WEBVTT
Kind: captions
Language: en
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namaskar and welcome to this course on accounting
and finance for civil engineers once again
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and this as you know is being hosted together
by iit kanpur and iit delhi
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so as far as the lecture today is concerned
this is the second in the series of economic
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decision making and in the last class we have
seen some of the methods used for this process
00:00:35.820 --> 00:00:41.580
as far as construction projects are concerned
so we will continue the discussion today and
00:00:41.580 --> 00:00:46.350
look at some of the methods that were remaining
from the list that i had shown you
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so you would recall that this was the list
that were shown to you out of pocket expenses
00:00:50.360 --> 00:00:55.680
and annual rate of return payback period and
so on right down to the incremental rate of
00:00:55.680 --> 00:01:01.280
return method and we have said that these
three methods do not consider the time value
00:01:01.280 --> 00:01:06.030
of money whereas these methods here they consider
the time value of money also
00:01:06.030 --> 00:01:13.030
we have continued our discussion and had completed
these four methods and what we were doing
00:01:13.030 --> 00:01:17.920
at that time was a part of this discussion
on present worth comparisons
00:01:17.920 --> 00:01:23.780
so let us resume our discussion from there
we were talking of npv which is the net present
00:01:23.780 --> 00:01:30.490
value and that is the sum of all cash flows
discounted to the present using the time value
00:01:30.490 --> 00:01:37.689
of money concept so what it effectively means
is that we have these different cash flows
00:01:37.689 --> 00:01:44.899
at different points in time and we want to
bring all of them back to time t is equal
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to zero and at the end of it sum this whole
thing if there are more than one cash flows
00:01:52.539 --> 00:01:59.539
so that sum is what is npv so mathematically
speaking we had looked at this formula that
00:01:59.539 --> 00:02:05.899
if we get all the cft’s at different points
in time we can use this one plus k to the
00:02:05.899 --> 00:02:13.960
power of t and we will get npv in this equation
cft is the cash flow at year t or time t n
00:02:13.960 --> 00:02:18.650
is the life of the project and k is the rate
of interest we have been using i but we could
00:02:18.650 --> 00:02:26.680
use k or any other variable and this basically
is an adaptation of the single payment present
00:02:26.680 --> 00:02:27.680
work factor
00:02:27.680 --> 00:02:33.250
so you can see that a single payment here
is being transferred to this point and that
00:02:33.250 --> 00:02:39.239
is what we had talked about when we had talked
of sppwf that is the single payment present
00:02:39.239 --> 00:02:46.989
work factor now what we are doing in addition
to that is just creating this sum of the different
00:02:46.989 --> 00:02:52.909
single payments so that is what is the concept
of npv and we have gone over it in the last
00:02:52.909 --> 00:02:54.799
class also
00:02:54.799 --> 00:02:59.270
so with this you would also recall this particular
chart where we had said that the present work
00:02:59.270 --> 00:03:04.390
concepts can be used for type one and type
two and type three kind of situations where
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type i deals with alternatives of equal lives
and type two deals with alternatives of unequal
00:03:10.370 --> 00:03:16.629
lives and alternatives with infinite lives
so now let us resume our discussion for alternatives
00:03:16.629 --> 00:03:19.500
with unequal lives
00:03:19.500 --> 00:03:24.880
in most real situations that is what it is
so alternatives may or may not have equal
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lives and most of the time you will find that
they do not have equal lives one equipment
00:03:29.629 --> 00:03:34.709
may last five years and other equipment may
last seven years so the two commonly used
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approaches when evaluating options with unequal
lives are the common multiple method and a
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study period method now what is the difference
between these two
00:03:46.819 --> 00:03:52.450
as far as the common multiple method is concerned
what we really do is re-evaluate the alternatives
00:03:52.450 --> 00:03:59.950
at the least common multiple of life periods
of alternatives as the coterminous life period
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for example if alternatives have a life period
of two and three years then they would be
00:04:05.269 --> 00:04:11.239
put to use for a period which is equal to
the lcm of these life periods and therefore
00:04:11.239 --> 00:04:18.100
six years which means that the alternatives
with two year life periods shall be replaced
