WEBVTT
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and welcome to this series of lectures on
principles of construction management once
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again and in the last few lecturers we have
been talking about project planning and scheduling
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and today we will concentrate on bar charts
and critical path on the previous discussions
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we have talked about different plans safety
plan quality plan materials plan time schedule
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a finance plan and so on which tell us that
all these resources and points need to be
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foreseen and today what we will do is concentrate
on planning from the point of view of time
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and scheduling now coming to construction
schedules what is the focus of the discussion
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today these can be drawn up either with the
time orientation or a resource orientation
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and what is the interest in drawing up these
schedules the basic interest is to find out
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and assign start and finish dates to various
activities in the project
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we have talked about the fact that a project
can be broken up into activities and we must
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know that if we have so many activities in
the project each of these activities when
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they can start and when they must start can
start depends on whenever the preceding activities
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has been completed and must start is basically
the point where it should be ensured that
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if they do not start by that time the project
is likely to be delayed so how do we arrive
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at these numbers or how do we arrive at the
earliest start and the latest start days for
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each activity is what the focus of our discussion
today is
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these schedules also serve as a basis for
matching and developing resource requirements
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with various activities overtime so we try
to figure out ok if this activity is going
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to be carried out and this particular point
in time we must ensure that all the resources
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required are available at that point in time
detailed construction schedules also serve
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as a means to monitor and control the progress
of projects and of course as a corollary to
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that they would also serve as a basis for
any mid course correction in strategy should
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that be required if there is an activity which
is likely to be delayed the top management
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has to be informed that this is the situation
and if a resource reallocation additional
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resources have been employed all that has
to be done and that is done on the basis of
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these construction schedules
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we have already discussed topics like the
work breakdown structures precedence relationships
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and constructing project networks in our earlier
discussion for simple cases and today the
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discussion is largely to look into some concepts
relating to time oriented scheduling more
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formally before we get started let us reiterate
some of the networks fundamentals once again
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as they are applied in construction project
management
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an activity is an item of work involving consumption
of resources and time and produces quantitative
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results they take place in between events
and have a very well defined start and a well
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defined end
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events indicate the start or time of one or
more activities an event does not consume
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any time nor resource then we had talked about
dummy activities is an activity which does
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not consume any resource or time and is used
to show logical inter dependency among different
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activities and a network is graphical representation
of activities in terms of nodes and arrows
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to show logical interdependency between them
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the networks help us better understand the
sequencing of activities give us various informations
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about the project such as the time required
for completion the activities that need to
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be monitored more closely the basis of plans
for material procurement and manpower deployment
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and as we will see later on the chance or
the probability of completing projects within
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a given time
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now we also talked about the two representations
of activities as far as network is concerned
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one is the activity on node which is shown
on here where activities a and b are shown
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at these nodes and this arrow basically says
that b can be done only after a has been completed
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and then there is this representation which
is activity on arrow which basically serves
