Goss, Brian C., M.D. [ARCHIVE] on Nostr: 📅 Original date posted:2013-11-08 📝 Original message:Peter, What is the ...
📅 Original date posted:2013-11-08
📝 Original message:Peter,
What is the propagation time within a pool? If my pool is made up of a ton
of fancy ASICs connected by 300 baud modems, how does that affect your
analysis (ie, Q for a mining pool is effectively a function of time as
well)?
Brian
P.S. I hope these are not ignorant questions; if they are, please feel free
to disregard!
Message: 1
Date: Thu, 7 Nov 2013 15:31:23 -0500
From: Peter Todd <pete at petertodd.org>
Subject: Re: [Bitcoin-development] On the optimal block size and why
transaction fees are 8 times too low (or transactions 8 times too
big)
To: Michael Gronager <gronager at ceptacle.com>
Cc: Bitcoin Dev <bitcoin-development at lists.sourceforge.net>
Message-ID: <20131107203123.GB3805 at petertodd.org>
Content-Type: text/plain; charset="us-ascii"
> Final conclusions is that the fee currently is too small and that
> there is no need to keep a maximum block size, the fork probability
> will automatically provide an incentive to not let block grows into
infinity.
Your definition of P_fork is inaccurate for a miner with non-negligable
hashing power - a miner will never fork themselves. Taking that into account
we have three outcomes:
1) The block propagates without any other miner finding a block.
2) During propagation another miner finds a block. (tie)
2.1) You win the tie by finding another block.
2.2) You lose the tie because someone else finds a block.
We will define t_prop as the time it takes for a block to propagate from you
to 100% of the hashing power, and as a simplifying assumption we will assume
that until t_prop has elapsed, 0% of the hashing power has the block, and
immedately after, 100% has the block. We will also define t_int, the average
interval between blocks. (600 seconds for Bitcoin) Finally, we will define Q
as the probability that you will find the next block.
The probabilities of the various outcomes:
1) 1 - (t_prop/t_int * (1-Q))
2) t_prop/t_int * (1-Q)
2.1) Q
2.2) 1-Q
Note that to simplify the equations we have not taking into account
propagation in our calculations for outcomes 2.1 or 2.2
Thus we can define P_fork taking into account Q:
P_fork(Q) = (t_prop/t_int * (1-Q))(1-Q) = t_pop/t_int * (1-Q)^2
Over the range 0 < Q < 0.5 the probability of a fork decreases approximately
linearly as your hashing power increases:
d/dq P_fork(Q) = 2(Q-1)
Q=0 -> d/dq P_fork(Q) = -2
Q=1/2 -> d/dq P_fork(Q) = -1
With our new, more accurate, P_fork(Q) function lets re-calculate the
break-even fee/KB using your original approach:
t_prop = t_0 + \alpha*S
E_fee = f*S
E(Q) = Q*(1 - P_fork(Q))*(E_bounty + E_fee)
E(Q) = Q*[1 - (t_0 + k*S)/t_int * (1-Q)^2]*(E_B + f*S)
d/dS E(Q) = Q*[ -2fSk/t_int*(1-Q)^2 - f*t_0/t_int*(1-Q)^2 + f -
E_b*k/t_int*(1-Q)^2 ]
Again, we want to choose the fee so that the more transactions we include
the more we earn, dE/dS > 0 We find the minimum fee to include a transaction
at all by setting S=0, thus we get:
d/dS E(Q, S=0) = Q*[ f - f*t_0/t_int*(1-Q)^2 - E_b*k/t_int*(1-Q)^2 ] > 0
f(1 - t_0/t_int*(1-Q)^2) > E_b*k/t_int*(1-Q)^2
f > [E_b*k/t_int(1-Q)^2] / [1 - t_0/t_int*(1-Q)^2]
f > [E_b*k*(1-Q)^2] / [t_int - t_0*(1-Q)^2]
With Q=0:
f > E_b*k / (t_int - t_0) ~ E_b*k/t_int
This is the same result you derived. However lets look at Q != 0:
df/dQ = 2*E_b*k * [t_int*(q-1)] / [t_int - t_0(q-1)^2]^2
With negligible latency we get:
df/dQ, t_0=0 = 2*E_b*k*(q-1)/t_int
So what does that mean? Well in the region 0 < q < 1/2, df/dQ is always
negative. In other words, as you get more hashing power, the fee/KB you can
charge and still break even decreases linearly because you will never orphan
yourself. Lets trythe same assumptions as your first analysis, based on the
work by Decker et al
Based on the work by Decker et al, lets try to calculate break-even fee/KB
for negligible, 10%, 25% and 40% hashing power:
t_0 = 10s
t_int = 600s
k = 80ms/kB
E_b = 25BTC