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three times and the one with three years of
life will be replaced twice
00:04:24.410 --> 00:04:28.770
it is assumed that the alternatives shall
be replaced after the service life with the
00:04:28.770 --> 00:04:35.300
same cost characteristics now this is a simplifying
assumption what is being basically said is
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that if we have an alternative which has a
service life of two years and another alternative
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which has a service life of three years one
two and three here and one and two here now
00:04:49.750 --> 00:04:55.590
what is being said is that if we go to a period
of six years then we would need to replace
00:04:55.590 --> 00:05:00.110
this equipment here go for another two years
here replace this here and then we will come
00:05:00.110 --> 00:05:01.290
to six
00:05:01.290 --> 00:05:06.760
whereas in this case if you replace this equipment
here and then we go to a cycle we will come
00:05:06.760 --> 00:05:12.720
to six so what is being assumed as far as
this last assumption is concerned that this
00:05:12.720 --> 00:05:18.190
replacement here and this replacement here
which is twice in the case of let us say alternative
00:05:18.190 --> 00:05:25.090
a and once in the case of alternative b that
will happen at the same cost characteristics
00:05:25.090 --> 00:05:30.250
as the original cost that is the original
cost here is what we are taking here
00:05:30.250 --> 00:05:35.280
please remember that it is not really necessary
to do that we can take any value that we want
00:05:35.280 --> 00:05:40.830
and so long as we are making the right kind
of corrections for that we can still make
00:05:40.830 --> 00:05:47.270
that comparison but in this particular method
that we are using we are not doing that so
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it is a simplifying assumption and under that
simplifying assumption we will go ahead and
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show illustrative calculations
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now coming to an example let assets a one
and a two have the capability of satisfactory
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performing the required function various details
of costs revenues and service lives are given
00:06:04.250 --> 00:06:10.470
in the table and if the required rate of return
is fifteen percent which as it is preferable
00:06:10.470 --> 00:06:17.310
using the common multiple method so now this
is the details of the two assets a one and
00:06:17.310 --> 00:06:18.310
a two
00:06:18.310 --> 00:06:21.940
so we have the initial cost which is a hundred
and sixty thousand and a hundred and fifteen
00:06:21.940 --> 00:06:26.430
thousand expected salvage value is twenty
thousand and we do not expect any salvage
00:06:26.430 --> 00:06:31.690
value here the service life let us say four
years and three years and the annual operating
00:06:31.690 --> 00:06:36.840
cost is zero here and twelve thousand five
hundred here so how do we go about doing this
00:06:36.840 --> 00:06:43.280
example let us try to calculate the npv that
is what our brief is
00:06:43.280 --> 00:06:47.470
that we want to calculate the npv of both
these options and they will be evaluated at
00:06:47.470 --> 00:06:53.710
a period of twelve years which is the lcm
of three and four so for the alternative a
00:06:53.710 --> 00:07:00.530
one this is our cash flow diagram so this
is what the initial investment is and this
00:07:00.530 --> 00:07:06.530
has a life of one two three four years and
again we will invest one lakh sixty thousand
00:07:06.530 --> 00:07:12.060
or hundred and sixty thousand go for another
four years here invest here and go for this
00:07:12.060 --> 00:07:16.350
and come to the coterminous value of twelve
years
00:07:16.350 --> 00:07:23.030
while we are doing this we remember that whereas
we will invest or will be required to invest
00:07:23.030 --> 00:07:28.400
hundred and sixty thousand in a new equipment
there will be a salvage value of twenty thousand
00:07:28.400 --> 00:07:35.610
as far as the old equipment is concerned and
please remember that there is no operation
00:07:35.610 --> 00:07:42.550
cost as for as alternative a one is concerned
so with this detail if you want to calculate
00:07:42.550 --> 00:07:48.170
the npv this is the formula that we discussed
in detail and we know how to do this
00:07:48.170 --> 00:07:55.440
so we transfer everything all these values
here to the present value which is this is
00:07:55.440 --> 00:08:00.320
my present value anyway this is all negative
so this hundred and sixty and hundred and
00:08:00.320 --> 00:08:05.280
sixty are both negative this is being charged
at fifteen percent here so one point one five
00:08:05.280 --> 00:08:09.770
to the power of four which is four years fh
in this case and eight years in this case
00:08:09.770 --> 00:08:16.210
so this becomes here these three terms reflect
what is the present value of this salvage
00:08:16.210 --> 00:08:17.210
