A PBKDF2 cost factor/iteration count of 12000 and 32-byte output means each candidate passphrase costs 12002 SHA256 blocks.
I can buy a crappy bitcoin miner which will do 2GH/s for about USD19.
Let's say we're going to use the Gawker leak as our dictionary. That's ~200,000 candidate passwords.
For a given user, I can therefore find their password (if it exists in the Gawker set) in 12002 * 200000 = 2.4GH SHA256 applications. That will take 1.2 seconds.
So for all 125 million eBay users, that's about 4 years. This work is trivially parallelisable, so buying more or faster hardware is brutally effective.
Note: there is obviously, and hopefully, a non-negligible probability that a user's password isn't in that set. Brute force of (say) the whole 8 printable-ASCII character password space would take longer but would be guaranteed to find to find about 50% (from Adobe leak) of user's passwords.
A PBKDF2 cost factor/iteration count of 12000 and 32-byte output means each candidate passphrase costs 12002 SHA256 blocks.
I can buy a crappy bitcoin miner which will do 2GH/s for about USD19.
Let's say we're going to use the Gawker leak as our dictionary. That's ~200,000 candidate passwords.
For a given user, I can therefore find their password (if it exists in the Gawker set) in 12002 * 200000 = 2.4GH SHA256 applications. That will take 1.2 seconds.
So for all 125 million eBay users, that's about 4 years. This work is trivially parallelisable, so buying more or faster hardware is brutally effective.
Note: there is obviously, and hopefully, a non-negligible probability that a user's password isn't in that set. Brute force of (say) the whole 8 printable-ASCII character password space would take longer but would be guaranteed to find to find about 50% (from Adobe leak) of user's passwords.