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PERT (Uncertain) Solution

The document provides information about a project with 8 activities (A-H) and calculates the probability of completing the project within 22, 21, and 25 weeks. It gives the optimistic, most probable, and pessimistic times for each activity along with formulas to calculate the expected time and variance. It identifies the critical path as A-D-F-H and calculates the total expected time as 22 weeks and total variance as 2.44. It then uses the standard normal distribution to determine the probability of completing within each of the given timeframes: 50% probability within 22 weeks, 26% probability within 21 weeks, and 97% probability within 25 weeks.

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Dominic Romero
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80 views2 pages

PERT (Uncertain) Solution

The document provides information about a project with 8 activities (A-H) and calculates the probability of completing the project within 22, 21, and 25 weeks. It gives the optimistic, most probable, and pessimistic times for each activity along with formulas to calculate the expected time and variance. It identifies the critical path as A-D-F-H and calculates the total expected time as 22 weeks and total variance as 2.44. It then uses the standard normal distribution to determine the probability of completing within each of the given timeframes: 50% probability within 22 weeks, 26% probability within 21 weeks, and 97% probability within 25 weeks.

Uploaded by

Dominic Romero
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as DOCX, PDF, TXT or read online on Scribd
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Given: (refer to Google forms and the picture below)

(Picture above was given thru Messenger)

Required:
Probability of the project to be completed within:
a. 22 weeks
b. 21 weeks
c. 25 weeks

Initial Formulas & Legend:

Let:
Expected Time = E
Total Expected Time = E(t)
Optimistic = x
Most Probable = y
Pessimistic = z
Variance = v
Total Variance = V2

E = (x + 4y + z) / 6 ; v = [(z – x) / 6]2

Solution:
*Table of Expected Time and Variance for each activity will make things easier
*Using the 2 formulas for each activity, the table below was made
*Immediate Predecessors were distinguished thru the picture in Given

Activity Immediate Expected Time Variance


Predecessor/s
A - 5 0.11
B - 3 0.03
C A 7 0.11
D A 6 0.44
E B 7 0.44
F D,E 3 0.11
G D,E 10 0.44
H C,F 8 1.78

*Using the table above and copy/pasting the picture in the given, the picture below is the updated
project network

C H
5
7 8
12 14
7
14 14
22
A 0
22
55 0
5 22
D F
6
5 3
5
11 11
11 11
14
14

B 0 E G
33 1 73 10
4 4
10 11
*As stated last meeting/s, the critical path is 11
A-D-F-H 12
21
22
*We compute for the total expected time and the total variance of the critical path

E(t) = EA + ED + EF + EH = 5 + 6 + 3 + 8 = 22

V2 = vA + vD + vF + vH = 0.11 + 0.44 + 0.11 + 1.78 = 2.44

z = [Time – E(t)] / V = (Time – 22) / √2.44

Requirements:

a. 22 weeks
z = (22 – 22) / √2.44 = 0 P(22 weeks) = 0.5000 + 0 = 0.5000 or 50%
b. 21 weeks
z = (21 – 22) / √2.44 = -0.64 P(21 weeks) = 0.5000 – 0.2389 = 0.2611 or 26%
c. 25 weeks
z = (25 – 22) / 2.44 = +1.92 P(25 weeks) = 0.5000 + 0.4726 = 0.9726 or 97%

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