0
$\begingroup$

I live in Argentina, and we have 30% anual rate of inflation.

I am paying a car also, in a plan of 84 payments. Each payment has a 27.7% of "Administrative Costs".

For example, for a pure fee (monthly) of $1807, I am paying $2306.

Current cost of car is $160,000. Each monthly fee is $160,000/84. (7 years).

Banks offer an annual rate of 23.4% of interest for monthly deposits.

My goal is to know, if I will be better depositing money in Bank than paying $500 of "Administrative Costs" each month.

But I don't know what formula to use in Excel, because, for example in the last part of the Excel, it gives me that for a $160,000 car, with 30% inflation, in 1 year it will cost $208,000, but the monthly calculation ends with $209933.87, so, I think I'm doing something wrong in the formula.

https://docs.zoho.com/sheet/published.do?rid=12cbm43aa83422b1c4c9eb327fdde40b43d1b

$\endgroup$
1
  • $\begingroup$ out of the topic $\endgroup$
    – mahavir
    Mar 6, 2014 at 16:59

1 Answer 1

1
$\begingroup$

For a single event of $30\%$ inflation, after one year the price goes from $160,000$ to $208,000$ as you say. If the inflation is $2.5\%$ per month and you apply it every month, in one year the $160,000$ goes to $160,000(1.025)^{12}\approx 215182.21$ The difference represents the effect of compound interest. Normally annual inflation should take account of the compound interest, so if it were truly $2.5\%$ every month the annual value would be $1.025^{12}\approx 34.49\%$. I didn't check your spreadsheet-without some explanation it is quite difficult.

$\endgroup$
2
  • $\begingroup$ So, how can I convert a 30% annual inflation into monthly inflation, and get the same result? $\endgroup$
    – JorgeeFG
    Mar 6, 2014 at 17:16
  • $\begingroup$ You are looking for $(1+x)^{12}=1.3, x=1.3^{(1/12)}-1, x\approx .0221$, so it is about $2.21\%$ per month to get $30\%$ per year. $\endgroup$ Mar 6, 2014 at 17:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .