Bitcoin Forum

R = 50 - (90% of F)

R is the reward for a block
F is the sum of all fees contained in that block.

Would this work?

What happens to the difference?

Sure, but what's the point? A smoother transition to fee based rewards? Looks to me like it's biased on inflation if that's the case.


Looks to me like he's talking about coinbase reward based off this math, meaning there is no difference left, just this much coins given to the solver.

If that is true, then 90% of all transaction fees simply cease to exist.  This is deflationary in it's own right.

Do you really think 50% of miners would vote for this?

No, no. ONLY the number of new coins printed changes. TX fees still rewarded to miners.

Not the way I understood it. From my perception, the coinbase reward is reduced by 90% of the redeemable transaction fees. This is only to determine the coinbase reward value. The transaction fees are still redeemed. Let's say it's a smoother way to go from coinbase reward to tx fee reward instead of just cutting BTC output in half at defined milestones. Destroying the transaction fees would be dangerous in too many ways to be implemented in such simple fashion.

Yes. If you get 0.1 BTC in fees in your block:

49.91 SHINY NEW BTC with no TX history
0.1 BTC in fees
for a total of 50.01.

Ok now let's discuss what's the point oO, cuz I can't fathom much of it.

This instead of the halving block rewards? Is this from the fear that fees won't provide enough incentive to mine?

I think it's mainly to skip the halving of coinbased reward milestones altogether. But then if the tx fee passes 55BTC before the 21mil target, you wouldn't be able to reach it, which isn't much of a problem I guess.

The determinism of the coinbase is a feature.  Excess supply is calculable.  What you propose would leave it subject to manipulation by altering the fees.  That would make a more volatile transaction processing market.