MESSAGE
DATE | 2014-12-06 |
FROM | Maneesh Kongara
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SUBJECT | Re: [LIU Comp Sci] Parity Bit Algebra
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From owner-learn-outgoing-at-mrbrklyn.com Sat Dec 6 21:50:06 2014 Return-Path: X-Original-To: archive-at-mrbrklyn.com Delivered-To: archive-at-mrbrklyn.com Received: by mrbrklyn.com (Postfix) id 8B40116007B; Sat, 6 Dec 2014 21:50:06 -0500 (EST) Delivered-To: learn-outgoing-at-mrbrklyn.com Received: by mrbrklyn.com (Postfix, from userid 28) id 7330816112F; Sat, 6 Dec 2014 21:50:06 -0500 (EST) Delivered-To: learn-at-mrbrklyn.com Received: from mail-qc0-f175.google.com (mail-qc0-f175.google.com [209.85.216.175]) by mrbrklyn.com (Postfix) with ESMTP id 8F13116007B for ; Sat, 6 Dec 2014 21:50:05 -0500 (EST) Received: by mail-qc0-f175.google.com with SMTP id b13so2464528qcw.6 for ; Sat, 06 Dec 2014 18:50:05 -0800 (PST) DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=gmail.com; s=20120113; h=mime-version:references:from:date:message-id:subject:to :content-type; bh=PSLCc9N+exp10h25hyt/mzgAU2c418m+FfSExlyQua4=; b=D1TzeRqcJovOtY/V9igV50KU5z4AfTeFGWuBx/e5ARIGyYsT58snfZFn4FOqW77/kJ qdmk45CoEjECiUXlDHtHJqVxH4xgEmMRF9A/M+mdJNJCeWC+IgiyCJpA3sM+umZWUz1h OVy2xYOXyNLenW9Dt3ktoT7LAOKUIKzOMUvTJgLL4BJNaldtPRCRYg+RM49/hZXkEeI2 uQlRlkyut6UizwFMHpYLD6OOTk/8eVPVdPQq310E9TmXomehCMZ79hxJ3W2M7FbeUBBO x71OLwrBz0CcBa1DLe77cIm9aGNBbTx/HU5TAJ6/FArKiME6HH7VVSbW6ELpv2OND9Xz Cskg== X-Received: by 10.140.89.200 with SMTP id v66mr38498485qgd.85.1417920605218; Sat, 06 Dec 2014 18:50:05 -0800 (PST) MIME-Version: 1.0 References: <54838832.6010205-at-panix.com> <54838C85.60705-at-panix.com> <5483BD3C.9020609-at-panix.com> <5483C042.2050104-at-panix.com> From: Maneesh Kongara Date: Sun, 07 Dec 2014 02:50:04 +0000 Message-ID: Subject: Re: [LIU Comp Sci] Parity Bit Algebra To: learn-at-mrbrklyn.com Content-Type: multipart/alternative; boundary=001a11c11b50fe6d3c0509975d80 Sender: owner-learn-at-mrbrklyn.com Precedence: bulk Reply-To: learn-at-mrbrklyn.com
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Have you seen the review questions?
On Sat, Dec 6, 2014, 9:49 PM Ruben Safir wrote:
> > > I can't even read the question > > > > On 12/07/2014 02:38 AM, Maneesh Kongara wrote: > > I've finished all other questions... I'm breaking my head with the last > > question. > > On Dec 6, 2014 9:36 PM, "Ruben Safir" wrote: > > > >> On 12/06/2014 11:09 PM, Maneesh Kongara wrote: > >>> Preliminary work as in answers to the other questions of the > assignment? > >>> On Dec 6, 2014 6:08 PM, "Ruben Safir" wrote: > >> I'm going to crash, how far did you get in the assingment. > >> > >> Call me is 20 minutes. I have one hour to do it :) > >> > >> > >>>> On 12/06/2014 10:52 PM, Maneesh Kongara wrote: > >>>>> Thanks so much for sharing Mr Rubin. Have you completed the > assignment? > >>>>> On Dec 6, 2014 5:49 PM, "Ruben Safir" wrote: > >>>> Not even looked at it outside of the library. I'm burning up trying > to > >>>> catch up. Do you have some preliminary work? > >>>> > >>>> I did give it throught. Part of the solution must have to do with the > >>>> Comp + 2 binary > >>>> > >>>> I can't even find my notes... > >>>> > >>>>>> This is from theidiscussion of Raid 3 Stallings Text which the slide > >>>>>> come from. It is related to the problem you asked about the parity > >> bit > >>>>>> calculation for the Hamming Code > >>>>>> > >>>>>> Not the methodology for logical calculation, it is straight math > >>>>>> > >>>>>> Note that X is XOR for whatever UFT reason > >>>>>> > >>>>>> > >>>>>> REDUNDANCY In the event of a drive failure, the parity drive is > >> accessed > >>>>>> and data is reconstructed from the remaining devices. Once the > failed > >>>>>> drive is replaced, the missing data can be restored on the new drive > >> and > >>>>>> operation resumed. > >>>>>> > >>>>>> Data reconstruction is simple. Consider an array of five drives in > >> which > >>>>>> X0 through X3 contain data and X4 is the parity disk.The parity for > >> the > >>>>>> ith bit is calculated as follows: > >>>>>> > >>>>>> X4(i) = X3(i) { X2(i) { X1(i) { X0(i) > >>>>>> > >>>>>> where { is exclusive-OR function. > >>>>>> > >>>>>> Suppose that drive X1 has failed. If we add X4(i) { X1(i) to both > >> sides > >>>>>> of the preceding equation, we get > >>>>>> > >>>>>> X1(i) = X4(i) { X3(i) { X2(i) { X0(i) > >>>>>> > >>>>>> > >>>>>> file:///home/ruben/william stallings - computer organization and > >>>>>> architecture designing for performance (8th edition).pdf > >>>>>> > >>>>>> Page 202 > >>>>>> > >>>>>> > >>>>>> > >>>>>> http://www.nylxs.com/images/raid3_parity_math.png > >>>>>> > >>>>>> Ruben > >>>>>> > >> > >
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Have you seen the review questions?