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that an activity a is represented as activity
a i j and i being the starting point of the
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activity and j being the end point of this
activity
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when we were talking about the example i had
added the activity b here to say and explain
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that now b is being represented by the activity
b j k where j represents also the starting
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point of b as it does the end point of a and
b off course can be undertaken only after
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a has been completed so we have done all this
before and in this course we will basically
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follow the activity on arrow representation
in our discussions
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so now coming to the scheduling techniques
there are three techniques which we will talk
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about as far as this course is concerned bar
charts the critical path method and the program
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evaluation and review technique some time
it is also called as project evaluation review
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technique it is the pert so now coming to
bar chart which is the first and most simple
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concept to understand these are graphical
representation of a project in which activities
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are shown on a real time scale
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these are shown as horizontal bars on a horizontal
time scale where the start and end locations
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of the bar coincide with the start and finish
dates of the activities the disadvantages
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we will see of bar charts is that they lack
informations about the logical dependence
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of activities and therefore it sometimes become
very difficult to figure out the criticality
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of an activity based on a bar charts or lets
look at example of a bar chart here we have
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activities like the contract negotiated contract
signed manufacturing schedules bill of materials
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short leave procurement material specifications
manufacturing plans and startup and this is
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the time on the horizontal axis which is given
in weeks after the go head and we find that
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the project can be completed or it scheduled
to be completed say nineteen weeks
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what this does not show is whether for example
this activity can be actually started only
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after this activity is completed or for that
matter whether this activity also has to be
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completed before this activity can be started
or not so this is the major disadvantage of
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this representation except that yes it gives
a very clear understanding that at a given
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point in time if you want to know what are
the activities that should have been completed
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it is very clear that as far as this line
that i have drawn is concerned at is point
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in time these activities should all have been
completed and this activity should be lets
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say fifty percent complete
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please also remember that this representation
is actually a schedule that emerges out of
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our planning in order for somebody to be able
to really ensure the deployment of the right
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kind of the resources so if we know in the
plan that this activity has been scheduled
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from this point to this point regardless of
its interdependencies and so on we know that
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whatever resources required for this particular
activity should be available at this point
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in time and may be free after this point in
time
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so this bar chart representation has its own
merits but of course it has its demerits so
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continuing with our discussion what is the
earliest starting time that is the e s t of
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an activity this is the time when the activity
can logically start given that all its preceding
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activities have been completed the earliest
finish time this is the earliest time at which
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the activity can be completed the earliest
finish time of an activity i j is basically
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the sum of the earliest starting time of that
activity and the duration that the activity
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takes
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so now if we take this as an example here
a is my activity a i j and d i j is the duration
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associated with that activity so the earliest
finish time of this activity i j is nothing
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but the sum of the earliest starting time
of the activity i j and the duration of the
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activity i j as its shown in this equation
as an illustrative example lets consider a
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project consisting of two activities a and
b which can start at the same time the durations
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of the activities a and b are four weeks and
seven weeks respectively now if you are required
