any simple solution to create an exponential distribution?

[bool]

Hello,

This may be a dumb question. I am trying to create a exponential distribution of a number of elements X distributed over y buckets rounded off to the nearest whole number.

So for example if x=2044 and y=9,

a nice distribution would look something like this:

bucket#1 =4
bucket#2 =8
bucket#3 =16
bucket#4 =32
bucket#5 =64
bucket#6 =128
bucket#7 =256
bucket#8 =512
bucket#9 =1024

Total of elements=2044

how can I easily acomplish this for let's say 1100 elements and 9 buckets. I feel like it's an easy formula but then there's the GROWTH function that can do it but I can't figure it out. Thanks!

[shg]

Don't you need to specify the initial value (4 in your example)?

[bool]

Sure the first value can be 2, 3, 4 , or 5... it really does not matter as long as the distribution adds to the number specified.

[MrShorty]

I find that I often must start with algebra/mathematics before I can get into Excel. Check this if it looks right:

1) y=x0+x1+x2+x3+...+x4 -- x(i+1)=2*x(i) Is the 2 fixed, or can it vary???
2) substitute into main sum: y=x0*2^0+x0*2^1+x0*x^2+...+x0*2^n Is n fixed, or is n variable??
3) divide both sides by x0: y/x0=sum(i=0 to n)(2^i)
4) It should be easy to show for powers of 2 that sum(i=0 to n)(2^i)=2^(n+1)-1. If the 2 is a variable, I do not know if this will hold for any "base".
5) substitute: y/x0=2^(n+1)-1. Assuming I am right to this point, then the only question is my mind is whether or not x0 is the unkown or n is the unkown to solve for. Or maybe the "base" 2 is the unkown?

If I have done that right, and my assumptions hold true, then the equation in step (5) should be easily solved for whichever quantity you need.

[bool]

I lost you. Not a math person here. Let's say we specify the number of elements X (2044 in my example). Which formula would take that number, round it off to produce results for Y (number of buckets) with a corresponding distribution. All numbers in the distribution rounded off, no decimals.

[MrShorty]

I lost you. Not a math person here.

That could be a problem, because you cannot check my derivation. So far, everything about my suggestions depends on the assumptions I made in coming up with the formula in step 5. You need to be able to understand this enough to be able to state whether or not I have understood this and made the right assumptions.

Assuming I have done it right, the answer to your question is in the equation in step 5 (edited here so that the buckets are numbered from 1 to 9 instead of 0 to 8).
y/x1=2^n-1
In this equation:
y= total sum (2044 in your case)
x1= the value in bucket 1 (4 in your case)
n= the number of buckets (9 in your case)
2044/4=2^(9)-1 (you should be able to verify this).

If you give it the total (y) and the value in the first bucket (x1), you should easily be able to solve for n. You may need to review your high school algebra on how to solve exponential equations (http://www.purplemath.com/modules/solvexpo2.htm is a good tutorial). You should get something like n=log(2044/4+1,2)

[bool]

That formula works. Thanks!