Count days but solutions can't land on holiday or weekend
 

Hi there,

I am trying count 15 days from 2/4/09, with the following caveats. I can't
count holiday's (2/16/09) and the solution can't land on a Saturday or
Sunday, if it does move it the following Monday.

Any help would be greatly appreciated.

Count days but solutions can't land on holiday or weekend
 

Consider using

=WORKDAY(D3,A1,H1:H7)

Where D3 is the start date, A1 contains the number of weekdays you want to
add to and H1:H7 contains a list of holidays you want excluded.

--
If this helps, please click the Yes button

Cheers,
Shane Devenshire


"NB292" wrote:

Hi there,

I am trying count 15 days from 2/4/09, with the following caveats. I can't
count holiday's (2/16/09) and the solution can't land on a Saturday or
Sunday, if it does move it the following Monday.

Any help would be greatly appreciated.

Count days but solutions can't land on holiday or weekend
 

Hi

general format
=workday(startdate, days, holidays)

Try placing the start date in a cell e.g. B1 and a list of the holiday dates
for the year in cells A1:A20
then
=Workday(B1,15,A1:A20)
will give your result
--
Regards
Roger Govier

"NB292" wrote in message
...
Hi there,

I am trying count 15 days from 2/4/09, with the following caveats. I can't
count holiday's (2/16/09) and the solution can't land on a Saturday or
Sunday, if it does move it the following Monday.

Any help would be greatly appreciated.

Count days but solutions can't land on holiday or weekend
 

Thanks Roger & Shane, not quite there yet though.

State date = 2/2/09, excel tells me 2/9/09 which is technically correct but
the answer I am looking for is 2/10/09. Because I must give an additional
business day if solution lands on a Saturday or Sunday.

Start date = 2/3/09, excel tells me 2/10/09, which is what I am looking.

Start date = 2/4/09, excel tells me 2/11/09, answer that I am look for is
2/10/09

-Nick

"Roger Govier" wrote:

Hi

general format
=workday(startdate, days, holidays)

Try placing the start date in a cell e.g. B1 and a list of the holiday dates
for the year in cells A1:A20
then
=Workday(B1,15,A1:A20)
will give your result
--
Regards
Roger Govier

"NB292" wrote in message
...
Hi there,

I am trying count 15 days from 2/4/09, with the following caveats. I can't
count holiday's (2/16/09) and the solution can't land on a Saturday or
Sunday, if it does move it the following Monday.

Any help would be greatly appreciated.

Count days but solutions can't land on holiday or weekend
 

Hi Nick

i don't think you can be entering the formula correctly.
Workday will never give you a date which is a Saturday or Sunday.
In this latest posting you are adding 7 days, not 15 as originally posted.
Using UK dates, and including just 16th Feb as a holiday, I get

Start Date Result Day
02/02/2009 11/02/2009 Wed
03/02/2009 12/02/2009 Thu
04/02/2009 13/02/2009 Fri
05/02/2009 17/02/2009 Tue
06/02/2009 18/02/2009 Wed
07/02/2009 18/02/2009 Wed
08/02/2009 18/02/2009 Wed
09/02/2009 19/02/2009 Thu
10/02/2009 20/02/2009 Fri
11/02/2009 23/02/2009 Mon

--
Regards
Roger Govier

"NB292" wrote in message
...
Thanks Roger & Shane, not quite there yet though.

State date = 2/2/09, excel tells me 2/9/09 which is technically correct
but
the answer I am looking for is 2/10/09. Because I must give an additional
business day if solution lands on a Saturday or Sunday.

Start date = 2/3/09, excel tells me 2/10/09, which is what I am looking.

Start date = 2/4/09, excel tells me 2/11/09, answer that I am look for is
2/10/09

-Nick

"Roger Govier" wrote:

Hi

general format
=workday(startdate, days, holidays)

Try placing the start date in a cell e.g. B1 and a list of the holiday
dates
for the year in cells A1:A20
then
=Workday(B1,15,A1:A20)
will give your result
--
Regards
Roger Govier

"NB292" wrote in message
...
Hi there,

I am trying count 15 days from 2/4/09, with the following caveats. I
can't
count holiday's (2/16/09) and the solution can't land on a Saturday or
Sunday, if it does move it the following Monday.

Any help would be greatly appreciated.