Thanks for that Biff, I never tested it properly.
--
Mike

When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.


"T. Valko" wrote:

=INDEX(A1:A5,MATCH(1,("x"=D1:D5)*(MIN(IF(E1:E50, E1:E5))=E1:E5),0))

That will return an error when the min number in the range is not associated
with "x" and is less than the min number that is associated with "x".

A......64
B......16
C..x..89
D......5
E..x...16

Based on that sample data the above formula returns #N/A. The correct result
should be E.

Try this (array entered):

=INDEX(A1:A5,MATCH(1,(D1:D5="x")*(E1:E5=MIN(IF(D1: D5="x",E1:E5))),0))

--
Biff
Microsoft Excel MVP


"Mike H" wrote in message
...
Hi,

Try this ARRAY formula

=INDEX(A1:A5,MATCH(1,("x"=D1:D5)*(MIN(IF(E1:E50,E 1:E5))=E1:E5),0))

This is an array formula which must be entered by pressing
CTRL+Shift+Enter
'and not just Enter. If you do it correctly then Excel will put curly
brackets
'around the formula {}. You can't type these yourself. If you edit the
formula
'you must enter it again with CTRL+Shift+Enter.

--
Mike

When competing hypotheses are otherwise equal, adopt the hypothesis that
introduces the fewest assumptions while still sufficiently answering the
question.


"jtfalk" wrote:

A B C D
E
First x
5
second x
0
third x
2
fourth x
2
fifth x
1

I am trying to get the A name with criteria of D and the lowest number.
This
list is about 100 items. So in the above case is would look through all
of D
for x's and the lowest E value which is 1 above. I was trying this:
=INDEX(A1:A100,MATCH(MIN(E1:E100),G2:G19,0)*(D1:D1 00="x"),0)
The problem is it looks at the zero and returns second




.