Ihab Saad – PDM Solved Example
AI: Summary ©
The speakers discuss various problems in construction management, including overlap and contiguous duration, and the network activities of various activities. They explain how the network works and the critical path, emphasizing the importance of the total float and the critical path in determining the optimal level of activity. They also discuss the importance of the critical path and how it will be different for each successor, emphasizing the importance of understanding the math in calculating PDM.
AI: Summary ©
Foreign welcome again to
problems in construction management, and today we're going
to talk about some PDM solved examples, presence Diagramming
Method, as we have discussed it in the previous lecture. So let's
have a quick look at this example. Here, very simple problem. As you
can see, the activities are represented by boxes. Each box has
the name of the activity and its duration, and then we have the
links, or the relationships between the activities. Some of
them are traditional, finish to start with no lag or overlap, like
here. Some of them are start to start with lag. Some of them are
finished to start with overlap, which is a negative lag, and so
on.
Now we're going to assume contiguous duration for the
activity, which means once the activity starts, it should keep
going without any interruption. As you may notice, here in this
network, we have activity B with an open end from the finish side.
So we're going to see. How does that affect the calculations? Now,
starting the calculations, we're going to start from the start side
of the activity A, which is the absolute left of the network.
And as we assume, the network's going to start on day zero.
So it's going to start on day zero, and as the duration is 12,
then it's going to end on day 12. Looking at the activities B, C and
D, they have a finish to start relationship with activity A. So
all of these activities are going to start on day 12. So and with
their respective durations. Activity B has eight days, so it's
going to finish on day 20. Activity C, 15 days ending on day
27 and activity D, 14 days ending on day 26
now looking at the following activities, activity e has only
one immediate predecessor, which is b. So from B, we're going to
take the 12 to get the start of activity e. It's a relationship of
start to start, but it has four days of lag. Therefore the start
of activity e is going to be 12, plus four days of lag here, which
is going to be 16, with a duration of 11 is going to end on day 27
for activity f, it has one immediate predecessor. C finished
to start with two days of overlap, which is the negative lag here,
which means activity f is going to start two days before the
completion of activity c. Therefore the expected start for f
is going to be 27 plus negative two, which is 27 minus two, that's
25 its duration is seven days. So it's going to be expected to
finish on day 32
activity G has only one immediate predecessor. D finished to start,
no lag, no overlap. Therefore it's going to start on day 26 with a
duration of 10 days, is going to end on day 36
going to activity h, it has two immediate predecessors, E and F.
And the dates coming from E are 27 and from F, 32 this is the forward
pass. We're moving forward. Therefore we're going to take the
larger of the two numbers. So for the start of activity, h is going
to be day 32 with the duration of 16 is going to end on day 48
activity I also has two immediate predecessors, but it's quite
interesting here, because the immediate predecessors link at two
different locations. So let's look at the dates coming to the start
of I from these two respective activities. From F, we have 32 no
lag, no overlap. So if it had only F as a predecessor, it should
start on day 32 but it has another predecessor, which is G. We
notice that the relationship with G is finish to finish. Look at the
connecting points on the relationship finish to finish with
five days of lag, which means I should finish five days after the
completion of G. Now let's calculate the two dates going to
the start and the finish of this activity. If we were going to use
32 from f. It would be 32 at the start, plus 18, that would give 50
at the end, at the end of activity i, if we were to take only
activity G, it would be 36 plus five, that would be 41 so the
larger of two numbers comes comes from activity f. Therefore we can
say that activity f is the one that drives activity I, therefore
we're going to take the 32
plus 18, that's 50.
Now looking at activity j, the last activity in the network. We
we have two immediate predecessors with the numbers 48 and 50.
Coming from H and I respectively. We're going to take the larger
number, since we are still in the forward pass, so it's going to be
50 plus nine, and the early finish of this network is day 59 now we
are done with the forward pass again. What did we do in the
forward pass? We moved from left to right, adding the durations,
adding the legs and adding the overlaps, basically plus negative
two, if you notice what I said here, plus negative two, which is
the same as minus two. So we just keep moving from left to right,
adding the durations and the legs and the overlaps until we reach
the end of the network, and that gives the early finish of the
network. Now we're going to start the trip backwards. So we're going
to start from the end of activity j and the early finish is 59 it's
going to be the same as the late finish. So we're going to drop
here, 59 minus nine. Now we're moving backwards, so we're going
to subtract, which gives 50 at the late start of activity J. The
first thing to notice is that the early start and the late start are
exactly the same. The early finish and the late finish are the same.
