WEBVTT
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and welcome once again to this series of lectures
on principles of construction management in
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the last couple of lectures we will talking
about planning scheduling resource management
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and today continuing from that discussion
we will be talking of what is called crashing
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of a network now what is crashing construction
projects involve considerable amounts of expenditure
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in terms of time and money activities in the
projects have to be scheduled such that they
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are completed with the minimum time and cost
in our earlier discussions we have seen that
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both time and cost are largely dependent on
the resources allocated to the individual
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activities in the project one of the important
features in network analysis applied to construction
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management like cpm is that the duration of
an activity can be reduced to a certain extent
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by increasing the resources assigned to it
now in this lecture lets look at the concept
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of reducing the project duration by reducing
the durations of critical activities in the
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project by increasing the resources allocated
to them what we had said last time and i would
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like to re trade that today there are two
kinds of activities lets say critical and
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noncritical by doing re allocation of resources
leveling resources we try to rationalize the
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resource deployment over the period of time
and that is done with non critical activities
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because the duration of the project is governed
by critical activities so now the discussion
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today is is there a possibility of reducing
the overall duration of the project now in
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order to reduce the duration of the project
obviously its not important to concentrate
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on the non critical activities obviously we
must concentrate on the critical activities
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meaning thereby that if
we want to reduce the duration of critical
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activities obviously they will be a requirement
or a commitment of additional resources which
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will cause additional expenses what has to
be seen is whether this additional expense
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of resources can be justified in terms of
the benefits that may crewed to the contractor
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by finishing the project early that is something
which is relevant because as far as the contractor
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is concerned the project cost has a direct
cost and indirect cost and so on so we have
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to see the overall perspective of balancing
the additional resources of balancing the
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expenditure in the additional resources and
the benefits accruing there from so that basically
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is the essence of what we will talk about
today and that is what is called crashing
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of network now what are the components of
total project cost one component is obviously
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direct cost which comprises of expenditure
that is specific to a project that is cost
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of materials labour and equipment indirect
cost includes expenditure like tariff for
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infrastructure facilities rent and so on now
cost time trade off is what we talk about
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really the analysis of the inter relationship
between the time and cost of a project in
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order to minimize its cost and duration is
called the cost time trade off this picture
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here is a schematic representation of what
goes on if we try to reduce the project duration
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about direct cost increases but our indirect
cost decreases so the total cost which is
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essentially the sum of the direct cost and
the indirect cost that is what has to be seen
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so it will reduce up to a certain point in
time and then start increasing again so we
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should be targeting at getting to this duration
of the project in other words by reducing
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the duration of individual activities the
total cost of the project can be reduced to
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an optimum level now let us technically defined
what is crashing expediting an activity to
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an earlier time by mobilizing more resources
committing more resources to it is known as
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crashing
now lets look at an example to explain this
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concept consider a project having six activities
a b c d e f and the network of the project
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along with the durations of the individual
activities is shown here the cost of crashing
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individual activities it could be in an units
its given in indian rupees here it shown in
