Forum Discussion
sum quotient
I was invited by Mr Kim for help. However, I am not so familiar with Financial Mathematics.
I created a column to create the quotients (C/(1+r)^t), where r is indicated in L1. Changing the value in L1 will give a different sum in L4. So, the user can guess the value of r.
There is a function called "XIRR" which seems to do this type of problem. But, I do not understand the setting.
i want to do the following with my numbers:
7.776/(1+r)^1+7.898/(1+r)^2+8.054/(1+r)^3+…+3.248/(1+r)^28 - 100
set equal to 0 and solve for r
for example i could imagine putting this:
=(E2/(1+I2)^B2+E3/(1+I2)^B3+E4/(1+I2)^B4 and so on subtract by 100 and then using the solver chaning the 0.2 (which are a random number now)
but im not sure..
what i want to do is:
(E2/(1+I2)^B2+E3/(1+I2)^B3+E4/(1+I2)^B4+E5/(1+I2)^B5+E6/(1+I2)^B6+E7/(1+I2)^B7+E8/(1+I2)^B8+E9/(1+I2)^B9+E10/(1+I2)^B10+E11/(1+I2)^B11+E12/(1+I2)^B12+E13/(1+I2)^B13+E14/(1+I2)^B14+E15/(1+I2)^B15+E16/(1+I2)^B16+E17/(1+I2)^B17+E18/(1+I2)^B18+E19/(1+I2)^B19+E20/(1+I2)^B20+E21/(1+I2)^B21+E22/(1+I2)^B22+E23/(1+I2)^B23+E24/(1+I2)^B24+E25/(1+I2)^B25+E26/(1+I2)^B26+E27/(1+I2)^B27+E28/(1+I2)^B28+E29/(1+I2)^B29)-100
set this equal to 0 and replace I2 by the value which gives me 0 for the equation. (whats an easy way to write it?)
- my apologies - this turns out to be very complex for me.....
solving for I2 is - there could be a circular reference in this..
maybe other people can solve this..
good luckI was invited by Mr Kim for help. However, I am not so familiar with Financial Mathematics.
I created a column to create the quotients (C/(1+r)^t), where r is indicated in L1. Changing the value in L1 will give a different sum in L4. So, the user can guess the value of r.
There is a function called "XIRR" which seems to do this type of problem. But, I do not understand the setting.
Mr. Chan
thank you for helping..
I think what Mr. Behler wanted is to determine first what the value of r is -- this is what's jolting my brain!
anyway I hope He can see something out of this...
many thanks and good health.
