Raid 6

Lee HLee H Member Posts: 1,135
in Server 70-290
Hi

This might be a daft question but if you dont know you have to ask

Is raid 6 covered in the server 2003 exam, how long has it been out, I dont know much abouit it really, is it being adopted by a lot of people at this moment in time

Lee H
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Comments

  • bertiebbertieb Member Posts: 1,031 ■■■■■■□□□□
    Nope.

    Raid 0/1/5 knowledge should be fine for the 70-290 exam.
    The trouble with quotes on the internet is that you can never tell if they are genuine - Abraham Lincoln
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  • dynamikdynamik Banned Posts: 12,312 ■■■■■■■■■□
    It's basically RAID-5 that allows two disks to fail. As bertieb says, it's not going to be on the exam. You're only tested on the software RAID options in Server 2003, so you won't have to worry about it.

    Wikipedia has a good write-up on the various levels and their features: http://en.wikipedia.org/wiki/Redundant_array_of_independent_disks#Standard_levels
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  • sprkymrksprkymrk Member Posts: 4,884 ■■■□□□□□□□
    dynamik wrote:
    It's basically RAID-5 that allows two disks to fail.

    Tsk tsk! If 2 disks fail in a RAID 5 you're done. You know that! :)


    Oh wait, you mean that's what RAID 6 is? Wow, you made my eyes pop out for a second there with that wording. icon_lol.gif
    All things are possible, only believe.
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  • Lee HLee H Member Posts: 1,135
    So it work in exactly the same way as Raid5 but with greater redundancy
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  • astorrsastorrs Member Posts: 3,139 ■■■■■■□□□□
    Lee H wrote:
    So it work in exactly the same way as Raid5 but with greater redundancy
    At a high level yes. At the "how it's actually done level" not even close. The math involved is significantly more complex than that of RAID-5 and the most common implementation (there are 4 different RAID-6 algorithms I know of) is based on Reed-Solomon encoding and involves Galois field algebra.

    sprkymrk, following our previous RAID-5 performance thread, the math required to calculate the write parity on RAID-6 is nuts and the rule of thumb usually says a penalty of 5-10% for sequential workloads and 25-35% for random workloads worse than RAID-5.
    twitter: @astorrs
    profile: linkedin.com/in/astorrs
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  • Lee HLee H Member Posts: 1,135
    the math required to calculate the write parity on RAID-6 is nuts and the rule of thumb usually says a penalty of 5-10% for sequential workloads and 25-35% for random workloads worse than RAID-5

    icon_eek.gif Straight over my head there buddy icon_eek.gif



    :D
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  • dynamikdynamik Banned Posts: 12,312 ■■■■■■■■■□
    This might fill you in a bit: http://www.techexams.net/forums/viewtopic.php?t=35676

    Astorrs, you can't be bothered to provide a link? Quit slacking! ;)
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  • astorrsastorrs Member Posts: 3,139 ■■■■■■□□□□
    dynamik wrote:
    Astorrs, you can't be bothered to provide a link? Quit slacking! ;)
    astorrs says to himself "must put TechExams away and get back to packing... flight leaves in a few hours... ugh...."
    twitter: @astorrs
    profile: linkedin.com/in/astorrs
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