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