value
00:08:17.210 --> 00:08:20.821
so one is to the power of four the other is
to the power of eight and third one is to
00:08:20.821 --> 00:08:29.390
the power of twelve so if we do this arithmetic
we find that this is our answer so basically
00:08:29.390 --> 00:08:36.050
as far as npv of a one is concerned which
has an initial investment of one sixty thousand
00:08:36.050 --> 00:08:42.880
salvage value of twenty thousand service life
of four years if we do this exercise thrice
00:08:42.880 --> 00:08:50.170
that is we get to twelve years then the npv
is two hundred and eighty two thousand seventy
00:08:50.170 --> 00:08:57.180
four rupees now let us try to do the same
exercise for asset a two and that is much
00:08:57.180 --> 00:08:58.850
simpler having done once
00:08:58.850 --> 00:09:02.750
we have an initial investment of one fifteen
thousand and this has to be repeated four
00:09:02.750 --> 00:09:09.079
times because we want to come to a co minus
terminus period of twelve years and this has
00:09:09.079 --> 00:09:13.910
only a service life of three years so every
three years we will have to replace the equipment
00:09:13.910 --> 00:09:19.819
and in this case there is a operating cost
of twelve thousand five hundred every year
00:09:19.819 --> 00:09:24.189
so what we need to do is to get all these
values that is this hundred fifteen thousand
00:09:24.189 --> 00:09:30.220
here and this twelve thousand five hundred
here back to the present value here
00:09:30.220 --> 00:09:35.050
please also remember that there is no salvage
value involved here so there are no plus term
00:09:35.050 --> 00:09:43.850
that we will get and as far as the formulae
that we use are concerned one will be a single
00:09:43.850 --> 00:09:50.560
payment present worth factor which could be
used to reduce these three values to the present
00:09:50.560 --> 00:09:55.610
worth which has been done here one point one
five to the power of three six and nine and
00:09:55.610 --> 00:10:01.500
here we will use equal payment present worth
factor which is basically the twelve thousand
00:10:01.500 --> 00:10:07.550
five hundred being spent for the entire period
of twelve years
00:10:07.550 --> 00:10:11.730
so this is the value here which is basically
the formula one plus i to the power of n minus
00:10:11.730 --> 00:10:18.220
one this is the formula we are using to reduce
this value to the present value and if you
00:10:18.220 --> 00:10:22.980
do this calculation with find that this turns
out to be three hundred and forty thousand
00:10:22.980 --> 00:10:29.499
seven seventy nine point seven so what really
happens is that since a one will be the chosen
00:10:29.499 --> 00:10:36.949
alternative because it is higher as far as
the npv is concerned than a two
00:10:36.949 --> 00:10:42.819
it is higher because it is less negative so
that is something which you must keep in mind
00:10:42.819 --> 00:10:48.180
because we were talking about negative values
so we are talking about negative values here
00:10:48.180 --> 00:10:54.850
the lesser negative value is actually more
harmful to us it is less profitable to us
00:10:54.850 --> 00:11:00.519
so in this case this is a more negative value
less profitable therefore choose alternative
00:11:00.519 --> 00:11:01.519
a one
00:11:01.519 --> 00:11:08.779
now let us come to the study period method
so instead of now trying to do the lcm we
00:11:08.779 --> 00:11:14.620
will try to identify a study period and then
try to do similar analysis so in this method
00:11:14.620 --> 00:11:19.439
a study period is chosen on the basis of the
length of the project the period is chosen
00:11:19.439 --> 00:11:25.139
in advance and only the cash flow within the
timespan are considered relevant the shortest
00:11:25.139 --> 00:11:29.100
life of the alternatives may also be chosen
as the study period
00:11:29.100 --> 00:11:35.579
so there are different options that we have
and i said right at the outset of this course
00:11:35.579 --> 00:11:41.110
that in several places we will talk about
what is the convention what are the options
00:11:41.110 --> 00:11:48.959
that are available to us and then somebody
as an engineer or as an accountant or as professional
00:11:48.959 --> 00:11:54.560
basically has to take a decision based on
the input available to him it is not really
00:11:54.560 --> 00:11:58.949
a matter of being absolutely wrong and absolutely
right
00:11:58.949 --> 00:12:05.899
there are cases where different methods may
give you different alternatives as profitable
00:12:05.899 --> 00:12:11.459
or advisable so we have to take a call and
try to say that well this method tells me
00:12:11.459 --> 00:12:17.130
this that method tells me something else now
what is really applicable in our case so that
00:12:17.130 --> 00:12:22.130
is what is the responsibility of an engineer
or a professional as far as construction management
00:12:22.130 --> 00:12:23.290
is concerned
00:12:23.290 --> 00:12:27.399
so moving forward to an illustrative example
which is basically the same example that we
00:12:27.399 --> 00:12:34.450
did just now using the lcm method so here
