What is Server MT Max and min MT serve

Hi all

What is MT Max and Min MT servers means in Siebel?
Why we need to use it. ?
Please give me an idea on this

Concerning
Remy

Here is an excerpt:

Our basic recommendation is to have 20 tasks by server of MT. (A server (Multi-Threaded) MT is equivalent to an operating system process).

Consider the expected number of concurrent users for the application in the application design phase. Although there may be a large number of potential users, a smaller number will be connected at one time. For the purposes of the following discussion, assume that there are 100 to 200 concurrent users.

There are a lot of variables to consider when you grant an object manager. This development should be discussed with Siebel Expert Services in the Production readiness review. Here are some variables to consider.

The number of concurrent users.
The duration of each user spends on the system during a session.
The session timeout.
The use of scripts.
Complexity of the application.
High usage time. (80% of users are online from 11:00 to 03:00)

...

The parameters used for this component object Siebel Manager are as follows:

Maximum task = 300
Maximum MT servers = 15
Minimum MT servers = 5

Maximum task represents the total number of concurrent users. Given that the expected number of concurrent users is between 100-200, that the parameter has been set to allow extra room for manoeuvre. This is important in order to allow the sessions which do not disconnect properly and so drag until the session timeout is reached. Session timeout can be set in the Web server and is usually set by default to 20 minutes.

Maximum MT servers is the total number of processes that can be run for this component. It is also used to calculate how many tasks can be performed on each process. The calculation is as follows:

Maximum task / Server MT = Maximum number of tasks per MT Server

300 / 15 = 20

As mentioned earlier, our basic recommendation is that the maximum task to MT Maximum servers report is 20. A widely configured or scripted application may require a smaller report.
Minimum server MT is used only when the Siebel server or component is started. This allows to determine how many process starting for this component. The only advantage it gives, it is that it reduces the initialization of the process time when a new MT server is required. In this example, 5 MT servers have been used because 100 users are expected. In this way, the system can handle 100 concurrent users before you have so that it starts a new process.

Tags: Oracle Applications

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    Site_1 user_1 RPT104 Activity_1 16 May 11 13.43.06.566193000 * May 16, 11 14.06.41.341612000 * ref_1 1125
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    Published by: MickyMick on June 7, 2011 03:21

    BobLilly wrote:
    Tabibitosan of Aketi method can be applied here (see {: identifier of the thread = 1005478})

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    Site_1 user_1 RPT104 Activity_2 2011-05-16 14.05.23.572211000 2011-05-16 14.06.01.058978000 ref_2 1125
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    Site_1 RPT104 Activity_3 2011-05-16 14.06.41.341612000 user_1 ref_1 14.06.45.341612000 2011-05-16 1125

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                   lead(DTStamp,1,DTStamp + interval '4' second) over(order by DTStamp) ENDTS,
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                     when     lag(Site) over(order by DTStamp)  = Site
                          and
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                          and
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                          and
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                          and
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                          and
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             from  t
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    Site_1 user_1 RPT104 Activity_1 16-MAY-11 02.06.01.058978000 PM     16-MAY-11 02.06.41.341612000 PM     ref_1       1125
    Site_1 user_1 RPT104 Activity_3 16-MAY-11 02.06.41.341612000 PM     16-MAY-11 02.06.53.530405000 PM     ref_1       1125
    
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  • in the header problem report the actual display Max and Min...

    Hi guys,.

    I have the following problem.

    1. I have a multi selection prompt where you take 2 years.

    2. I have the following columns. Year, month, Exp, calc YEAR (max), calc (min)

    3. I need to get the months max and min across all lines. So far, I tried MAX, MIN, RMAX and RMIN without success. However, it is not picking up along all lines. For example,.

    2001 1 100050 2001 2001
    .....
    2002 1 244005 2002 2002

    I'm sure it has to do something with my aggregation in the computed column, but not quite sure where exactly.

    c max (max (year)) and min (min (year))

    :)

  • scanning question (Min, Max and Min return)

    I have a tunable laser that is supposed to make a sweep of the Min to Max and then back to min.

    For some reason, it seems to be an additional step after the peak wavelength as shown below.

    Strange is that it seems to happen only when I put a wavelength that has 1 decimal place or 4 decimals.

    I'm relatively new to labVIEW, Please HELP? any advice would be helpful.

    This looks like a problem of accuracy point.  Each language beats in it.

    What you need to do is to calculate how many steps to the top must be made and then make comparisons on the step number (an integer).

