Thomas Voegtlin [ARCHIVE] on Nostr: ๐ Original date posted:2015-05-10 ๐ Original message:Le 08/05/2015 22:33, Mark ...
๐
Original date posted:2015-05-10
๐ Original message:Le 08/05/2015 22:33, Mark Friedenbach a รฉcrit :
> * For each block, the miner is allowed to select a different difficulty
> (nBits) within a certain range, e.g. +/- 25% of the expected difficulty,
> and this miner-selected difficulty is used for the proof of work check. In
> addition to adjusting the hashcash target, selecting a different difficulty
> also raises or lowers the maximum block size for that block by a function
> of the difference in difficulty. So increasing the difficulty of the block
> by an additional 25% raises the block limit for that block from 100% of the
> current limit to 125%, and lowering the difficulty by 10% would also lower
> the maximum block size for that block from 100% to 90% of the current
> limit. For simplicity I will assume a linear identity transform as the
> function, but a quadratic or other function with compounding marginal cost
> may be preferred.
>
Sorry but I fail to see how a linear identity transform between block
size and difficulty would work.
The miner's reward for finding a block is the sum of subsidy and fees:
R = S + F
The probability that the miner will find a block over a time interval is
inversely proportional to the difficulty D:
P = K / D
where K is a constant that depends on the miner's hashrate. The expected
reward of the miner is:
E = P * R
Consider that the miner chooses a new difficulty:
D' = D(1 + x).
With a linear identity transform between block size and difficulty, the
miner will be allowed to collect fees from a block of size: S'=S(1+x)
In the best case, collected will be proportional to block size:
F' = F(1+x)
Thus we get:
E' = P' * R' = K/(D(1+x)) * (S + F(1+x))
E' = E - x/(1+x) * S * K / D
So with this linear identity transform, increasing block size never
increases the miners gain. As long as the subsidy exists, the best
strategy for miners is to reduce block size (i.e. to choose x<0).
๐ Original message:Le 08/05/2015 22:33, Mark Friedenbach a รฉcrit :
> * For each block, the miner is allowed to select a different difficulty
> (nBits) within a certain range, e.g. +/- 25% of the expected difficulty,
> and this miner-selected difficulty is used for the proof of work check. In
> addition to adjusting the hashcash target, selecting a different difficulty
> also raises or lowers the maximum block size for that block by a function
> of the difference in difficulty. So increasing the difficulty of the block
> by an additional 25% raises the block limit for that block from 100% of the
> current limit to 125%, and lowering the difficulty by 10% would also lower
> the maximum block size for that block from 100% to 90% of the current
> limit. For simplicity I will assume a linear identity transform as the
> function, but a quadratic or other function with compounding marginal cost
> may be preferred.
>
Sorry but I fail to see how a linear identity transform between block
size and difficulty would work.
The miner's reward for finding a block is the sum of subsidy and fees:
R = S + F
The probability that the miner will find a block over a time interval is
inversely proportional to the difficulty D:
P = K / D
where K is a constant that depends on the miner's hashrate. The expected
reward of the miner is:
E = P * R
Consider that the miner chooses a new difficulty:
D' = D(1 + x).
With a linear identity transform between block size and difficulty, the
miner will be allowed to collect fees from a block of size: S'=S(1+x)
In the best case, collected will be proportional to block size:
F' = F(1+x)
Thus we get:
E' = P' * R' = K/(D(1+x)) * (S + F(1+x))
E' = E - x/(1+x) * S * K / D
So with this linear identity transform, increasing block size never
increases the miners gain. As long as the subsidy exists, the best
strategy for miners is to reduce block size (i.e. to choose x<0).