A project is described by the simple activity network shown at the right, consis

Question

A project is described by the simple activity network shown at the right, consisting of drected branches representing activities and nodes representing the beginning/termination of activities. The duration in days of two activities A, B and C in the construction project are denoted as TA, TB and TC, whose probability mass functions (pmfs) are given below. Assume that TA, TB and TC are statistically independent. 3. Activity A Activity C Activity B Start day 0 Aclivity Network 2 6 7 Pr(TA D) 32.14% 14.29% 3.57% 0.00% 3.57% 14.29% 32.14% Pr(T8-D) 26.28% 16.33% 9.95% 7.15% 7.91% 12.25% 20.15% Pr(TCzD) 20.41% 18.37% 16.33% 14.29% 12.24% 10.20% 8.16% Activities A and B are scheduled to start simultaneously at day 0. Node (a) represents the completion of both activities A and B, and node (b) represents the completion of the project. Activity C can start only when both A and B are completed. Determine and plot the pmf of the time T required to complete the project. Compute its mean, its variance and the most likely time required to complete the project

Explanation / Answer

Answer :

the functions of Aand B

A ~ N(50,10), so mu = 50 and sigma = sqrt(10)Then, P(A <= 60) = P(Z <= (60-50)/sqrt(10)) = P(Z <= sqrt(10)) = from Excel, normsdist(sqrt(10)) =0.999217298870999

B ~ N(45,15) so P(B <= 60) = P(Z <= (60-45)/sqrt(15)) = P(Z <= sqrt(15)) = from Excel, normsdist(sqrt(15)) = 0.999946244411635

Then, P(max(A,B) <= 60 = 0.999217298870999 * 0.999946244411635 =0.999163585357194

This is the answer to a.

(Note that if the variances 10 and 15 are supposed to be 10^2 and 15^2 respectively, then P(max(a,b) <= 60) = (P(Z <= 1))^2 = 0.841344746068543^2 = 0.707860981737141; this becomes a much more difficult problem to deal with.)

There is a 15 delay of max(A,B) >= 60. (There would also be some probability that one of A and B is greater than 75, which would cause more delay; however, given the large z's, this is extremely unlikely; P(z >= 2sqrt(15) = 5*10^-15, and P(z >= 2.5 sqrt(10) is much less than that).

Then, if C starts on time, P(C <= 150-60) = P(z <= (90-80)/sqrt(25)) = P(Z <= 2) = normsdist(2) =0.977249868051821mIf C starts late, P(C <= 150-60-15) = P(z <= (75-80)/sqrt(25)) = P(Z <= -1) = normsdist(-1) =0.158655253931457

Then, the probability that the project is completed on time is

P(max(a,b) <= 60)*P(C <= 90) + P(max(a,b) >= 60)*P(C <= 75) =0.999163585357194 * 0.977249868051821 + (1-0.999163585357194)*0.158655253931457 =0.976565183530048

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