Maintaining iCAP Compliance on HP Integrity & HP 9000 Mid-range & Superdome Servers

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Details And Example Scenario
• If a system is out of compliance (as described in the section “iCAP boot time compliance
enforcement”), then the iCAP software will enforce compliance on the next reboot of any
hard partition. This may have a significant impact on what the customer might be able to use
in production as illustrated by the example below.
• Example scenario: Consider a complex with two nPars (nPar 1, nPar 2). In this example we
assume that the complex has 8 Cores with Permanent Usage Rights and 8 Cores without
Usage Rights. Both the nPars have 4 Active Cores and 4 Inactive Cores each. nPar 1 is
running a vPar which is configured to use a minimum of 4 cores. Since the number of Inactive
cores is equal to Cores without Usage Rights the complex is compliant and the TiCAP balance
will not be decremented. This state is shown in the figure below.
• Now consider the case where nPar 2 is at EFI (or BCH) boot prompt for more than 12 hours.
The iCAP software in nPar 1 will assume that all cores in nPar 2 are active. In this state the
number of active cores in the complex will be 12 (4 + 8). Since there are 8 cores without
usage rights for the complex there should be 8 or more inactive cores. Currently there are only
4 inactive cores in the complex because of which the complex is out of compliance. If nPar 2
remains at EFI (or BCH) prompt for more than 12 hours TiCAP will be consumed for 4 cores
every 30 minutes. This may lead to a negative TiCAP balance on Complex 1. The new state of
cores in Complex 1 will be: # of Inactive Cores = 4; # of Active Cores = 4 + 8 (assumed) =
12. This state is shown in the figure below.
• If nPar 2 were to remain at the EFI (or BCH) prompt and nPar 1 were to be rebooted in nPar
mode then the iCAP software will enforce compliance and activate only as many cores as
nPar 1
4 Active Cores
4 Inactive Cores
vPar with minimum 4
cores
nPar 2
(at EFI or BCH
prompt)
8 Active cores
Complex 1
Total # of cores in Complex 1 = 16
# of Cores with Permanent Usage Rights = 8
# of Cores without Usage Rights (CWuR) = 8
# of Inactive Cores = 4
# of Active Cores = 4 + 8 (assumed) = 12
nPar 1
4 Active Cores
4 Inactive Cores
vPar with minimum 4
cores
nPar 2
4 Active Cores
4 Inactive Cores
Complex 1
Total # of cores in Complex 1 = 16
# of Cores with Permanent Usage Rights = 8
# of Cores without Usage Rights (CWuR) = 8
# of Inactive Cores = 4 + 4 = 8
# of Active Cores = 4 + 4 = 8