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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.

Please help me to get this.I'm struggling for this for a long while. Any help will be appreciated. Thanks in advance.

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3 Answers 3

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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.

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  • 10^5 * 1.06^2 = 112,360, not 111,641.
    – RonJohn
    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.
    – RonJohn
    Jan 18, 2018 at 14:23
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    @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).
    – RonJohn
    Jan 18, 2018 at 16:21
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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

See http://financeformulas.net/Real_Rate_of_Return.html

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also https://money.stackexchange.com/a/56847/11768

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

Additional note by RonJohn

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.

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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.

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