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CPM Network Computation

Activity
Depends on t
AOA AON
1-2 A ─ 3
2-3 B A 4

Spring 2008, Activity on Node 1
King Saud University Dr. Khalid Al-Gahtani

AOA A ct ivit y N a m e

1 3
A
B

3
4
2 A ct ivit y D urat io n

AON
3 4

A B


Drawing CPM Networks
• Prerequisites: Before drawing a CPM
network, we must have:
– List of all activities comprising the project
– Order of precedence of each activity
– Duration estimate of each activity


Example:
Depends on Duration (Day)
Activity
(Immediate Predecessor(s)) (Time to perform)
a ─ 14
b ─ 3
c ─ 7
d a, b 4
e b, c 10


Activity on arrow Solution

a
d

b D um m y

c
e


Activity-on-node Solution
a

d

ST A RT b F IN IS H

e

c


Class work#1:
Draw AON Network for the fowling project:

Activity Depends upon Activity Depends upon
A G F
B A I F
F A J H
H A K I, J, F
C B, A L G, D, E
D B M K
E C N L, M


“Activity on Node”
• Nodes = Activities
• Links = Precedence Relationships
• Dummy activities are not required

ES t ES
Activity Name
LS TF LF

Example
Activity Depends on Duration (Days)
A 5
B A 15
C A 10
D B 15
E B, C 10
F D, E 5


AON
ES t EF

N am e

B D LS TF LF

F FIN ISH
ST A R T A

C E


AON
ES t EF

N am e
5 15 20 20 15 35

B D LS TF LF

5 20 20 35

35 5 40 40 0 40
0 0 0 0 5 5
F FIN ISH
ST A R T A

35 40 40 40
0 0 0 5

5 10 15 20 10 30

C E

15 25 25 35


Constraints with Lead/lag time


Finish-to-Start (FSij):
• FSij is equal to the minimum number of
time units that must transpire from the
completion of the predecessor (i) prior to
the start of the successor (j).
• The time between the finish of one activity
and the start of its successor is called
“Lag”.


Finish-to-Start (FSij):
• If the relationship is not listed on the
dependency arrow, FS is assumed with
Lag= 0.
• Example: a planner may wish to have an
activity of removing formwork from a new
building component follow the concrete
pour by some pre-defined lag period to
allow setting.


Start-to-Start (SSij):
• SSij is equal to the minimum number of
time units that must be complete on the
preceding activity (i) prior to the start of the
successor (j).


Start-to-Start (SSij):
• “Lag” is always applied to SS relation.
• Example: parallel in starting of Installing
and Finishing the walls activity of 100
rooms on a project must be 10 days
difference (SS=10 days).
– You don’t have to wait installing 100 wall’s
room to start doing the finishing work.


Finish-to-Finish (FFij):
• FFij is equal to the minimum number of
time units that must remain to be
completed on the successor (j) after the
completion of the predecessor (i).
• It is applied as productivity control.


Finish-to-Finish (FFij):
• The finish date of Installing and Finishing
walls’ activity of 100 rooms on a project
must have 10 days difference in order to
control productivity (FF=10 days).
– In this example, the productivity of installing
the walls’ activity might be less than finishing
the rooms’ activity.
• “Lag” is always applied to FF relation as
buffer between the two activities.

Start-to-Finish (SFij):
• SFij is equal to the minimum number of
time units that must transpire from the
start of the predecessor (i) to the
completion of the successor (j).
• It is applied also for controlling productivity


Start-to-Finish (SFij):
Example: The start date of Installing 100
rooms’ wall’s activity and the finish date of
Finishing same walls’ activity of a project
must maintain 30 days difference to
control productivity (SF=30 days).
“Lag” is always applied to SF relation as
buffer between the two activities.
• It is not recommended to use by planner.

Start-to-Start and Finish-to-Finish
(ZZij):
• ZZij is a combination of two constraints.
i.e., a start-to-start and finish-to-finish
relationship. It is written with the SSij time
units first, followed by the FFij time units.
• These two relations are used combined to
maintain buffer between the start and
finish of two activities.


Forward Pass Computations
Initial T im e
E Fi F S ij
E S j = M ax all i ESi SS ij
E Fi F F ij Dj
ESi SF ij Dj

EFj = ESj + Dj


Backward Pass Computation
T erm inal T im e
LS j F S ij
L F i = M in all j LF j F Fij
LSi SS ij Di
LF j SFij Di

LSi = LFi Dj


Example
2 2 6
B FS 4 C SF 5 D
SS 3

FF 1
SF 5
4 1 3 6
A F SF 4 L E

5 1 2 ES D EF
K G FS 4 H A ctivity
LS F LF


Example Solution
4 2 6 10 2 12 9 6 15
B FS 4 C SF 5 D
5 1 7 SS 3 11 1 13 10 1 16

FF 1
SF 5
0 4 4 4 1 5 9 3 12 16 6 22
A F SF 4 L E
0 0 4 10 6 11 11 2 14 16 0 22

ES D EF
4 5 9 9 1 10 14 2 16 A ctivity
K G FS 4 H LS F LF
4 0 9 9 0 10 14 0 16


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