it might be possible to do this as a conditional formula (=if) otherwise a
macro might be needed.

the thing i'm unsure about is this formula looks at the cells beneath it and
keeps looking until a result is returned, i'm not sure if that can be done as
a conditional formula.

if you really want to know what this is for send me a pm and i'll tell you.


if m3 = 0 then ""
if l3 = i then ""
if m3 = 1 and if l3 = "c" and if i3 > 0

and if e4 – d3 < -.0066 then result = 1
write result in n3
else go to row 5

if e5 – d3 < -.0066 then result = 1

if d5 >= d3 then result = 0

i seriously doubt this will happen but in case it does
if e5 – d3 < -.0066 and if d5 > d3 then result = "ambiguous"
else go to next row
keep going down the rows (it should not take any more than 3 or 4) until
either


if i3 < 0
and if e4 – d3 > .0066 then result = 1
write result in n3
else go to row 5

if e5 – d3 > .0066 then result = 1
if d5 < d3 then result = 0

must loop through all the rows, there 8800 of them

any thoughts on this?

"kylefoley2000" wrote:

> it might be possible to do this as a conditional formula (=if) otherwise a
> macro might be needed.
>
> the thing i'm unsure about is this formula looks at the cells beneath it and
> keeps looking until a result is returned, i'm not sure if that can be done as
> a conditional formula.
>
> if you really want to know what this is for send me a pm and i'll tell you.
>
>
> if m3 = 0 then ""
> if l3 = i then ""
> if m3 = 1 and if l3 = "c" and if i3 > 0
>
> and if e4 – d3 < -.0066 then result = 1
> write result in n3
> else go to row 5
>
> if e5 – d3 < -.0066 then result = 1
>
> if d5 >= d3 then result = 0
>
> i seriously doubt this will happen but in case it does
> if e5 – d3 < -.0066 and if d5 > d3 then result = "ambiguous"
> else go to next row
> keep going down the rows (it should not take any more than 3 or 4) until
> either
>
>
> if i3 < 0
> and if e4 – d3 > .0066 then result = 1
> write result in n3
> else go to row 5
>
> if e5 – d3 > .0066 then result = 1
> if d5 < d3 then result = 0
>
> must loop through all the rows, there 8800 of them

It's hard for anybody to have an idea because your statement of the problem
is long on minute points but short on a concise statement of what you are
trying to do. Help us help you by being clear and concise about what you are
trying to accomplish.


"kylefoley2000" wrote:

> any thoughts on this?
>
> "kylefoley2000" wrote:
>
> > it might be possible to do this as a conditional formula (=if) otherwise a
> > macro might be needed.
> >
> > the thing i'm unsure about is this formula looks at the cells beneath it and
> > keeps looking until a result is returned, i'm not sure if that can be done as
> > a conditional formula.
> >
> > if you really want to know what this is for send me a pm and i'll tell you.
> >
> >
> > if m3 = 0 then ""
> > if l3 = i then ""
> > if m3 = 1 and if l3 = "c" and if i3 > 0
> >
> > and if e4 – d3 < -.0066 then result = 1
> > write result in n3
> > else go to row 5
> >
> > if e5 – d3 < -.0066 then result = 1
> >
> > if d5 >= d3 then result = 0
> >
> > i seriously doubt this will happen but in case it does
> > if e5 – d3 < -.0066 and if d5 > d3 then result = "ambiguous"
> > else go to next row
> > keep going down the rows (it should not take any more than 3 or 4) until
> > either
> >
> >
> > if i3 < 0
> > and if e4 – d3 > .0066 then result = 1
> > write result in n3
> > else go to row 5
> >
> > if e5 – d3 > .0066 then result = 1
> > if d5 < d3 then result = 0
> >
> > must loop through all the rows, there 8800 of them