Lines Matching defs:loadavg
206 * 90% of l_estcpu usage in (5 * loadavg) seconds
208 * We wish to decay away 90% of l_estcpu in (5 * loadavg) seconds. That is, we
210 * for (i = 0; i < (5 * loadavg); i++)
214 * for all values of loadavg.
217 * decay ** (5 * loadavg) ~= .1
220 * decay = (2 * loadavg) / (2 * loadavg + 1)
226 * decay ** (5 * loadavg) ~= .1
229 * b = 2 * loadavg
234 * 1) Given [factor ** (5 * loadavg) =~ .1], prove [factor == b/(b+1)].
235 * 2) Given [b/(b+1) ** power =~ .1], prove [power == (5 * loadavg)].
250 * factor ** (5 * loadavg) =~ 0.1
251 * => ln(factor) =~ -2.30 / (5 * loadavg)
252 * => factor =~ exp(-1 / ((5 / 2.30) * loadavg))
253 * =~ exp(-1 / (2 * loadavg))
257 * =~ (2 * loadavg) / ((2 * loadavg) + 1)
264 * => power =~ 4.60 * loadavg + 2.30
265 * => power =~ 5 * loadavg
267 * Conclusion: decay = (2 * loadavg) / (2 * loadavg + 1)
271 #define loadfactor(loadavg) (2 * (loadavg))
423 * in (5 * loadavg) seconds. This causes the system to favor processes which