 1: $1,722; Years 2-4: $1,983/year - Salvage value received in 4 years: $5,804 #### **Cash Flow Diagram:** - Year 0: +$11,048 (if sold), or -$11,048 (opportunity cost if kept) - Year 1: -$1,722 (O&M) - Years 2-4: -$1,983 (O&M per year) - Year 4: +$5,804 (salvage) #### **Calculate Present Worth (PW):** - O&M (Year 1): $1,722 / (1.11)^1 = $1,550.45 - O&M (Year 2): $1,983 / (1.11)^2 = $1,608.77 - O&M (Year 3): $1,983 / (1.11)^3 = $1,449.27 - O&M (Year 4): $1,983 / (1.11)^4 = $1,305.65 - Salvage in Year 4: $5,804 / (1.11)^4 = $5,804 / 1.5181 = $3,823.29 Now sum up: \[ PW = -11,048 - 1,550.45 - 1,608.77 - 1,449.27 - 1,305.65 + 3,823.29 \] \[ PW = -11,048 - (1,550.45 + 1,608.77 + 1,449.27 + 1,305.65) + 3,823.29 \] \[ PW = -11,048 - 5,914.14 + 3,823.29 \] \[ PW = -11,048 - 2,090.85 \] \[ PW = -13,138.85 \] #### **B. Calculate EUAC for Used Machine** EUAC formula: \[ EUAC = PW \times \frac{A/P, i, n} \] Where \( A/P, i, n \) is the capital recovery factor: \[ A/P = \frac{i(1+i)^n}{(1+i)^n - 1} \] For \( i = 11\% \), \( n = 4 \): \[ (1+i)^n = 1.11^4 = 1.5181 \] \[ A/P = \frac{0.11 \times 1.5181}{1.5181 - 1} = \frac{0.16699}{0.5181} = 0.3224 \] \[ EUAC_{used} = -13,138.85 \times 0.3224 = -4,238.98 \] (Take absolute value for cost, as EUAC is a cost.) --- ### Step 2: **New Machine** #### **A. Find Present Worth (PW) of Costs if Bought** - Initial Cost: $12,799 (Year 0) - O&M: Year 1: $1,360; Years 2-4: $1,592/year - Salvage Value: $7,060 (Year 4) O&M: - Year 1: $1,360 / 1.11 = $1,225.23 - Year 2: $1,592 / (1.11)^2 = $1,292.87 - Year 3: $1,592 / (1.11)^3 = $1,164.74 - Year 4: $1,592 / (1.11)^4 = $1,049.02 Salvage: - $7,060 / 1.5181 = $4,648.98 \[ PW = -12,799 - 1,225.23 - 1,292.87 - 1,164.74 - 1,049.02 + 4,648.98 \] \[ PW = -12,799 - (1,225.23 + 1,292.87 + 1,164.74 + 1,049.02) + 4,648.98 \] \[ PW = -12,799 - 4,731.86 + 4,648.98 \] \[ PW = -12,799 - 82.88 \] \[ PW = -12,881.88 \] #### **B. Calculate EUAC for New Machine** \[ EUAC_{new} = -12,881.88 \times 0.3224 = -4,153.91 \] --- ## 3. **Compare and Select the Optimal Option** - **Used Machine EUAC:** $4,238.98 - **New Machine EUAC:** $4,153.91 ### **Conclusion:** The lowest EUAC is for the **new machine**. --- ## **Final Answer** \[ \boxed{4153.91} \] --- ### **Summary Table** | Option | EUAC | |---------------|-----------| | Used Machine | 4238.98 | | New Machine | 4153.91 | --- ### **Final Answer (No commas, no currency, no spaces, rounded to two decimals):** **4153.91**