DigitalVolcano Software Support

I suspect that most people want to know "When I have eliminated all duplicates, what size am I left with? Will it all fit on drive x?"

Currently, my scan result tells me:
Total size of all files: 21 TB
Files have duplicates: 16.1 TB

So 21 - 16.1 = 5.9 TB of unique single files that I can't delete.
But for the duplicates, :
I might have two 8 TB files, so after de-dupe I will be left with 5.9 + 8 TB = 13.9 TB
or might be four 2 TB files, so after de-dupe I will be left with 5.9 + 2TB = 7.9 TB (So, yay! It can all fit on my 8TB drive!)

(In reality, my dupes are 8 million files, but I simplify here.)

So my request is for the scan result to include the size of the source duplicates. Something like:

Total size of all files: 21 TB
Unique single files: 5.9 TB
Duplicate Files: 16.1 TB* consisting of:
- Source files: 2 TB
- Duplicate copies: 14.1 TB**

Size after eliminating all duplicate copies: 7.9 TB

* (for me, this one is a useless measure but needs to be kept because many programs state it this way and surely people will compare the results of different programs.)
** (i.e. a clear statement of how much wasted space I have. What I will gain by eliminating duplicates.)

What do you think?
What statistics do you want to see?

It's a good idea and could be useful for 'Exact content' matching (Regular mode).
On the other modes the duplicates within the groups may be wildly different sizes and it's not for the program to decide which copies are to be removed. Perhaps a minimum/maximum saving could be appropriate here.

Yes. That would work. In exact mode the min and max would be the same.

Thank you for all of your replies. I won't reply thanks in every post, so that there's not more msgs for you to review.