Quote from: DumbFruit on June 09, 2015, 03:02:25 PM

Ok, so my post was nuked again (My computers fault) so this will be short and sweet. I have been away, so I apologize for the delay;

Suppose there is a big miner with 33% of the hashpower, and there are 67 small miners with 1% of the hashpower. Because of the Law of Big Numbers and simplicity, I'll assume that they all mine blocks in a predictable order;
BigMiner, SmallMiner[1], SmallMiner[2], BigMiner, SmallMiner[3], SmallMiner[4], .. , BigMiner, SmallMiner[66], SmallMiner[67]

I'll also assume that the rollover fee will be distributed over the next 100 blocks.

BigMiner can assume that it can recover 33% of the lost reward on a future block. So if the BigMiner can mine a block that exceeds T while simultaneously getting a reward that is better than a block less than T, given that his reward is reward - penalty + .33*penalty, then it's at a net positive. Whereas a small miner could only do reward - penalty + 0.01*penalty. What is further problematic for the SmallMiner is that if they mine anything with a rollover fee, the BigMiner will get 33% of the reward whereas they will be lucky to get 1% (Not only is the reward less for himself, but any rolling over reward is more beneficial to his larger competitor(s)). In this way, the BigMiner has a competitive advantage over SmallMiners.

I tried to come up with specific numbers that would show the attack on your algorithm, but I wasn't sure what values you were using to get your graph¹, so I think it would end up being a straw man anyway. The bottom line is that penalties or rewards rolling over hurts the SmallMiner more than the BigMiner. SmallMiners being less able to recover over future blocks.

By "penalties or rewards" I mean that it doesn't make a difference whether you are rewarding the current larger block and rolling a bonus over to future blocks, or if you are penalizing a big block and then passing the difference to future blocks.

I didn't see a response to that problem, sorry if I missed it.

¹http://i.imgur.com/EZvlJq7.png

Edit: T was defined as the target block size.

See a worked example at https://bitcointalk.org/index.php?topic=1078521.msg11557115#msg11557115. I use the penalty function described in the OP - f(x) = max(x-T,0)^2 / (T*(2T-x)). And an earlier one with made up numbers but additional analysis.

You need to be thinking in terms of reward per block. Of course the 33% miner should get about x33 the rewards of a 1% miner, simply because he's bigger. What we don't want is that the big miner would have more than x33 the rewards (superlinear gain). What I've shown is that the big miner actually gets less than x33 the total rewards of a small miner, or in other words, that the reward per block is smaller for the big miner.

Every miner gets, per block, (tx fees - penalty) + collection. The rollover pool is the same for all miners so collection is a roughly constant value. But (tx fees - penalty) differs per block, and is lower for the big miner. The big miner purposefully builds supersized blocks, because increasing the pool works to his advantage - but it works to the advantage of the small miners, too.

Put another way, the small miners don't reclaim much of their own penalty, but they reclaim the penalty of the big miners.

If you still disagree, please either
1. Explain why we shouldn't be looking at reward per block, or
2. Explain what is wrong with the example (the one with real numbers) that shows the reward per block is smaller for big miners.

Quote from: tacotime on June 09, 2015, 04:56:28 PM

Quote from: Meni Rosenfeld on June 09, 2015, 11:56:42 AM

Furthermore, if the reward is (1 - penalty) * (minted coins) + tx fees, then in the future when the minted coins go down to 0, there is no penalty at all. That's the opposite of future-proof.

Monero is permanently inflationary, so there will always be an impact of the penalty.

Ok. I was primarily refuting the implied claim that this formula would make sense when applied to Bitcoin.