UBC derived configuration examples

User Beancounters Definition /proc/user_beancounters /proc/bc/ General information Units of measurement VSwap Parameters description Primary parameters numproc, numtcpsock, numothersock, vmguarpages Secondary parameters kmemsize, tcpsndbuf, tcprcvbuf, othersockbuf, dgramrcvbuf, oomguarpages, privvmpages Auxiliary parameters lockedpages, shmpages, physpages, numfile, numflock, numpty, numsiginfo, dcachesize, numiptent, swappages Internals User pages accounting RSS fractions accounting On-demand accounting UBC consistency Consistency formulae System-wide configuration vzubc(8) Configuration examples Basic Derived Intermediate configurations Tables List of parameters Parameter properties Consistency Config examples

This table shows examples of configurations derived from the basic examples.

Contents

Derived UBC configurations

Name	A	B	2A	0.5A+0.5B
Power	1GB/200	1GB/60	1GB/100	1GB/92
Helper values
total mem	4986KB	16000KB	9972KB	10493KB
total kmem	1536KB	3712KB	3072KB	2624KB
total umem	3450KB	12288KB	6900KB	7869KB
avnumproc	15	40	30	27
Primary parameters
numproc	40	65	80	52
numtcpsock	40	80	80	60
numothersock	40	80	80	60
Secondary parameters
kmemsize	851,968	2,359,296	1,703,936	1,605,632
tcpsndbuf	262,144	524,288	524,288	393,216
tcprcvbuf	262,144	524,288	524,288	393,216
othersockbuf	163,840	327,680	327,680	245,760
dgramrcvbuf	32,768	65,536	65,536	49,152
Auxiliary parameters
shmpages	512	1,024	1,024	768
privvmpages	3,450	9,216	6,900	6,333
numfile	512	1,280	1,024	896
dcachesize	196,608	524,288	393,216	360,448

Explanation

The columns A and B are exactly the already considered examples A and B.

2A

The next column labelled “2A” was produced by multiplying all the numbers from example A by 2. This configuration is good for half the number of containers plan A was created for, that is, this new configuration is for 100 containers on a system with 1GB of RAM.

0.5A+0.5B

The next column “0.5A+0.5B” was produced by multiplying the numbers from example A by 0.5 and adding the numbers from example B multiplied by 0.5. The resulting confguration is a mean confguration between A and B, being computed as (A + B)/2. It is good for

${\frac {1}{0.5/200+0.5/60}}\approx 92$

containers on a system with 1GB of RAM.

Other configurations

A lot of other configurations can be produced the same way, by multiplying by a constant (and summing, if necessary) the values from existing configurations.

Configurations produced by multiplying existing configurations by a number greater than 1 will be consistent if the original configurations were consistent. Configurations produced by multiplying by a number less than 1 may or may not be consistent. See Intermediate UBC configurations for more details about producing new configurations basing on the existing ones.

Caution: do not subtract the values of one configuration from others.