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Diffstat (limited to 'lib/gitlab/ci/pipeline_duration.rb')
| -rw-r--r-- | lib/gitlab/ci/pipeline_duration.rb | 141 |
1 files changed, 0 insertions, 141 deletions
diff --git a/lib/gitlab/ci/pipeline_duration.rb b/lib/gitlab/ci/pipeline_duration.rb deleted file mode 100644 index 3208cc2bef6..00000000000 --- a/lib/gitlab/ci/pipeline_duration.rb +++ /dev/null @@ -1,141 +0,0 @@ -module Gitlab - module Ci - # # 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 PipelineDuration - 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 |