also the table is exactly the same and we
00:12:34.450 --> 00:12:39.149
are considering the study period to be three
years
00:12:39.149 --> 00:12:44.800
so if we consider the study period to be three
years in that case the relevant cash flow
00:12:44.800 --> 00:12:49.709
diagram as far as a one is concerned is just
as much there is no operational expenditures
00:12:49.709 --> 00:12:55.960
there is no salvage value because the service
life of this equipment is four years so and
00:12:55.960 --> 00:13:03.689
we have said that only the cash flows or transactions
or whatever it is happening within the study
00:13:03.689 --> 00:13:08.300
life is being considered so since the study
life is only three years this is all that
00:13:08.300 --> 00:13:11.819
matters as far as we are concerned for alternative
a one
00:13:11.819 --> 00:13:18.180
so the present worth of this alternative obviously
simply hundred sixty thousand and when it
00:13:18.180 --> 00:13:26.569
comes to asset a two the cash flow diagram
is one fifteen thousand and an operating expense
00:13:26.569 --> 00:13:31.519
of twelve thousand five hundred every year
so this gives me a net present worth of one
00:13:31.519 --> 00:13:40.100
fifteen thousand and the same old formula
for eppwf and we get the net present worth
00:13:40.100 --> 00:13:43.400
of minus hundred and forty four thousand five
forty eight
00:13:43.400 --> 00:13:51.110
so now what happens alternative a two should
be chosen as its npv is higher than that of
00:13:51.110 --> 00:13:58.360
a one note that the result in this case is
different than that obtained from the common
00:13:58.360 --> 00:14:03.029
multiple method and that is exactly what i
had mentioned just now that these different
00:14:03.029 --> 00:14:09.600
methods may lead to different recommendations
because each of them have inherent limitations
00:14:09.600 --> 00:14:10.829
and assumptions
00:14:10.829 --> 00:14:18.490
so these methods just give you different tools
which as an engineer or as an professional
00:14:18.490 --> 00:14:23.610
you have to use with judgement and that is
exactly what is written here that though this
00:14:23.610 --> 00:14:29.399
may not happen every time the example is illustrative
of the idea that depending on the evaluating
00:14:29.399 --> 00:14:34.990
method different results may be obtained for
the same set of alternatives or the same set
00:14:34.990 --> 00:14:42.420
of data and professional judgement is required
when making a final decision
00:14:42.420 --> 00:14:48.879
moving forward we go to alternatives with
infinite lives so the alternative with infinite
00:14:48.879 --> 00:14:55.410
lives are evaluated using what is called the
capitalised equivalent method that is the
00:14:55.410 --> 00:15:02.379
ce method now the ce or the capitalised equivalent
is a single amount determined at time zero
00:15:02.379 --> 00:15:07.730
which at a given rate of interest will be
equivalent to the net difference of receipts
00:15:07.730 --> 00:15:12.369
and disbursements if the given cash flow pattern
is repeated in perpetuity
00:15:12.369 --> 00:15:18.740
so basically let us go back and try to see
what we have already done what we had done
00:15:18.740 --> 00:15:26.059
when we consider the eppwf that is equal payment
present worth factor we were trying to find
00:15:26.059 --> 00:15:35.759
out what is the p if this a which is an instalment
kind of a payment being repeated all the time
00:15:35.759 --> 00:15:41.051
for a long period of time in perpetuity then
what we had said was that this is the formula
00:15:41.051 --> 00:15:43.899
that can be used or this is the factor that
can be used
00:15:43.899 --> 00:15:51.310
but please remember that if the n becomes
very large then this factor simply reduces
00:15:51.310 --> 00:15:56.959
to one upon i now how that happens i am going
to show you i would also like to draw your
00:15:56.959 --> 00:16:02.769
attention to the fact that we use this formula
just now when we were trying to determine
00:16:02.769 --> 00:16:09.600
the npv for operating costs twelve thousand
five hundred or whatever that number was incurred
00:16:09.600 --> 00:16:14.269
every year for a period of three years for
a period of six years and so on and we try
00:16:14.269 --> 00:16:16.480
to use precisely this formula
00:16:16.480 --> 00:16:23.040
so now this is something which we must clearly
understand so let us try to go back to the
00:16:23.040 --> 00:16:29.769
basics and try to understand that if we are
considering the whole concept of interest
00:16:29.769 --> 00:16:38.589
rate and a period which is n years then the
first thing that comes to mind is that what
00:16:38.589 --> 00:16:48.509
happens if i have a unit value here how much
will it grow to after a period of n years
00:16:48.509 --> 00:16:54.449
if the rate of interest applicable is i that
we know from the compounding interest formula
00:16:54.449 --> 00:16:57.589
that this will be nothing but one plus i to
the power of n
00:16:57.589 --> 00:17:05.319
so basically if this was one this value will
be higher than one that is how we are normally
00:17:05.319 --> 00:17:11.030
