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Microsoft (MCSE, MCSD, MOUS, MCAD) > SQL server exams > Raid 5

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Author Raid 5
naeem
Member




Registered: Aug 2001
Location: Islamabad
Country: Pakistan
State:
Certifications: MCSD
Working on: MCDBA

Total Posts: 53
Post Raid 5

While implementing RAID 5 , u need 3 hard disk at least. at each hard disk along with data some parity informations r written too.in case of any one of the hard disk failure , other 2 hard disk r able to recover its data.

now my question is
---- what kind of parity informations is?
I mean what is its structure ?

Is it some kind of backup. if this is backup then it means there will huge amount of redundatnt data (excessive amont of data repetition)

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Vidra
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M




Registered: May 2001
Location: Rustenburg
Country: South Africa
State:
Certifications: MCSE NT 4.0,MCSE 2000, A+, Net+, MCP+I, MCP, SQL 2000(Admin)
Working on: MCDBA, MCSD, Server+, I-Net+, MOUS

Total Posts: 27
Lightbulb Parity

Here Is A Parity Table (Sort OF)

(Parity)
0 + 0 = 1
0 + 1 = 0
1 + 0 = 0
1 + 1 = 1

hope this helps!?

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colinbun
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Registered: Nov 2001
Location:
Country: United Kingdom
State:
Certifications: MCDBA
Working on:

Total Posts: 2
Re: Raid 5

You are absolutely right about the amount of disk used in raid 5. For example, if you have only 3 disks, data is written to disks 1 & 2 to start with, with parity written to disk3. For the next read, disk 1 might have the parity data, then disk 2 and so on. As you can see, you have just lost a third of your disk size.

The whole formula is

lost space = 1 / number of disks

ie. a third for 3, a quarter for 4 and so on.

Cheers,
Colin

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