    Also, why use the node formula?  A structure matter would take care of the logic.

  • Max and min function

    I have a special request, I'm looking for the max and, possibly, of the minutes of the date that the order was made.

    EMP_ID Full_Name Date
    1 Jim Smith 02/01/2011
    1 Jim Smith 13/01/2011
    1 Jim Smith 15/01/2011
    2 Susan Morgan 20/02/2011
    2 Susan Morgan 22/02/2011
    2 Susan Morgan 25/02/2011
    2 Susan Morgan 28/02/2011

    So, if I have 2 tables Emp and joined by say Emp_id orders and I need the elements following result set

    EMP_ID Full_Name Date max (date) min (date)
    1 Jim Smith 1/2/2011 1/15/2011 1/2/2011
    1 Jim Smith 1/13/2011 1/15/2011 1/2/2011
    1 Jim Smith 1/15/2011 1/15/2011 1/2/2011
    2 Susan Morgan 2/20/2011 2/28/2011 2/20/2011
    2 Susan Morgan 2/22/2011 2/28/2011 2/20/2011
    2 Susan Morgan 2/25/2011 2/28/2011 2/20/2011
    2 Susan Morgan 2/28/2011 2/28/2011 2/20/2011


    It is a simplified query that my orginal query has more columns that I have to choose and more tables. I don't know if I can use a select group on because I would have to force all the columns and that would be just grap date unique columns max since I bring together down the line

    Hello

    Oracle_Rookie wrote:
    ... I don't know if I can use a select group on because I would have to force all the columns and that would be just grap date unique columns max since I bring together down the line

    Yes, probably do not want to use GROUP BY. Use more Analytics MIN and MAX fucntions. For example:

    SELECT       deptno
    ,       ename
    ,       hiredate
    ,       MAX (hiredate) OVER (PARTITION BY deptno)     AS last_in_dept
    FROM       scott.emp
    ORDER BY  deptno
    ,            hiredate
    ;
    

    Output:

    `   DEPTNO ENAME      HIREDATE  LAST_IN_D
    ---------- ---------- --------- ---------
            10 CLARK      09-JUN-81 23-JAN-82
            10 KING       17-NOV-81 23-JAN-82
            10 MILLER     23-JAN-82 23-JAN-82
    
            20 SMITH      17-DEC-80 23-MAY-87
            20 JONES      02-APR-81 23-MAY-87
            20 FORD       03-DEC-81 23-MAY-87
            20 SCOTT      19-APR-87 23-MAY-87
            20 ADAMS      23-MAY-87 23-MAY-87
    
            30 ALLEN      20-FEB-81 03-DEC-81
            30 WARD       22-FEB-81 03-DEC-81
            30 BLAKE      01-MAY-81 03-DEC-81
            30 TURNER     08-SEP-81 03-DEC-81
            30 MARTIN     28-SEP-81 03-DEC-81
            30 JAMES      03-DEC-81 03-DEC-81
    

    Most of the aggregates fucntions have analytical counterparts, which can give you the same results without collapsing the entire down to one line per result group. The analytical PARTITION BY clause matches the GROUP BY clause in an aggregate query.

    I hope that answers your question.
    If not, post a small example of data (CREATE TABLE and only relevant columns, INSERT statements) for all tables and also post the results desired from these data.
    Make the sample data as simple as you can, so it has the same problem that occurs in your actual query.
    Explain, using specific examples, how you get these results from these data.
    Always tell what version of Oracle you are using.

    Published by: Frank Kulash, March 15, 2011 12:50

  • Max and Min of successive summits with a different amplitude

    Hello

    I have a signal having different successive Max and Minima.ie each pic + / is having a different value.

    I wanted to get successive + / maximum values of the. CSV file.

    I'm done with the chart from .csv file, but I am not able to get values of crete please help.

    What function can I use?

    How about waveform peaks?

    Check it out. Heres where it is:
    Programming > Waveform > Analog Waveform > waveform measures > waveform monitoring

    This is peaks and valleys (maxs and mn).

    It will then display locations (x-value) and amplitue (value y) point.

  • AVG, max and min calculated from form data

    I have two tables. One with the raw data and the other with summary data. There is a form which the raw data to column came in, and then click submit, I use a cfloop function to get the data in the same column in the table of raw data. I can't find an easy way to calculate the max, min and average values for the data that are in the raw data table that can be put in the summary data table. The data are customized according to the date and place.

    This is the code to get the raw data into the table.