assuming given the fact that i is positive
so if i is positive n is positive then this
00:17:11.030 --> 00:17:16.569
one will grow to so much if we have a different
value than one we have a hundred we know that
00:17:16.569 --> 00:17:21.680
this has to be simply multiplied by hundred
so this is basically just the factor that
00:17:21.680 --> 00:17:30.450
we have already discussed earlier the converse
of this problem if we have one here at the
00:17:30.450 --> 00:17:38.970
end of n years with a rate of interest applicable
being high then what is the present worth
00:17:38.970 --> 00:17:40.720
of this one
00:17:40.720 --> 00:17:45.700
this is precisely what we have been talking
about as the single payment present worth
00:17:45.700 --> 00:17:50.620
factor and if you look at the tables those
will tell you that this value will obviously
00:17:50.620 --> 00:17:57.450
be always less than one so if you want a certain
value here that can be simply multiplied by
00:17:57.450 --> 00:18:05.280
that factor and we get the present worth now
coming to the eppwd kind of model this is
00:18:05.280 --> 00:18:11.620
what we are doing so if this was unit what
we are calling a
00:18:11.620 --> 00:18:20.020
if a was a unit quantity and this was happening
n times at a rate of interest of i then the
00:18:20.020 --> 00:18:26.310
present worth factor is one plus i times to
the power of n minus one upon i times one
00:18:26.310 --> 00:18:33.960
plus i to the power of n now this is how these
factors are coming about we have to understand
00:18:33.960 --> 00:18:41.660
what is actually happening now if simple arithmetic
tells us that if we break this up into one
00:18:41.660 --> 00:18:47.020
plus i to the power of n minus one and this
is going to be here at both sides one plus
00:18:47.020 --> 00:18:51.830
i to the power of n here and i times one plus
i to the power of n here
00:18:51.830 --> 00:18:56.800
if n tends to be infinity very large then
the denominator goes to a large number and
00:18:56.800 --> 00:19:01.381
this term goes to zero and in this case tha
fa as far as the first term is concerned they
00:19:01.381 --> 00:19:06.980
cancel out and your factor simply becomes
one upon i and this is what we had mentioned
00:19:06.980 --> 00:19:12.590
in the previous slide so it simply becomes
one upon i let us try to do a simple simulation
00:19:12.590 --> 00:19:22.420
and try to find out that if my a was one the
rate of interest was ten and the n was also
00:19:22.420 --> 00:19:23.420
ten
00:19:23.420 --> 00:19:29.820
we try to find out what is the present worth
or the present worth factor as far as this
00:19:29.820 --> 00:19:35.130
is concerned if we use the actual formula
we will find that this is nothing but one
00:19:35.130 --> 00:19:43.180
plus one to the power of ten minus one divided
by point one times one point one to the power
00:19:43.180 --> 00:19:50.930
of ten and this will turn out to be six point
one three eight now if we were to use the
00:19:50.930 --> 00:19:58.920
formula that it is one upon i then n is equal
to ten then one upon i is actually ten
00:19:58.920 --> 00:20:05.340
so we have this is the difference between
the rigorous analysis and the approximate
00:20:05.340 --> 00:20:11.410
analysis when we taken n is equal to ten and
taken it to be large number so now instead
00:20:11.410 --> 00:20:17.340
of ten if this n was to be made let us say
twenty then what happens the numbers become
00:20:17.340 --> 00:20:23.220
one point one to the power of twenty minus
one upon point one into one point one to the
00:20:23.220 --> 00:20:28.860
power of twenty and then the number turns
out to be eight point one five four
00:20:28.860 --> 00:20:33.590
you can do this arithmetic and check and you
can see that this is becoming closer to this
00:20:33.590 --> 00:20:39.290
value now instead of twenty if we take this
number to be forty so we see that we are gradually
00:20:39.290 --> 00:20:45.430
increasing this and to reach a large number
then what happens is that this turns out to
00:20:45.430 --> 00:20:50.400
be one point one to the power of forty minus
one divide by point one times one point one
00:20:50.400 --> 00:20:56.070
to the power of forty and this will turn out
to be nine point seven seven
00:20:56.070 --> 00:21:03.860
so gradually as n is being increased we find
that the final answer as far as this is concerned
00:21:03.860 --> 00:21:09.710
is indeed tending towards this ten value and
that is what simple calculus tells us this
00:21:09.710 --> 00:21:16.630
is what this particular term or this breakup
of the terms tells us so just keep that in
00:21:16.630 --> 00:21:24.010
mind that these formulae have their own limitations
if you want to use simple calculations they
00:21:24.010 --> 00:21:30.950
will always be approximate but yes we can
always use them as first guesses so now having
00:21:30.950 --> 00:21:34.290
done that let us try to look at an example
and try to see how it works
00:21:34.290 --> 00:21:42.060
let us consider design a of a check dam which
costs fifty lakhs to construct and will cost
00:21:42.060 --> 00:21:47.440
rupees seven point five lakhs per year to
