I have a simple pivot table with Dollar sales in it by Product. Can I
calculate, in the Pivot Table itself, a RANK calculation to show where each
individual Products ranks?

Can anyone help me.

Thanks KEN

there is an index option under show values as in the field setting.
you ca also take a look at this: http://support.microsoft.com/kb/151515


wroote:

I have a simple pivot table with Dollar sales in it by Product. Can I
calculate, in the Pivot Table itself, a RANK calculation to show where each
individual Products ranks?

Can anyone help me.

Thanks KEN

Andrea,

I've tried Index as suggested but this does not provide the RANK sequence
which I want. Any other suggestions?

KEN

"Andrea" wrote:

there is an index option under show values as in the field setting.
you ca also take a look at this: http://support.microsoft.com/kb/151515


wroote:

I have a simple pivot table with Dollar sales in it by Product. Can I
calculate, in the Pivot Table itself, a RANK calculation to show where each
individual Products ranks?

Can anyone help me.

Thanks KEN

Depending on the use you have to do, there is one workaround not very much
usefull: it does not work well if you have two identical amount and is fixed.

In the Pivot sort the product in ascending or descending order.
Add a calculated field call it rank and add this formula =1.
Drag the rank in the data area and summarize as running total.

whenever you add a new product or anything changes in the source data always
sort the pivot table as you did at step 1 after refreshing



"Ken" wrote:

Andrea,

I've tried Index as suggested but this does not provide the RANK sequence
which I want. Any other suggestions?

KEN

"Andrea" wrote:

there is an index option under show values as in the field setting.
you ca also take a look at this: http://support.microsoft.com/kb/151515


wroote:

I have a simple pivot table with Dollar sales in it by Product. Can I
calculate, in the Pivot Table itself, a RANK calculation to show where each
individual Products ranks?

Can anyone help me.

Thanks KEN