Optimistic time, most likely time and pessimistic times for the activities of a network in the given figure are written above their arrows. If the contractual obligation time for the project is 75, the latest occurrence time for the event 2, is
A. 20
B. 25
C. 35
D. 15
Answer: Option B
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The normal time required for the completion of project in the above problem is
A. 9 days
B. 13 days
C. 14 days
D. 19 days
A. $$\frac{{{{\text{t}}_{\text{o}}} + 3{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{2}$$
B. $$\frac{{{{\text{t}}_{\text{o}}} + 3{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{3}$$
C. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{4}$$
D. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{5}$$
E. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{6}$$
A construction schedule is prepared after collecting
A. Number of operations
B. Output of labour
C. Output of machinery
D. All the above
A. 3.5 and $$\frac{5}{6}$$
B. 5 and $$\frac{{25}}{{36}}$$
C. 3.5 and $$\frac{{25}}{{36}}$$
D. 4 and $$\frac{5}{6}$$
Latest occurrence time calculated from backward pass method.
Therefore TL of event 4 is 75 given;
Now( 75- (TE of 3-4) - (TE of2-3)),
TE =( opt time + 4 *most lik time + pess time) /6.
75 - (15) - (35) = 25 i.e TL of event 2 i.e asked