    < CFLOOP INDEX = "onerow" FROM = "1" TO = "#form.numrows #" >
    < CFQUERY NAME = "insertrawdata" DATASOURCE = 'test' >
    INSERT INTO test.dbo.rawdata (locationid, date, data)
    VALUES (' #form.locaitonid # ', #DateValue #, ' #Form ['data' & onerow] #")
    < / CFQUERY >
    < / CFLOOP >

    OK, well, do it in 2 queries then...


    Select max (data) as max, min (data) as min, avg (data) like avg test.dbo.rawdata where locationid = ' #form.locationid # ' and date = #datevalue # group by locationid, date


    insert into test.dbo.summarydata (locationid, date, max, min, avg)
    values (' #form.locationid #', #datevalue #,)
    (#getSummaryData.max #, #getSummaryData.min #, #getSummaryData.avg #)

  • Create a table based on Max and Min

    Hello

    I found a thread fairly similar, but he has not answered my question.

    I'm building a Vi that simulates a cyclone running at high volumes of different admission. I need to generate a series of volumes of admission based on a user of max, min and interval/step input. These data must be used for the calculations and the x scale of a chart. What is the best way to do it? I could have the user specify each data point, but that only works if the user wants to only like 10 data points. I need to be able to say, start at 0 and 1 to 900. I know how to do this in matlab, but I can't understand it in labview.

    Thanks in advance for any advice.

    It's still there. Even back in LabVIEW 7.1, the range of functions had a search capability. Have you tried it?

    7.1 if you want as the inputs to be the stage of start-up/stop/then you can do something like this (the ramp model VI icon will be different for 7.1):

  • Help requests on max and min

    HI friends,

    referring to the sample data below, can you please help me GET the supplied output.

    WITH tt (ID, OS, prev_os, START_DT, NEXT_DT) AS

    (SELECT 1, 'a', TO_DATE (September 10, 2014 00:00:00 ',' DD/MM/YYYY HH24:MI:SS'), (NULL), TO_DATE('09/29/2014 00:00:00', 'MM/DD/YYYY HH24:MI:SS') OF double UNION ALL to_char)

    SELECT 1, 'a', TO_DATE (30 September 2014 00:00:00 ',' DD/MM/YYYY HH24:MI:SS'), 'a', TO_DATE('10/09/2014 00:00:00', 'MM/DD/YYYY HH24:MI:SS') FROM dual UNION ALL

    SELECT 1, 'b', TO_DATE (October 10, 2014 00:00:00 ',' DD/MM/YYYY HH24:MI:SS'), 'a', TO_DATE('10/19/2014 00:00:00', 'MM/DD/YYYY HH24:MI:SS') FROM dual UNION ALL

    SELECT 1, 'a', TO_DATE (20 October 2014 00:00:00 ',' DD/MM/YYYY HH24:MI:SS'), 'b', NULL FROM dual)

    SELECT * FROM tt

    1one10/09/201429/09/2014
    1one30/09/2014one09/10/2014
    1b10/10/2014one19/10/2014
    1one20/10/2014b

    Out put:

    ID os strt_dt end_dt
    1one
    10/09/2014
    09/10/2014
    1b
    10/10/2014
    19/10/2014
    1one
    20/10/2014

    Hello

    866088 wrote:

    OKI,

    I'm currently implementing scd2

    is there are several consecutive admissions for an id, operating system after a sort on the start date, then I should have only one entry for this id, os with min (start_dt) and max (end_dt)

    What is scd2?

    There are several different things that might mean by "consecutive".  What do you mean exactly?

    Maybe you want something like this:

    WITH got_grp_id AS

    (

    SELECT id, os, start_dt, next_dt

    ROW_NUMBER () OVER (PARTITION BY id ORDER BY start_dt)

    -ROW_NUMBER () OVER (PARTITION BY id, os ORDER BY start_dt) AS grp_id

    TT

    )

    SELECT id, os

    MIN (start_dt) AS min_start_dt

    MAX (next_dt) AS max_next_dt

    OF got_grp_id

    GROUP BY id, the os, the grp_id

    ORDER BY id, min_start_dt

    ;

    Output:

    ID O MIN_START_D MAX_NEXT_DT

    ---------- - ----------- -----------

    1 one 10-Sep-2014, October 9, 2014

    1 b 10 October 2014 October 19, 2014

    1 to 20 October 2014

    For an explanation of the technique of Difference sets used here, see

    Analytical question lag and lead

    or

    Re: Ranking of queries

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