What are the hardware requirements for installing Directory and Resource Administrator 6.x? (NETIQKB312)

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  • 02-Feb-2007
  • 19-Jun-2007

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What are the hardware requirements for installing Directory and Resource Administrator 6.x?

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Directory and Resource Administrator 6.x

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Hardware Requirements

Element

Requirements

CPU

Intel Pentium III computer, 450MHz or higher.

RAM

512MB recommended for the Administration server. The amount of memory required depends on the number of managed accounts. For more information, see "Calculating Memory Usage" below.

Drive Space

200MB

Other Considerations

Application Event Log

    The Administration products record entries in the Windows 2000 application event log. You may need to increase the size of your application event log.

Registry

    The Administration products require about 5MB of disk space for the security data stored in the Windows 2000 registry.

Domain Files

    To improve performance, the Administration server caches trusted domain information in domain files stored on the Administration server computer. If there is insufficient disk space the Administration server logs warning messages in the event log and will not be able to load information from some trusted domains. These domains are marked Incomplete in the right pane of the MMC interface

    Domain files vary in size based on the number of user accounts, groups, OUs, and contacts in a domain. For example, a domain with 470 computers, 65 groups, and 3,000 users with the average user in 5 groups yields a domain file of 320KB. A domain with 10,000 users yields a domain file of 1MB.


Calculating Memory Usage

The amount of memory used by Directory and Resource Administrator varies depending on the individual enterprise structure. The following equations are meant to assist in calculating the average memory the Administration products would use. Frequent calls for large amounts of information could cause a burst of memory usage, resulting in a higher actual average.


Memory Calculation Equations

The following variables are used in equating cache memory usage:

  • AvgName - Average bytes in name (user, computer, group, contact, or OU)
  • AvgDesc - Average bytes in description (user, computer, group, contact, or OU)
  • AvgSAM - Average bytes in SAM account name (user, computer, or group)
  • AvgUPN - Average bytes in User Principal Name (user)


The following list shows the equations for calculating the memory used by the Administration products:

  • Number of users x (54 + AvgName + AvgDesc + AvgSAM + AvgUPN) = User bytes
  • Number of computers x (32 + AvgName + AvgDesc + AvgSAM) = Computer bytes
  • Number of groups x (52 + AvgName + AvgDesc + AvgSAM) = Group bytes
  • Number of contacts x (32 + AvgName + AvgDesc) = Contact bytes
  • Number of OUs x (70 + AvgName + AvgDesc) = OU bytes
  • User bytes + Computer bytes + Group bytes + Contact bytes + OU bytes = Total memory used


Example Memory Calculation

If the managed domains contain 3000 users with an average name of 12 bytes, an average description of 20 bytes, an average SAM account name of 8 bytes, and an average User Principal Name of 1.
1 bytes, the total bytes of memory used for user accounts is calculated as follows:


    3000 (54 + 12 + 20 + 8 + 11) = 315000


If the managed domains contain 200 computers with an average name of 12 bytes, an average description of 25 bytes, and an average SAM account name of 8 bytes, the total bytes of memory used for computer accounts calculated as follows:

    200 (32 + 12 +25 + 8) = 154000


If the managed domains contain 700 groups with an average name of 8 bytes, an average description of 25 bytes, and an average SAM account name of 8 bytes, the total bytes of memory used for group accounts is calculated as follows:

    700 (52 + 8 + 25 + 8) = 65100


If the managed domains contain 300 contacts with an average name of 14 bytes and an average description of 25 bytes, the total bytes of memory used for contact accounts is calculated as follows:

    300 (32 + 14 + 25) = 21300


If the managed domains contain 100 OUs with an average name of 8 bytes and an average description of 12 bytes, the total bytes of memory used for OU accounts is calculated as follows:

    100 (70 + 8 + 12) = 9000


The total number of bytes calculated by adding the totals for each object type. The total amount of memory used in the example, therefore, is calculated as follows:

    315000 + 154000 + 65100 + 21300 + 9000 = 564400 (564KB)

For more information regarding software requirements for installing Directory and Resource Administrator 6.0x, see article NETIQKB297.

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Additional Information

Formerly known as NETIQKB312