MESSAGE
DATE | 2014-12-06 |
FROM | Ruben Safir
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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 22:19:45 2014 Return-Path: X-Original-To: archive-at-mrbrklyn.com Delivered-To: archive-at-mrbrklyn.com Received: by mrbrklyn.com (Postfix) id B2FCC16112C; Sat, 6 Dec 2014 22:19:45 -0500 (EST) Delivered-To: learn-outgoing-at-mrbrklyn.com Received: by mrbrklyn.com (Postfix, from userid 28) id 96C4D161130; Sat, 6 Dec 2014 22:19:45 -0500 (EST) Delivered-To: learn-at-mrbrklyn.com Received: from mailbackend.panix.com (mailbackend.panix.com [166.84.1.89]) by mrbrklyn.com (Postfix) with ESMTP id C2A5516112C for ; Sat, 6 Dec 2014 22:19:44 -0500 (EST) Received: from [10.0.0.42] (unknown [96.57.23.82]) by mailbackend.panix.com (Postfix) with ESMTPSA id 8D2F0137AF for ; Sat, 6 Dec 2014 22:19:44 -0500 (EST) Message-ID: <5483C776.6060505-at-panix.com> Date: Sun, 07 Dec 2014 03:20:22 +0000 From: Ruben Safir User-Agent: Mozilla/5.0 (X11; Linux x86_64; rv:31.0) Gecko/20100101 Thunderbird/31.2.0 MIME-Version: 1.0 To: learn-at-mrbrklyn.com Subject: Re: [LIU Comp Sci] Parity Bit Algebra References: <54838832.6010205-at-panix.com> <54838C85.60705-at-panix.com> <5483BD3C.9020609-at-panix.com> <5483C042.2050104-at-panix.com> <5483C287.3030707-at-panix.com> In-Reply-To: Content-Type: text/plain; charset=utf-8 Content-Transfer-Encoding: 7bit Sender: owner-learn-at-mrbrklyn.com Precedence: bulk Reply-To: learn-at-mrbrklyn.com
Nuh?
So post and give me credits in your HW assignment since I gave you the key to the solution 3 days ago
Ruben
On 12/07/2014 03:02 AM, Maneesh Kongara wrote: > Damn! I get it now!!! > On Dec 6, 2014 10:00 PM, "Maneesh Kongara" wrote: > >> It gives us the desired answer? >> On Dec 6, 2014 9:58 PM, "Ruben Safir" wrote: >> >>> On 12/07/2014 02:50 AM, Maneesh Kongara wrote: >>>> Have you seen the review questions? >>> >>> What happens when you subtract a comp 2 number again and again >>> >>> >>>> 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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