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queue runner: attempt at slightly smarter scheduling criteria

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Instead of just going for "whatever is the oldest build we know of",
use the following first:

- Is the step more constrained? If so, schedule it first to avoid
  filling up "more desirable" build slots with less constrained builds.

- Does the step have more dependents? If so, schedule it first to try
  and maximize open parallelism and breadth of scheduling options.
This commit is contained in:
Pierre Bourdon 2024-04-21 17:36:16 +02:00
parent ee1a7a7813
commit b8d03adaf4
Signed by untrusted user: delroth
GPG key ID: 6FB80DCD84DA0F1C
1 changed files with 6 additions and 0 deletions
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6
src/hydra-queue-runner/dispatcher.cc
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@ -133,6 +133,8 @@ system_time State::doDispatch()
comparator is a partial ordering (see MachineInfo). */
int highestGlobalPriority;
int highestLocalPriority;
size_t numRequiredSystemFeatures;
size_t numRevDeps;
BuildID lowestBuildID;
StepInfo(Step::ptr step, Step::State & step_) : step(step)
@ -141,6 +143,8 @@ system_time State::doDispatch()
lowestShareUsed = std::min(lowestShareUsed, jobset->shareUsed());
highestGlobalPriority = step_.highestGlobalPriority;
highestLocalPriority = step_.highestLocalPriority;
numRequiredSystemFeatures = step->requiredSystemFeatures.size();
numRevDeps = step_.rdeps.size();
lowestBuildID = step_.lowestBuildID;
}
};
@ -193,6 +197,8 @@ system_time State::doDispatch()
a.highestGlobalPriority != b.highestGlobalPriority ? a.highestGlobalPriority > b.highestGlobalPriority :
a.lowestShareUsed != b.lowestShareUsed ? a.lowestShareUsed < b.lowestShareUsed :
a.highestLocalPriority != b.highestLocalPriority ? a.highestLocalPriority > b.highestLocalPriority :
a.numRequiredSystemFeatures != b.numRequiredSystemFeatures ? a.numRequiredSystemFeatures > b.numRequiredSystemFeatures :
a.numRevDeps != b.numRevDeps ? a.numRevDeps > b.numRevDeps :
a.lowestBuildID < b.lowestBuildID;
});