The document describes determining the critical path of a project using network analysis. It lists the steps as: 1) drawing the network diagram with activities and relationships, 2) calculating earliest and latest start/finish times, 3) determining total float for each activity, 4) identifying critical activities with zero total float, and 5) calculating the project duration. An example project is provided to demonstrate the process of identifying the critical path and activities with zero total float to determine the project completion time.

1. PROJECT MANAGEMENT AND
NETWORK MODEL

2. The procedure of determining the critical path
 Step 1. List all activities and then draw arrow (network) diagram.
 Step2. Indicate the normal time (tij) for each activity ( i,j) above the
arrow which is deterministic.
 Step 3. Calculate the earliest start, the earliest finish, the latest finish
and latest start times for each event.
 Step 4. Indicate the various times computed above on the arrow
diagram.
 Step 5. Determine the total float for each activity by taking the
difference between the earliest start and the latest start time.
 Step 6. Identify the critical activities and connect them by double line
arrows.
 Step 7. Calculate the project duration.

3. CONT’D
Example:
 A small maintenance project consists of the following jobs
whose precedence relationship is given below.
Required:
A. Draw an arrow diagram representing the project.
B. Find the total float for each activity.
C. Find the critical path and the total project duration
Jobs 1-2 1-3 2-3 2-5 3-4 3-6 4-5 4-6 5-6 6-7
Duration (
days)
15 15 3 5 8 12 1 14 3 14

4. SOLUTION
A. THE NETWORK DIAGRAM THAT REPRESENTS THE PROJECT IS AS FOLLOWS

5. CONT’D
 The paths of the network are:
 Path: 1 – 2 –5 -6 -7=>37days
 Path: 1 – 2 – 3– 4 – 5 – 6– 7 =>44days
 Path: 1 – 2 – 3– 4 – 6– 7 =>54days- Critical Path Length
 Path: 1 – 3– 4– 5 – 6 – 7 =>41days
 Path: 1 – 3– 4 – 6 – 7 =>51days
 Path: 1 – 3–6 – 7 =>41days
B. The total float for each activity
 To determine the total float first the earliest start and finish; late
start and finish should be computed.
 This calls for the forward pass and backward pass computation.

6. CONT’D
 Forward pass calculation
 Finding ES and EF times involves a “forward pass” through the
network.
 The method associated with finding ES and EF is called
forward pass method.
 ES: the earliest time the activity can start, assuming all
preceding activities start as early as possible.
 EF: The earliest time the activity can finish.
EF = ES + t
 Esj is given by:
 ESj= Max (ESj, ti-j)
 Where Esi is the earliest time and tij is the normal time for the
activity (i,j).

7.  ES1=0
 ES2= ES1 + t1-5= 0 + 15=15
 ES3= Max (ES 2 + t2-3, Es1 + 13)
 = Max (15 +3, 0+15) = 18
 ES4= ES3+ t34 = 18+8 = 26
 ES5 = Max (ES2 + 25, ES4 + t4-5)
 = Max (15 +5, 26+1) = 27
 ES6 = Max (ES3+t3-6, ES4+ t4-6, ES5+t5-6)
 = Max (18+12, 26+24, 27+3)
 = 40
 ES7 = ES6 + t6-7 = 40 +14 = 54

8. CONT’D
 Backward pass Calculation
 Finding LS and LF times involves a “backward pass” through
the network.
 The method of computing LS and LF is called backward pass
method.
 LS, the latest time the activity can start and not delay the
project.
 LF, the latest time the activity can finish and not delay the
project
 Computation of the latest starting and finishing times is aided
by the use two rules.
 LS = LF -t
 The latest starting (LS) time for each activity is equal to its
latest finishing (LF) time minus its expected duration