operate and maintain whereas design b costs
00:21:47.440 --> 00:21:52.570
seventy five lakhs to build and will require
five lakhs per year to operate and maintain
00:21:52.570 --> 00:21:56.660
now both installations may be considered to
be as well as permanent and if the minimum
00:21:56.660 --> 00:22:01.900
required rate of return is five percent which
alternative should be preferred so let us
00:22:01.900 --> 00:22:04.340
see how it works
00:22:04.340 --> 00:22:09.630
this basically is the cash flow diagram as
far as design a is concerned some kind of
00:22:09.630 --> 00:22:14.110
cost in the beginning fifty lakhs and then
seven lakhs fifty thousand seven and a half
00:22:14.110 --> 00:22:24.460
lakhs going on forever cash flow diagram for
b is seventy five lakhs and only five lakhs
00:22:24.460 --> 00:22:31.870
going on forever now how do we compare the
two npv of design a is fifty lakhs
00:22:31.870 --> 00:22:37.780
and this seven point five divided by point
zero five which is basically the factor of
00:22:37.780 --> 00:22:45.880
one over i and we get the npv is two crores
that is two hundred lakhs as far as design
00:22:45.880 --> 00:22:50.750
b is concerned you do the same approach and
we get the npv to be one point seven five
00:22:50.750 --> 00:22:59.030
crores so the npv of design b is higher than
that of design a and hence select design b
00:22:59.030 --> 00:23:04.750
so that is how it works when we are talking
of alternatives which have a long life or
00:23:04.750 --> 00:23:06.420
an infinite life
00:23:06.420 --> 00:23:11.720
now coming to a very different discussion
let us try to understand and look more closely
00:23:11.720 --> 00:23:17.360
at the relationship between npv and the rate
of interest
00:23:17.360 --> 00:23:22.951
an illustrative example which will make things
clear if we were looking at a cash flow diagram
00:23:22.951 --> 00:23:27.260
like this you would recall that this is coming
from one of the previous examples that we
00:23:27.260 --> 00:23:32.960
have already done now calculate the npv of
the cash flow at an interest of fifteen percent
00:23:32.960 --> 00:23:39.860
throughout now if we were to do that this
is our cash flow diagram from the table
00:23:39.860 --> 00:23:44.711
our two hundred coming from here this is my
two hundred here and this thirty five is all
00:23:44.711 --> 00:23:51.270
over the place here and this thirty seven
is coming here we cannot use the equal payment
00:23:51.270 --> 00:23:56.320
present worth factor because of this thirty
seven sitting here instead of thirty five
00:23:56.320 --> 00:23:57.970
so we can do any which way we want
00:23:57.970 --> 00:24:03.300
and so long as we do this npv calculation
at fifteen percent what will happen is that
00:24:03.300 --> 00:24:08.530
the present factors you have to calculate
or see from the table and come to the present
00:24:08.530 --> 00:24:13.480
values which we have already done in the previous
example when we were talking earlier and we
00:24:13.480 --> 00:24:21.750
will find that the net present worth is minus
forty three thousand six hundred and ten
00:24:21.750 --> 00:24:26.860
if we would repeat this exercise now with
an interest rate of ten percent the whole
00:24:26.860 --> 00:24:32.840
process remains the same and we will find
that the net present worth has now become
00:24:32.840 --> 00:24:37.840
minus twelve thousand three eighty different
from minus forty three thousand six hundred
00:24:37.840 --> 00:24:44.340
when we use the interest rate of fifteen percent
so by changing the interest rate from fifteen
00:24:44.340 --> 00:24:51.530
to ten the npv has gone from minus forty three
six hundred to minus twelve hundred three
00:24:51.530 --> 00:24:54.510
eighty with the kind of data that we had
00:24:54.510 --> 00:24:59.590
now if we were to further to do this exercise
once again at five percent then what we will
00:24:59.590 --> 00:25:05.410
find is that the present value has gone to
a plus value that is plus two thousand one
00:25:05.410 --> 00:25:14.260
seventy and this clearly shows that the npv
is very strongly related to the rate of interest
00:25:14.260 --> 00:25:21.660
that we use and depending on the interest
rate the values could be plus or minus
00:25:21.660 --> 00:25:28.690
so obviously the interest rate at which the
npv becomes zero is known as the internal
00:25:28.690 --> 00:25:34.950
rate of return now this is a concept which
is used very often in construction economics
00:25:34.950 --> 00:25:40.450
and decision making and that is why we must
be very clear about this the internal rate
00:25:40.450 --> 00:25:46.400
of return is that rate of interest at which
the npv is zero
00:25:46.400 --> 00:25:52.260
geometrically speaking this is what it is
so you remember that at fifteen percent our
00:25:52.260 --> 00:25:59.760
npv was minus forty three at ten percent it
was minus twelve at five percent it became
00:25:59.760 --> 00:26:08.511
plus two something of that nature so basically
as we reduced the interest rate the npv moved
00:26:08.511 --> 00:26:11.940
in the positive direction so there is a negative
relation between these two
