# How rate of interest affects inflation rate [duplicate]

Suppose i invest 100000 today.and suppose rate of interest is 12% and inflation rate is 6% . so what amount i will get after 2 year at the rate of 12% and inflation of 6%.

In other word you can say,how the rate of interest affects inflation rate.

Inflation doesn’t affect your nominal investment return at all, only its spending power. If you invest \$100,000 for two years at 12% per year, you will have \$125,440, whatever the inflation rate is. If the inflation rate for the things you want to buy (which may be different from the headline inflation rate) was 6% per year, then that \$125,440 will only buy you as much as \$111,641 would have at the time you made the initial investment.

• `10^5 * 1.06^2 = 112,360`, not `111,641`. Jan 18, 2018 at 11:03
• @RonJohn see Chris Degnen's answer for why 112,360 is wrong.
– ssn
Jan 18, 2018 at 11:15
• @ssn `10^5 * 1.06^2 = 112,360` most certainly is mathematically correct as to what 10^5 is after two years of inflation. `111,641` is the answer to a different problem: `1.12^2 / 1.06^2 = 1.11641153`. Jan 18, 2018 at 14:23
• @RonJohn No 111,641 is the answer to this problem. 112,360 is how much you should have in `t + 2` to equal 100,000 today. Yes the math is correct isolated - but that is the answer to a different problem.
– ssn
Jan 18, 2018 at 15:48
• @ssn you seem to be assuming I don't know the different between `10^5 * 1.06^2` (which is 112,360) and `10^5 * 1.12^2/1.06^2` (which is 111,641). Jan 18, 2018 at 16:21

With principal, interest rate and inflation

``````p = 100000
r = 0.12
i = 0.06
``````

after two years you have

``````p (1 + r)^2 = 125440.00
``````

However, accounting for inflation, in today's value that is

``````125440/(1 + i)^2 =  111641.15
``````

This is the same as adjusting for inflation after each compounding period, which would be necessary if there were intervening cash flows.

``````year1 =     p (1 + r)/(1 + i) = 105660.38
year2 = year1 (1 + r)/(1 + i) = 111641.15
``````

The quick and dirty method you may find mentioned elsewhere is

``````p (1 + (r - i))^2 = 112360.00
``````

but that's just lazy and wrong.

It is more rigorous to use an extra step calculating `x`

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

p (1 + (r - x))^2 = 111641.15
``````

The Q&D method is wrong because `(1 + (r - i))^2 where r=12% and i=6%` reduces to `1.06^2`, whereas `1.12^A` grows faster than `1.06^A`.

The balance of your investment investment account will grow at 12% APR no matter what the rate of inflation.

This means that after two years the account value will be 1.122 = 1.2544 times more than the starting balance.

But... since the value of money has -- in some economist's delusionally uniform world -- decreased by 6%/annum for two years, you now require 1.062 = 1.1236 times more than the starting balance just to be able to buy what you did two years ago.

You'd think that because 12% is 2x as large as 6% that you'd have 2x as much money every year. Not so, because compounding is a power function while ratios are linear.

The formula for the yearly percentage difference between a 12% investment and 6% inflation is calculated by `((1.12^A)-1)/((1.06^A)-1) - 2` where A is the number of years since the start.

This is derived from:

``````1.12^A/1.06^A
``````

Correct: I've not mentioned your 100000 starting amount, because it's irrelevant. Multiply your starting balance (whatever it's size, whether 20 or 80 or 10^5 or 4.2*10^72) by 1.06^A and 1.12^A to get the values for each year `A`. And naturally, the `.12` and `.06` can be changed to any number you'd like depending on the investment and inflation rates.