Re: working out the number of customers over ten year period
- From: "windandwaves" <nfrancken@xxxxxxxxx>
- Date: 31 Jan 2007 17:46:04 -0800
On Feb 1, 1:27 pm, matt271829-n...@xxxxxxxxxxx wrote:
On Jan 31, 4:24 am, "windandwaves" <nfranc...@xxxxxxxxx> wrote:
On Jan 31, 3:36 pm, matt271829-n...@xxxxxxxxxxx wrote:
On Jan 31, 1:53 am, "windandwaves" <nfranc...@xxxxxxxxx> wrote:
Hi Folk
I am putting together a business plan and I am trying to work out the
number of customers over a ten year cycle.
My assumptions are as follows:
I will add 2 new customers per month. On average, a customer will
make 3 purchases per month and after each purchase the likelyhood is
60% that they will make another purchase.
Can this be worked out using a fairly simple formula (for excel) ...
OR
... is just far too complicated? In this case then how should I do
it instead? I want a situation where some new customers come on board
each month and some fall off. It will also be likely that people who
have purchase for a while are more likely to purchase forever, but
that is just making things TOOO complicated I think.
Are you trying to calculate the number of customers per month? I.e.,
every month that a customer makes a purchase they are counted as a
customer for that month (but not for other months when they don't make
a purchase)? Is the number of purchases they make in a month relevant?
The sentence "on average, a customer will make 3 purchases per month
and after each purchase the likelihood is 60% that they will make
another purchase" does not really make sense. This implies an ever
diminishing likelihood of continuing to purchase, so you are bound to
reach a point when they can't possibly be averaging 3 purchases per
month. Do you mean 3 purchases in the *first* month perhaps?
It's all kind of confusing...- Hide quoted text -
- Show quoted text -
Hi
THis is the way it works I think.
1. on average a customer makes three purchases a month.
2. independent of that, after each purchase, there is a 60% likelihood
that the customer will make another purchase. Their frequency is not
affected, just their likelihood to come back at all.
Now, to make the picture complete,
3. with each purchase, the 60% increases just a little bit (i.e. if a
person has been purchasing for fifty years, then it is highly unlikely
that they will NOT come back the next time).
Does that make more sense now?
Not entirely. If after, each purchase, the probability of a return
purchase is 60%, then the average number of puchases that a customer
will *ever* make is 1 + 0.6 + 0.6^2 + 0.6^3 ... continued
indefinitely, which equals 2.5. Therefore as far as I can see the
average cannot be 3 in the first month, or, by restarting the argument
should they continue being a customer, in any subsequent month.
It might be easier simply to assign a probability that someone who is
a customer in one month will remain a customer the next month. Then
multiply the expected customers by 3 to get expected purchases. Then
you would not have the awkward need to reconcile the return-purchase
probability with the requirement of an average of three purchases per
month.
Assuming that this monthly return probability is 60% you would have
something like this:
Month 1
Expected customers = 2
Month 2
Expected customers = New customers + Customers retained from previous
month
= 2 + 0.6*2 = 3.2
Month 3
Expected customers = New customers + Customers retained from previous
month
= 2 + 0.6 * 3.2 = 3.92
Etc.
To adjust the return probability according to how long the customer
had been around would need a little more work.- Hide quoted text -
Yes, that is great, lets keep it like that....
Thanks a million!
.
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