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| This page describes how guarantees for resources can be implemented. | | This page describes how guarantees for resources can be implemented. |
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− | = How to guarantee a guarantee =
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− | It's not obvious at the first glance, but ''there are only two ways of how a guarantee can be provided'':
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− | # reserve desired amount in advance
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− | # limit consumers to keep some amount free
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− | The first way has the followong disadvantages:
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− | ; Reservation is impossible for certain resources
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− | : such as CPU time, disk or network bandwidth and similar can not be just reserved as their amount instantly increases;
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− | ; Reserved amount is essentially a limit, but much more strict
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− | : cutting off X megabytes from RAM implies that all the rest groups are limited in their RAM consumption;
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− | ; Reservation reduces containers density
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− | : if one wants to run some identical containers, each requiring 100Mb on 1Gb system, reservations can be done for only 10 containers, and starting the 11th is impossible.
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− | On the other hand, ''limiting'' of containers can provide guarantees for them as well.
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| = Providing a guarantee through limiting = | | = Providing a guarantee through limiting = |
Revision as of 12:36, 23 May 2011
This page describes how guarantees for resources can be implemented.
comment4,
Providing a guarantee through limiting
The idea of getting a guarantee is simple:
if any group
requires a
units of resource from
units available then limiting all the rest groups with
units provides a desired guarantee
For groups in the system this implies solving a linear equation set to get limits like this:
In a matrix form this looks like
where
and thus the solution is
Skipping boring calculations, the reverse matrix is
This solutions looks huge, but the vector is calculated in time:
void calculate_limits(int N, int *g, int *l)
{
int sum;
int i;
if (N == 1) {
l[0] = R;
return;
}
sum = 0;
for (i = 0; i < N; i++)
sum += R - g[i];
for (i = 0; i < N; i++)
l[i] = (sum - (R - g[i]) - (N - 2) * (R - g[i]))/(N - 1);
}
Disadvantages of this approach
This approach has only one disadvantage: O(n) time needed to start a new container.