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the table given below now this is the table
determine which activity should be crushed
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first in order to reduce the project duration
so suppose we have this project has which
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activities and these of the durations lets
say that is given in days and this table gives
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us the cost of crashing for one day so if
we want to carry out activity a in three days
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it is possible even though it is four days
here it is possible to do it in three days
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provided we are able to mobilize additional
resources of four thousand rupees similarly
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activity e instead of six can be done in five
days provided we are willing to spend another
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thousand five hundred rupees we must remember
that at the end of it it is not possible for
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any activity to be infinitely crashed that
is we cannot say that if we are willing to
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increase the commitment of resources indefinitely
that time can be reduced indefinitely this
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cannot be done there may be minimum times
for any of these activities beyond which no
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matter how we crush it it will not work so
there is a minimum time which is required
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there is a reasonable time which we are willing
to gave and beyond that reasonable time yes
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it is possible to reduce it to that minimum
time by putting in additional resources so
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this is the back drop of carrying out this
exercise so once we have this information
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listed with us which is the cost of crashing
for one day for the different activities obviously
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it make sense to begin with the activity which
requires the minimum amount of cost so we
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should do that activity first which involves
minimum additional commitment of resources
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the second part of it the discussion is that
is that activity critical will really making
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that commitment help as far as reducing the
project duration is concerned that will happen
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only when the activities critical so the first
thing to do perhaps is to find out which are
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the critical activities and then try to see
what are the resources required if you want
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to crash those activities so lets move forward
and we do the network analysis we try to find
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out that the paths in the network a d b e
and c f these are the three paths in this
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network comprising of activities a d on one
side b e on the other and c f is the third
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and the critical path is b e this takes six
and six twelve days this takes ten days and
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this is nine and we are looking at the maximum
of this and therefore the critical path is
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b e which is twelve days long so this project
with the given information can be completed
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in twelve days at the certain cost now lets
try to see how we will use this information
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on crashing the first question to be asked
is which activity should be crashed from our
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analysis given here the duration of the project
is governed by the critical path b e and since
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the cost to crash e by one day is lower than
that of b activity e should be crashed first
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so we are looking at e is critical and b is
critical to crash b requires two thousand
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to crash e it requires thousand five hundred
and therefore it make sense to crash e first
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and then see how things change only to reiterate
even if any of the non critical activities
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a d c and f is crashed the project duration
will still remain twelve days and it makes
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no sense to do that now let us try to formally
define what is called the cost slope of an
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activity the extra cost in expediting an activity
to reduce its duration by unit time is called
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the cost slope of that activity that is given
by crash cost minus normal cost divided by
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normal time minus crash time that is the cost
slope in fact in this table when the cost
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of crashing for one day was given what was
essentially given is the cost slope of all
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these activities because what we mean by this
is that this is the cost that is incurred
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for crashing that activity by a single day
and obviously from the given set of activities
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the activity that has the minimum cost slope
should be trashed first
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lets consider two activities a and b the normal
durations the crash durations normal cost
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and crash cost of a and b are given in the
following table and if you are ask the simple