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to construct a network of the project and
prepare the bar chart if the activities are
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to start at the earliest times
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now coming to the solution of this problem
we create an activity one two which is a and
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on top of that we write duration of this activity
lets say four weeks and then we create an
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activity b which is one three and write the
duration seven which is also the duration
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in weeks so and as far as completion of the
project is concerned we create this dummy
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activity two three and we will complete this
project when we have reached node three node
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three represents the completion of activity
b and also activity a through this dummy activity
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now lets try to carry out the calculations
that we said earlier this zero here represents
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the starting point of the activity a which
is the earliest that can start the project
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we have seen that the earliest finishing time
of an activity i j is e s t of i j plus d
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i j and therefore the earliest that we can
reach node two and that is the completion
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of the activity a is four because as far as
a is concerned its earliest starting point
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is zero the duration of the activity is four
and this becomes four
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similarly for b it can starts at zero because
its an independent activity and we can reach
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node three at seven that is zero plus seven
so what is the minimum time with the project
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can be completed intimately we know that this
project can now be completed in seven weeks
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so what becomes the bar chart representation
this is how it can be represented we start
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both these activities a and b at zero a gets
completed at four and b gets completed at
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seven
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so obviously since i have said that the activities
will start at their earliest starting point
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a has been started at zero we will very soon
see that a need not start at zero and thats
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why we have to be careful an activity can
start and it must start and that is the difference
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which i had pointed out even at the outset
of todays discussion
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so it is very clear that though a can start
at zero it need not start at zero however
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b must start at zero in order that the project
is not delayed so now lets continue this discussion
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little further this is the latest time at
which the activity must be completed to avoid
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the delay in the project completion and latest
start time of the activity is the latest time
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when an activity must be started in order
to avoid delays
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mathematically speaking the latest starting
time of an activity i j is represented as
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latest finishing time of that activity minus
the duration of that activity so if we look
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at this example once again the latest starting
time of an activity i j is the latest finishing
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time of this activity minus the duration of
this activity so continuing with our illustrative
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example where we had these two activities
a and b taking four weeks and seven weeks
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respectively a and b being independent to
start with this is the network we had drawn
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and now what we are trying to do is to determine
or is to see how the situation changes when
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it comes to the determination of the latest
start times
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we know that the project can be completed
in seven weeks b being the activity which
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is governing it and that duration is given
to be seven weeks now because of the dummy
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activity node two will also be seven and if
we use this equation try to find out what
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is the latest start times for these activities
i j which is in this case is a and b a being
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one two and b being one three so as far as
a is concerned the latest finish time of a
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is seven the duration is four and therefore
the latest a must start is three
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similarly as far as b is concerned it must
start at zero because the latest finishing
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time for b is seven and its duration is also
seven so seven minus seven gives you zero
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here and seven minus four gives you three
so this is how we evaluate the latest starting
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times of an activity so please note that as
far as the first discussion is concerned basically
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we went from left to right and this part now
is gone from right to left so this understanding
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is going to come in hand even we go for the
