Civil Engineering - Construction Management - Discussion

Discussion Forum : Construction Management - Section 2 (Q.No. 33)
33.
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
20
25
35
15
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
17 comments Page 2 of 2.

Kajal Tomar said:   7 years ago
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.
(6)

Pranab said:   9 years ago
It is considering 1-2, the contractual obligation 75 and latest occurrence is 2. It is considering latest occurrence 1-2.

{PT + (4 x MLT) + OT}/6.
i.e {15+(4x28)+21}/6 which is equal to 24.66.

But, contractual obligation 75,
From event 1-2, 2-3, and 3-4 total 3 duration time,
so, time duration to reach 1-2 is 75/3 = 25,
25>24.66.
Due to contractual obligation, the time will be the greater one
Ans = 25.
(5)

Amjad said:   10 years ago
Then what is the contractual obligation 75 and latest occurrence is 2?

Amjad said:   10 years ago
Then what is the contractual obligation 75 and latest occurrence is 2?

Amjad said:   10 years ago
Then what is the contractual obligation 75 and latest occurrence is 2?

Rituraj said:   10 years ago
Is this is possible that pessimistic time can be less than optimistic time? Please answer this.

Naveen B said:   1 decade ago
{Op time + (4 * Most lik time) + pess time}/6.

i.e {15+(4*28)+21}/6 which is equal to 24.6666 = 25.
(1)


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