Formula Help

neil.p.b

Hi,

I was wondering if anyone knew if there was a formula for the following kind of problem:

I have a sum of money, say 10000. Each year for ten years this is reduced by 5%. So at the end of year one there is 9500 left, year 2 9025, etc. What i want to know is over the ten years what is the average amount of money. Is there any formula that will make it easier to calculate this without having to individually reduce the amount by 5% each year and then averaging all ten year end totals?

TIA

Neil

You can multiply r^y by your initial amount, where r is the percentage rate (5%) as a fraction (5% is 19/20), y is the number of years you'd like to calculate it over, and ^ symbolises "to the power of."

So, in your case, r^y would be (19/20)^10, which is 0.598737, and if you multiply that by your initial amount you get 5987.3693, rounded to two decimal places is 5987.37.

neil.p.b

gvn wrote: »

You can multiply r^y by your initial amount, where r is the percentage rate (5%) as a fraction (5% is 19/20), y is the number of years you'd like to calculate it over, and ^ symbolises "to the power of."

So, in your case, r^y would be (19/20)^10, which is 0.598737, and if you multiply that by your initial amount you get 5987.3693, rounded to two decimal places is 5987.37.

GVN,

Thanks for the reply. The problem with the above is it gets only gets the year 10 value. What i'm looking for is the average over the 10 years.

Neil

neil.p.b

neil.p.b wrote: »

GVN,

Thanks for the reply. The problem with the above is it gets only gets the year 10 value. What i'm looking for is the average over the 10 years.

Neil

Oh right, I understand what you're saying now; I misunderstood you originally.

You want to find the average reduction? i.e. ("reduction year 1" + "reduction year 2" + ... + "reduction year 10") / 10.

If that's the case, then would the average reduction not just be something like:

[(Initial amount) - (r^y)]/y

So, (10000 - 5987.37)/10, which is 401.26, meaning the amount is reduced, on average, by 401.26 each year.

Perhaps I'm being overly simplistic or misunderstanding you again. If so, hopefully an accountant will spot this and help you out.

Edit: scratch the above, I think I misunderstood you (yet again!).

764dak

Multiplier = [r(r+1)^n + (r+1)^n - 1]/[r*(n+1)]
In this case r= -0.05 and n = 10

average = 10000*Multiplier = 7840.00 to 2 d.p.

This is where I got it from:

http://www.wolframalpha.com/input/?i=sum+of+%281%2Br%29%5En+from+n%3D0+to+n

You basically sum (1+r)^n from 0 to n. Wolfram Alpha got that formula for the sum. After that you divide it by n+1 to get the average. The link only shows the sum.

neil.p.b

764dak - perfect, exactly what i was looking for. Cheers.

GVN - thanks for your help anyway. I probably could have worded the problem better!