GATE Past Year Questions: Network Models | Industrial Engineering - Mechanical Engineering PDF Download

For 4 – 6 Free float = (Ej – Ei) – dij = (8 – 2) – 4
= 6 – 4 = 2
Project duration , E = 13
therefore, (2, 13)

If duration of activity F has changed to 10 days, critical path remains the same and project duration will increase to 19 days.

As activity F is only in the critical path.

∴ Critical path will remain same i.e. A-D-F-H

Project duration changes to 19 days.

Path QSUW = Q + S + U + W = 4 + 5 + 5 +10 = 24 days.

d_{ij =} length of direction from node i to node j.

S_j  = length of the shortest path from P to node j.

There will be two possible paths to reach from P to G.
1. By reaching (P to Q) and then (Q to G)
2. By reaching (P to R) and then (R to G)

∴ S_G = min {S_O + d_OG, S_R + d_RG}

As the network has three paths A-B, C-D, and E-F. they can be drawn as:

The network has three paths
A - B = 7 + 5 = 12 days
C - D = 6 + 6 = 12 days
E - F = 8 + 4 = 12 days
Here we can see that all three paths are critical.
If we crash any one activity in either A-B, C-D, and E-F by one day, the project duration will remain the same.
Similarly, if we crash any two activities from paths A-B, C-D, and E-F by one day, the project duration will still be the same.
To reduce the project duration by 1 day, we need to crash 3 activities from all critical paths.

∴ the number of activities to be crashed to decrease the project duration by 1 day = 3.

Project completed = Activity C + Activity F + Activity K + Activity M
⇒ 4 + 9 + 3 + 8 = 24

Beta distribution extended to the Maximum & Minimum & gives strict definition for the mean & variance

EST = Earliest start time LFT = Latest Finish Time \ Path along which EST & LFT is equal are called critical path Critical path duration is 18 days.

⇒ a – c – e – f
⇒ 30 + 60 + 45 + 20 = 155 days

Dummy activities often have a zero completion time & are used to represents precendence relationship.