Three schedule activities of 10 days duration each need to be complete before their outputs can be integrated.
Activity 1 & 2 both have a 90% probability of achieving the estimated duration of 10 days.
Activity 3 has an 80% probability of achieving the 10 days.
The three activities are in parallel with no cross dependencies, what is the probability of the integration activity starting on schedule?
Possible solution #1
There is a 10% probability of the start being delayed by Activity 1 overrunning.
There is a 10% probability of the start being delayed by Activity 2 overrunning.
There is a 20% probability of the start being delayed by Activity 3 overrunning.
Therefore in aggregate there is a 40% probability of the start being delayed meaning there is a 60% probability of the integration activity starting on time.
Possible solution #2
The three activities are in parallel and the start of the integration is dependent on all 3 activities achieving their target duration. The probability of a ‘fair coin toss’ landing on heads 3 times in a row is 0.5 x 0.5 x 0.5 = 0.125 (an independent series)
Therefore the probability of the three activities achieving ‘on time’ completion as opposed to ‘late’ completion should be 0.9 x 0.9 x 0.8 = 0.648 or a 64.8% probability of the integration activity starting on time.
Which of these probabilities are correct?
The more usual project scheduling situation where activities 1, 2 and 3 are joined ‘Finish-to-Start’ in series (an interdependent series). Is there any way of determining the probability of activity 4 starting on time from the information provided or are range estimates needed to deal with the probability of the activities finishing early as well as late?
There is a correct answer and an explanation – see the next post
(its too long for a comment)