down in our discussion today and if we represent
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this as a bar chart with the activities being
shown at the latest start times what it will
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show is like this that yes b will start at
zero finish at seven but a can be started
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at seven now to ensure that a also gets completed
at seven it must be ensured that a must start
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at three
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so in the previous discussion we had said
that a can start at zero yes but it must start
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at three given that its duration is four and
the fact that its finishing time cannot exceed
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seven so now coming formally to what is called
a forward pass the forward pass moves from
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the start node towards the finish node and
basically calculates the earliest occurrences
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of all the events we have illustrated this
with a very small network of three nodes and
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we take up some more illustrative examples
later on and a forward pass is done to calculate
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the earliest project completion and when an
activity can start
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while carrying out the forward pass at a given
node j we need to check all the arrows which
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meet at j and take the maximum of the early
finish times of all those activities meeting
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at j as the earliest occurrence time of the
event j this is something which must be clearly
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understood that if there is a node j where
several activities are coming together so
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what we are talking of as far as the node
j is concerned there is an activity let us
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say a b and c which is i one j i two j and
i three j the earliest that the node j can
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be reached is basically the maximum of e i
depending on whether we are talking of i one
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i two or i three plus that respective durations
of that activities
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so in this case it will be the maximum of
e i one plus d i one j e i two plus d i two
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j e i three plus d i three j so the maximum
of this will be the e of j so now if we take
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an illustrative example based on the information
given in the following table about the activities
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in the project determine the minimum time
required to complete the project you will
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recall that this example has also have been
done earlier we have four activities in the
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project with durations being five six four
and three the predecessors that is a and b
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do not require any preceding activity they
can be started on their own c requires both
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a and b to be completed d requires only b
to be completed
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so if we want to draw the network first we
draw this network two three being a dummy
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activity as discussed earlier now coming to
evaluating the minimum project duration through
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the forward pass we look at this network once
again a b c and d with the respective durations
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plotted we start at zero reach node two at
zero plus six we reach node three from this
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route at zero plus five but please note that
this node three also has an incoming arrow
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from two so we have to see when can we reach
node three from two and that is six plus zero
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because this being a dummy activity does not
consume any resource and therefore this becomes
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six so the value itself is also six because
this is the maximum of zero plus five and
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zero plus six which in this case is five and
six the maximum of this is six so node three
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is reached at six
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similarly coming to node four from this side
that is completion of activity c the time
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where it can be reached is six plus four which
is equal to ten and from this side it is six
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plus three which is equal to nine so the maximum
of these two values is ten so the minimum
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time required to complete this project is
ten days
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now let us take another illustrative example
where the project has five activities which
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is given in weeks three four four three and
two are durations of the activities a b c
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d and e and given these interdependences we
have already drawn this network diagram and
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now proceed to evaluate the minimum project
duration for this project so given this information
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we obviously start at zero what is the earliest
that we can reach node two is zero plus three
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what is the earliest we reach node three is
zero plus four and as far as the node four
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is concerned it is the maximum of four plus
zero and three plus zero which is the maximum
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of three and four and this node is therefore
