How to understand buffer swaps

In SAP training materials, it says that buffer swaps occur when buffer content is deleted from the buffer and is replaced by other content. It's no relation at all to the swapping mechanism on OS level.
I am confused about the source and target of buffer swaps. For example, when buffer swaps occur on program buffer(PXA), where will the data in program buffer be put, physical memory?, and where does new data come from? SAP buffers locate on share memory, belong to virtual memory, but virtual memory is the summary of physical memory and OS swap space(or OS paging file). I just remember that extended memory resides on physical memory, don't know whether sap buffers reside on physical too.
For swapping on OS level, I think it's page exchange between physical memory and swap space which locates on disk. Is it right?
Please advise on how to understand buffer swaps and OS swapping. Thanks so much.
James

Hi James,
Plz find the link  I hope it might help you,
http://help.sap.com/saphelp_nw70/helpdata/EN/c4/3a6dbb505211d189550000e829fbbd/frameset.htm
as the question you have asked is concentrating in
depth and this can be understood if you have a look
at the following link.
regards,
Manjula.

Similar Messages

  • How to Understand AWR Report

    Hi
    How to understand AWR report?

    Hi,
    first of all, you need to understand things that this report is about -- like wait events, DB time, CPU time, etc. If you are familiar with these things, the rest is easy.
    There are also some articles about reading AWR reports in my blog, including some examples: http://savvinov.com/awr/
    Best regards,
    Nikolay

  • How to understand STATPACKS report ?

    Hi,
    How to understand STATPACKS report ? Some documents or link ?
    Thank you.

    http://www.akadia.com/services/ora_statspack_survival_guide.html
    http://download-uk.oracle.com/docs/cd/B10501_01/server.920/a96533/statspac.htm
    http://jonathanlewis.wordpress.com/2008/02/18/analysing-statspack/
    and many..many more....!!!!
    Grrod reading...
    Sim

  • How to reduce buffer busy waits, session hanging due to buffer busy waits

    Hi,
    How to reduce buffer busy waits, session hanging due to buffer busy waits.
    Thanks,
    Sathis.

    When I see through enterprise manager I see lot of
    tables with buffer busy waits.
    Is there any way by table name we can check the
    blocks info.
    The simple way is to look at the SQL statement and corresponding table name?
    P1=file#, P2=block#. You can extract segment name(table or index) using this info.
    Query v$bh like following:
    SQL> select file#, block#, class#, objd from v$bh where file# = P1 and block# = P2;
    SQL> select object_name from all_objects where object_id = <objd>;See following doc:
    http://download-west.oracle.com/docs/cd/B19306_01/server.102/b14237/dynviews_1051.htm
    Or you can dump block:
    SQL> alter system dump datafie <P1> block <P2>;Some excerpts from block dump:
    scn: 0x07df.17e70782 seq: 0x01 flg: 0x04 tail: 0x07822301
    frmt: 0x02 chkval: 0x61d0 type: 0x23=PAGETABLE SEGMENT HEADER
    Map Header:: next  0x00000000  #extents: 1    obj#: 55881  flag: 0x10000000>
    Can we do something at table level that will reduce
    the waits.
    Yes, some methods are known. But before thinking of that, you must verify which block class and which access are involved.
    Typo... always. :(
    Message was edited by:
    Dion_Cho

  • How to understand the way OBIEE implements outer joins?

    Hello guys
    I have a few scenarios where I have to implement outer joins and inner joins between dimensions and facts..
    There are 2 ways as far as I know that allows me to implement outer joins in BMM layer.
    1, Join the Logical dim table A to Logical Fact table B using outer join. The modeling will look like the below:
    A Dim ---outer------B fact -----------inner----C Dim
    By this design, the query that selects columns from all 3 tables will look at this:
    select columns from ((B fact inner joins C on key1 = key2) left outer joins on A dim) on key3=key4)..
    2, Join logical dim table A to fact table B inside the LTS of fact B by mapping the LTS B to dim A using outer join. then join fact B to C:
    B fact (mapped to fact B outer join Dim A) ------------inner -------C dim
    By this design, the query that selects columns from all 3 tables will look at this instead:
    select columns from C, B left outer join A on key1=key2 where key3=key4
    Comparing these 2 queries, the first one seems to do inner joins first and then outer join the result set to Dim A, the second query seems to outer join Dim A first and the result set inner joins to Dim C..
    I ran the same report using these 2 different designs, and the data comes out very different. The report of the first query is much smaller than the report of the second query...
    Can anyone help me understand how OBIEE understands outer joins? The second query is so far giving the right result, however, I can't get rid of the outer join in the query even if not selecting columns from Dim A, which is impacting the performance of other reports without Dim A.
    Your inputs will be greatly appreciated
    Thanks
    Edited by: user7276913 on Apr 20, 2010 9:31 AM
    Edited by: user7276913 on Apr 20, 2010 9:31 AM