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question as to which of these should be crashed
first the analysis would basically b that
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a has a normal duration of ten and a crash
duration of seven b has twelve and eight so
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these are essentially the minimum times that
is required for the activities a and b beyond
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this it cannot be crashed now the crash cost
which is the cost at the crash duration this
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is not the cost slope in the previous example
what was given was crashing per day here we
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are giving the crash cost that is the cost
at the crash duration and the normal cost
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is given here
so if you want to calculate the cost slope
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of these activities we have to carry out the
small calculation crash cost minus normal
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cost divided by normal time and crash time
difference and we get one thousand here and
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seven fifty here this one thousand here is
nothing but eight thousand minus five thousand
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divided by ten minus seven which is equal
to three thousand divided by three and thats
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equal to a thousand similarly the cost slope
for activity b is seven fifty given this data
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activity b cost slope is lower and therefore
if it is required activity b needs to be crash
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first so having done the simple example we
can lay down the protocol or the procedure
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to be followed for crashing a network the
first step is identify the critical path or
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paths of the network and the corresponding
activities determine the project cost find
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the cost slope of critical activities rank
the critical activities in ascending order
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of the cost slope and crash the activities
in the critical path as per the ranking activities
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can be crashed only till their respective
crash durations obviously we cannot try to
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crash them beyond that crash duration calculate
the revised project cost repeat steps one
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to five till the optimum cost and duration
have been obtained remember that what we are
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ask to do is repeat steps one to five what
is one one is to find out the critical path
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having done the critical path first or having
identified the critical path first but having
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change the durations of those activities by
crashing and so on it is likely that some
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other activities of some other paths would
become critical and therefore as we crash
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activities in a stepwise manner its important
that at each step we check if other activities
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have become critical or not because if they
have then the cost slope there also will become
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important this something which you have to
keep in mind as we do an actual example and
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thats what we will do we will consider a project
having seven activities as shown below for
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that dependencies normal durations and crash
durations of activities in days and the event
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times are shown in the network so this number
here is t n that is the normal time this is
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the crash time which is t c
so with this information about the crash time
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and the normal times given for all the activities
we will proceed to find out what activities
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to crash what not to crash and so on and we
are also given the information that the indirect
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cost as far as the project is concerned is
six thousand rupees per day so this table
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here in addition to the normal times and the
crash times it gives you the information about
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the normal cost and the crash cost for all
the activities so if we look at a for example
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the normal times for three and two as i explained
in the previous slide and the crash cost is
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seven thousand and five thousand so that analysis
we can do for all the activities and what
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we proceed to do is the first step which is
identification of the critical path and activities
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now this we are already familiar with and
i am not going to spend time on this we find
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that the normal duration of the project is
sixteen days and one two four five six is
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the critical path that is our critical path
runs like this one two four five and six so
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the activities a c e n g are critical so there
is activity a c e and g these are the critical
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activities its not a very difficult network
i am sure you can find out that there are
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other paths but they are not critical so now
as far as the project cost is concerned at
