Ihab Saad – PDM Solved Example

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.