Quote from: NewLiberty on June 04, 2015, 05:59:23 PM

Quote from: molecular on June 04, 2015, 05:52:03 PM

Quote from: NewLiberty on June 04, 2015, 04:32:01 PM

Can you explain a bit about the mechanism wherein the miner pays into the rollover pool, and why that is different from the 'original proposal'? It is not obvious why this dictinction makes a difference. It seems to still incentivize out of chain payments to miners for transaction inclusion regardless of whether it is paid by the miner, or deducted from the miner's reward, both are dependent on the fees in the block (which aren't there in out of block payments schemes).

I think the penalty is not dependent on the fee:

Quote from: Meni Rosenfeld on June 02, 2015, 08:16:17 PM

The miner of a large block must pay a penalty that depends on the block's size.

So it still has to be paid regardless of wether or not the tx fee payment was on- or off-chain.

This is penalty then ONLY taken from the coinbase reward and not also TX fees?
How would this method survive the successive halvings (much less the eventual cessation)?

Quote from: Meni Rosenfeld on June 02, 2015, 08:16:17 PM

The penalty will be deducted from the funds he collects in the generation transaction

I read that as "from reward, tx fees and rollover pool input".

Quote from: Meni Rosenfeld on June 02, 2015, 08:16:17 PM

If the penalty exceeds the miner's income of tx fees + minted coins + his share of the current rollover pool, the block is invalid.

I stumbled accross this problem, too and I remember it was adressed upthread, not very far from OP. I just searched, but couldn't find it and I also can't remember why (or wether) this wasn't a problem. I think I wasn't convinced or didn't think it through.

But thinking about that a bit, assuming 0 block reward is reached. A block with no transactions in it can always be mined because it doesn't carry any penalty at all. As soon as there's a penalty to be payed there need to be some transactions in the block and the fees they pay must exceed the penalty (otherwise the block is invalid). Is this a problem? I'm not sure, seems to me that depends on the particularities of the penalty function...