# Can someone explain this for me please

• Jan 5th 2010, 06:43 AM
RhysGM
Can someone explain this for me please
$\displaystyle LV= \frac{mr}{(1+i)} + \frac{mr^2}{(1+i)^2} + \frac{mr^3}{(1+i)^3} + . . . = m (\frac{r}{1+i-r})$

What does this mean an how was the last bit created???
• Jan 5th 2010, 07:37 AM
Drexel28
Quote:

Originally Posted by RhysGM
$\displaystyle LV= \frac{mr}{(1+i)} + \frac{mr^2}{(1+i)^2} + \frac{mr^3}{(1+i)^3} + . . . = m (\frac{r}{1+i-r})$

What does this mean an how was the last bit created???

Let $\displaystyle S=1+r+\cdots+r^n$ then $\displaystyle rS=r+r^2+\cdots+r^n+r^{n+1}$ so then $\displaystyle rS-S=r^{n+1}-1\implies S=\frac{r^{n+1}-1}{r-1}$ assuming the $\displaystyle r$ in your case (identify it) has the quality $\displaystyle |r|<1$ take the limit as $\displaystyle n\to\infty$
• Jan 5th 2010, 07:38 AM
RhysGM
If;

$\displaystyle LV = \frac{mr}{(1+i)} + \frac{mr^2}{(1+i)^2} + \frac{mr^3}{(1+i)^3} + . . . + \frac{mr^n}{(1+i)^n}$

then if we sum the arithmetic sequence is this true;

$\displaystyle LV = \frac{(\frac{mr}{(1+i)}+\frac{mr^n}{(1+i)^n})n}{2}$

$\displaystyle LV = \frac{n(\frac{mr(1+i)^n+mr^n(1+i)}{(1+i)(1+i)})}{2 }$
• Jan 5th 2010, 07:40 AM
Drexel28
Quote:

Originally Posted by RhysGM
If;

$\displaystyle LV = \frac{mr}{(1+i)} + \frac{mr^2}{(1+i)^2} + \frac{mr^3}{(1+i)^3} + . . . + \frac{mr^n}{(1+i)^n}$

then if we sum the arithmetic sequence is this true;

$\displaystyle LV = \frac{(\frac{mr}{(1+i)}+\frac{mr^n}{(1+i)^n})n}{2}$

$\displaystyle LV = \frac{n(\frac{mr(1+i)^n+mr^n(1+i)}{(1+i)(1+i)})}{2 }$

Your $\displaystyle LV$ looks like an infinite sum. Try just $\displaystyle Lv=\frac{m}{1-\frac{r}{1+i}}$
• Jan 5th 2010, 07:59 AM
RhysGM
Thanks for you help, I just like to explain a few things; I have a book that gives me this formula

Quote:

Originally Posted by RhysGM
$\displaystyle LV= \frac{mr}{(1+i)} + \frac{mr^2}{(1+i)^2} + \frac{mr^3}{(1+i)^3} + . . . = m (\frac{r}{1+i-r})$

which cacluates the life time value of a customer where;

m = the margin that customer makes in a year
r = the annual rention rate of your customer
i = future discount factor due to interest and external ecconomic factors

I don't understand how they got to the last bit;

I understand the sequence and it makes sense logically as to why it works, but how does that apply to the final bit.

Quote:

Originally Posted by RhysGM
If;

$\displaystyle LV = \frac{mr}{(1+i)} + \frac{mr^2}{(1+i)^2} + \frac{mr^3}{(1+i)^3} + . . . + \frac{mr^n}{(1+i)^n}$

then if we sum the arithmetic sequence is this true;

$\displaystyle LV = \frac{(\frac{mr}{(1+i)}+\frac{mr^n}{(1+i)^n})n}{2}$

$\displaystyle LV = \frac{n(\frac{mr(1+i)^n+mr^n(1+i)}{(1+i)(1+i)})}{2 }$

This is my attempt to sum the sequence, however I can't simplify it to get the same result as the book.

All I'm trying to do is understand why the book is giving me a particular formula.
• Jan 5th 2010, 08:08 AM
Drexel28
Quote:

Originally Posted by RhysGM
Thanks for you help, I just like to explain a few things; I have a book that gives me this formula

which cacluates the life time value of a customer where;

m = the margin that customer makes in a year
r = the annual rention rate of your customer
i = future discount factor due to interest and external ecconomic factors

I don't understand how they got to the last bit;

I understand the sequence and it makes sense logically as to why it works, but how does that apply to the final bit.

This is my attempt to sum the sequence, however I can't simplify it to get the same result as the book.

All I'm trying to do is understand why the book is giving me a particular formula.

It appears as though you are really attempting to calculate $\displaystyle \sum_{n=1}^{{\color{red}\infty}}\frac{mr^n}{(1+i)^ n}=m\sum_{n=1}^{{\color{red}\infty}}\left(\frac{r} {1+i}\right)^n$ and, forgive my complete ignorance of business related affairs (:/), I assume that $\displaystyle 0<\frac{r}{1+i}<1$ in which case we see (by our above calculations that) $\displaystyle m\sum_{n=1}^{{\color{red}\infty}}\left(\frac{r}{1+ i}\right)^n$$\displaystyle =m\lim_{n\to{\color{red}\infty}}\frac{\left(\frac{ r}{1+i}\right)^{n+1}-1}{\frac{r}{i+1}-1}$ and since by assumption $\displaystyle 0<\frac{r}{1+i}<1$ we see that $\displaystyle \left(\frac{r}{1+i}\right)^n\to0$ so that $\displaystyle LV=\frac{m}{1-\frac{r}{1+i}}=\frac{m(1+i)}{1+i-r}$
• Jan 5th 2010, 01:23 PM
RhysGM
Ok, lets simplify this for my little brain and put a little business sense to it so I can understand it better.

r in the above formulas is the retention rate of a customer base, for example let us assume a bank (I don't work for a bank so no comments about bankers please) has 1,000 customers and their retention rate year on year is 70%, ie they will loose 30% of their customer base each year. Assuming no one opens a new account. So in year 2 they will have 700 customers, in year 3; 490, year 4; 343 and so forth. Theorectically for infinity, realistically until you have 1 customer left.

If we add all those percentages together;

70% + 49% + 34.3% + ...

this is the same as 70% over 1-70% = 2.333

Why??? How can that be proved?

The m is how much money that customer makes you per year, so if that is £50 for example this would be;

£50 + £50 x 70% + £50 x 49% + £50 x 34.3% ...

The result of this is how much money one customer will make you over their lifetime with the bank.

The i in the equation is to do with interest rates, so £50 today will not be worth £50 next year and this is just a factor that can be ignored.

The result will give a value for all new current accounts in that year for their lifetime. This is handy to know as it tells you want you can spend per customer to try and retain them including offering any incentives etc. Anyway I digress.

I want to know why;

£50 + £50 x 70% + £50 x 49% + £50 x 34.3% ...

is the same as;

£50 x 0.7/0.3