four
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similarly when we try to do the node five
we are looking at three plus three four plus
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four and four plus two that is we are looking
at the maximum of six six and eight and this
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is the eight so the minimum time required
to complete this project is eight weeks
20:33.530 --> 20:40.510
so now lets try to find out what are the latest
starting point or the latest starting times
20:40.510 --> 20:46.210
of each of these activities through the process
called the backward pass now the backward
20:46.210 --> 20:52.430
pass moves from the finish node towards the
start node and basically calculates the latest
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occurrence times of all these events the backward
pass is done to calculate the time at which
20:59.000 --> 21:06.830
a given activity must start to avoid project
delay like we did in the case of a forward
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pass where we tried to find out earliest that
we can reach a node now we are finding out
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the latest that we should reach a node
21:14.780 --> 21:19.080
while carrying out the backward pass at a
given node i we must check all the arrows
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that emerge from i and take the minimum of
the latest starting times of all those activities
21:25.370 --> 21:32.470
succeeding i as the latest occurrence for
that event i mathematically speaking the l
21:32.470 --> 21:42.170
i can be written as the minimum of the l j
minus all the d i js where d i js are the
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durations of these activities i and j so as
is given in this example given that three
21:48.730 --> 22:03.090
activities lets say a b and c which are i
j one i j two and i j three then the latest
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that we are reaching lets say nodes j one
j two j three is l j one l j two and l j three
22:10.870 --> 22:18.860
the latest we must reach here is basically
the minimum of l j one minus d i j one l j
22:18.860 --> 22:28.160
two minus d i j two and l j three minus d
i j three now lets do an illustrative example
22:28.160 --> 22:33.230
performing this backward pass we will use
the same example that we did earlier where
22:33.230 --> 22:39.010
we did the forward pass and there is a network
that we have drawn and we had found that this
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project can be completed in ten days
22:42.270 --> 22:47.280
now we are trying to find out what is the
latest time that these activities a b c and
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d must start now going back from this we find
that as far as the activity c is concerned
22:54.290 --> 23:05.220
this must start at ten minus four which is
six similarly as for as b is concerned it
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must start at ten minus three which is seven
and six minus zero which is six so the minimum
23:14.410 --> 23:23.080
of this gives us six so we go back and find
out the latest we must be at one that is six
23:23.080 --> 23:29.630
minus six and six minus five and that obviously
is the minimum of zero and one and that gives
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us zero we must be at node one at zero we
must be at node two at six node three is also
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at six and we will reach node four at ten
so this is how we now calculated the latest
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starting times for each of those nodes
23:50.060 --> 23:57.620
lets do another example a project which has
six activities a b c d e f which are in a
23:57.620 --> 24:02.400
manner that a b and c are interdependent and
inter related to the extent that c can be
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done if b is completed and b can be done if
a is completed on an independent route f can
24:07.870 --> 24:14.790
be done when e is completed and e can be done
when d is completed and a and d can be started
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independently so if we have this information
given to us with the duration of the activities
24:19.970 --> 24:26.070
also listed here on the arrows lets try to
find out the earliest start time and the latest
24:26.070 --> 24:27.070
start times
24:27.070 --> 24:32.830
we start with zero here here also we reach
six as far as nodes two and three are concerned
24:32.830 --> 24:39.180
because both these activities are taking six
units of time and we reach four at fourteen
24:39.180 --> 24:46.350
which is six plus eight and we reach node
five at ten which is six plus four and we
24:46.350 --> 24:54.950
reach node six which is twenty one because
that is the maximum of fourteen plus seven
24:54.950 --> 24:59.790
and ten plus three so this is twenty one this
is thirteen and the maximum of this being
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twenty one we reach node six at twenty one
which now becomes our minimum project duration
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so if we want to plot this in a bar chart
with the activities starting at their earliest
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starting times we have activity a which can
start at zero and goes up to six activity
25:20.870 --> 25:28.100
b which can start only after a has been completed
it starts at six and goes up to fourteen activity
25:28.100 --> 25:33.910
c which can start after b has completed starting
at fourteen going up to twenty one and on
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the independent route the other route which
is d e and f the activities are d e and f
25:43.120 --> 25:49.050
we see that these activities d e and f have
been positioned in a manner that they will
25:49.050 --> 25:57.400
start on their e s t which is the earliest
starting time that these activities can start
25:57.400 --> 26:03.440
so continuing this example lets do the backward
pass analysis now we start with twenty one
26:03.440 --> 26:10.110
we must reach node five at eighteen because
this eighteen is twenty one minus three so
26:10.110 --> 26:17.920
nearly we must reach node four at fourteen
which is twenty one minus seven we must reach
26:17.920 --> 26:24.740
node three at fourteen which is eighteen minus
four node two at six which is fourteen minus
26:24.740 --> 26:33.200
eight and node one at zero now given this
