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Consider a project consisting of four activities A, B, C, and D.
a) A and B, the first activities of the project, can be started simultaneously.
b) C can be started only after A is completed.
c) D can be started only after B is completed
Suppose the activity times for the activities are A = 2 weeks, B = 2 weeks, C = 2 weeks, D = 2 weeks.

a) How long does the project take to complete?
b) We ask the engineer working on activity A to factor in uncertainty in activity times and to provide better estimates of the activity times. The engineers tell us that activity A takes 1 week 50% of the time, and 3 weeks 50% of the time. How long does the project now take to complete (on average)?
c) We now also ask the engineer working on activity B to factor in uncertainty in activity times and to provide better estimates of the activity times. The engineer now tell us that activity B takes 1 week 50% of the time, and 3 weeks 50% of the time. How long does the project now take to complete (on average)? [ASSUME THAT BOTH ACTIVITY A AND ACTIVITY B ARE UNCERTAIN]

(a) is 4 weeks
however i am confused about b and c wouldn't the time average out to 2 weeks so both for a and b the time to complete a project would again be 4 weeks, this answers is sketchy though because it seems too easy to get
my logic .5(1)+ .5(3) = 2 average, 2 + 2 = 4

notamathwiz
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  • I believe you are correct about (b). However, consider that (c) now involves four possibilities. What is the completion time required in each case, and what is the probability of each case occurring? (The expected completion time is not 4 weeks again...) – colormegone May 01 '13 at 01:58
  • would it then be .5(1)+ .5(3) + .5(1)+ .5(3) = 4, so 4 + 2 = 6? – notamathwiz May 01 '13 at 02:07
  • The probabilities of all the cases can't add up to 2. What is the chance of A taking just one week and B also taking just one week? What are the other possibilities, and what is the probabilities of each (A - 1 wk., B - 3 wks.; A - 3 wks. , B - 1 wk. ; etc.)? – colormegone May 01 '13 at 02:12
  • ok so in the case that it a and b will both me 1 wk is 1/8
    in the case that it a and b will both me 1 wk is 4/8
    – notamathwiz May 01 '13 at 02:24
  • If the outcomes of tasks A and B are independent of one another, then the probability that A takes one week and B also takes one week is $\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ . The contribution of this to the sum for the expected time is then $0.25 \cdot (1+2) $ , since tasks C and D will need 2 weeks after the one week in which A and B have been completed. Now what do the other cases look like? – colormegone May 01 '13 at 02:29
  • ok so case A 3 wk B 3 wk is (.5)(.5)=.25 The contribution of this to the sum for the expected time is then (.25)(3+2)= 5/4, case A 1 wk B 3 wk is (.5)(.5)=.25 The contribution of this to the sum for the expected time is then (.25(1+2)+.25(3+2)) = 2, case A 3 wk B 1 wk is (.5)(.5)=.25 The contribution of this to the sum for the expected time is then (.25(3+2)+.25(1+2)) = 2 (5/4 sorry that's what i meant) – notamathwiz May 01 '13 at 02:40
  • That would be 5/4. Now consider the two remaining cases, add everything up, and you're there. – colormegone May 01 '13 at 02:42
  • adding everything up gives me 6 weeks is this correct? – notamathwiz May 01 '13 at 02:47

1 Answers1

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One way in which part (c) can be evaluated is to use a "tree diagram" of the possible outcomes (which I'm not going to attempt to render right now using TeX...):

Task A could take either 1 week (0.5 probability) or 3 weeks (also 0.5 prob.). Independently of that (we assume), task B has the same probabilities for such outcomes. So a "tree" looks something like:

                          1 week (0.5) --- 0.25 prob.
  1 week (0.5) ---  B --
                          3 weeks (0.5) --- 0.25 prob.

A --
1 week (0.5) --- 0.25 prob. 3 weeks (0.5) --- B -- 3 weeks (0.5) --- 0.25 prob.

Since tasks A and B are performed concurrently, the probability that both are done after just one week is 0.25. Otherwise one or the other or both take three weeks, which occurs with a probability of 0.25 + 0.25 + 0.25 = 0.75 .

Now tasks C and D each take two weeks, but cannot start until A and/or B are complete. So if A and B are both done in a week, C and D can both get started and everything will finish in 3 weeks. There's a 0.25 probability of that happening. Otherwise, at least one of A and B will not be done for 3 weeks, which means at least one of C and D will need another 2 weeks after that. In those cases, the project will need (3 + 2) = 5 weeks; there's a 0.75 probability of that.

Hence, the expected time to complete the project with the uncertainities for tasks A and B considered is $ \ 0.25 \cdot 3 \ + \ 0.75 \cdot 5 \ = \ 0.75 \ + \ 3.75 \ = \ 4.5 $ weeks.

colormegone
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