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From: Mark Greene <greenemj@gmail.com>
Date: Wed, 12 May 2010 08:45:42 -0400
Message-ID: <AANLkTikZaSit6m4kkwfGODztPEOdbrZxBE-eBvIynLrc@mail.gmail.com>
Subject: Re: replication impact on write throughput
To: user@cassandra.apache.org
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>>If the replication factor is 2, then everything is written twice. So
>>your throughput is cut in half.

throughput of new inserts is cut in half right? I think I was thinking about
capacity in more general terms from the node's perspective. The node has the
ability to write so many operations per second whether that be a replicated
key or not and you could potentially take that number and multiply it by the
number of nodes in the ring to give you your write capacity per second.

On Tue, May 11, 2010 at 10:13 PM, Paul Prescod <paul@prescod.net> wrote:

> On Tue, May 11, 2010 at 5:56 PM, Mark Greene <greenemj@gmail.com> wrote:
> > I was under the impression from what I've seen talked about on this list
> > (perhaps I'm wrong here) that given the write throughput of one node in a
> > cluster (again assuming each node has a given throughput and the same
> > config) that you would simply multiply that throughput by the number of
> > nodes you had, giving you the total throughput for the entire ring (like
> you
> > said this is somewhat artificial). The main benefit being that adding
> > capacity was as simple as adding more nodes to the ring with no
> degradation.
> > So by your math, 100 nodes with each node getting 5k wps, I would assume
> the
> > total capacity is 500k wps. But perhaps I've misunderstood some key
> > concepts. Still a novice myself ;-)
>
> If the replication factor is 2, then everything is written twice. So
> your throughput is cut in half.
>
>  Paul Prescod
>

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<div><br></div>&gt;&gt;If the replication factor is 2, then everything is w=
ritten twice. So<br>&gt;&gt;your throughput is cut in half.<div><br></div><=
div>throughput of new inserts is cut in half right? I think I was thinking =
about capacity in more general terms from the node&#39;s perspective. The n=
ode has the ability to write so many operations per second whether that be =
a replicated key or not and you could potentially take that number and mult=
iply it by the number of nodes in the ring to give you your write capacity =
per second.=A0<br>

<br><div class=3D"gmail_quote">On Tue, May 11, 2010 at 10:13 PM, Paul Presc=
od <span dir=3D"ltr">&lt;<a href=3D"mailto:paul@prescod.net">paul@prescod.n=
et</a>&gt;</span> wrote:<br><blockquote class=3D"gmail_quote" style=3D"marg=
in:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;">

<div class=3D"im">On Tue, May 11, 2010 at 5:56 PM, Mark Greene &lt;<a href=
=3D"mailto:greenemj@gmail.com">greenemj@gmail.com</a>&gt; wrote:<br>
&gt; I was under the impression from what I&#39;ve seen talked about on thi=
s list<br>
&gt; (perhaps I&#39;m wrong here) that given the write throughput of one no=
de in a<br>
&gt; cluster (again assuming each node has a given throughput and the same<=
br>
&gt; config) that you would simply multiply that throughput by the number o=
f<br>
&gt; nodes you had, giving you the total throughput for the entire ring (li=
ke you<br>
&gt; said this is somewhat=A0artificial). The main benefit being that addin=
g<br>
&gt; capacity was as simple as adding more nodes to the ring with no degrad=
ation.<br>
&gt; So by your math, 100 nodes with each node getting 5k wps, I would assu=
me the<br>
&gt; total capacity is 500k wps. But perhaps I&#39;ve misunderstood some ke=
y<br>
&gt; concepts. Still a novice myself ;-)<br>
<br>
</div>If the replication factor is 2, then everything is written twice. So<=
br>
your throughput is cut in half.<br>
<font color=3D"#888888"><br>
=A0Paul Prescod<br>
</font></blockquote></div><br></div>

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