Hi Claus,

I see what you are doing.

Suppose we set first 4 rows of days as 1,2,3,4 and w(1), w(2), w(3) and w(4) as 10,20,30,40 respectively and compare the results with the manually generated output.

Your formula gives

f(1) = 10
f(2) = 20
f(3) = 41.03
f(4) = 60.12

Unless I forgot how to do math, :) here is the manually done computation :

When :

n = 1, no f(1) is calculated as upper limit tends to 0. No number exists?
n = 2, f(2) = 10 x e^(-1/4) = 7.7880
n = 3, f(3) = [10 x e^(-2/4)] + [20 x e^(-1/4)] = 21.6413
n = 4, f(4) = [10 x e^(-3/4)] + [20 x e^(-2/4)] + [30 x e^(-1/4)] = 40.2183


Do you see where I'm going...?


On Thursday, April 26, 2018 at 7:21:50 AM UTC+1, Claus Busch wrote:
Hi Ron,

Am Wed, 25 Apr 2018 13:46:11 -0700 (PDT) schrieb Ron George:

Thanks a lot. I put the "C2-1" bit there make it understand that the upper limit of sigma function is n-1 where n is the day on any row. If n = 3, C2-1 = 1.

Also, do you think my formula as a whole is correct?

my understanding for your formula is a bit different.
Have a look:
https://1drv.ms/x/s!AqMiGBK2qniTgcxMl0ftNOJuVY6Duw


Regards
Claus B.
--
Windows10
Office 2016