information we can again plot the bar chart
26:33.200 --> 26:41.550
for these activities and we find that this
is how our activities c and f both must be
26:41.550 --> 26:46.600
completed at twenty one and therefore as per
the activity c is concerned given the fact
26:46.600 --> 26:52.780
that it takes seven units of time it must
start at fourteen so fourteen becomes the
26:52.780 --> 27:00.980
latest starting point for activity c similarly
for f eighteen becomes the latest starting
27:00.980 --> 27:06.540
time because it takes three units of time
and we do not want a delay beyond twenty one
27:06.540 --> 27:14.820
so carrying this discussion forward we have
the plot for activities b e a and d so this
27:14.820 --> 27:22.150
set here is now plotted with the condition
that they will start at their latest starting
27:22.150 --> 27:28.680
times so now we obviously have a difference
between the earliest starting times and the
27:28.680 --> 27:33.490
latest starting times for some activities
in some activities there is no difference
27:33.490 --> 27:39.461
between the earliest that they can start and
the earliest that they must start and that
27:39.461 --> 27:43.660
is what we need to investigate little further
and that is what we planned to do in this
27:43.660 --> 27:50.990
slide if we use this representation which
is basically the earliest that these activities
27:50.990 --> 27:57.300
can be started against that the nodes can
be reached and so on and this representation
27:57.300 --> 28:02.050
which is the latest that these activities
can be started finished the nodes that the
28:02.050 --> 28:04.840
node we are interested in and so on
28:04.840 --> 28:09.340
so now if we combine these two representations
we get the representation which is shown at
28:09.340 --> 28:14.350
the bottom here what we have done and that
is conventionally done also if we look at
28:14.350 --> 28:22.090
different books in the subject we have created
this box which has two entries this entry
28:22.090 --> 28:29.630
is taken from here which is basically represents
the earliest that this node can be reached
28:29.630 --> 28:35.950
it also represents the earliest of that activities
eliminating from that node can be started
28:35.950 --> 28:41.870
this digit however has been borrowed from
the figure on the right here and this shows
28:41.870 --> 28:48.230
the latest this node must be reached and also
the latest starting time of the activities
28:48.230 --> 28:55.460
eliminating from that node so now if we see
this route the a b c route we find that the
28:55.460 --> 29:02.710
earliest and the latest starting times are
all the same where as if we go through this
29:02.710 --> 29:08.550
route there is a difference in these values
and this is something which we investigate
29:08.550 --> 29:11.400
a little more in subsequent slides
29:11.400 --> 29:15.920
now before we do that let us look at this
bar chart representation side by side this
29:15.920 --> 29:20.550
has been drawn on the assumption that the
activities are started at the earliest that
29:20.550 --> 29:25.980
they can be started whereas this has been
drawn on the basis that the activities are
29:25.980 --> 29:30.990
started at the last movement that is they
must be started at that time less the project
29:30.990 --> 29:38.580
is delayed what we see is that there is some
flexibility in activities d e and f is starting
29:38.580 --> 29:43.370
and finishing times whereas activities a b
and c do not have any such flexibility and
29:43.370 --> 29:47.360
thats what we have discussed in the previous
slide and what have we also conclude from
29:47.360 --> 29:54.160
that is that the project duration is primarily
governed by activities a b and c so as far
29:54.160 --> 30:01.660
as this route is concerned maybe we can do
d as it is but shift e a little bit f a little
30:01.660 --> 30:07.480
bit so long as the finishing dates of these
activities are preserved
30:07.480 --> 30:13.850
taking this concept formally we have the concept
of total float the total float of an activity
30:13.850 --> 30:20.010
is the amount of time by which that start
of an activity may be delayed without causing
30:20.010 --> 30:25.880
a delay in the project so what we see for
example here is that this activity can be
30:25.880 --> 30:31.630
started here but must be started here so there
is this duration which is being called as
30:31.630 --> 30:34.140
float
30:34.140 --> 30:40.730
similarly this activity can be started here
but must be started here so there will be
30:40.730 --> 30:48.120
this duration which also float so e the float
is associated with an activity to reiterate
30:48.120 --> 30:53.350
it is the amount of time by which the start
of an activity may be delayed without causing
30:53.350 --> 31:00.910
a delayed to the project completion mathematically
the total float of an activity i j can be
31:00.910 --> 31:09.410
given as l s t of i j minus the e s t of i
j or l f t of i j minus e f t of i j or l
31:09.410 --> 31:17.450
j minus l i minus d i j so i am leaving it
to you to ensure that the arithmetic suggested
31:17.450 --> 31:22.760
here is correct but the principle is something
which i have already explained fairly clearly
31:22.760 --> 31:28.560
i hope that total float of an activity is
the amount of time by which the start of that
31:28.560 --> 31:33.190
activity may be delayed without causing a
delay in the project completion so moving
31:33.190 --> 31:40.670
forward float can be evaluated either using
occurrence times of activities or events and
31:40.670 --> 31:46.100
if the total float of an activity is zero
then the activity is known as the critical
31:46.100 --> 31:52.900
activity for the network that is what does
float zero mean float zero means that the
31:52.900 --> 31:59.060
activity must be started at the time when
it can be started there is absolutely no rule
31:59.060 --> 32:06.140
for a delay there is no way that we can allow
any delay in that activity for other activity
32:06.140 --> 32:11.770
is yes its possible that the can start and
must start dates are slightly different
32:11.770 --> 32:19.030
so now coming to the critical path method
it is the most widely used scheduling technique
32:19.030 --> 32:24.580
and the path represents the series of activities
which should not be delayed for timely completion
32:24.580 --> 32:30.410
of project we can imagine that a project or
a network representing the project can have
32:30.410 --> 32:36.540
different paths through which the starting
point and the end points are connected now
32:36.540 --> 32:43.090
each path has a set up activities so the critical
path represents that series of activities
32:43.090 --> 32:47.360
