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  1. #1
    Join Date
    Feb 2003
    Posts
    60

    Unanswered: Size of virtual memorysegments

    Hello,

    OS = HP11i
    IDS= 7.10.UD3

    SHMVIRTSIZE 131072
    SHMADD 131072
    SHMTOTAL 0


    Block size = 2048 bytes



    # onstat -g seg
    INFORMIX-OnLine Version 7.10.UD3 -- On-Line -- Up 03:31:42 -- 694560 KbytesSegment Summary: (resident segments are not locked)id key addr size ovhd class blkused blkfree
    ...
    ...
    1381386250 e68f5000 134217728 2652 V 1299 15085


    If I take the size of V segment :
    134217728 pages = 67108864 KB = 65536 MB

    It is far from the value of SHMVIRTSIZE ....


    And if I take
    blkused + blkfree * 2048 = 16384 * 2048 = 32768 KB = 32 MB.


    Once again, pretty different figures ....


    So my question is :
    Is there any trick, or ain't I doing the right sum or what ?


    Cheers.

  2. #2
    Join Date
    Feb 2003
    Posts
    60

    Mistake

    Opps, sorry, I wrote :

    If I take the size of V segment :
    134217728 pages = 67108864 KB = 65536 MB

    It is far from the value of SHMVIRTSIZE ....


    In fact , the size in correct here, because in bytes ...

    134217728 bytes = 131072 KB

    which is well the value of SHMVIRTSIZE


    The problem is with bock used and block free

  3. #3
    Join Date
    Aug 2002
    Location
    Bonn/Germany
    Posts
    152
    (1299+15085)*8192 = 134217728

    These are 8K-Blocks, they have nothing to do
    with the page size.

    BTW 7.10 is very very old and no longer supported
    by Informix. Using such a version in a production
    environment is unwarrantable.

    Best regards

    Eric
    --
    IT-Consulting Herber
    WWW: http://www.herber-consulting.de
    Email: eric@herber-consulting.de

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  4. #4
    Join Date
    Feb 2003
    Posts
    60

    thanks

    Thanks Eric

  5. #5
    Join Date
    Apr 2003
    Posts
    1

    Re: Mistake

    When you have,

    size : 134217728

    blkused : 1299
    blkfree : 15085

    blkused and blkfree are the number of 8Ko memory fragment and not 2 Ko pages.

    So, ( 1299 + 15085 ) * 8 Ko = 132 Mo ... it's OK !

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