Therefore, what's the total float of activity J? Basically it's
zero, which means activity j is critical. And this is something
that we can conclude, and we can expect, usually in a network like
this, starting with one activity and ending with one activity,
usually the first and the last activities are going to be
critical, since there's going to be one continuous path linking
these two activities from beginning to end, therefore
they're going to fall on that critical path, or that longest
path in the network. Now we have to move back and see which other
activities are going to be critical and where's the critical
path going to be. So this 50 is going to be transferred, basically
to the finish of activity h and the finish of activity i, and then
we're going to subtract the durations, and we have here 50
minus 16, that's 34 at the late start of H, and 50 minus 18,
that's 32 at the late start of i, we notice again for activity I,
the dates are exactly the same at both sides. So early start, late
start, early finish and late finish, the early and late dates
are pretty much the same. Therefore this is another critical
activity. But looking at activity h, we notice that the late finish
is different from the early finish and the late start is the
different different from the late from the early start, which means
this is going to be what we call the total float. Total float is
late minus early, either from the start or the finish for the time
being. So late finish minus early finish, or late start minus early
start. So the total float for activity h is going to be two
days. Now, remember the sad method that I told you about when we try
to calculate the free float of activity h we said that the free
float of an activity is going to be its total float minus the total
float of its immediate successor, or the largest total float of any
of its immediate successors if it has more than one immediate
successor. So looking here, this is an activity that has two days
of total float with only one immediate successor which has zero
days of total float. Therefore, the free float for activity h is
going to be exactly the same as its total float, which is also
equal to two days activity I of course, since it's critical, and
by default, any critical activity has zero total float and zero free
float as well. Now let's move back going to, for example, activity e,
we're going to find that it has only one immediate successor,
therefore that 34 is going to be transferred here as is, because
there's no lag or overlap.
And for activity f, which has two immediate successors. We have 34
coming from H, 32 coming from I remember, this is the backward
pass. So we take the smaller number. Therefore we're going to
take, here the 32 so let's look at the numbers here. For example, we
have 50 minus five, which is the lag. So it's going to end on day
45 minus the duration 10 is going to start on day 35 and here we
have the 32 which is the smaller of the two numbers, 32 and 3432
minus seven. That's 25 and here we're going to have 34 and 2334
minus 11. That's 23
as you notice, we are moving one column at a time. We're not moving
to the whole beginning of the network. We're taking it one
column at a time.
All right now if we move back to activity D, for example, it has
only one immediate success.
Minus nine, which gives 30 at the beginning. But here from E, we
have 3033 plus negative two, which gives 31 larger than the 30. So e
is driving G, therefore it's going to be 31 and nine is 40 activity.
H is going to be here we have again, 33 and 30. We're going to
take the larger number, the 33 plus 841,
again, looking at activity, I we have numbers coming from both
sides. If we look at G, is going to be 40 plus 12, that's 52 if we
look at H, it's going to be 41 plus nine. That's 50 at the end.
So at the end we have from G 52 from H 50 we're going to take the
larger number. So here we're going to have 40 and 52
that's the forward pass. Very simple, very straightforward. We
add the numbers as we are moving from left to right now starting
the backward pass again, the late finish is going to be the same as
the early finish. So it's going to be 52 minus 12. That's 40 again.
Network starts with one activity ends with one activity. We should
expect the first and the last activities to be on the critical
path. Now think about it for a second. Where is the longest path?
Which path gives the largest numbers. Someone might get
confused and say, well, H is 41 and g is 40 so h is longer. No,
that's a trick, because h goes to the end, but G goes to the
beginning. And we found out that G is the one that drives I,
therefore g must be the one that's critical. So it's going to be
here. We're gonna have 52 minus nine. That's 43 minus 835,
and here we're gonna have
if we follow that same,
same path.
Here we have the 40 and the 31
now going to activity e, we have, what's the number coming to E,
from G and H, from H, what do we have? We have 35
from G. What do we have? 31 No, it's 31 minus negative two, which
is 31 which is 31 plus two, therefore it's 33 so 33 from G and
35 from H, this is backward pass we take the smaller number. So
we're going to take the 33 minus nine, that's 24 so obviously the
critical path is going to be I, G, E, and then tracing backward.
We're not sure yet whether it's going to be B or C. We're gonna
see in a minute. Here, we're gonna have 40 minus six going to the
end. We did not take a number from the beginning, because we don't
have a relationship here. So 40 minus six, that's 34 minus 14,
that's 20 now here at C, we have 24 coming from E,
and we have 35 minus seven, which would be a 28 coming to the end of
C, we're going to take the smaller number. So here we have 24 and at
B, also we're going to have 24 and 13. So obviously B is going to be
the critical activity. Moving back to activity. A, here we have 11.
And from C, we have nine. Again, no, it's nine minus negative
three, which is nine plus three. Therefore the number coming from C
is 12. From B is 11. We take the smaller number, so it's going to
be 11 and zero. And here's our critical path.
Looking at the total and free floats again, for activity D,
obviously we have a total float of one,
its immediate successor is critical. So according to the SAT
method, the free float for D is also one.
Activity H has a total float of two, immediate successor,
critical, therefore the free float is also two. Activity f has total
float of two. Immediate successor has also a total float of two. So
the free float here is going to be zero.
Activity C has a total float of one,
immediate successors. One is critical. One has a total float of
two. We're going to take the largest of the total floats of the
immediate successors. So one minus two gives us negative one. We
agreed before that we cannot have a negative free float. So whenever
you get a negative free float, just put zero, and that would be
the free float. So the free float for activity c is zero, its total
float is one.
As you can see, the calculations are not that hard at all. I hope
that these two examples illustrate the calculations for PDM. The only
thing that you have to pay attention to is, where are the
relationship points starting and finishing?
Is it a finish to start? Is it a start to start? Is it a finish to
finish? Do we have any lags? Do we have an overlaps? And how are
these going to be factored in our calculations? And again, apart
from that, is just very simple math. One of the very common
mistakes that I usually see on assignments, on exams, is errors
in that very simple math like 19 plus 14 equals 34 or 35 or
something like that. Take your time with these initial
calculations, because this is what makes most of the mistakes on
these problems. The math, as you can see, is very simple and very
straightforward. I'll see you later in another lecture and in
another example.
You.