MESSAGE
| DATE | 2014-12-06 |
| FROM | Maneesh Kongara
|
| SUBJECT | Re: [LIU Comp Sci] Parity Bit Algebra
|
From owner-learn-outgoing-at-mrbrklyn.com Sat Dec 6 21:38:07 2014
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Date: Sat, 6 Dec 2014 21:38:04 -0500
Message-ID:
Subject: Re: [LIU Comp Sci] Parity Bit Algebra
From: Maneesh Kongara
To: learn-at-mrbrklyn.com
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--089e0129503e07aca50509973362
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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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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> wrote:= 0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex">On 12/06/2014 11:09=
PM, Maneesh Kongara wrote:
> Preliminary work as in answers to the other questions of the assignmen=
t?
> On Dec 6, 2014 6:08 PM, "Ruben Safir" <mrbrklyn-at-panix.com">mrbrklyn-at-panix.com> wrote:
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 completed the as=
signment?
>>> On Dec 6, 2014 5:49 PM, "Ruben Safir" <"mailto:mrbrklyn-at-panix.com">mrbrklyn-at-panix.com> wrote:
>> Not even looked at it outside of the library.=C2=A0 I'm burnin=
g 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 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.=C2=A0 It is related to the problem you asked ab=
out the parity bit
>>>> calculation for the Hamming Code
>>>>
>>>> Not the methodology for logical calculation, it is straigh=
t math
>>>>
>>>> Note that X is XOR for whatever UFT reason
>>>>
>>>>
>>>> REDUNDANCY In the event of a drive failure, the parity dri=
ve 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 d=
rives in which
>>>> X0 through X3 contain data and X4 is the parity disk.The p=
arity 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 organizati=
on and
>>>> architecture designing for performance (8th edition).pdfr>
>>>>
>>>> Page 202
>>>>
>>>>
>>>>
>>>> ng" target=3D"_blank">http://www.nylxs.com/images/raid3_parity_math.png=
>>>>
>>>> Ruben
>>>>
>>
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