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this point in time assuming that the activities
will take place at their normal durations
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will have the total cost which is the sum
of all the activities that is activities a
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plus b plus c and so on write up to g the
sum of the normal cost will give us the direct
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cost of the project and that in this case
turns out to be sixty thousand inr
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as far as the indirect cost is concerned we
have the information that it is six thousand
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rupees per day and sixteen being our duration
of the project the total indirect cost is
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ninety six thousand and we have the total
cost as the sum of this plus this which is
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hundred and fifty six thousand inr this of
course is this normal cost of the project
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now lets try to move forward and try to do
the crashing business we know that activities
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a c e n g are the critical activities from
the information given the t n and the t c
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as for as the activity g is concerned is both
the same which means that this activity simply
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cannot be crashed it will take that amount
of time as per as other activities are concerned
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a c and e they can be crashed by different
amounts e can be crashed by one day c can
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be crashed by two days and e can be crashed
by three days provided we are willing to put
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in a certain amount of additional resources
and that is where we have to find out the
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cost slope the cost slope for each of these
activities is two thousand four thousand and
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three thousand calculated by dividing the
difference in cost by the difference in time
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so once that is done we know that if we rank
these activities a c and e a has the minimum
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slope then it comes to e followed by c so
since a has the minimum cost slope lets try
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to crash activity a by one day and this requires
an additional cost in terms of direct cost
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which is equal to the cost slope of activity
a but it also results in a saving of one day
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which is equal to the indirect cost of the
project itself
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which means that as far as the revised cost
calculation is concerned the revised project
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cost becomes hundred and fifty six thousand
plus two thousand which is the cost slope
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of activity a minus six thousand which is
the indirect cost for one day and that is
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hundred fifty two thousand which is lower
than hundred and fifty six so basically we
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have a table like this where we say that if
you are completing the project in sixteen
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days which was the normal duration the direct
cost was sixty indirect cost was ninety six
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and the total was hundred and fifty six in
thousands now we have crashed the project
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by changing the duration of activity a two
fifteen in the direct cost and indirect cost
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the sum of this is hundred and fifty two which
is four thousand lower than the previous total
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cost so this obviously is a better solution
but now can we find an even better solution
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is whats the next step but before that we
need to modify our network to check if there
17:20.300 --> 17:25.500
are any changes in the critical activities
so if we redraw this network and say that
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instead of thee this activities been completed
in two days we find that there is no change
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in the critical path in the network which
continues to be having activities a c e and
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g even though the project duration is come
down to fifteen days and the cost has come
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down to hundred and fifty two thousand so
lets go to the next step which we did last
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time that is the cost slope analysis we find
that even though a has the minimum slope but
17:52.830 --> 17:57.750
it cannot be crushed any further because it
has already reached the crash duration so
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our choice is limited to c and e and between
these two obviously we must crash e by one
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day and see what happens to the direct indirect
and the total costs
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so that analysis is shown here given that
the cost slope of activity e is three thousand
18:17.130 --> 18:23.340
the revised project cost becomes hundred and
fifty two plus three minus six thousand and
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we have a lower number thats the hundred forty
nine thousand which means that we revise the
18:30.240 --> 18:35.530
table that for shown last time in this from
now that from sixteen to fifteen and now to
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fourteen the direct cost has increased from
sixty to sixty two and now to sixty five but
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the indirect cost has reduced by six thousand
