Sorry,

I checked myself and A25 = the Start weeknumber in the current month
and B25 = the end weeknumber in the month.
=TRUNC(((B22-DATE(YEAR(B22),1,0))+6)/7)

A1 is the Current date =Today()

Regards
Dylan

"Niek Otten" wrote in message
...
Just a few questions:

What does B25 have to do with this?
Where do you test for 1.5?
Can't you use the WEEKNUM() function?

--
Kind regards,

Niek Otten
Microsoft MVP - Excel



"dd" <dd.dd wrote in message
...
| Following on from the post titled "How do I count the number of even weeks
| in the current month."
| I am taking the number of even weeks per month and assigning payments
| against each even week in the month.
|
| Since there are a maximum of three even weeks per month:
| For the first even week I assign the maximum value I will receive for that
| month.
| If the middle of the month has not been reached I assign the remaining
value
| of £100
| If it is none of the above (after the middle of the month) then I assign
the
| value of £0
|
| I have the following
|
=IF(TRUNC(((A1-DATE(YEAR(A1),1,0))+6)/7)=A25,A28*100,IF(ROUND(B25-TRUNC(((A1-DATE(YEAR(A1),1,0))+6)/7)/2,-1)=1,100,0))
|
| Broken down:
| If the present week number (A1) = the start of month week number (A25)
| Then multiply the number of even weeks in this month X £100
|
| If the present week number (A1) divided by two = 1.5
| Then the value is £100
|
| Or else the Value is £0
|
| Checking it the start of this month is week 4
| The end is week 8
|
| When I try changing my cell values this celldoes not update. I remain at
"0"
|
|
|
|