On Sa= t, Dec 6, 2014, 9:49 PM=C2=A0Ruben Safir < ix.com">mrbrklyn-at-panix.com> wrote: te" style=3D"margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"= >
I can't even read the question
On 12/07/2014 02:38 AM, Maneesh Kongara wrote:
> I've finished all other questions... I'm breaking my head with= the last
> question.
> On Dec 6, 2014 9:36 PM, "Ruben Safir" <mrbrklyn-at-panix.com" target=3D"_blank">mrbrklyn-at-panix.com> wrote:
>
>> On 12/06/2014 11:09 PM, Maneesh Kongara wrote:
>>> Preliminary work as in answers to the other questions of the a= ssignment?
>>> On Dec 6, 2014 6:08 PM, "Ruben Safir" <"mailto:mrbrklyn-at-panix.com" target=3D"_blank">mrbrklyn-at-panix.com> wr= ote:
>> I'm going to crash, how far did you get in the assingment.
>>
>> Call me is 20 minutes.=C2=A0 I have one hour to do it :)
>>
>>
>>>> On 12/06/2014 10:52 PM, Maneesh Kongara wrote:
>>>>> Thanks so much for sharing Mr Rubin. Have you complete= d the assignment?
>>>>> On Dec 6, 2014 5:49 PM, "Ruben Safir" < href=3D"mailto:mrbrklyn-at-panix.com" target=3D"_blank">mrbrklyn-at-panix.com>> wrote:
>>>> Not even looked at it outside of the library.=C2=A0 I'= m burning up trying to
>>>> catch up.=C2=A0 Do you have some preliminary work?
>>>>
>>>> I did give it throught.=C2=A0 Part of the solution must ha= ve to do with the
>>>> Comp + 2 binary
>>>>
>>>> I can't even find my notes...
>>>>
>>>>>> This is from theidiscussion of Raid 3 Stallings Te= xt which the slide
>>>>>> come from.=C2=A0 It is related to the problem you = asked about the parity
>> bit
>>>>>> calculation for the Hamming Code
>>>>>>
>>>>>> Not the methodology for logical calculation, it is= straight math
>>>>>>
>>>>>> Note that X is XOR for whatever UFT reason
>>>>>>
>>>>>>
>>>>>> REDUNDANCY In the event of a drive failure, the pa= rity drive is
>> accessed
>>>>>> and data is reconstructed from the remaining devic= es. Once the failed
>>>>>> drive is replaced, the missing data can be restore= d on the new drive
>> and
>>>>>> operation resumed.
>>>>>>
>>>>>> Data reconstruction is simple. Consider an array o= f five drives in
>> which
>>>>>> X0 through X3 contain data and X4 is the parity di= sk.The parity for
>> the
>>>>>> ith bit is calculated as follows:
>>>>>>
>>>>>> X4(i) =3D X3(i) { X2(i) { X1(i) { X0(i)
>>>>>>
>>>>>> where { is exclusive-OR function.
>>>>>>
>>>>>> Suppose that drive X1 has failed. If we add X4(i) = { X1(i) to both
>> sides
>>>>>> of the preceding equation, we get
>>>>>>
>>>>>> X1(i) =3D X4(i) { X3(i) { X2(i) { X0(i)
>>>>>>
>>>>>>
>>>>>> file:///home/ruben/william stallings - computer or= ganization and
>>>>>> architecture designing for performance (8th editio= n).pdf
>>>>>>
>>>>>> Page 202
>>>>>>
>>>>>>
>>>>>>
>>>>>> y_math.png" target=3D"_blank">http://www.nylxs.com/images/raid3_pari= ty_math.png
>>>>>>
>>>>>> Ruben
>>>>>>
>>
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