Imposing Lag Durations in Critical Path

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Do curso por Columbia University

Construction Scheduling

519 ratings

Columbia University

Construction Scheduling

519 ratings

Curso 2 de 4 em Specialization Construction Management

Learners will discover the key project scheduling techniques and procedures including; how to create a network diagram, how to define the importance of the critical path in a project network, and defining project activities float. Also covered are the fundamentals of Bar Charts, Precedence Diagrams, Activity on Arrow, PERT, Range Estimating, and linear project operations and the line of balance.

Na lição

Critical Path

Professor Odeh discusses the importance of critical path in this module including what it is and why it is important. Learners will identify a critical path in a project and how to determine a critical path in different relationships.

Introduction to Critical Path5:44

Critical Path in a Project Schedule10:02

Determining the Number of Critical Paths in a Project2:09

Imposing Lag Durations in Critical Path12:14

Determining Critical Path in a Start to Start Relationship7:26

Determining Critical Path in a Finish to Finish Relationship6:57

Conheça os instrutores

Ibrahim Odeh, Ph.D., MBA
Instructor, Department of Civil Engineering and Engineering Mechanics, Columbia University
Director of Research and Founder, Global Leaders in Construction Management

Let's take it a step further.

We just finished Example 2 here with the following activities,

predecessors and duration, and

we came up with the following activity on node diagram.

If I want to impose a little bit of a lag duration between several activities

on this diagram, it will look like this, as another example.

So, activity D if you notice here, we added a relationship between it and

activity, a finish to start but with a lag of 4 days.

That's mean, after I finish with activity A,

I'm going to wait 4 days before I start with activity D.

In addition, I added two other lag times a new project.

One is the relationship between Activity F and one of its predecessors, not both.

Activity F, as you can see here, has two predecessors, B and C

and only B will have a lag time between B and F of 3 days.

As we can see in the last column of the table.

Moreover, the last activity of the project I has two predecessors and

I also added one lag of 1 day with one of its predecessor

which is activity H and activity G, there is no lag.

Just a traditional relationship of finish to start between I and G.

So how we want to draw this?

It will be exactly the same from a logic

point of view, like, example number 2, we'll not change anything, but

the calculations will be changing based on the lag, and

when we calculate the forward, and the backward paths calculations.

So, let's quickly [INAUDIBLE] diagram,

so we have A, we have B here,

C B,

all finish to start relationships.

And then we have E here, F and G.

It's exactly the same as the previous example but we want to build on it.

So we have a relationship between B and E.

And F has two predecessors between B and C.

And G has two predecessors of C and D.

So the last two activities, we have activity H and

has two predecessors of E and F.

And the last activity of I,

With two predecessors of H and G.

From the table, in example number 3 with the lags, we highlighted three lags.

One of the lags is between A and D of 4 days.

And, another lag between B and F of 3 days.

And the last lag time we inforced

between H and I with the 1 day.

So in this case, let's do the calculations of identifying the early start or

the forward pass calculations for

early start time and early finish time of all the activities.

Let's quickly highlight

the duration of each activity to 5,

9, 3, 6, 4, 2.

And we have 4 here and 2 here.

So the early start as we remember, we start with 0 + 2 = 2.

We have the early start for B and C to be 2.

So once we had a lag time, by definition,

that means we will wait 4 days after the finish of activity A,

which is in day number 2, to start its successor

which will be in 2 + 4 in day number 6.

So the early start day for

activity D will be equal to early finish date of its predecessor plus

the lag time inforced between the relationship between both activities.

And then we continue with the forward past calculations.

From 7 here, 11, 9.

And then we will go to E here, we have 7.

The same thing, the relation applies between activity B and activity F.

So in this case, the relation to find the early start date for activity F,

we have in this case two predecessors, B and C.

So, the early start date for activity F will be either

7 + 3 which is equal 10 or 11.

And, as I explained to you before, it will be the maximum number

between the early finish dates of all the predecessors

of that activity which, in this example, activity F.

So in this case, we have here early start date of 11 for activity F.

And if you want to think about it, without mathematical

approach, when we finish activity B, and

wait also at least 3 days, before we start activity F.

If there is no relationship with C and F, we still can have this starts right away.

However there is a condition we have in our project

that activity F will not start until both activities B and C finish.

So in this case we take the last activity from the we'll finish the last.

That will impose the early start to the early finish between the successor and

the predecessor.

The same concept as we just explained before but no lags between 11 and 9.

We have 11 for Activity G here.

Which will have 15.

We'll have 15, 13 and

then we have the maximum between 13 and

15 which is 15 here, 19.

And the same concept will goes for Activity I.

The early start date for activity I will be either 13,

or 19 + 1, the maximum will be 20.

And 20 + 2 will be 22.

So in this case, based on the information of adding

the lag time on these three relationships, between AD, BF,

H I, it causes the total project duration to jump 22 days.

That being said, let's go do the backward pass calculations.

Which as I spend the last day of the last activity

will go to be the late finish date and you go the same calculations but

backward to 20.

So in this case, when we try to find that late finish

date of activity that has a relation with its successors,

with the lag instead of adding 1,

we'll be subtracting 1 from here to have 19.

And then going to be the same concept of 15 here.

There will be 20, 18.

It goes here to 18 and 15.

So, it goes here 15, 15, 11,

And we'll go back to the, to 9 here.

And we will stop by activity B and

try to think before we move with the solution,

what will be the late finish date for activity B,

based on the two successors of E and F with a lag of 3 between B and F.

That will be either 9 or 11- 3, which is 8.

The minimum, because we're going backward, not the maximum, would be then the 8.

So that being said for activity C then,

we have the 11 and the 18, so

we'll take the 11, 11- 9, 2.

8- 5, 3.

And then the minimum between 3,

2 and 15- 4 which is 11.

Then that will be the number 2.

So we'll take the late start of activity C

to be the late finish for activity A.

And by that, we have the solution having activity on node with

the same example of example number 2 but with lag times as expand here.

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