for everyday of reduction that were able to
18:46.060 --> 18:52.890
achieve and we are now working at hundred
and forty nine the issue is can we do more
18:52.890 --> 18:59.050
lets try to now identify if there are any
changes in the network if you do this we still
18:59.050 --> 19:04.880
find that the critical path is a c and g and
we are free to go ahead and look at the cost
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slope of the critical activities so we find
that we can still crash activities and four
19:11.860 --> 19:19.080
can still become two now this is the revise
network after this revision we have crashed
19:19.080 --> 19:27.160
a from three to two and now we have crashed
e to four days and the project duration has
19:27.160 --> 19:32.810
become fourteen and the cost is hundred forty
nine and if we do the network analysis at
19:32.810 --> 19:37.810
this point in time we still find that the
critical activities are a c e and g so we
19:37.810 --> 19:46.710
go back to this table which gives us the cost
slopes and we try to crash e by one more day
19:46.710 --> 19:51.300
once we do that we come to a total cost of
hundred forty six hundred forty nine was the
19:51.300 --> 19:58.090
previous cost at the cost slope of activity
e and subtract the indirect cost so once we
19:58.090 --> 20:03.870
come to one forty six we go back to this table
and we find that this one forty six is still
20:03.870 --> 20:12.260
lower than the previous iteration so we perhaps
have still some room for doing more crashing
20:12.260 --> 20:17.500
so we come back to this picture here now we
are begin with the project duration of thirteen
20:17.500 --> 20:24.670
and the durations of the activities have already
been adjusted here we now find that the critical
20:24.670 --> 20:35.940
paths are a c e g and a b f g so a a c e g
is not the only critical path there is another
20:35.940 --> 20:43.630
critical path a b f and g so activities b
and f have also now become critical so if
20:43.630 --> 20:48.650
we bring it to thirteen days now if we want
to reduce the project by one more day we have
20:48.650 --> 20:55.040
to ensure that we also account for the cost
slopes in activities b and f and then try
20:55.040 --> 21:00.410
to see if we can get a better cost than a
hundred forty six thousand and order to do
21:00.410 --> 21:06.650
that we come to this table we are we have
now included activities b and f earlier we
21:06.650 --> 21:11.670
are working only we c and
now we are working with b and f of course
21:11.670 --> 21:17.400
a and g are out of the reckoning having reached
the crash durations in the first iteration
21:17.400 --> 21:23.540
itself now if we look at these cost slopes
we cannot really proceed as simply as we did
21:23.540 --> 21:29.240
in the previous case and the reason is the
following there are two critical paths and
21:29.240 --> 21:34.980
the activities along both paths have to be
crashed simultaneously by one day and only
21:34.980 --> 21:40.660
then the total duration of the project will
reduce from thirteen to twelve what i am trying
21:40.660 --> 21:46.830
to tell you is that if you look at this project
the way it is shown there is this path here
21:46.830 --> 21:54.230
which is c and e there is this path here which
is and f it makes no sense to just crash an
21:54.230 --> 22:00.910
activity whether it is c or e based on its
cost slope because then this path here we
22:00.910 --> 22:06.430
still remain critical and we will not get
any reduction in the project time
22:06.430 --> 22:11.280
so in order to get a reduction in project
time we have to find a combination it could
22:11.280 --> 22:18.260
be b and c which is this and this it could
be b and e it could be f and c it could be
22:18.260 --> 22:25.750
f and e so any of these four combinations
if the cost slope that is the additional cost
22:25.750 --> 22:32.250
that is incurred in changing the duration
of the activities by one day the sum of that
22:32.250 --> 22:38.560
has to be considered and thats what has been
done here so b and c gives a total cost slope
22:38.560 --> 22:42.710
of six thousand b and e gets five thousand
five thousand five hundred here and four thousand
22:42.710 --> 22:50.530
five hundred here and this obviously then
ranks highest that is the cost slope of crashing
22:50.530 --> 22:54.500
f and e together is four thousand five hundred
rupees
22:54.500 --> 23:01.390
so if we do that then we will be able to reduce
the project by one day in other words what
23:01.390 --> 23:06.130
we will do is will have this hundred forty
six thousand we will add to it not the cost
23:06.130 --> 23:11.830
slope of an activity but two activities that
is e and f and subtract the indirect cost
23:11.830 --> 23:17.170
for one day and we get a number hundred forty
four thousand five hundred which is still
23:17.170 --> 23:22.340
better than a hundred forty six thousand so
if we look at this table now from sixteen
23:22.340 --> 23:28.610
days which we started we have come down to
twelve days and we have reduced the total
23:28.610 --> 23:33.850
cost from one fifty six to one forty four
point five even though we had to increase
23:33.850 --> 23:40.490
our direct cost from sixty to seventy two
and a half so the network after crashing e
23:40.490 --> 23:46.290
and f by one day looks like this and you would
now like to find out if it is possible to
23:46.290 --> 23:51.230
crash it any further of course the critical
path continue to be having activities a c
23:51.230 --> 23:59.930
e g and a b f g only this activity still continues
to be non critical as far as this project
23:59.930 --> 24:02.800
is concerned
now if you look at this table to carry out
24:02.800 --> 24:11.330
the cost slope calculations we find that activities
a e f and g have all reached their critical
24:11.330 --> 24:18.010
times and therefore the only possibility that
we have for crashing is activities b and c
24:18.010 --> 24:22.640
and since there are two critical paths activities
along both the paths have to be crashed simultaneously
24:22.640 --> 24:28.040
by one day and only then we will get a reduction
from twelve to eleven now keeping this in
24:28.040 --> 24:35.190
mind b and c combination has a total cost
slope of six thousand please note that activities
24:35.190 --> 24:41.740
a e and f cannot be crashed any further and
in that case the revised cost becomes hundred
24:41.740 --> 24:49.620
forty four point five plus six which is the
