(AOA) and (AON) Network construction and critical path calculations (using four methods)
This document provides an example of constructing an activity on arrow (AOA) network and activity on node (AON) network for a project with multiple activities. It shows the steps to calculate early start, early finish, late start, late finish, total float, and critical path for identifying the longest duration of project completion. The critical path is identified as A → C → G → J, with an overall project completion time of 27 days.
(AOA) and (AON) Network construction and critical path calculations (using four methods)
1.
Project Management
(AOA) and(AON) Network Construction and
Critical Path Calculations
(Using Four Methods)
Dr. Mahmoud Abbas Mahmoud
Assistant Professor
2.
(AOA) and (AON)Network Construction and
Critical Path Calculations
(Using Four Methods)
EXAMPLE:
Construct the (AOA) and (AON) networks for the project whose activities,
their durations and their precedence relationships are as given in the table,
also find:
i) [ES, EF, LS, LF, Slack] for all activities.
ii) Critical path(s).
iii) Project completion time.
Activity
Predecessor
Activity
Duration
(days)
A - 2
B A 8
C A 10
D B 6
E B 3
F C 3
G C 7
H D, F 5
I E 2
J G 8
3.
Solution (1):
(AOA) network
Activity
Path
(i- j)
Duration
(days)
ES EF LS LF
TF
(Slack)
Critical
A (1 – 2) 2 0 2 0 2 0* Yes
B (2 – 3) 8 2 10 8 16 6 No
C (2 – 4) 10 2 12 2 12 0* Yes
D (3 – 5) 6 10 16 16 22 6 No
E (3 – 6) 3 10 13 22 25 12 No
F (4 – 5) 3 12 15 19 22 7 No
G (4 – 7) 7 12 19 12 19 0* Yes
H (5 – 8) 5 16 21 22 27 6 No
I (6 – 8) 2 13 15 25 27 12 No
J (7 – 8) 8 19 27 19 27 0* Yes
Therefore:
Critical path: A --- C --- G --- J
Project completion time = 27 days
4.
Solution (2):
(AOA) network(calculations in node)
Activity
Path
(i - j)
Duration
(days)
ES EF LS LF
TF
(Slack)
Critical
A (1 – 2) 2 0 2 0 2 0* Yes
B (2 – 3) 8 2 10 8 16 6 No
C (2 – 4) 10 2 12 2 12 0* Yes
D (3 – 5) 6 10 16 16 22 6 No
E (3 – 6) 3 10 13 22 25 12 No
F (4 – 5) 3 12 15 19 22 7 No
G (4 – 7) 7 12 19 12 19 0* Yes
H (5 – 8) 5 16 21 22 27 6 No
I (6 – 8) 2 13 15 25 27 12 No
J (7 – 8) 8 19 27 19 27 0* Yes
Note: (LC = LF) & (EC = EF)
Therefore:
Critical path: A --- C --- G --- J
Project completion time = 27 days
5.
Solution (3):
(AOA) network(calculations on arrow)
Activity
Path
(i - j)
Duration
(days)
ES EF LS LF
TF
(Slack)
Critical
A (1 – 2) 2 0 2 0 2 0* Yes
B (2 – 3) 8 2 10 8 16 6 No
C (2 – 4) 10 2 12 2 12 0* Yes
D (3 – 5) 6 10 16 16 22 6 No
E (3 – 6) 3 10 13 22 25 12 No
F (4 – 5) 3 12 15 19 22 7 No
G (4 – 7) 7 12 19 12 19 0* Yes
H (5 – 8) 5 16 21 22 27 6 No
I (6 – 8) 2 13 15 25 27 12 No
J (7 – 8) 8 19 27 19 27 0* Yes
Therefore:
Critical path: A --- C --- G --- J
Project completion time = 27 days
6.
Solution (4):
(AON) network
Activity
Path
(i- j)
Duration
(days)
ES EF LS LF
TF
(Slack)
Critical
A (1 – 2) 2 0 2 0 2 0* Yes
B (2 – 3) 8 2 10 8 16 6 No
C (2 – 4) 10 2 12 2 12 0* Yes
D (3 – 5) 6 10 16 16 22 6 No
E (3 – 6) 3 10 13 22 25 12 No
F (4 – 5) 3 12 15 19 22 7 No
G (4 – 7) 7 12 19 12 19 0* Yes
H (5 – 8) 5 16 21 22 27 6 No
I (6 – 8) 2 13 15 25 27 12 No
J (7 – 8) 8 19 27 19 27 0* Yes
Therefore:
Critical path: A --- C --- G --- J
Project completion time = 27 days