=MIN(IF(BF3:IV3=D3/2,BF3:IV3))

As always Harlan, concise and to the point.<bg
--

Regards,

RD
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Please keep all correspondence within the Group, so all may benefit !
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"Harlan Grove" wrote in message
...
"Ragdyer" wrote...
....
Anyway, will a zero work for both error traps:

=IF(OR(MIN($BF3:$IV3)=0,MIN(BF3:IV3)<D3/2),0,
SMALL(BF3:IV3,COUNTIF(BF3:IV3,"<"&D3*0.5)+1))
....

You don't need to calculate the minimum twice. If MIN(...) were 0,
MIN(...) would also be < D3/2 if D3 0. Since D3 should be positive,
better to make the test OR(D3<=0,MIN(BF3:IV3)<D3/2).

However, this is a whole lot of unnecessary calculation for something
that should be solved with a simple array formula.

=MIN(IF(BF3:IV3=D3/2,BF3:IV3))

If there are values in BF3:IV3 = D3/2, this would return the smallest
of them; otherwise, it'll return 0. And it's much clearer in addition
to being much more efficient and much shorter.

And, FTHOI, returning the LARGEST value in BF3:IV3 if there were no
values in BF3:IV3 = D3/2 could be done using

=LARGE(BF3:IV3,MAX(COUNTIF(BF3:IV3,"="&D3/2),1))