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From: Sergio Demian Lerner <sergio.d.lerner@gmail.com>
Date: Tue, 9 May 2017 16:15:32 -0300
Message-ID: <CAKzdR-pN0B-wjhPh3n6Jw7L6wsz9yQ=kyVX7NYx5ACZC3bg8OA@mail.gmail.com>
To: Matt Corallo <lf-lists@mattcorallo.com>
Content-Type: multipart/alternative; boundary=001a114670905d9ba1054f1c300a
Cc: bitcoin-dev <bitcoin-dev@lists.linuxfoundation.org>
Subject: Re: [bitcoin-dev] Some real-world results about the current Segwit
	Discount
Precedence: list

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Let n be the non-segwit bytes. Let the seg/noseg ratio be 1.7.

Segwit with 75% discount: (let WITNESS_SCALE_FACTOR=4)
n*WITNESS_SCALE_FACTOR+n*1.7 = 4,000,000
Then n=4,000,000 / 5.7 = 701K
Average block size = 701K*(1+1.7)=1.8 Mbytes
Maximum block size = 4 MBytes

Segwit with 50% discount + 2MB HF: (let WITNESS_SCALE_FACTOR=2)
n*2+n*1.7 = 4,000,000
n = 4,000,000/ 3.7 = 1.08M
Average block size = 1.08M*(1+1.7)=2.9 Mbytes
Maximum block size = 4 MBytes

The capacity of Segwit(50%)+2MbHF is 50% more than Segwit, and the maximum
block size is the same.


On Tue, May 9, 2017 at 3:58 PM, Sergio Demian Lerner <
sergio.d.lerner@gmail.com> wrote:

>
>>
>> You suggested "If the maximum block weight is set to 2.7M, each byte of
>> non-witness block costs 1.7", but these numbers dont work out - setting
>> the discount to 1.7 gets you a maximum block size of 1.7MB (in a soft
>> fork), not 2.7MB.
>
>
> Yes. In a soft-fork is true.
> I was thinking about what a HF could do to optimize the balance, and I
> forgot I was in the context of a SF.
>
>
>

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<div dir=3D"ltr">Let n be the non-segwit bytes. Let the seg/noseg ratio be =
1.7.=C2=A0<div><br></div><div>Segwit with 75% discount: (let WITNESS_SCALE_=
FACTOR=3D4)</div><div></div><div>n*WITNESS_SCALE_FACTOR+n*1.7 =3D 4,000,000=
</div><div>Then n=3D4,000,000 / 5.7 =3D 701K</div><div>Average block size =
=3D 701K*(1+1.7)=3D1.8 Mbytes</div><div>Maximum block size =3D 4 MBytes<br>=
</div><div><br></div><div>Segwit with 50% discount + 2MB HF: (let WITNESS_S=
CALE_FACTOR=3D2)</div><div><div>n*2+n*1.7 =3D 4,000,000<br></div></div><div=
>n =3D 4,000,000/ 3.7 =3D 1.08M</div><div><div>Average block size =3D 1.08M=
*(1+1.7)=3D2.9 Mbytes</div></div><div><div>Maximum block size =3D 4 MBytes<=
br></div></div><div><br></div><div>The capacity of Segwit(50%)+2MbHF is 50%=
 more than Segwit, and the maximum block size is the same.=C2=A0</div><div>=
<br></div></div><div class=3D"gmail_extra"><br><div class=3D"gmail_quote">O=
n Tue, May 9, 2017 at 3:58 PM, Sergio Demian Lerner <span dir=3D"ltr">&lt;<=
a href=3D"mailto:sergio.d.lerner@gmail.com" target=3D"_blank">sergio.d.lern=
er@gmail.com</a>&gt;</span> wrote:<br><blockquote class=3D"gmail_quote" sty=
le=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><div d=
ir=3D"ltr"><div class=3D"gmail_extra"><div class=3D"gmail_quote"><span clas=
s=3D""><blockquote class=3D"gmail_quote" style=3D"margin:0px 0px 0px 0.8ex;=
border-left:1px solid rgb(204,204,204);padding-left:1ex"><br>
<br>
You suggested &quot;If the maximum block weight is set to 2.7M, each byte o=
f<br>
non-witness block costs 1.7&quot;, but these numbers dont work out - settin=
g<br>
the discount to 1.7 gets you a maximum block size of 1.7MB (in a soft<br>
fork), not 2.7MB. </blockquote><div><br></div></span><div>Yes. In a soft-fo=
rk is true.</div><div>I was thinking about what a HF could do to optimize t=
he balance, and I forgot I was in the context of a SF.</div><div><br></div>=
<div>=C2=A0</div></div></div></div>
</blockquote></div><br></div>

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