Anthony,


On Thu, Dec 17, 2015 at 6:55 PM, Anthony Towns via bitcoin-dev <
bitcoin-dev@lists.linuxfoundation.org> wrote:

> On Thu, Dec 17, 2015 at 04:51:19PM +0100, sickpig--- via bitcoin-dev wrote:
> > On Thu, Dec 17, 2015 at 2:09 PM, Jorge Timón wrote:
> > > Unless I'm missing something, 2 mb x4 = 8mb, so bip102 + SW is already
> > > equivalent to the 2-4-8 "compromise" proposal [...]
> > isn't SegWit gain ~75%? hence 2mb x 1.75 = 3.5.
>
> Segwit as proposed gives a 75% *discount* to witness data with the
> same limit, so at a 1MB limit, that might give you (eg) 2.05MB made up
> of 650kB of base block data plus 1.4MB of witness data; where 650kB +
> 1.4MB/4 = 1MB at the 1MB limit; or 4.1MB made up of 1.3MB of base plus
> 2.8MB of witness, for 1.3MB+2.8MB/4 = 2MB at a 2MB limit.
>
> > 4x is theoric gain you get in case of 2-2 multisig txs.
>
> With segregated witness, 2-2 multisig transactions are made up of 94B
> of base data, plus about 214B of witness data; discounting the witness
> data by 75% gives 94+214/4=148 bytes. That compares to about 301B for
> a 2-2 multisig transaction with P2SH rather than segwit, and 301/148
> gives about a 2.03x gain, not a 4x gain. A 2.05x gain is what I assumed
> to get the numbers above.
>
> You get further improvements with, eg, 3-of-3 multisig, but to get
> the full, theoretical 4x gain you'd need a fairly degenerate looking
> transaction.
>
> Pay to public key hash with segwit lets you move about half the
> transaction data into the witness, giving about a 1.6x improvement by
> my count (eg 1.6MB = 800kB of base data plus 800kB of witness data,
> where 800kB+800kB/4=1MB), so I think a gain of between 1.6 and 2.0 is
> a reasonable expectation to have for the proposed segwit scheme overall.
>
>
many thanks for the explanation.

so it should be fair to say that BIP 102 + SW would bring a gain between
2*1.6 and 2*2.

Just for the sake of simplicity if we take the middle of the interval we
could say
that BIP102 + SW will bring us a max block (virtual) size equal to 1MB * 2
* 1.8 = 3.6

Is it right?