# Embroidery Machine Cost Analysis: Model B vs Model S G Fashions is considering two embroidery machine models. This analysis compares their costs over their respective useful lives, discounted at a Minimum Attractive Rate of Return (MARR) of 18% per year. --- ## 1. Input Data Summary | Model | Initial Cost (R) | Energy Cost/year (R) | Useful Life (yrs) | Salvage Value (R) | Maintenance Cost Pattern | |----------|------------------|----------------------|-------------------|-------------------|----------------------------------------------| | B | 110,000 | 6,000 | 10 | 0 | 2,500 in Year 1, then +1,200/year thereafter | | S | 150,000 | 4,500 | 15 | 15,000 | 1,700 in Year 7, then +800/year thereafter | --- ## 2. Maintenance Cost Calculations ### Model B - Year 1: R2,500 - Year 2: R2,500 + R1,200 = R3,700 - Year 3: R3,700 + R1,200 = R4,900 - Continue incrementing by R1,200 each year up to Year 10. | Year | Maintenance Cost (R) | |------|---------------------| | 1 | 2,500 | | 2 | 3,700 | | 3 | 4,900 | | 4 | 6,100 | | 5 | 7,300 | | 6 | 8,500 | | 7 | 9,700 | | 8 | 10,900 | | 9 | 12,100 | | 10 | 13,300 | --- ### Model S - Years 1–6: R0/year (no maintenance) - Year 7: R1,700 - Year 8: R1,700 + R800 = R2,500 - Year 9: R2,500 + R800 = R3,300 - Continue incrementing by R800 each year up to Year 15. | Year | Maintenance Cost (R) | |------|---------------------| | 1–6 | 0 | | 7 | 1,700 | | 8 | 2,500 | | 9 | 3,300 | | 10 | 4,100 | | 11 | 4,900 | | 12 | 5,700 | | 13 | 6,500 | | 14 | 7,300 | | 15 | 8,100 | --- ## 3. Present Worth (PW) Calculation Approach \[ PW = C_0 + \sum_{n=1}^{N} \frac{C_n}{(1+i)^n} - \frac{S}{(1+i)^N} \] Where: - \( C_0 \): Initial Cost - \( C_n \): Annual Costs (energy + maintenance) - \( i \): MARR (0.18) - \( S \): Salvage Value (if any) - \( N \): Useful Life We need to find the Present Worth (PW) for both models. --- ## 4. Present Worth Calculation ### Model B **Initial cost:** R110,000 **Energy cost:** R6,000/year for 10 years **Maintenance:** As calculated above **Salvage value:** R0 **MARR:** 18% **Useful Life:** 10 years #### a. Energy Cost Present Worth \[ PW_{energy,B} = 6,000 \times \frac{1 - (1+0.18)^{-10}}{0.18} \] \[ PW_{energy,B} = 6,000 \times 4.947 = 29,682 \text{ (rounded to nearest R)} \] #### b. Maintenance Cost Present Worth Each year's cost discounted: \[ PW_{maint,B} = \sum_{n=1}^{10} \frac{C_n}{(1.18)^n} \] | Year | Cost (R) | Discount Factor (1.18^n) | Present Value (R) | |------|----------|------------------------|-------------------| | 1 | 2,500 | 1.18 | 2,119 | | 2 | 3,700 | 1.3924 | 2,659 | | 3 | 4,900 | 1.6420 | 2,984 | | 4 | 6,100 | 1.9375 | 3,149 | | 5 | 7,300 | 2.2872 | 3,193 | | 6 | 8,500 | 2.6989 | 3,150 | | 7 | 9,700 | 3.1857 | 3,045 | | 8 | 10,900 | 3.7571 | 2,902 | | 9 | 12,100 | 4.4334 | 2,730 | | 10 | 13,300 | 5.2334 | 2,541 | **Total Present Worth (Maintenance):** \[ PW_{maint,B} \approx 28,472 \] #### c. Total Present Worth (Model B) \[ PW_B = 110,000 + 29,682 + 28,472 = 168,154 \] --- ### Model S **Initial cost:** R150,000 **Energy cost:** R4,500/year for 15 years **Maintenance:** As above **Salvage value:** R15,000 at year 15 **MARR:** 18% **Useful Life:** 15 years #### a. Energy Cost Present Worth \[ PW_{energy,S} = 4,500 \times \frac{1 - (1+0.18)^{-15}}{0.18} \] \[ PW_{energy,S} = 4,500 \times 5.556 = 25,002 \] #### b. Maintenance Cost Present Worth | Year | Cost (R) | Discount Factor (1.18^n) | Present Value (R) | |------|----------|------------------------|-------------------| | 1–6 | 0 | - | 0 | | 7 | 1,700 | 3.1857 | 534 | | 8 | 2,500 | 3.7571 | 666 | | 9 | 3,300 | 4.4334 | 744 | | 10 | 4,100 | 5.2334 | 783 | | 11 | 4,900 | 6.1746 | 794 | | 12 | 5,700 | 7.0870 | 805 | | 13 | 6,500 | 8.3637 | 777 | | 14 | 7,300 | 9.8702 | 740 | | 15 | 8,100 | 11.6508 | 695 | **Sum (Maintenance PW):** \[ PW_{maint,S} \approx 6,538 \] #### c. Salvage Value Present Worth \[ PW_{salvage,S} = \frac{15,000}{(1.18)^{15}} \approx \frac{15,000}{14.14} = 1,061 \] #### d. Total Present Worth (Model S) \[ PW_S = 150,000 + 25,002 + 6,538 - 1,061 = 180,479 \] --- ## 5. Decision | Model | Present Worth (R) | |----------|------------------| | B | 168,154 | | S | 180,479 | - **Lower present worth is better (lower cost).** - **Model B** is the more economical option. --- ## 6. Conclusion **G Fashions should buy Model B**. It has a lower present worth cost (R168,154) compared to Model S (R180,479), when all costs and the time value of money at 18% MARR are considered. --- **Note:** All calculations are rounded; if absolute precision is required, use a finance calculator or spreadsheet for discounting each cash flow. **Assumptions:** Yearly costs occur at year-end, and all values are in South African Rand (R).