# Step-by-Step Solution ## **Given Data** ### **Old Car** - Market value now: $400 - Maintenance: $800/year - Fuel efficiency: 10 miles/gallon - Annual miles: 15,000 miles/year - Gas price: $3.50/gallon ### **New Car** - Purchase price: $8,000 (paid cash) - Maintenance: Negligible (due to warranty) - Fuel efficiency: 30 miles/gallon - Resale value after 2 years: $5,000 ### **Other Info** - Analysis period: 2 years - **MARR (discount rate): 15%** - Ignore taxes --- ## **Step 1: Calculate Costs for Each Option** ### **Old Car (Keep for 2 Years)** #### **a. Maintenance (Present Worth)** - $800/year for 2 years #### **b. Fuel Cost** - Annual fuel usage = 15,000 miles / 10 mpg = **1,500 gallons/year** - Annual fuel cost = 1,500 gallons × $3.50 = **$5,250/year** #### **c. Salvage Value** - Can be sold for $400 now (if not replaced), but if kept, assume negligible value after 2 years. #### **d. Total Costs (Cash Flows)** | Year | Item | Amount ($) | |------|--------------|------------| | | (Salvage value if sold) | | | 1 | Maintenance | -800 | | 1 | Fuel | -5,250 | | 2 | Maintenance | -800 | | 2 | Fuel | -5,250 | | 2 | Salvage Value| | --- ### **New Car (Buy Now, Sell in 2 Years)** #### **a. Initial Cost** - $8,000 (Year , cash outflow) #### **b. Maintenance** - Negligible (warranty covers for 2 years) #### **c. Fuel Cost** - Annual fuel usage = 15,000 / 30 = 500 gallons/year - Annual fuel cost = 500 × $3.50 = **$1,750/year** #### **d. Resale Value** - Sell for $5,000 at end of Year 2 #### **e. Total Costs (Cash Flows)** | Year | Item | Amount ($) | |------|--------------|------------| | | Purchase | -8,000 | | 1 | Fuel | -1,750 | | 2 | Fuel | -1,750 | | 2 | Salvage Value| +5,000 | --- ## **Step 2: Calculate Present Worth (PW) of Each Option** ### **Discount Factors** - Year 1: \( \frac{1}{(1+.15)^1} = .8696 \) - Year 2: \( \frac{1}{(1+.15)^2} = .7561 \) --- ### **Old Car: Present Worth** #### **PW = PV(Maintenance) + PV(Fuel)** #### **Year 1** - Maintenance: $800 × .8696 = $695.68 - Fuel: $5,250 × .8696 = $4,561.40 #### **Year 2** - Maintenance: $800 × .7561 = $604.88 - Fuel: $5,250 × .7561 = $3,969.53 #### **Total PW (Old Car)** \[ PW_{old} = \$695.68 + \$4,561.40 + \$604.88 + \$3,969.53 = \$9,831.49 \] (Salvage value at Year 2 is assumed $$) --- ### **New Car: Present Worth** #### **Year ** - Purchase: $8,000 #### **Year 1** - Fuel: $1,750 × .8696 = $1,522.80 #### **Year 2** - Fuel: $1,750 × .7561 = $1,323.18 - Salvage: $5,000 × .7561 = $3,780.50 (inflow) \[ PW_{new} = \$8,000 + \$1,522.80 + \$1,323.18 - \$3,780.50 = \$7,065.48 \] --- ## **Step 3: Decision and Final Answer** \[ PW_{old} = \$9,831.49 \\ PW_{new} = \$7,065.48 \] **Since the Present Worth (cost) of the new car is lower, you should REPLACE the old car with the new one.** --- ## **Summary Table** | Option | Present Worth (\$) | |-----------|-------------------| | Old Car | 9,831.49 | | New Car | 7,065.48 | --- ## **Assumptions Stated** - No resale value for old car after 2 years. - Negligible maintenance for the new car due to warranty. - Ignore income taxes. - All cash flows are at year-end except initial purchase/sale. - Analysis period is strictly 2 years. --- ## **Final Answer** > **You should replace the old car with the new car. The present worth of costs for the new car ($7,065.48) is less than that of keeping the old car ($9,831.49) over the two-year period at a 15% MARR.**

# MARR Definition and Subject Context ## **MARR Definition** MARR stands for **Minimum Attractive Rate of Return**. It is the lowest rate of return on an investment that an investor is willing to accept before undertaking the investment. This rate reflects the opportunity cost of capital and is used in capital budgeting to evaluate the attractiveness of an investment project. ## **Subject Context** The question relates to **Engineering Economy** or **Financial Engineering**. It involves analyzing the costs and benefits of two investment options (an old car versus a new car) using present worth calculations over a specified time period. This analysis helps in making informed financial decisions based on cash flows, maintenance costs, and salvage values.