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Given a Job which needs to be completed within $30\ min$ for the $99\%$ of the times it's executed and within $20\ min$ for the $95\%$ of the times it's executed.

The Job completion depends on the following 7 tasks:

  1. 1 x Task $A $
  2. 1 x Task $B$
  3. 5 x Task $C$

Task $A, B, C$ need to be completed sequentially, but the 5 iterations for Task $C$ can be started in parallel i.e. Task $B$ gets started when Task $A$ completes and all 5 Task $C$ iterations start together as soon as task $B$ completes.

It's also estimated that Task $A$ and $B$ will take $10\%$ of the time each and $80\%$ of the time will be spent completeting $1$ iteration of Task $C$. Running $5$ iterations in parallel for Task $C$ is assumed to take same amount of time as running only $1$ iteration of Task $C$.

Given the above requirement how can we calculate the $99\%$ and $95\%$ excpected times for Task $A, B, C$ so that the Job requirment of $99\% = 30\ min$ and $95\%=20\ min$ are still met?

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    $\begingroup$ We need more information to answer this question. How are the execution times of tasks A,B,C distributed? If there is no uncertainty in the execution times of the tasks you describe then the tasks will finish within a given time T either with probability 0 or 1. So no point is asking about 99% or 95% probabilities within time T. If there is uncertainty, then you have to tell us the probability distributions. $\endgroup$ – Thanassis Mar 3 '18 at 4:21
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    $\begingroup$ Try to quantify from your given information: Do you mean $E[A]:E[B]:E[C] = 1:1:8$? And for parallel task $C_i$ I expect the complete time $C = \max\{C_i\}$, and the distribution should be different from individual $C_i$ unless they are some how dependent / constants. And you want to ask what is the expected values $E[A], E[B], E[C]$ such that $\Pr\{A + B + C < 30\} = 99\%$ and $\Pr\{A + B + C < 25\} = 95\%$. Just like the comment above, you do not have enough information to determine all these - you need more assumptions. $\endgroup$ – BGM Mar 3 '18 at 8:46
  • $\begingroup$ The problem doesn't state this explicitly: can we assume tA95 (task A 95% time) = tB95 = (1/8) * tC95, and tA99 (task A 99% time) = tB99 = (1/8) tC99 ? And can we assume the C task times for any given run are independent of each other? In other words, if two C tasks are running, the probability that BOTH will complete in tC95 minutes or less is 0.95 * 0.95 = 0.9025 ? $\endgroup$ – Brian Trial Mar 9 '18 at 19:55
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When you take $20\ min$ to do a the job, $2\ min$ where dedicated to task $A$, $2\ min$ to task $B$ and $18\ min$ to task $C$. Do the same for $30\ min$.

Expected times for $A$ and $B$ will be: $99\%=3\ min$ and $95\%=2\ min$.

Expected times for $C$ (the $5$ iterations count as $1$) will be: $99\%=24\ min$ and $95\%=18\ min$.

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    $\begingroup$ I don't think this works purely by calculating the percentage of time. The probability of completion of Job = probability of task A and B and C to complete. If we need to achieve 95% probability for the Job then the probabilities of task need to be much higher than 95% because at .95*.95*.95 = 0.85, ie. you can guarantee the job to complete only in 85% of the time. $\endgroup$ – newbie Jun 12 '15 at 0:43
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    $\begingroup$ "Expected times for C (the 5 iterations count as 1 ) will be: 99%=24 min and 95%=18 min " I think you meant 95% = 16 min. And I don't agree with the "5 iterations count as 1". If that is the case, why does the problem even bother to mention 5 C tasks running in parallel? It is only worth mentioning if they run independently of each other. The problem wording is confusing. $\endgroup$ – Brian Trial Mar 9 '18 at 17:30

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