Create Formula - Add # of Networkdays to a specific start date to find an end date

Thread view:

Enable EMail Alerts

Start New Thread

Thread rating:

bdicarlo1@yahoo.com - 08 Jan 2007 23:31 GMT

I need to create a formula where I add a set number of Networkdays to a
start date. Example:

Start Date = 1/2/07
Number of Networkdays = 21
End Date = Is calculated

If Start = 1/2/07 and 21 Networkdays are added, what is the retured end
date????

I can only find example of Networkdays where I would be providing the
Start and End Dates, and it will calculate the Networkdays. I can't
find example where End Date is calculated based on the number of
networkdays from start date.

Can anyone help with a valid formula for this?

Thanks!!

Reply to this Message

Dave Peterson - 08 Jan 2007 23:36 GMT

Look for =workday() in Excel's help.
(Also part of the analysis toolpak)

> I need to create a formula where I add a set number of Networkdays to a
> start date. Example:
[quoted text clipped - 14 lines]
>
> Thanks!!

Signature

Dave Peterson

Reply to this Message

Jerry - 09 Jan 2007 03:23 GMT

Dave,

It sounds as though you may need to calculate workdays in each month. The
excel function is =eomonth(start_date,no of months to calculate days on).
For example eomonth(1/2/7,0) returns 1/31/7. So in your formula you would
have =networkdays(1/2/7,eomonth(1/2/7,0)) would return 22 days.
Substituting a 1-12 in the last parameter for eomonth would count the days
in all months through the number of months past january.

This isn't the final answer you are looking for but you could experiment
with it.
the eomonth function is in the analysis pak.

> Look for =workday() in Excel's help.
> (Also part of the analysis toolpak)
[quoted text clipped - 17 lines]
>>
>> Thanks!!

Reply to this Message

Dave Peterson - 09 Jan 2007 12:40 GMT

First, it wasn't me who asked the question.

Second, it sounds more like an =workday() question (well, to me anyway).

> Dave,
>
[quoted text clipped - 34 lines]
> >
> > Dave Peterson

Signature

Dave Peterson

Reply to this Message