Totally IFless

=(G6+MAX(0,G6-40)*0.5)*(35+(B6>=DATE(2006,6,1))*2)

--

HTH

Bob Phillips

(replace xxxx in the email address with gmail if mailing direct)

"Dave F" <DaveF@discussions.microsoft.com> wrote in message
news:C7686F83-E19D-46B6-8A8E-C3806A29CB84@microsoft.com...
> Wow, Bob, that's impressive.
>
> Will have to try it out ,thanks.
>
> Dave
>
> "Bob Phillips" wrote:
>
> > How about this
> >
> > =(MIN(G6,40)+MAX(0,G6-40)*1.5)*(IF(B6<DATE(2006,6,1),35,37))
> >
> >
> > or
> >
> > =(G6+MAX(0,G6-40)*0.5)*(IF(B6<DATE(2006,6,1),35,37))
> >
> > --
> >
> > HTH
> >
> > Bob Phillips
> >
> > (replace xxxx in the email address with gmail if mailing direct)
> >
> > "Dave F" <DaveF@discussions.microsoft.com> wrote in message
> > news:F87FB881-8C23-488E-B630-A483B823FB99@microsoft.com...
> > > Here's the formula:
> > >
> > >
> >
=IF(B6<DATE(2006,6,1),IF(G6<=40,G6*35,(40*35+((G6-40)*35*1.5))),IF(G6<=40,G6
> > *$F$2,(40*37+((G6-40)*$F$2*1.5))))
> > >
> > > This formula calculates correctly, so that's not the issue. I'm
wondering
> > > if there is a clearer (more concise) way to accomplish the calculation
I'm
> > > trying to do, without using helper columns.
> > >
> > > Here's the scenario: for dates prior to 6/1/2006 the client's billing
rate
> > > is $35/hr, with 1.5x for any hours over 40 hrs per week, and, for all
> > dates
> > > after 6/1/2006 the client's billing rate is $37 per hour with 1.5x for
any
> > > hours over 40 hours per week.
> > >
> > > Any ideas?
> >
> >
> >