which should not be delayed for timely accomplishment
of the project so in the previous example
32:47.360 --> 32:55.490
a b c constituted one path d e f constituted
another path and we find that a b c is the
32:55.490 --> 33:01.330
critical path indeed it is the longest path
in the network representing the minimum time
33:01.330 --> 33:05.920
required to complete the project that is what
we have illustrated in the example that we
33:05.920 --> 33:10.660
just completed and the duration of activities
is considered to be deterministic
33:10.660 --> 33:16.370
now what we have assumed in our discussion
all the time is that the duration of each
33:16.370 --> 33:22.280
of these activities is clearly known we know
that an activity a will be completed in six
33:22.280 --> 33:28.640
weeks we know that the activity b is completed
in ten weeks and we have already seen an example
33:28.640 --> 33:35.910
as to how these estimates are drawn so once
this value is known to us we can use the critical
33:35.910 --> 33:41.360
path method the way we have illustrated in
this discussion and find out what are the
33:41.360 --> 33:47.060
activities that need to be watched more carefully
and which are the activities that determine
33:47.060 --> 33:52.560
the time of completion of the project and
for which of the activities we can afford
33:52.560 --> 33:59.320
to be a little more relaxed or we can do resource
allocation in a manner that they need not
33:59.320 --> 34:01.960
take the same amount of time
34:01.960 --> 34:08.540
we can possibly allow an extension of times
on non critical activities because they do
34:08.540 --> 34:15.320
not affect the project duration this is the
luxury which we should not take with the activities
34:15.320 --> 34:19.929
lying on the critical path so as already stated
activities on the critical path have total
34:19.929 --> 34:25.190
float as zero so now lets do an illustrative
example computing the critical path for a
34:25.190 --> 34:32.620
project which is shown here so in this case
there are activities a b c d e f and g and
34:32.620 --> 34:36.409
the interdependence is also on along with
the durations is given here and what we are
34:36.409 --> 34:40.760
required to find out is the activities in
the critical paths the minimum time that the
34:40.760 --> 34:45.370
project can be completed and the total floats
of all these activities
34:45.370 --> 34:50.440
what i have not done here is tried to make
the table where we are said that there are
34:50.440 --> 34:56.570
activity a b c d e and so on and what do they
depend on so i am leaving to you to complete
34:56.570 --> 35:03.590
this table because this table is actually
the starting point for creating this network
35:03.590 --> 35:06.990
so because it is an illustrative example and
we have done some of these things before i
35:06.990 --> 35:11.390
have not done this but in this case we find
a does not depend on anything b does not depend
35:11.390 --> 35:17.920
on anything c depends on a and so on this
is something that which you can do as a homework
35:17.920 --> 35:22.730
and as far as we have concerned lets move
forward and do the computations
35:22.730 --> 35:28.900
so now the first step is to find out the earliest
that we can reach the different nodes using
35:28.900 --> 35:34.790
this equation that we already discussed in
detailed before we start zero here we get
35:34.790 --> 35:43.380
to three which is zero plus three for activity
a node three zero plus two node four is three
35:43.380 --> 35:51.090
plus three node five is the maximum of two
plus five and three plus seven which is ten
35:51.090 --> 35:59.740
and node six is the maximum of six plus six
and ten plus five so this fifteen represents
35:59.740 --> 36:05.040
the earliest that the project can be completed
in whatever units it is weeks months days
36:05.040 --> 36:06.250
whatever
36:06.250 --> 36:14.090
so i am leaving this computation of the latest
starting times as an assignment now taking
36:14.090 --> 36:20.870
this information of fifteen days or weeks
units of time being the minimum duration of
36:20.870 --> 36:27.990
the project and translating it through the
backward pass to the latest starting times
36:27.990 --> 36:33.280
of activities we are not doing the calculations
here and i am leaving it out to you to verify
36:33.280 --> 36:40.920
these if these are numbers that we get what
we see is that in the case of nodes two nodes
36:40.920 --> 36:49.750
five and of course node six and node one these
numbers are the same so what is our critical
36:49.750 --> 36:54.440
paths it is basically a d and g
36:54.440 --> 37:01.450
now what we do is we calculate or tabulate
the earliest and the latest event times for
37:01.450 --> 37:07.190
all the events all these activities and the
d i js are given here earliest starting time
37:07.190 --> 37:15.220
the latest starting time if we do all these
we find that the critical path is a d g and
37:15.220 --> 37:22.100
this what we have already done that is activities
a d and g lie on the critical path and these
37:22.100 --> 37:29.800
are the floats which are associated with activities
b c e and f
37:29.800 --> 37:36.760
so this becomes our final network that a d
and g are on the critical path and activities
37:36.760 --> 37:48.390
b e c and f have some float that is their
earliest starting dates are not that critical
37:48.390 --> 37:56.240
there can be a small gap or there can be a
resource reallocation such that they can be
37:56.240 --> 38:01.490
moved around a little bit with this we come
to an end of our discussion today and there
38:01.490 --> 38:05.520
are so many reference books so many texts
that can help you understand this concept
38:05.520 --> 38:11.010
of bar charts the critical path method and
calculating the critical path and i am sure
38:11.010 --> 38:17.150
you can do that what we will do next is to
handle a little bit of the uncertainty that
38:17.150 --> 38:23.040
is associated with the completion times of
each of these activities in this discussion
38:23.040 --> 38:30.200
today i have also pointed out that we have
assumed at time for each activity is known
38:30.200 --> 38:37.790
to us now what happens if that time has an
distribution in terms of the probability of
38:37.790 --> 38:42.940
completion we may have an activity which can
be completed in five days it may also take
38:42.940 --> 38:50.140
ten days so once that uncertainty is brought
in how do we handle this uncertainty in our
38:50.140 --> 38:55.210
approach is what is the subject for the next
discussion i look forward to see you at that
38:55.210 --> 38:56.210
time
38:56.210 --> 38:56.570
thank you