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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_i}
requires a
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G_i}
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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A^{-1} = \frac {1} {N - 1} \left( A - (N - 2) E \right) }
This solutions looks huge, but the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L}
vector is calculated in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle O(N)}
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.