Q=0 -> f = 0.0033 BTC/kB
Q=0.1 -> f = 0.0027 BTC/kB
Q=0.25 -> f = 0.0018 BTC/kB
Q=0.40 -> f = 0.0012 BTC/kB
Let's assume every miner is directly peered with every other miner, each of
those connections is 1MB/s, and somehow there's no latency at all:
k = 1mS/kB
Q=0 -> f = 0.000042 BTC/kB
Q=0.1 -> f = 0.000034 BTC/kB
Q=0.25 -> f = 0.000023 BTC/kB
Q=0.40 -> f = 0.000015 BTC/kB
Regardless of how you play around with the parameters, being a larger miner
has a significant advantage because you can charge lower fees for your
transactions and therefor earn more money. But it gets even more ugly when
you take into account that maybe a guy with 0.1% hashing power can't afford
the high bandwidth, low-latency, internet connection that the larger pool
has:
k = 10mS/kB, t_0=5s, Q=0.01 -> 0.000411 BTC/KB k = 1mS/kB, t_0=1s, Q=0.15
-> 0.000030 BTC/KB
So the 1% pool has an internet connection capable of 100kB/s to each peer,
taking 5s to reach all the hashing power. The 15% pool can do 1MB/s to each
peer, taking 1s to reach all the hashing power. This small different means
that the 1% pool needs to charge 13.7x more per KB for their transactions to
break even! It's a disaster for decentralization.
Businesses live and die on percentage points, let alone orders of magnitude
differences in cost, and I haven't even taken into account second-order
effects like the perverse incentives to publish your blocks to only a
minority of hashing power.(1)
This problem is inherent to the fundemental design of Bitcoin:
regardless of what the blocksize is, or how fast the network is, the current
Bitcoin consensus protocol rewards larger mining pools with lower costs per
KB to include transactions. It's a fundemental issue. An unlimited blocksize
will make the problem even worse by increasing fixed costs, but keeping the
blocksize at 1MB forever doesn't solve the underlying problem either as the
inflation subsidy becomes less important and fees more important.
1)
http://www.mail-archive.com/bitcoin-development@lists.sourceforge.net/msg032
00.html
--
'peter'[:-1]@petertodd.org
00000000000000054eeccf3ac454892457bf4919d78efb275efd2ddd1a920c99
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📝 Original message:Peter,
What is the propagation time within a pool? If my pool is made up of a ton
of fancy ASICs connected by 300 baud modems, how does that affect your
analysis (ie, Q for a mining pool is effectively a function of time as
well)?
Brian
P.S. I hope these are not ignorant questions; if they are, please feel free
to disregard!
Message: 1
Date: Thu, 7 Nov 2013 15:31:23 -0500
From: Peter Todd <pete at petertodd.org>
Subject: Re: [Bitcoin-development] On the optimal block size and why
transaction fees are 8 times too low (or transactions 8 times too
big)
To: Michael Gronager <gronager at ceptacle.com>
Cc: Bitcoin Dev <bitcoin-development at lists.sourceforge.net>
Message-ID: <20131107203123.GB3805 at petertodd.org>
Content-Type: text/plain; charset="us-ascii"
> Final conclusions is that the fee currently is too small and that
> there is no need to keep a maximum block size, the fork probability
> will automatically provide an incentive to not let block grows into
infinity.
Your definition of P_fork is inaccurate for a miner with non-negligable
hashing power - a miner will never fork themselves. Taking that into account
we have three outcomes:
1) The block propagates without any other miner finding a block.
2) During propagation another miner finds a block. (tie)
2.1) You win the tie by finding another block.
2.2) You lose the tie because someone else finds a block.
We will define t_prop as the time it takes for a block to propagate from you
to 100% of the hashing power, and as a simplifying assumption we will assume
that until t_prop has elapsed, 0% of the hashing power has the block, and
immedately after, 100% has the block. We will also define t_int, the average
interval between blocks. (600 seconds for Bitcoin) Finally, we will define Q
as the probability that you will find the next block.
