# How do we include inflation in our calculations? [closed]

How do we include inflation in our compound interest calculations? E.g. if we have current principal of 1000$ and the interest rate is 3% after 10 years we have 1344$ (used this calculator)
But if for this exercise we wanted to take inflation into account let's say 2% how would that be part of our formula?

Update:
I know that the number including the inflation is 1102$. I don't know exactly how to do the calculations to get the 1102$. Getting the 1334$ is straightforward but I am confused on how to include inflation to get the 1102$

• Hi Jim, welcome to Quant.SE! You would still have \$1334, it's just worth less in real terms. To get the amount in real terms you can substract the inflation from the interest rate, this gives \$1105 for me. However, questions of this type are considered too basic here. Dec 25, 2015 at 17:04
• @BobJansen:Is there a more appropriate SE for my question then?
– Jim
Dec 27, 2015 at 11:26
• I don't know, but I think the information here should be sufficient. You might have made a rounding error in your last comment to @Brumder, see this calculation and note that $1000 \times 1.03^{10} = 1344 > 1334$. Dec 27, 2015 at 11:39

This calculator does not include inflation in whatever interest rate you specify (I checked). Usually, the rate quoted by banks is the nominal interest rate, which is simply how much your capital will appreciate with inflation (e.g. higher inflation would yield higher returns). It does not take into account purchasing power and is calculated as follows:

Nominal Rate = (1 + Real Interest Rate)(1 + Inflation Rate) - 1


In terms of trying to "take inflation into account" I'm not totally sure what you mean. If you mean factor inflation into the equation so that the purchasing power of the future value is quoted in today's dollar amount, the simplest way would be to lower the rate you entered into the calculator by expected future inflation (assuming you're using the commonly cited nominal rate or using the interest rate as a proxy for total return). You could also divide the future value from the calculator by:

(1 + expected inflation)^n

• What I mean by "take inflation into account" is that if I have an amount and the interest rate I can just do the sum over the years to figure out the final amount (in my example 1344$after 10 years with 3% interest rate). What I want is how do I deduct the inflation from this 1344$?
– Jim
Dec 24, 2015 at 20:03
• You're not giving me any clues behind your assumption for a 3% interest rate and what the FV of 1334 equals. I'll assume, again, that you're thinking the interest rate specified does not account for inflation. Therefore, if you want to factor in inflation of 2% annually, just adjust the interest rate down 2%. You'll get an inflation-adjusted final value. Dec 24, 2015 at 20:15
• Yes the interest rate of 3% does not include inflation. I was just saying that having 1000$ starting sum after 10 years with 3% interest we get 1334$. This calculation/rate did not include inflation. So if the interest rate was 3% over these years and the inflation rate was 2% how do we calculate what we get after the 10 years? I think it is 1334 - X (where X is the inflation errosion). So how can I calculate this X?
– Jim
Dec 24, 2015 at 20:34
• That's exactly what I wrote at the end of my answer. Take 1334 & divide it by (1 + expected inflation)^n. YOU put the interest rate in, so my entire point earlier was that I'm assuming YOU did not include inflation expectations in your rate. Simple answer now: (1334) / (1 + expected inflation %)^n. Dec 24, 2015 at 20:41
• I see what you mean. The issue I have with this is that because I know that the number after calculating inflation is 1102 (I know the number but did not know how it is derived) I can't get this by using the formula 1334/((1.02)^10)
– Jim
Dec 24, 2015 at 22:50

Further to a post here, you can appreciate by the interest rate and depreciate by the inflation rate at the same time like this:

principal       p = 1000
interest rate   r = 0.03
inflation       i = 0.02
number of years n = 10

p (1 + r)^n (1 + i)^-n = 1102.48


The calculation can be simplified with a factor x:

x = i (1 + r)/(1 + i) = 0.0201961

p (1 + (r - x))^n = 1102.48


These calculations give the future value in 10 years appreciated by interest at 3% but depreciated by inflation at 2%.

However, it differs from the example here, which - like Brumder's initial answer - counts inflation as a appreciating factor.

• Yes 1102 is the number that I know is the result but did not know how to derive. What is the intuitive explanation of the formula p (1 + r)^n (1 + i)^-n (the post you linked to is helpful but does only a generalization)
– Jim
Dec 24, 2015 at 22:53
• Inflation is the increase in the price of goods, meaning today's dollar buys less in the future. In your example the discount to adjust for ten years of inflation is is calculated by $1000/(1 + 0.02)^10 =$820.35; the value of \$1000 in ten years. The other component is straightforward accumulation of interest: (1 + 0.03)^10. Dec 25, 2015 at 0:18
• But if I subtract the (1000- 820.35 = 179.65) from the 1334 we don't get the 1102.
– Jim
Dec 25, 2015 at 11:41