&=
#13;
=
‹#›
![](3D"master04_image003.gif")
![](3D"slide0025_image004.gif")
A
5 WEEKS
B
10 WEEKS
C
12 WEEKS
D
8 WEEKS
E
6 WEEKS
ACTIVITY<=
/b>
(t)
ES
EF
FL
LS
LF
We have a network with five activities: A, B, C, D and E. The =
=
activities have different durations =
that are
shown in the figure. For example, activity A has a duration of 5 weeks.
The figure in the lower left corner shows the different values that a=
re =
calculated in the network.
First, we calculate early positions for all of the activities. These =
=
show the earliest start points (ES) =
and
earliest finish points (EF).
0
5
5
15
5
17
17
25
25
31
![](3D"slide0023_image005.gif")
A
5 WEEKS
B
10 WEEKS
C
12 WEEKS
D
8 WEEKS
E
6 WEEKS
ACTIVITY<=
/b>
(t)
ES
EF
FL
LS
LF
F=
irst,
we take the starting po=
int in
activity A and assume that it starts in the timepoint 0.
The earliest start point (ES) for A is 0, and we can directly find th=
e earliest finish (EF) by adding to it=
the
duration of A:
0 +
5 =3D 5
By looking at the network, we see that the earliest time point for =
activity A gives the earliest time p=
oint
that both activities B and C can start on.&nbs=
p;
Activity B has a duration of 10 weeks and its’ earliest finish =
is
in week 15.
5 +
10 =3D 15
Activity C has a duration of 12 weeks and its’ earliest finish =
is
in week 17.
5 +
12 =3D 17
The earliest time point that activity D can start on is the maximum o=
f =
the earliest finish values for B and=
C.
Therefore, activity D starts on week 17.
The
duration of activity D is 8 weeks, so its earliest finish is 25.
17
+ 8 =3D 25
Activity
E therefore has an earliest start point in week 25.
The earliest finish for activity E is week 31 because the duration of=
the activity is 6 weeks.
=
25 +
6 =3D 31
0
5
7
17
5
17
17
25
25
31
![](3D"slide0024_image006.gif")
A
5 WEEKS
0
5
B
10 WEEKS
5
15
C
12 WEEKS
5
17
D
8 WEEKS
17
25
E
6 WEEKS
25
31
ACTIVITY<=
/b>
(t)
ES
EF
FL
LS
LF
Now we can calculate backwards in the network to find the latest =
position for all activities. This me=
ans that
we will calculate the latest start points and latest finish points.
The finish date of the project is not given, so we assume that the =
latest finish point is equal to the =
earliest
finish point. Thus, we set the latest finish equal to 31.
Now
we can subtract to find the latest start for all activities.
For activity E, which has a duration of 6 weeks, t=
he
latest start is week
25.
31
- 6 =3D 25
Additionally, by looking at the network structure we can see that =
activity D at has a latest finish po=
int in
week 25.
Activity D has a duration of 8 weeks, which gives us a latest start <=
/span>point of 17.
25
- 8 =3D 17
Since 17 is the latest time point at which activity D can start, both=
=
activities B and C should be finishe=
d by
this time. If either activit=
y <=
span
lang=3DNO-BOK style=3D'font-size:58%'>B or C are finished later, the entire
project does not finish in week 31.
Activity
B has a duration of 10 weeks and the latest start point is 7.
<=
/div>
17
- 10 =3D 7
Activity C has a duration of 12 weeks and has a latest start point of=
5.
17 -
12 =3D 5
Now we have two values for activity A that can give us the latest =
finish. We should choose the lowest =
value,
which is value 5.
The
latest start for activity A is 0 and the duration is 5.
5
- 5 =3D 0
For
activity A, the float becomes 0.
5
- 5 =3D 0
0
2
0
0
0
![](3D"slide0026_image007.gif")
A
5 WEEKS
0
5
0
5
B
10 WEEKS
5
15
7
17
C
12 WEEKS
5
17
5
17
D
8 WEEKS
17
25
17
25
E
6 WEEKS
25
31
25
31
ACTIVITY<=
/b>
(t)
ES
EF
FL
LS
LF
Now we will calculate the float of the different activities in the =
network. We find the floats by looki=
ng at
the differences between the latest finish points (LF) and the earliest fin=
ish
points (EF).
For
activity E, the float becomes 0.
31
- 31 =3D 0
For
activity D, the float becomes 0.
25
- 25 =3D 0
For
activity C, the float becomes 0.
17
- 17 =3D 0
For
activity B, the float becomes 2.
17
- 15 =3D 2
![](3D"slide0028_image008.gif")
0
2
0
0
0
A
5 WEEKS
0
5
0
5
B
10 WEEKS
5
15
7
17
C
12 WEEKS
5
17
5
17
D
8 WEEKS
17
25
17
25
E
6 WEEKS
25
31
25
31
ACTIVITY<=
/b>
(t)
ES
EF
FL
LS
LF
This means that all activities in the network are critical, except =
activity B. Therefore, we find a cha=
in of
critical activities A, C , D and E, which form a critical path in the network.&=
#13;
At the project start at time 0, the project will h=
ave a
total duration of 31
weeks.
![](3D"slide0028_image009.gif")
&=
#13;
&=
#13;
=
‹#›
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