The probabilities of the various outcomes:
1) 1 - (t_prop/t_int * (1-Q))
2) t_prop/t_int * (1-Q)
2.1) Q
2.2) 1-Q
Note that to simplify the equations we have not taking into account
propagation in our calculations for outcomes 2.1 or 2.2
Thus we can define P_fork taking into account Q:
P_fork(Q) = (t_prop/t_int * (1-Q))(1-Q) = t_pop/t_int * (1-Q)^2
Over the range 0 < Q < 0.5 the probability of a fork decreases approximately
linearly as your hashing power increases:
d/dq P_fork(Q) = 2(Q-1)
Q=0 -> d/dq P_fork(Q) = -2
Q=1/2 -> d/dq P_fork(Q) = -1
With our new, more accurate, P_fork(Q) function lets re-calculate the
break-even fee/KB using your original approach:
t_prop = t_0 + \alpha*S
E_fee = f*S
E(Q) = Q*(1 - P_fork(Q))*(E_bounty + E_fee)
E(Q) = Q*[1 - (t_0 + k*S)/t_int * (1-Q)^2]*(E_B + f*S)
d/dS E(Q) = Q*[ -2fSk/t_int*(1-Q)^2 - f*t_0/t_int*(1-Q)^2 + f -
E_b*k/t_int*(1-Q)^2 ]
Again, we want to choose the fee so that the more transactions we include
the more we earn, dE/dS > 0 We find the minimum fee to include a transaction
at all by setting S=0, thus we get:
d/dS E(Q, S=0) = Q*[ f - f*t_0/t_int*(1-Q)^2 - E_b*k/t_int*(1-Q)^2 ] > 0
f(1 - t_0/t_int*(1-Q)^2) > E_b*k/t_int*(1-Q)^2
f > [E_b*k/t_int(1-Q)^2] / [1 - t_0/t_int*(1-Q)^2]
f > [E_b*k*(1-Q)^2] / [t_int - t_0*(1-Q)^2]
With Q=0:
f > E_b*k / (t_int - t_0) ~ E_b*k/t_int
This is the same result you derived. However lets look at Q != 0:
df/dQ = 2*E_b*k * [t_int*(q-1)] / [t_int - t_0(q-1)^2]^2
With negligible latency we get:
df/dQ, t_0=0 = 2*E_b*k*(q-1)/t_int
So what does that mean? Well in the region 0 < q < 1/2, df/dQ is always
negative. In other words, as you get more hashing power, the fee/KB you can
charge and still break even decreases linearly because you will never orphan
yourself. Lets trythe same assumptions as your first analysis, based on the
work by Decker et al
Based on the work by Decker et al, lets try to calculate break-even fee/KB
for negligible, 10%, 25% and 40% hashing power:
t_0 = 10s
t_int = 600s
k = 80ms/kB
E_b = 25BTC
Q=0 -> f = 0.0033 BTC/kB
Q=0.1 -> f = 0.0027 BTC/kB
Q=0.25 -> f = 0.0018 BTC/kB
Q=0.40 -> f = 0.0012 BTC/kB
Let's assume every miner is directly peered with every other miner, each of
those connections is 1MB/s, and somehow there's no latency at all:
k = 1mS/kB
Q=0 -> f = 0.000042 BTC/kB
Q=0.1 -> f = 0.000034 BTC/kB
Q=0.25 -> f = 0.000023 BTC/kB
Q=0.40 -> f = 0.000015 BTC/kB
Regardless of how you play around with the parameters, being a larger miner
has a significant advantage because you can charge lower fees for your
transactions and therefor earn more money. But it gets even more ugly when
you take into account that maybe a guy with 0.1% hashing power can't afford
the high bandwidth, low-latency, internet connection that the larger pool
has:
k = 10mS/kB, t_0=5s, Q=0.01 -> 0.000411 BTC/KB k = 1mS/kB, t_0=1s, Q=0.15
-> 0.000030 BTC/KB
So the 1% pool has an internet connection capable of 100kB/s to each peer,
taking 5s to reach all the hashing power. The 15% pool can do 1MB/s to each
peer, taking 1s to reach all the hashing power. This small different means
that the 1% pool needs to charge 13.7x more per KB for their transactions to
break even! It's a disaster for decentralization.
Businesses live and die on percentage points, let alone orders of magnitude
differences in cost, and I haven't even taken into account second-order
effects like the perverse incentives to publish your blocks to only a
minority of hashing power.(1)
This problem is inherent to the fundemental design of Bitcoin:
regardless of what the blocksize is, or how fast the network is, the current
Bitcoin consensus protocol rewards larger mining pools with lower costs per
KB to include transactions. It's a fundemental issue. An unlimited blocksize
will make the problem even worse by increasing fixed costs, but keeping the
blocksize at 1MB forever doesn't solve the underlying problem either as the
inflation subsidy becomes less important and fees more important.
1)
http://www.mail-archive.com/bitcoin-development@lists.sourceforge.net/msg032
00.html
--
'peter'[:-1]@petertodd.org
00000000000000054eeccf3ac454892457bf4919d78efb275efd2ddd1a920c99
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