00:26:11.940 --> 00:26:18.450
iss the npv if so if we plot the npv versus
the discount rate or the interest rate there
00:26:18.450 --> 00:26:26.960
will be a point where the npv will be zero
and that is precisely what is called the internal
00:26:26.960 --> 00:26:32.960
rate of return so i would encourage you to
do more problems by varying the interest rates
00:26:32.960 --> 00:26:37.660
and try to determine the irr’s for those
cases as well
00:26:37.660 --> 00:26:43.380
so more formally irr is the discount rate
at which the npv of the cash flow is zero
00:26:43.380 --> 00:26:48.860
and this is what we have already done and
how do we calculate the npv we already know
00:26:48.860 --> 00:26:54.130
that this is how we are going to calculate
the npv except that in npv we had said that
00:26:54.130 --> 00:26:58.070
we will take this term here to be i and calculate
the npv
00:26:58.070 --> 00:27:05.440
the zero was not coming to the picture but
if this i gets replaced by the irr then the
00:27:05.440 --> 00:27:11.950
npv is zero so that is precisely what the
definition is and this is calculated using
00:27:11.950 --> 00:27:18.770
a trial and error procedure obviously an alternative
with a higher irr is preferable as far as
00:27:18.770 --> 00:27:20.390
a company is concerned
00:27:20.390 --> 00:27:24.150
now beginning this point there is a four or
five step procedure which tells you how to
00:27:24.150 --> 00:27:29.100
calculate the irr let us quickly go through
that in principle the irr can be determined
00:27:29.100 --> 00:27:33.590
by equating the net present worth of the cash
flow to zero that is setting the difference
00:27:33.590 --> 00:27:38.940
of the benefits and cost of the present worth
to zero as shown in this equation here and
00:27:38.940 --> 00:27:45.220
the first step is to assume a trial rate of
return at i percent counting the cost to be
00:27:45.220 --> 00:27:50.220
negative and the income to be positive find
the equivalent present worth of all costs
00:27:50.220 --> 00:27:52.220
and income
00:27:52.220 --> 00:27:57.490
the third step would be if the equivalent
net worth is positive then the income from
00:27:57.490 --> 00:28:01.880
the investment is worth more than the cost
of the investment and the actual percentage
00:28:01.880 --> 00:28:09.040
of interest or return is higher than the trial
rate and vice versa so adjust the rate according
00:28:09.040 --> 00:28:15.790
to the result that you get in step three and
go back to step two until one value of i is
00:28:15.790 --> 00:28:21.980
found that results in a positive equivalent
net worth and the other higher value of i
00:28:21.980 --> 00:28:26.860
is found with a negative equivalent net worth
and solve by interpolation
00:28:26.860 --> 00:28:31.220
so that is what we had actually done from
the previous example we had fifteen then ten
00:28:31.220 --> 00:28:37.610
and then five and we found that when we transition
from ten to five there was the transition
00:28:37.610 --> 00:28:42.310
from a negative value to a positive value
so we obviously know that the irr is somewhere
00:28:42.310 --> 00:28:48.702
between five and ten we can do a closer iteration
once again and determine the actual irr which
00:28:48.702 --> 00:28:54.720
i am not really doing as far as this particular
class is concerned but we will do a very simple
00:28:54.720 --> 00:28:59.000
illustrative example of calculating the irr
with this cash flow shown here
00:28:59.000 --> 00:29:05.640
there is a minus hundred to begin with and
then plus twenty thirty twenty forty and forty
00:29:05.640 --> 00:29:07.910
which is represented like this
00:29:07.910 --> 00:29:15.460
and now if we try to calculate the npv at
an i of ten percent we find that this plus
00:29:15.460 --> 00:29:22.070
ten point one six if we increase the rate
of interest to be fifteen percent we find
00:29:22.070 --> 00:29:26.910
that it is minus four point zero two and if
you are happy with that we can do a linear
00:29:26.910 --> 00:29:33.960
interpolation and find that the irr is between
ten and fifteen thirteen point one five percent
00:29:33.960 --> 00:29:39.710
if you want to be more accurate than doing
linear interpolation between these two numbers
00:29:39.710 --> 00:29:44.790
sure enough go ahead and try to do it with
twelve percent fourteen percent and then try
00:29:44.790 --> 00:29:51.010
to get a number which will possibly be slightly
better approximation of the irr now this completes
00:29:51.010 --> 00:29:57.800
my discussion as far as irr is concerned the
internal rate of return is concerned let us
00:29:57.800 --> 00:30:04.530
move further and try to discuss a little bit
about future worth comparison
00:30:04.530 --> 00:30:10.110
so far what we had done was the present worth
that is all expenses were being brought to
00:30:10.110 --> 00:30:17.300
a present worth discounting it using the concept
of interest rate and then coming up with a
00:30:17.300 --> 00:30:21.460
present worth trying to do the analysis based
on present worth what is the present worth
00:30:21.460 --> 00:30:27.260
positive negative less negative more po negative
