Author

Topic: Self-regulating (in|de)flation (Read 1308 times)

full member
Activity: 182
Merit: 100
June 02, 2011, 10:10:35 PM
#12
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.
legendary
Activity: 3738
Merit: 1360
Armory Developer
June 02, 2011, 09:35:54 PM
#11
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.
full member
Activity: 154
Merit: 100
June 02, 2011, 09:30:33 PM
#10
This instead of the halving block rewards? Is this from the fear that fees won't provide enough incentive to mine?
legendary
Activity: 3738
Merit: 1360
Armory Developer
June 02, 2011, 09:25:00 PM
#9
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.
sr. member
Activity: 364
Merit: 250
June 02, 2011, 09:22:43 PM
#8

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.

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.
legendary
Activity: 3738
Merit: 1360
Armory Developer
June 02, 2011, 09:13:29 PM
#7

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.

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.
sr. member
Activity: 364
Merit: 250
June 02, 2011, 09:11:29 PM
#6

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.

No, no. ONLY the number of new coins printed changes. TX fees still rewarded to miners.
hero member
Activity: 812
Merit: 1001
-
June 02, 2011, 09:07:46 PM
#5
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?


Do you really think 50% of miners would vote for this?
legendary
Activity: 1708
Merit: 1010
June 02, 2011, 09:07:32 PM
#4

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.
legendary
Activity: 3738
Merit: 1360
Armory Developer
June 02, 2011, 09:05:42 PM
#3
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?


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.

Quote
What happens to the difference?

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.
legendary
Activity: 1708
Merit: 1010
June 02, 2011, 09:02:36 PM
#2
What happens to the difference?
sr. member
Activity: 364
Merit: 250
June 02, 2011, 08:13:50 PM
#1
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?
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