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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
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