    Xcode is the IDE.
    Objective-C is the language typically used.
    There's lots of getting started stuff at https://developer.apple.com

  • How to understand the Error message of the connection of OPC server

    hi guys, one issue need your help
    when I run the NI OPC servers, one error message will appear in the process information.
    Just like as followed:
    ==================================================​=========================
    2011-6-23   11:50:05   Default User   NI OPC Servers   Stopping Simulator device driver.
    2011-6-23   11:50:05   Default User   NI OPC Servers   Closing project D:\National Instruments\Shared\NI OPC Servers\projects\simdemo.opf
    2011-6-23   11:50:25   Default User   NI OPC Servers   NI OPC Servers
    2011-6-23   11:50:25   Default User   NI OPC Servers   Unable to load driver DLL 'D:\National Instruments\Shared\NI OPC Servers\drivers\dataforth_isolynx_u.dll'
    more messages
    ==================================================​===========================
    how to understand "2011-6-23   11:50:25   Default User   NI OPC Servers   Unable to load driver DLL 'D:\National Instruments\Shared\NI OPC Servers\drivers\dataforth_isolynx_u.dll'??????
    Thanks

    Hi!
    I have the same problem when opening the OPC Server.
    Unfortunately I have no solution, but I think it has something to do with incorrect Certificates and is more of a Windows Problem.
    If anybody can help, please let us know!
    Best Regards
    Martin

  • How to understand the STAD relevant for RFC

    Hello performance experts.
    I have question about how to understand the STAD relevant for RFC.
    I want to know how to calculate the response time of action of "Push save buttion"
    which call function module by RFC.
    In our system, the screen is created using web dynpro and after push the button,
    function module for application will be called.
    When I push Save button, 2 STAD line is created.
    One is for function code " SAVE" and the other is for "RFC".
    Line for "SAVE" include RFC as client, and RFC line has RFC info as server (also client)
    I wonder how I should calculate the response time of action"Save" button in following case.
    Should I add the 2 response time?
    1) 5,867 ms +   5,797 m  is collect?
    2) 5,867 ms  include RFC time, so no need to sum the value and 5,867 ms is collect??
    3)  590 ms  +   5,797 ms is collect?
    4) Or other calculation is correct ?                                                   
    1) STAD for function code:    
    CPU time                     578 ms    
    RFC+CPIC time              5,280 ms                                                  
    Total time in workprocs      594 ms
    Response time            5,867 ms                                                                 
    Processing time              590 ms    
    Load time                      1 ms               
    Wait            5,273  ms
    Roll (in+wait) time        5,274 ms
    as Client
    Number    Connections                              1
               Destinations                             1
               Users                                    1
               Calls                                    2
    Time      Calling                              5,280   ms
               Remote execution                    10,527   ms
               Idle                                11,110   ms
    Data      Sent                               135,836   Bytes
               Received                               620   Bytes
    2) STAD for RFC
    CPU time                   1,406 ms
    RFC+CPIC time                  9 ms
    Total time in workprocs    5,080 ms
    Response time            5,797 ms
    Processing time            2,132 ms
    Load time                     50 ms
    Roll (in+wait) time          718 ms
    Database request time      2,882 ms
    Enqueue time                  16 ms
    Roll time
    Wait              717  ms
    as Client
    Number    Connections                              1
                      Destinations                             1
                         Users                                    1
                           Calls                                    4
    Time      Calling                                  9   ms
              Remote execution                         1   ms
              Idle                                     0   ms
    Data      Sent                                   686   Bytes
                 Received                             1,360   Bytes
    as Server
    Number    Connections                              1
              Destinations                             1
              Users                                    1
              Calls                                    2
    Time      Calling                              5,272   ms
              Remote execution                     5,269   ms
              Idle                                        11,118   ms
    Data      Sent                                   620   Bytes
              Received                           135,760   Bytes
    Thanks for your cooperation in advance.
    Keiichiro

