141 lines
4.4 KiB
Ruby
141 lines
4.4 KiB
Ruby
module Gitlab
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module Ci
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# # Introduction - total running time
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#
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# The problem this module is trying to solve is finding the total running
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# time amongst all the jobs, excluding retries and pending (queue) time.
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# We could reduce this problem down to finding the union of periods.
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#
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# So each job would be represented as a `Period`, which consists of
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# `Period#first` as when the job started and `Period#last` as when the
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# job was finished. A simple example here would be:
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#
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# * A (1, 3)
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# * B (2, 4)
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# * C (6, 7)
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#
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# Here A begins from 1, and ends to 3. B begins from 2, and ends to 4.
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# C begins from 6, and ends to 7. Visually it could be viewed as:
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#
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# 0 1 2 3 4 5 6 7
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# AAAAAAA
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# BBBBBBB
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# CCCC
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#
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# The union of A, B, and C would be (1, 4) and (6, 7), therefore the
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# total running time should be:
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#
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# (4 - 1) + (7 - 6) => 4
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#
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# # The Algorithm
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#
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# The algorithm used here for union would be described as follow.
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# First we make sure that all periods are sorted by `Period#first`.
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# Then we try to merge periods by iterating through the first period
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# to the last period. The goal would be merging all overlapped periods
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# so that in the end all the periods are discrete. When all periods
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# are discrete, we're free to just sum all the periods to get real
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# running time.
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#
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# Here we begin from A, and compare it to B. We could find that
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# before A ends, B already started. That is `B.first <= A.last`
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# that is `2 <= 3` which means A and B are overlapping!
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#
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# When we found that two periods are overlapping, we would need to merge
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# them into a new period and disregard the old periods. To make a new
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# period, we take `A.first` as the new first because remember? we sorted
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# them, so `A.first` must be smaller or equal to `B.first`. And we take
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# `[A.last, B.last].max` as the new last because we want whoever ended
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# later. This could be broken into two cases:
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#
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# 0 1 2 3 4
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# AAAAAAA
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# BBBBBBB
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#
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# Or:
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#
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# 0 1 2 3 4
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# AAAAAAAAAA
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# BBBB
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#
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# So that we need to take whoever ends later. Back to our example,
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# after merging and discard A and B it could be visually viewed as:
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#
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# 0 1 2 3 4 5 6 7
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# DDDDDDDDDD
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# CCCC
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#
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# Now we could go on and compare the newly created D and the old C.
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# We could figure out that D and C are not overlapping by checking
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# `C.first <= D.last` is `false`. Therefore we need to keep both C
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# and D. The example would end here because there are no more jobs.
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#
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# After having the union of all periods, we just need to sum the length
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# of all periods to get total time.
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#
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# (4 - 1) + (7 - 6) => 4
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#
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# That is 4 is the answer in the example.
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module PipelineDuration
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extend self
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Period = Struct.new(:first, :last) do
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def duration
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last - first
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end
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end
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def from_pipeline(pipeline)
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status = %w[success failed running canceled]
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builds = pipeline.builds.latest.
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where(status: status).where.not(started_at: nil).order(:started_at)
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from_builds(builds)
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end
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def from_builds(builds)
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now = Time.now
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periods = builds.map do |b|
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Period.new(b.started_at, b.finished_at || now)
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end
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from_periods(periods)
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end
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# periods should be sorted by `first`
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def from_periods(periods)
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process_duration(process_periods(periods))
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end
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private
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def process_periods(periods)
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return periods if periods.empty?
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periods.drop(1).inject([periods.first]) do |result, current|
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previous = result.last
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if overlap?(previous, current)
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result[-1] = merge(previous, current)
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result
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else
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result << current
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end
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end
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end
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def overlap?(previous, current)
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current.first <= previous.last
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end
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def merge(previous, current)
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Period.new(previous.first, [previous.last, current.last].max)
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end
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def process_duration(periods)
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periods.sum(&:duration)
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end
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end
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end
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end
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