Answer to Question #172309 in Operations Research for Bernice maina

Question

Find the critical path of the following network using the EST/LSTs.

Activity Preceding Activity Duration (days)

1 - 4

2 1 7

3 1 5

4 1 6

5 2 2

6 3 3

7 5 5

8 2, 6 11

9 7, 8 7

10 3 4

11 4 3

12 9, 10, 11 4

(b). Calculate the floats of the network diagram in part (a).

(c ). The standard deviations of the activities on the critical path in part (a) are: 1, 2, 1.5, 3,

2.5 and 3 respectively. Based on these values calculate the probability of achieving a

scheduled time of 40 days for the project duration.

Answer 1

(a)

EST for incident 4: actions C, E and H unite with C for 0 + 9 = 9 days, E for 7 + 9 = 16 days and H for 8 + 7 = 15 days. As a result, we take the maximum, which is 16.

(b)

LFT for incident 2: events E and F stem from it coming regressive from incident 4, the LFT for 2 is 16 – 9 = 7 plus, from incident 6, the LFT for 2 is 23.5 – 5 = 18.5. As a result, we take the lowermost, which is 7.

We find the actions 1, 2, 4, 5, 6 and 7 are holding EST = LFT. They are serious actions, and the double-line shots revealed in the system epitomize Critical Path with events A, E, I, J and L; total project time being 28 days, TE is 28 days.

(c)

Get the critical path's total sum is 1, 2, 1.5, 3, 2.5, and 3, which is 13.

Divide 13 by the number of intervals to get the median interval, which is 13 divided by six, leading to 2.16. Going to the nearest ten, we get 2.20.

Thus, the probability of 2.2 is 2.2/100, which is 0.022

Getting the probability of 40 days, 0.022*40=0.88

0.88*100=80%