Is it more economical to buy an automobile for $29,000 cash or to pay $4,500 down and $3,000 at the end of each quarter for 2 years, if the money is worth 8% compounded quarterly? what is the cash equivalent savings by picking the better plan?
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To determine the more economical option, we need to compare the total cost of each plan and consider the time value of money.
1. Calculating the cost of paying in cash:
The automobile costs $29,000. Since it is paid in cash, there are no additional charges or interest fees associated with this option. Therefore, the total cost remains $29,000.
2. Calculating the cost of paying in installments:
Considering that the installments are made at the end of each quarter for two years, there will be a total of 8 payments ($3,000 x 8 = $24,000) plus the initial down payment of $4,500. This results in a total cost of $28,500 when paying in installments.
Next, we need to consider the time value of money, which takes into account the interest earned or charged over time.
3. Calculating the cash equivalent savings:
First, we need to calculate the future value (FV) of the installment payments at the end of two years using the compound interest formula:
FV = PV * (1 + r/n)^(n*t)
Where:
PV = Present value (installment amount)
r = Annual interest rate (8%)
n = Number of compounding periods per year (4 quarters)
t = Number of years (2)
FV = $3,000 * (1 + 0.08/4)^(4*2)
FV = $3,000 * (1 + 0.02)^8
FV = $3,000 * (1.02)^8
FV ≈ $3,000 * 1.1723
FV ≈ $3,516.90
Now that we have the future value of the installment payments, we can find the equivalent cash savings by calculating the present value (PV) of this future value:
PV = FV / (1 + r/n)^(n*t)
PV = $3,516.90 / (1 + 0.08/4)^(4*2)
PV = $3,516.90 / (1 + 0.02)^8
PV = $3,516.90 / (1.02)^8
PV ≈ $3,516.90 / 1.1723
PV ≈ $2,999.78
Therefore, the cash equivalent savings by choosing the installment plan is approximately $2,999.78 - $28,500 = $1,500.22.
Thus, by selecting the installment plan, you would save approximately $1,500.22 compared to paying in cash upfront.