Sorry, I was hoping for someone to respond so I went to another catagory.

I tried your suggestion and am getting a value but it is not the one I was
hoping for. Below is the formula.

SUMPRODUCT(($D$4:$D$10="DS")*($G$4:$H$100))+SUMPR ODUCT(($D$4:$D$10="DS")*($J$4:$K$10))

The results are that "DS" occurs 2x on D5 and D10 and H5=1 and H10=2 and
J10=2.

I am trying to cound the number of times the exceptions are "0" for each
loan that person processed. Each row is a loan. The errors are defined in a
range of G4:H10 and J4:K10. I then will divide the number of occurences per
the number of loans that "DS" processed - which in the case above should be
2/2 = 100% Error rate. Therefore, if they processed 5 loans and only 2 loans
had Compliance exceptions it would be 2/5=40% error rate.
--
Deb


"Peo Sjoblom" wrote:

Please don't post more than once, stay in the original thread.
The regulars will find your post

As I posted to your other post you cannot use that function with ranges that
are not equal in size

You can try


=SUMPRODUCT((D4:D10="CR")*(MOD(COLUMN(G4:K10),3)0 )*(G4:K100))


or


=SUMPRODUCT((D4:D13="CR")*(G4:H130))+SUMPRODUCT(( D4:D13="CR")*(J4:K130))

--


Regards,


Peo Sjoblom

"Deb" wrote in message
...
=COUNTIFS(G$4:H$10,"0",D$4:D$10,"CR")+COUNTIFS(J$ 4:K$10,"0",D$4:D$10,"CR")

The above calculation does not work. Can anyone help me? I need to count
the number of exceptions in a range of columns skipping column "I" if the
persons initials in a range in column "D" equals their initials.

Thanks.
--
Deb