# Step-by-Step Solution Let's determine which machine to buy using **Annual Worth (AW)** analysis at a **MARR of 12%**. We'll convert all costs to equivalent uniform annual costs over each machine's life. --- ## **Given Data** | | Machine X | Machine Y | |---|---|---| | Initial Cost | $12,000 | $15,000 | | Annual Labour Cost | $4,000 in year 1, increasing $200/year | $3,500 in year 1, increasing $250/year | | Annual Maintenance Cost | $400 in year 1, increasing $100/year | $350 in year 1, increasing $150/year | | Overhaul (every 2 years) | $2,100 | $2,100 | | Salvage Value | $1,500 | $3,000 | | Life (years) | 8 | 6 | | MARR | 12% | 12% | --- ## **Step 1: Calculate Present Worth (PW) of Each Cost Component** ### **A. Initial Cost** Already present value: - Machine X: $12,000 - Machine Y: $15,000 --- ### **B. Salvage Value (PW)** \[ PW_{SV} = \text{SV} \times (P/F, i, n) \] Where: - \(\text{SV}\) = Salvage Value - \(i = 12\%\) - \(n = \) machine life #### Machine X: \[ PW_{SV,X} = 1,500 \times (P/F, 12\%, 8) = 1,500 \times (1+0.12)^{-8} = 1,500 \times 0.4039 = \$606 \] #### Machine Y: \[ PW_{SV,Y} = 3,000 \times (P/F, 12\%, 6) = 3,000 \times (1+0.12)^{-6} = 3,000 \times 0.5066 = \$1,520 \] --- ### **C. Annual Labour Cost (PW of Arithmetic Gradient Series)** Labour cost is a gradient series: - \(A_1\) = first year cost - \(G\) = annual increase PW for gradient series over \(n\) years: \[ PW = A_1(P/A, i, n) + G(P/G, i, n) \] Where: - \(P/A\) = present worth factor for uniform series - \(P/G\) = present worth factor for gradient series #### Find Factors: - \((P/A, 12\%, 8) = \frac{1-(1+0.12)^{-8}}{0.12} = 4.968\) - \((P/A, 12\%, 6) = 4.111\) - \((P/G, 12\%, 8) = \frac{1-(1+0.12)^{-8}}{0.12^2} - \frac{8}{0.12} = 21.15 - 66.67 = -45.52\) (But let's use the correct formula: \((P/G, i, n) = \frac{1}{i} \left[\frac{1-(1+i)^{-n}}{i} - n(1+i)^{-n}\right]\)) Let’s calculate \((P/G, 12\%, 8)\): \[ (P/G, 12\%, 8) = \frac{1}{0.12}\left[\frac{1 - (1.12)^{-8}}{0.12} - 8 \times (1.12)^{-8}\right] \] \[ = 8.333 \left[4.968 - 8 \times 0.4039 \right] = 8.333 \left[4.968 - 3.231\right] = 8.333 \times 1.737 = 14.475 \] Similarly for 6 years: \[ (P/G, 12\%, 6) = \frac{1}{0.12}\left[\frac{1 - (1.12)^{-6}}{0.12} - 6 \times (1.12)^{-6}\right] = 8.333 \left[4.111 - 6 \times 0.5066 \right] = 8.333 \left[4.111 - 3.04\right] = 8.333 \times 1.071 = 8.924 \] #### Machine X (Labour): - \(A_1 = 4,000\), \(G = 200\), \(n = 8\) \[ PW_{Lab,X} = 4,000 \times 4.968 + 200 \times 14.475 = 19,872 + 2,895 = \$22,767 \] #### Machine Y (Labour): - \(A_1 = 3,500\), \(G = 250\), \(n = 6\) \[ PW_{Lab,Y} = 3,500 \times 4.111 + 250 \times 8.924 = 14,388.5 + 2,231 = \$16,619.5 \] --- ### **D. Annual Maintenance Cost (PW of Arithmetic Gradient Series)** #### Machine X (Maintenance): - \(A_1 = 400\), \(G = 100\), \(n = 8\) \[ PW_{Maint,X} = 400 \times 4.968 + 100 \times 14.475 = 1,987.2 + 1,447.5 = \$3,434.7 \] #### Machine Y (Maintenance): - \(A_1 = 350\), \(G = 150\), \(n = 6\) \[ PW_{Maint,Y} = 350 \times 4.111 + 150 \times 8.924 = 1,438.85 + 1,338.6 = \$2,777.5 \] --- ### **E. Overhaul Every 2 Years (PW of Recurring Cost)** PW of $2,100 every 2 years: - For Machine X (\(n = 8\)): at years 2, 4, 6, 8 - For Machine Y (\(n = 6\)): at years 2, 4, 6 \[ PW = 2,100 \times [(P/F, 12\%, 2) + (P/F, 12\%, 4) + (P/F, 12\%, 6) + (P/F, 12\%, 8)] \] First, calculate \((P/F, 12\%, n)\): - Year 2: \((1.12)^{-2} = 0.7972\) - Year 4: \((1.12)^{-4} = 0.6355\) - Year 6: \((1.12)^{-6} = 0.5066\) - Year 8: \((1.12)^{-8} = 0.4039\) #### Machine X: \[ PW_{Ovhl,X} = 2,100 \times (0.7972 + 0.6355 + 0.5066 + 0.4039) = 2,100 \times 2.3432 = \$4,920.7 \] #### Machine Y (only to year 6): \[ PW_{Ovhl,Y} = 2,100 \times (0.7972 + 0.6355 + 0.5066) = 2,100 \times 1.9393 = \$4,072.5 \] --- ## **Step 2: Find Net Present Worth (NPW) for Each Machine** ### **Machine X** \[ NPW_X = -12,000 - 22,767 - 3,434.7 - 4,920.7 + 606 = -12,000 - 22,767 - 3,434.7 - 4,920.7 + 606 = -43,122.4 \] ### **Machine Y** \[ NPW_Y = -15,000 - 16,619.5 - 2,777.5 - 4,072.5 + 1,520 = -15,000 - 16,619.5 - 2,777.5 - 4,072.5 + 1,520 = -36,949.5 \] --- ## **Step 3: Convert NPW to Equivalent Uniform Annual Worth (AW)** \[ AW = NPW \times (A/P, i, n) \] Where \((A/P, 12\%, 8) = 0.1760,\ (A/P, 12\%, 6) = 0.2432\) ### **Machine X** (\(n = 8\)): \[ AW_X = -43,122.4 \times 0.1760 = -\$7,593 \] ### **Machine Y** (\(n = 6\)): \[ AW_Y = -36,949.5 \times 0.2432 = -\$8,989 \] --- ## **Step 4: Decision** - **AW for Machine X = \(-\$7,593\)** - **AW for Machine Y = \(-\$8,989\)** Since we want the **least negative (highest) annual worth**, **Machine X** is preferred. --- # **Final Answer** > **Machine X should be purchased, because it has a higher (less negative) annual worth at a 12% MARR.** --- If you need to see any step in more detail (e.g., the arithmetic gradient formulas), let me know!