Subject: Re: [Bitcoin-development] On the optimal block size and why transaction fees are 8 times too low (or transactions 8 times too big)

From: Peter Todd <[email-masked]>
Date: Thu, 7 Nov 2013 15:31:23 -0500

> 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/[email-masked]/msg03200.html

-- 
'peter'[:-1]@petertodd.org
00000000000000054eeccf3ac454892457bf4919d78efb275efd2ddd1a920c99