RE: RAID

From: Satar Naghshineh <Satar.Naghshineh_at_irvine.mellesgriot.com>
Date: Mon, 30 Oct 2000 10:30:42 -0800
Message-Id: <10665.120585@fatcity.com>

This message is in MIME format. Since your mail reader does not understand this format, some or all of this message may not be legible.

------_=_NextPart_001_01C0429F.82907180
Content-Type: text/plain;

charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable

That I understand. You were using "spindle" to mean "Logical device". = When I
see the word "spindle", I think of the physical spindle of a hard = drive. So,
in a RAID5 configuration, you are using multiple spindles (Which was my initial statement).

Regards,
Satar

> -----Original Message-----
> From: Ron Rogers [SMTP:RROGERS_at_galottery.org]
> Sent: Monday, October 30, 2000 2:55 PM
> To: Multiple recipients of list ORACLE-L
> Subject: RE: RAID

>=20

> Satar,
> Maybe I'm not explaining it clear enough or in terms we both =
understand.
> I don't think that it has to do with the lunar phases of the moon and =
the
> tides of the world.
> If you use the RAID 5 configuration of drives, the drives are =
treated as
> one drive. As an example: if you have 5 -10 GIG drives used =
seperately as
> individual drives you can place the 5 drives where ever you want and =
on
> what controller you want. You have the full benefit of the 5 =
independent
> drives. If you use the drives as a RAID5 configuration you in effect =
have
> only 1 drive to use. The RAID5 configured drive. The total capacity =
of the
> "drive" will be 40 GIG, not 50 GIG, because of the parity algorythem =
that
> uses one of the drives. If you lose one of the physical drives in the
> RAID5 configuration the values on that drive can be durived from the
> values on the other 4 drives.
> As example: if the 5 drives have the values 1, 2, 3, 4, 10 stored on =
them
> in the same stripe then you can calculate the missing value on the =
drive
> stripe that was defective. 1+2+3+4=3D10. 10 being the "parity" =
information.
> Losing the first drive and calculating the value of the data stored =
is
> 10-(2+3+4) =3D1.
> The RAID5 configuration is treated as one logical device (spindle) =
not 5
> logical devices.
> HTH
> ROR =AA=BF=AA

>=20
>=20
>=20
>=20

------_=_NextPart_001_01C0429F.82907180
Content-Type: text/html;

charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2//EN">
<HTML>
<HEAD>
<META HTTP-EQUIV=3D"Content-Type" CONTENT=3D"text/html; =

charset=3Diso-8859-1">
<META NAME=3D"Generator" CONTENT=3D"MS Exchange Server version = 5.5.2650.12">

<TITLE>RE: RAID</TITLE>
</HEAD>
<BODY>

That I understand. = You were using "spindle" to mean "Logical device". = When I see the word "spindle", I think of the physical = spindle of a hard drive. So, in a RAID5 configuration, you are using = multiple spindles (Which was my initial statement).

Regards, Satar

</P>
<UL>
<P><FONT SIZE=3D1 FACE=3D"Arial">-----Original Message-----</FONT>
<BR><B><FONT SIZE=3D1 FACE=3D"Arial">From:&nbsp;&nbsp;</FONT></B> <FONT =

SIZE=3D1 FACE=3D"Arial">Ron Rogers [SMTP:RROGERS_at_galottery.org] Sent:   Monday, October 30, 2000 2:55 PM To:     Multiple recipients of list ORACLE-L Subject:       = RE: RAID

Satar,   Maybe I'm not explaining it = clear enough or in terms we both understand. I don't think that it has = to do with the lunar phases of the moon and the tides of the = world.

If you use the RAID 5 = configuration of drives, the drives are treated as one drive. As an = example: if you have 5 -10 GIG drives used seperately as individual = drives you can place the 5 drives where ever you want and on what = controller you want. You have the full benefit of the 5 independent = drives. If you use the drives as a RAID5 configuration you in effect = have only 1 drive to use. The RAID5 configured drive. The total = capacity of the "drive" will be 40 GIG, not 50 GIG, because = of the parity algorythem that uses one of the drives. If you lose one = of the physical drives in the RAID5 configuration the values on that = drive can be durived from the values on the other 4 drives.

As example: if the 5 drives have = the values 1, 2, 3, 4, 10 stored on them in the same stripe then you = can calculate the missing value on the drive stripe that was defective. = 1+2+3+4=3D10. 10 being the "parity" information. Losing the = first drive and calculating the value of the data stored is 10-(2+3+4) = =3D1.

The RAID5 configuration is = treated as one logical device (spindle) not 5 logical devices. HTH ROR =AA=BF=AA Received on Mon Oct 30 2000 - 12:30:42 CST