    Hi Keiichiro,
    I wonder how I should calculate the response time of action"Save" button in following case.
    Should I add the 2 response time?
    1) 5,867 ms +   5,797 m  is collect?
    2) 5,867 ms  include RFC time, so no need to sum the value and 5,867 ms is collect??
    3)  590 ms  +   5,797 ms is collect?
    4) Or other calculation is correct ?
    It depends of what you mean with response time of action "Save" button.
    If you mean ALL time, the value is 5.867 because the local time from save was 594ms, the "external" RFC time was 5280ms.
    If you want separete analysis, the 594ms is your option since the 5280ms is in other part...
    Also, analysing the other part there's another other part which represents a good part of processing time:
    Database request time      2,882 ms
    Hope this help youl.
    Regards, Fernando Da Ró

  • How to caluculate buffer size for import/exports in oracle

    Hi DBAS,
    how to caluculate buffer size for import exports,here is the formula but how can we use if there is a 100 of tables and datatypes,is the any query to find out rows_in_array ,maximum_row_size total schema and datbase
    buffer_size = rows_in_array maximum_row_size*
    Thanks,
    tmadugula

    http://download.oracle.com/docs/cd/B10501_01/server.920/a96652/ch01.htm
    Search "buffer"

  • How to understand on hand in user guide and table MTL_ONHAND_QUANTITIES

    version:12.1.1
    IS on hand in user guide from table MTL_ONHAND_QUANTITIES?
    How to understand on hand in user guide and table MTL_ONHAND_QUANTITIES?
    http://etrm.oracle.com/pls/et1211d9/etrm_fndnav.show_object?n_appid=401&n_tabid=52063&c_type=TABLE
    What does "Not implemented in this database" mean?

    What the page is saying that MTL_ONHAND_QUANTITIES is not a table. It is a view.
    The main table underlying the view is mtl_onhand_quantities_detail.
    After you perform a transaction, the data in this table (and view) is updated automatically.
    You don't have to do anything.
    Sandeep Gandhi

  • How to calculate #Buffer Gets    # Exec           Buffer Gets/Exec

    Hi,
    How to calculate #Buffer Gets,# Execution time,Buffer Gets/Exec for a sql query?

    Nirmal
    You can find out these statistics from two places
    1) using SQL_TRACE (10046 trace) and then TKPROF (or Autotrace in SQL*Plus)
    2) or looking at V$SQL which records the cost assigned to each SQL statement since the statement was first cached.
    If you use Statspack or AWR, you can see the difference between two points in time, so you can calculate the cost for a period of time.
    See Using SQL_Trace and TKPROF
    http://download-uk.oracle.com/docs/cd/B10501_01/server.920/a96533/sqltrace.htm#1018
    and Using Statspack
    http://download-uk.oracle.com/docs/cd/B10501_01/server.920/a96533/statspac.htm#34837
    (see the 10g documentation equivalents if necessary).
    Remember, ratios (eg gets/exec) aren't always very helpful. You're best off concentrating on those operations which take longest (ie where there is the most potential to save time). See (eg) www.hotsos.com, www.oraperf.com, and others to identify effective performance methodologies.
    HTH
    Regards Nigel

  • [q] how to understand the 'order' of a list?

    hi,everyone
    there's a test program:
    import java.util.*;
    public class testList
    void test()
    LinkedList ll= new LinkedList();
    ll.add("3");
    ll.add("1");
    ll.add("2");
    ll.add("5");
    ListIterator li = ll.listIterator();
    while (li.hasNext())
    System.out.println(li.next());
    public static void main(String[] args)
    testList tl = new testList();
    tl.test();
    As we know, order is the most important feature of a List.
    But this program's result is " 3 1 2 5 " which is not a sorted list!
    How we understand the phrase 'order' ?
    And , is there any method in Collection which can sort the elements in one collection (list or set)?
    thanx a lot!

    As we know, order is the most important feature of a
    List. Yes, and it preserves the order you have used to put elements in the List. :)
    But this program's result is " 3 1 2 5 " which is not
    a sorted list!Well, what order to you actually like? "1 2 3 5"? Why not "5 3 2 1"? And for more complicated objects "natural" order may be even less obvious. And never forget that List may contain objects that belong to different classes - for example Integer, String and Date - how they are to be ordered? (Of course, such usage is not recommended).
    How we understand the phrase 'order' ?
    And , is there any method in Collection which can
    sort the elements in one collection (list or set)?There are sort-methods in java.util.Collections class.