9. CONT’D
 2.For nodes with one leaving arrows: LF for arrows
entering that node equals the LS of the leaving arrow.
 For nodes with multiple leaving arrows: LF for arrows
entering that node equals the smallest LS of leaving
arrows.
 LFi is given by:
 LFi = Min ( LFj-ti-j) where LFj is the latest finish time for the
event j.
 LF7 = 54
 LF6= LF7 = t6-7 = 54-14 = 40
 LF5= LS6-t5-6= 40-3 = 37
 LF4 = Min (LS5-t4-5, LS6-t4-6)
 = Min (37-1, 40-14) = 26


10.  LF3 = Min (LF4-t3-4, LF6-t3-5)
 = Min (26-8, 40-12) = 18
 LF2 = Min (LF5 - t2-5, LF3-t2-3)
 = Min (37-5, 18-3) = 15
 LF1 = Min (LF3-t1-3, LF2-t1-2)
 = Min (18-15, 15-15)=0
 The following table gives the summarized calculation for
critical path and total float.

11. CONT’D
Activity
Normal
time
Earliest Latest
Total float LFj-
ESj or LFi-ESi
Start Finisgh Start Finish
ESi ESj LSi LFj
1-2 15 0 15 0 15 0
1-3 15 0 15 3 18 3
2-3 3 15 18 15 18 0
2-5 5 15 20 32 37 17
3-4 8 18 26 18 26 0
3-6 12 18 30 28 40 10
4-5 1 26 40 26 40 0
5-6 3 27 30 37 40 10
6-7 14 40 54 40 54 0

12. CONT’D
c)The Critical Path
 From the above table we observe that the activities 1-2, 2-
3, 3-4, 4-6, 6-7 are the critical activities.
 The critical path is given by, 1-2-3-4-6-7
 The total project completion time (duration)is given by 54
days.
Project Evaluation and Review Technique (PERT)
 PERT is a time-oriented technique designated to cater for
projects where it is not possible to estimate the exact
duration of the activity.
 It uses statistical theory to estimate how long a project is
likely to last, and the probability of completing the project
by a particular date.
 It helps to address situations which require a probabilistic
approach

13. CONT’D
 The probabilistic approach involves three times estimates
for each activity instead of one:
 1) Optimistic time: - The length of time required under
optimum conditions.
 It is represented be type represented by the letter “a”.
 2) Pessimistic time: - The amount of time that will be
required under the worst conditions.
 It is represented by the letter “b”.
 3) Most – likely time: - the most probable amount of time
required.
 It is represented by the letter “m”.

14. CONT’D
The β– distribution is commonly used to describe the inherent variability in time estimates.
te
m b
a
0
Activity
start
Optimistic
time Most-likelytime
(Mode)
Pessimistic time

15. CONT’D
 β – Distribution can be symmetrical or skewed to either
the right or left according to the nature of the activity.
 The mean and variance of the β-distribution can be found
from the three time estimates.
 β – Distribution is unimodal with a high concentration of
probability surrounding the most-likely time estimate.
 The important issue in network analysis is the average of
expected time for each activity, te, and the variance of
each activity time,δ_i^2.

16. CONT’D


17. CONT’D


18. CONT’D
Activity Optimistic
time
Most likely
time
Pessimistic
time
1-2 3 5 7
2-3 4 8 12
3-4 2 4 6
3-5 3 6 9
3-6 3 6 9
4-5 2 4 6
4-6 1 3 6
5-6 2 5 8
6-7 2 5 14
7-8 4 7 10

19. CONT’D
 Given the above information, compute the following:
A. Find the expected duration and variance of each
activity
B. Find the expected duration of the project.
C. Calculate the variance and standard deviation of
project length and what is the probability that the
project will be completed:
i.at least 4 days earlier than expected?
ii.no more than 4 days later than expected?
C. If the project due date is 41 days what is the probability
of meeting the due date.
Solution for a) and b)

