{"id":42,"date":"2014-12-16T13:25:00","date_gmt":"2014-12-16T12:25:00","guid":{"rendered":"https:\/\/caipirinha.spdns.org\/wp\/delay-propagation-in-parallel-task-chain\/"},"modified":"2022-10-09T14:38:51","modified_gmt":"2022-10-09T12:38:51","slug":"delay-propagation-in-parallel-task-chain","status":"publish","type":"post","link":"https:\/\/caipirinha.spdns.org\/wp\/?p=42","title":{"rendered":"Delay Propagation in a Parallel Task Chain"},"content":{"rendered":"

<\/span>Problem Statement<\/span><\/h2>\n

Similar to the last blog, we now want to examine how delays in parallel tasks propagate if we have a discrete and non-binary probabilistic pattern of delay for each involved task. We will assume to have reached our milestone if each task has been completed. We take the same example tasks as yesterday, but now, they are in parallel:<\/span>
\n\"3<\/span><\/span><\/p>\n

Source Data<\/span><\/h2>\n

<\/span>In our example, we have 3 tasks, named “A”, “B” and “C”, and their respective durations is distributed according to the table and the graph below:<\/span><\/p>\n\n\n\n\n\n\n
Duration<\/span><\/b><\/td>\n0 d<\/span><\/b><\/td>\n1 d<\/span><\/b><\/td>\n2 d<\/span><\/b><\/td>\n3 d<\/span><\/b><\/td>\n4 d<\/span><\/b><\/td>\n5 d<\/span><\/b><\/td>\n6 d<\/span><\/b><\/td>\n7 d<\/span><\/b><\/td>\n8 d<\/span><\/b><\/td>\n9 d<\/span><\/b><\/td>\n10 d<\/span><\/b><\/td>\n11 d<\/span><\/b><\/td>\n12 d<\/span><\/b><\/td>\n13 d<\/span><\/b><\/td>\n14 d<\/span><\/b><\/td>\n15 d<\/span><\/b><\/td>\n16 d<\/span><\/b><\/td>\n17 d<\/span><\/b><\/td>\n18 d<\/span><\/b><\/td>\n19 d<\/span><\/b><\/td>\n20 d<\/span><\/b><\/td>\n <\/td>\n<\/tr>\n
A<\/span><\/b><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n5%<\/span><\/td>\n10%<\/span><\/td>\n25%<\/span><\/td>\n20%<\/span><\/td>\n15%<\/span><\/td>\n10%<\/span><\/td>\n10%<\/span><\/td>\n5%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n100%<\/span><\/td>\n<\/tr>\n
B<\/span><\/b><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n40%<\/span><\/td>\n30%<\/span><\/td>\n20%<\/span><\/td>\n10%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n100%<\/span><\/td>\n<\/tr>\n
C<\/span><\/b><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n50%<\/span><\/td>\n50%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n0%<\/span><\/td>\n100%<\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

 \"Probability<\/span>As you can see, there is not a fixed duration for each task as we have that in traditional project management. Rather than that, there are several possible outcomes for each task with respect to the real duration, and the probability of the outcome is distributed as in the graph above. The area below each graph will sum up to 1 (100%) for each task.<\/span>
\nIf we assume these distributions, how does the probability distribution of our milestone outcome look like?<\/span>
\n<\/span><\/p>\n

Solution<\/span><\/h2>\n

If we consider our milestone to be reached when all tasks have been completed, that is, when A, B, and C have been completed, then it becomes obvious that the distribition of task C has no influence at all. Even after 8 days when C has been completed with 100% certainty, there is no chance that either A or B will have started ever. The earliest day when our milestone could happen is on day 11 because this is the first day when there is a slight chance that A could be completed.<\/span>
\nOn the other side, on day 19, all tasks have been completed with 100% certainty, and hence by then, the milestone will have been reached for sure.<\/span>
\nThe correct mathematical approach in order to compute the resulting distribution is, for 2 tasks:<\/span>
\n \"Resulting<\/span>
\nIn order to facilitate our work and enable us to compute many parallel tasks, we use a C program named <\/span>parallel.c<\/span><\/b><\/a> which processes an input file<\/a><\/b> of probability distributions in a certain format:<\/span><\/p>\n