Question

Using the following information about Project A, its activities and predecessor activities below: Activity Predecessors A в B

The following table contains the pessimistic, likely, and optimistic estimates for how long activities are estimated to take

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Answer #1

(e) time estimate (TE) = (TO + 4TM + TP)/ 6

Where TO = OPTIMISTIC TIME

TM = likely time

TP= PESSIMISTIC time

(Te To + 4to Activity (TP) Possibisha L Optiroiste Expected time 725 4.83 3.1667 ó co w 8.6667 in & ww one of of w 3.8333 5.5(f) standard deviation = (TP- TO)/6

VARIANCE = Square of standard deviation

Where TP= pessimistic time

TO= optimistic time

To +4 tot TE EER to Ap-to]. go [In] Likely Te To +4 totle Espectedt To ptimiste Activity Pelisha Vasiloxe Stodos Delah 0.8333

(g)

glos verting TE to cobole istegess:- Activity TE в з co E H ) ( ..--.0 لا لا ی د

(H) (ì)(j)(k)

coorliqq pote Caitical pall diagons Cafter forward f bookward pass ES: 17 LF: 1411 F-6 ES: 23 : 23 C90 TO G-7 AS @ B-3 @ GB O

(l) Total float or slack comes zero only for activity in critical path. Hence our answer in (h) is confirmed.

The critical path are A-B-D-E-G-H and A-B-C-F-G-H

(J) Standard deviation of project = square root of variance of expected time of project

Variance of expected time = sum of variances of critical path activities

= 0.694+ 0.25+ 5.4444+0.6944+1.7778+3.3611

= 12.2213 (for critical path A-B-D-E-G-H)

Standard deviation of project= square root of variance

= 3.4959

For critical path A-B-C-FG-H , standard deviation = 3.036

Note: time limitation for answering last part

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