  • How to understand this statement?

    How to understand this statement?
    What is a zero-parameter public constructor
    <key-partitioning/class-name>: Specifies the name of a class that implements the com.tangosol.net.partition.KeyPartitioningStrategy interface. This implementation must have a zero-parameter public constructor.

    It is a public constructor that doesn't take any parameters. See the following simple class.
    public class Platypus
      public Platypus() {} //public constructor with no arguments
    }

  • How to understand the hierarchy structure of 0PLANT io?

    how to understand the hierarchy structure of 0PLANT io ?
    i can understand the hierarchy structure of PRODUCT io,
    well, i hope someone could explain the hierarchy structure of 0PLANT io
    thanks.

    Hi
    You can use following hierarchy tables
    RSTHIERNODE - Texts of Non-Postable Hierarchy Nodes
    RSEHIERNODE - Master Data: Hierarchy Nodes that Cannot Be Posted To
    RSMHIERNODE - Master data: Hierarchy nodes that cannot be posted to
    RSHIEDIR - Hierarchy Catalog
    RSHIEDIRT - Hierarchy directory texts
    RSREQHIER - Data Request hierarchy
    RSROLEHIERARCHY - Role hierarchy
    In addition to these have a look at K table & I table for 0PLANT.
    K table - Hierarchy SID table
    I table - Hierarchy structure table
    Hope this will help..!!
    Thanks,
    Vikrant

  • How sharepoint understand when user requests for web applications by their DNS names

    HI
    I configured Alternate access mapping in my sharepoint farm for default ,intranet zones
    and spt farm has two web front end servers and they load balancing by F5 device
    in WFE servers there are different web applications are running on different ports
    so here I want to know how load balancing works, load balancing configured in F5 device.
    when users request a webapplication from browser (ex http://cms) where this request will go
    1)when I ping cms and other web applicaations  it returns me a loadbalancer  server IP  for all web applications;
    ping cms : it returns 10.xxx.0.80 , same ip returns when I ping for other web app
    but ex CMS web application run on the 10.xxx.1.26:81 port in sharepoint server
    2) and these sharepoint web applications running on different ports in sharepoint  web servers , so here  how sharepoint understand when user requests for web applications by their DNS names
    http://cms and http://products  etc
    adil

    I'm not sure if the F5 can add a port number (I'm not an expert on load balancers).  But in general if you design the SharePoint site to run on port 81 then you need to have port 81 appended to the request or it won't work.  http://cms in your
    example would take you to http://cms:80 not http://cms:81.  But in general DNS will resolve the address back to the F5 load balancer.  The load balancer will look at the header of the HTTP request (which contains the original address you requested)
    and forward the request to the appropriate web front end IP address.  If your web front end is using one IP address for multiple sites then those sites need to be differntiated by using a custom port like 81 (which must be included in the original request)
    or because a host header was bound to the web application when it was created.  If they are running on different port numbers then the request must include the port number by the time it gets to the SharePoint server.
    Paul Stork SharePoint Server MVP
    Principal Architect: Blue Chip Consulting Group
    Blog: http://dontpapanic.com/blog
    Twitter: Follow @pstork
    Please remember to mark your question as "answered" if this solves your problem.

  • Program Buffer swap

    Hi Techies
    In our Production servers i found that Program buffer swap are more around 10 thousand swaps
    I have restarted the server twice but still after two days again the swap are increased ( only the program buffer swaps are more) and moreover hitratio is 99% and no performance issue .
    abap/buffersize                  512000
    what shall i do shall increase the program buffer size to reduce the swaps
    Thank You
    Haroon
    Edited by: virtualharoon on Aug 14, 2009 7:15 AM

    Hi techies
    thanks for the reply techies
    I have configured the logon load balancing, i have 10 application servers ,As per suggestions i will increase the paramater to 600MB. will that be oK
    Will this parameter has any dependencies(abap/buffersize)
    Awaiting for the reply
    ENV
    SAP R/3 4.6B
    Oracle
    Memory : 50 GB

Maybe you are looking for