slope of the crashed activities and minus
24:49.620 --> 24:54.471
six thousand which is the indirect cost for
one day and we find that there is no change
24:54.471 --> 24:59.910
in the cost of the project so what we have
reached now is a plato here hundred forty
24:59.910 --> 25:02.660
four point five to hundred forty four point
five
25:02.660 --> 25:09.160
and even though we have reduced the project
duration by one day in other words we would
25:09.160 --> 25:14.780
still like to crash the project because for
the same cost we are able to complete the
25:14.780 --> 25:20.110
project one day earlier so this is definitely
a better option than the previous one that
25:20.110 --> 25:24.630
is trying to complete the project in twelve
days at the same cost and we basically say
25:24.630 --> 25:29.090
that the project cost is hundred and forty
four point five thousand with these critical
25:29.090 --> 25:34.100
paths please also take note that once we have
reduced the project duration to eleven days
25:34.100 --> 25:41.270
d also has become critical two plus four plus
five is eleven two plus three plus one plus
25:41.270 --> 25:44.560
five is eleven two plus two plus two plus
five is eleven
25:44.560 --> 25:49.320
so all the activities basically as far as
this network is concerned are now critical
25:49.320 --> 25:53.450
activities and there is no float in these
activities we can complete the project in
25:53.450 --> 25:57.970
eleven days at the cost of a hundred forty
four point five thousand but of course we
25:57.970 --> 26:03.060
we should try to examine whether or not we
would like to crash the project further if
26:03.060 --> 26:07.860
the activities can still be carried out in
a shorter time for that we go back to this
26:07.860 --> 26:14.310
table we now find b c and d are the activities
which are candidates for crashing and we find
26:14.310 --> 26:21.130
that yes there is a scope because you we are
still working with three two and four s t
26:21.130 --> 26:29.490
p for the activities b c and d which is higher
than there crash durations and these are our
26:29.490 --> 26:35.220
cost slopes with this information we try to
find out what is the actual cost in terms
26:35.220 --> 26:39.710
of crashing the network by one day which would
involve all the paths
26:39.710 --> 26:46.060
now again the discussion that the activities
along all the paths have to be crashed simultaneously
26:46.060 --> 26:51.350
and only then that duration of the project
can be reduced from eleven to ten this takes
26:51.350 --> 27:00.810
us to the discussion that what are the possibilities
b c and d have all to be crashed by one day
27:00.810 --> 27:05.790
and only then we will be able to get the project
to ten days what is the cost of crashing all
27:05.790 --> 27:11.460
these three activities the cost is ten thousand
now obviously this ten thousand is greater
27:11.460 --> 27:17.510
than the saving of six thousand and therefore
we expect the cost be higher lets do that
27:17.510 --> 27:22.180
and formally close this discussion hundred
and forty four point five plus ten thousand
27:22.180 --> 27:28.090
minus six thousand gives you a higher total
cost then what you had when you are completing
27:28.090 --> 27:30.250
the project in eleven days
27:30.250 --> 27:35.380
so this is a project cost in eleven days this
the project cost in ten days so if you put
27:35.380 --> 27:40.080
it back in a table we find that a hundred
forty four point five has now become a hundred
27:40.080 --> 27:47.220
forty eight point five and therefore it is
a case that crashing may be stopped at this
27:47.220 --> 27:52.690
point because we have already reached the
optimum we have now a started increasing the
27:52.690 --> 27:59.490
cost of the project even though the activities
are still crash able that is theoretically
27:59.490 --> 28:06.150
if we want we can complete the activities
in a shorter duration then what is being planned
28:06.150 --> 28:13.060
but that will involve cost to the extent that
the total cost of project will increase we
28:13.060 --> 28:18.740
must also remember that in this model what
we have done is we have assumed that the cost
28:18.740 --> 28:23.390
slope is constant from one day to another
it is always a possibility that there is an
28:23.390 --> 28:30.210
activity which from six days to five days
cost some x but if you try to crash it from
28:30.210 --> 28:36.590
five to four it might cost two x this is the
kind of thing which we have not done in this
28:36.590 --> 28:39.050
discussion
what we have assumed it i that whether you
28:39.050 --> 28:43.490
are crashing the activity from six to five
or five to four of possibly four to three
28:43.490 --> 28:50.430
the crash cost is the same or the cost slope
is still the same that is something which
28:50.430 --> 28:56.030
we can do as an exercise try to change the
cost slope and see how it works but the algorithm
28:56.030 --> 29:01.610
that has been laid out that is we are trying
to understand re calculate the cost slopes
29:01.610 --> 29:06.280
that enables us to take care of any change
in the cost slopes should that information
29:06.280 --> 29:11.990
be made available to us now with this here
is the summary of a our discussion the projects
29:11.990 --> 29:16.110
shall be completed in eleven days with the
cost of hundred forty four point five and
29:16.110 --> 29:21.940
the benefit has been that we have been able
to crash from one fifty six to one forty four
29:21.940 --> 29:27.180
point five and reduce the duration by five
days we shall we started with sixteen and
29:27.180 --> 29:29.960
now we are able to complete the project in
eleven
29:29.960 --> 29:35.210
so its a win win situation its a shorter project
duration and a cheaper cost so in addition
29:35.210 --> 29:40.430
to the cost and time recoveries the early
completion of the project has also got the
29:40.430 --> 29:45.600
following benefits recovery of deal is early
availability of resources for other project
29:45.600 --> 29:50.230
receive early completion bonuses if thats
applicable for a project avoidance of seasonal
29:50.230 --> 29:55.570
issues which may affect productivity at a
later point in time improve project cash flows
29:55.570 --> 30:01.460
so with this we come to an end of our discussion
today and as usual this is the list of references
30:01.460 --> 30:05.850
which might help you understand the subject
a little better and i look forward to seeing
30:05.850 --> 30:13.190
you at another lecture
thank you