DFT2

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DFT2

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January 7th, 2009 02:00

RAID 5 3+1 or 7+1?

Hi,

I'd like to know the pro and cons of choosing R5 3+1 over 7+1, in terms of performance, when operating normally, and when reconstructing?

I also noticed the 400 GB 10K drive is way faster than the 300 10K drive, is this true in real life?

Thanks a lot for your future answers!
  • Quincy561

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    Posted January 7th, 2009 05:00

    7+1 will generally have better performance because each LUN is spread over 8 disks instead of 4. Most systems have workloads which are skewed across LUNs, so this can be significant. Rebuild times are not really that different for either, just that the 7+1 touches more disks in a rebuild scenario.

    Each different drive brand and type have slightly different performance characteristics. I'm not sure how you are "testing" this 400GB drive you are talking about, but maybe it is better for this workload that the other drive you tested.
  • xe2sdc

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    Posted January 7th, 2009 02:00

    I usually choose 3+1 or 7+1 depending on Read/Write ratio.
    If you have little writes (up to 10%), go raid7+1 and enjoy wonderful read performances (you are reading from 8 different drives)
    If you have 70/30 you can't use 7+1 and have to go with 3+1.
    If you have more then 35% of writes, choose mirror.

    Obviously total iops DO make difference .. 30% of 100 iops means 30 writes per second.. 5% of 1000 iops means 50 writes per second ;-)
    Since it's hard to predict actual iops I prefer to forget about iops while choosing protection ;-)
    However someone else may have different rules .. So please share them :D

    Please note that you always write in cache. But as soon as you fill the cache with writes (with tracks waiting to be destaged) you can easily have troubles (performances impact on your hosts). That's why I don't considere cache wile choosing actual protection (I prefer to considere "the worst case").
  • DFT2

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    Posted January 7th, 2009 04:00

    OK, so, to sum this up, these two type of RAID deliver approximatly the same performance, but reconstruction is longer, and you must add 8 drives for an upgrade, instead of 4?

    And for the 400 GB drive? Is this drive really special compared to the others?
  • Quincy561

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    Posted January 7th, 2009 04:00

    Stefano, There really is no difference in the write overhead for 3+1 over 7+1, except that if you get a specific capacity, 7+1 is likely to have fewer disks to work with overall.

    3+1 and 7+1 must both do 4 disk IOs to perform a random write. Both can do optimized writes for sequential IO, with 7+1 being better than 3+1 because it does the parity for 7 data writes. So given the same number of physical drives in the system, 7+1 would generally perform better. They both have similar rebuild characteristics, however 7+1 is statistically more likely to suffer a 2nd drive failure in a raid group.
  • DFT2

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    Posted January 7th, 2009 05:00

    I just looked at the DMX 4 spec, or data sheet. The one with the drives, and I saw the 400 Go drive was much better.
  • xe2sdc

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    Posted January 7th, 2009 05:00

    Chapeau .. I can only learn from you!! :D
  • dynamox

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    Posted January 8th, 2009 04:00


    3+1 and 7+1 must both do 4 disk IOs to perform a
    random write. Both can do optimized writes for
    sequential IO, with 7+1 being better than 3+1 because
    it does the parity for 7 data writes. So given the
    same number of physical drives in the system, 7+1
    would generally perform better.


    Quincy,

    would you please explain why 3+1 and 7+1 do 4 disk io ? I see 3+1 doing that ..but how about 7+1 ?

    Also i would think that 7+1 would have worth write performance because it has to calculate parity against 8 drives versus only 4 ?

    Hopefully i am not hijacking too much :)
  • Quincy561

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    Posted January 8th, 2009 06:00

    Copied this from a page on the Microsoft site.
    http://www.microsoft.com/technet/prodtechnol/windows2000serv/reskit/core/fndb_con_dwtx.mspx?mfr=true

    The parity stripe is the exclusive OR (XOR) of all the data values for the data stripes in the stripe. If no disks in the RAID-5 volume have failed, the new parity for a write can be calculated without having to read the corresponding stripes from the other data disks. Thus, only two disks are involved in a write operation

    So to do a random write on RAID5 3+1 or 7+1, we read the old data, xor with the new data, write the new data, take the interim xor result and xor that with the old parity and then write the new parity. Two disk reads and two disk writes.