20. Paths Activity to tm tp
Ze
To+4tm+
tp
6
Expected
path
length
(E the
path)
S2
1-2-3-5-
6-7-8
1-2 3 5 7 5 37 4/9
2-3 4 8 12 8 16/9
3-5 3 6 9 6 1
5-6 2 5 8 5 1
6-7 2 5 14 6 4
7-8 4 7 10 7 1
1-2-3-4-
5-6-7-8
1-2 3 5 7 5 39
Critical
path
4/9
2-3 4 8 12 8 16/9
3-4 2 4 6 4 4/9
4-5 2 4 6 4 4/9
5-6 2 5 8 5 1

21. CONT’D
Path Activity (
2
path)  Path
1-2-3-5-6-7-8 1-2
83/9
2-3
3-5
5-6
6-7
7-8
1-2-3-4-5-6-7-8 1-2
82/9
Project variance
2-3
3-4
4-5
5-6
6-7
7-8
1-2-3-4-6-7-8 1-2
285/
36
2-3
3-4
4-6
6-7
7-8
1-2-3-6-7-8 1-2
74/9
2-3
3-6
6-7
7-8
9
/
83 9
/
82 36
/
2859
/
74

22. CONT’D


23. CONT’D


24. CONT’D


25. darla
/smb
s/vit
25
NETWORK EXAMPLE
Illustration of network analysis of a minor redesign of a product and its associated
packaging.
The key question is: How long will it take to complete this project ?

26. darla
/smb
s/vit
26
FORWARD PASS
 Earliest Start Time (ES)
 earliest time an activity can start
 ES = maximum EF of immediate predecessors
 Earliest finish time (EF)
 earliest time an activity can finish
 earliest start time plus activity time
EF= ES + t
Latest Start Time (LS)
Latest time an activity can start without delaying critical path time
LS= LF - t
Latest finish time (LF)
latest time an activity can be completed without delaying critical path time
LS = minimum LS of immediate predecessors
Backward Pass

27. EXAMPLE 2
SUPPOSE A CIVIL ENGINEERING FIRM HAS TO BID FOR THE CONSTRUCTION OF A DAM. THE ACTIVITIES
AND TIME ESTIMATES ARE GIVEN BELOW. DO THE NEXT REQUIREMENTS.
Activity a m b
1-2 (a) 1 3 4
1-3 (b) 2 4 6
1-4 (c) 2 3 5
2-5 (d) 3 4 5
3-6 (e) 3 5 7
4- 7 (f) 5 7 9
5-8 (g) 2 3 6
6-8 (h) 4 6 8
7-8 (I) 3 4 6

28. CONT……….
REQUIRED:
a) Draw the project network and identify all the paths.
b) Compute the expected time for each activity and the expected
duration for each path.
c) Determine the expected project length.
d) Identify the critical path and critical activities.
e) Compute the variance for each activity and the variance for
path .

29. SOLUTION
A)

30. CONT…..
Solution
B)
Path Activity
Times
a m b te = a+4m+b
6
Path total
1-2-5-8 a
b
c
1 3 4
2 4 6
2 3 5
2.83
4.00
3.17
10.00
1-3-6-8 d
e
f
3 4 5
3 5 7
5 7 9
4.00
5.00
7.00
16.00
1-4-7-8 g
h
i
2 3 6
4 6 8
3 4 6
3.33
6.00
4.17
13.5

31. CONT…..
C) The critical path is path 1-3-6-8, since it has the longest expected
duration.
Path
Activity
Times
a m b δ=activity=(b-a)2
36
δ2 path δ path
1-2-5-8 a
b
c
1 3 4
2 4 6
2 3 5
(4-1)2/36=9/36
(6-2)2/36=16/36
(5-2)2/36=9/36
34/36=0.944 0.97
1-3-6-8 d
e
f
3 4 5
3 5 7
5 7 9
(5-3)2/36=4/36
(7-3)2/36=16/36
(9-5)2/36=16/36
36/36=1.00 1.00
1-4-7-8 g
h
i
2 3 6
4 6 8
3 4 6
(6-2)2/36=16/36
(8-4)2/36=16/36
(6-3)2/36=9/36
41/36=1.139 1.07