determining the irr and so on
00:30:27.260 --> 00:30:35.010
but now what we will do is do the future worth
the future worth is calculated for each component
00:30:35.010 --> 00:30:42.300
once again and frequently used in cases when
the owner wants to estimate the net worth
00:30:42.300 --> 00:30:48.590
at some future date such as planning for retirement
and so on and comparison and evaluation seems
00:30:48.590 --> 00:30:54.720
to be more meaningful as it provides some
insight into the future receipts
00:30:54.720 --> 00:30:59.490
as an illustrative example if you look at
the data that is given here the owner of a
00:30:59.490 --> 00:31:06.110
plant at x decides to set up a new plant at
another place y and is considering two alternatives
00:31:06.110 --> 00:31:11.910
either to construct a new plant or remodel
an old facility which is available either
00:31:11.910 --> 00:31:16.970
way the company will be able to start its
operations only after three years the timing
00:31:16.970 --> 00:31:22.050
and costs of various components in the two
options are as given in the table shown here
00:31:22.050 --> 00:31:26.340
if the interest rate is taken to be eight
percent which alternative setting up the new
00:31:26.340 --> 00:31:32.890
plant or remodelling an existing plant result
in an lower equivalent cost when the firm
00:31:32.890 --> 00:31:38.750
begins production at the end of the third
year so we will try to do this with the future
00:31:38.750 --> 00:31:43.480
worth concept so this is the data which is
given to us
00:31:43.480 --> 00:31:48.730
as far as constructing a new plant is concerned
there is buying the land design and the initial
00:31:48.730 --> 00:31:52.610
construction cost and the main construction
cost and the setting up of equipment when
00:31:52.610 --> 00:31:57.550
it comes to remodelling purchase of factory
initial remodelling the main remodelling and
00:31:57.550 --> 00:32:05.480
construction costs and setting up the equipment
so we can reduce both these alternatives to
00:32:05.480 --> 00:32:06.740
cash flow diagrams
00:32:06.740 --> 00:32:12.570
that is this is my cash flow diagram for option
a this is the first expenditure the second
00:32:12.570 --> 00:32:19.610
expenditure the third and the fourth expenditure
in terms of purchase of land initial construction
00:32:19.610 --> 00:32:25.840
and design main construction and finally the
setting up of equipment similarly for option
00:32:25.840 --> 00:32:37.100
b this is the cash flow diagram now we want
to do this analysis using future worth future
00:32:37.100 --> 00:32:43.600
worth of option a is given like this that
is we are talking of the future worth of this
00:32:43.600 --> 00:32:48.510
fifteen at an interest rate of eight percent
after three years
00:32:48.510 --> 00:32:55.310
so we are trying to reduce everything to this
point in time which is three years hence so
00:32:55.310 --> 00:33:02.400
therefore this fifteen three years hence this
two point five two years hence this ten one
00:33:02.400 --> 00:33:10.610
year hence and this five as it is so if we
do this exercise we will find that the future
00:33:10.610 --> 00:33:19.230
worth is minus thirty seven point six so this
is the total expenditure in future worth terms
00:33:19.230 --> 00:33:25.990
similarly if you calculate the future worth
of option b which is ten for a period of three
00:33:25.990 --> 00:33:31.570
years that is this ten to a period of three
years this three point five for a period of
00:33:31.570 --> 00:33:37.840
two years this twelve here for a period of
one year and finally this five point five
00:33:37.840 --> 00:33:44.990
as it is so if that is what we do we find
that the future worth of option b is minus
00:33:44.990 --> 00:33:46.250
thirty five point one
00:33:46.250 --> 00:33:53.350
now which of these do we choose the equivalent
cost of remodelling the old facility is projected
00:33:53.350 --> 00:34:00.800
to be having a smaller future cost and hence
that is preferable so that is how we do future
00:34:00.800 --> 00:34:07.740
worth kind of analysis as far as evaluation
of options is concerned so now coming to the
00:34:07.740 --> 00:34:12.200
last concept of the day the equivalent annual
charge which is the eac
00:34:12.200 --> 00:34:17.369
the eac distributes the present value of the
project equally over the life of the project
00:34:17.369 --> 00:34:25.119
and there we find that we can use this formula
now this if you notice carefully is just the
00:34:25.119 --> 00:34:34.559
converse of the eppwf that is the equal payment
present worth factor so pv is the present
00:34:34.559 --> 00:34:36.700
value of the project n is the life of the
project
00:34:36.700 --> 00:34:41.429
k is the discount rate or the interest rate
which we have been using i and k interchangeable
00:34:41.429 --> 00:34:47.329
and a is the annual charge nothing but the
capital recovery factors so if you go back
00:34:47.329 --> 00:34:53.389
to those tables which we showed you and try
to tell you how those tables can be used you
00:34:53.389 --> 00:34:58.749
will find that sometimes it is referred to
as the capital recovery factor
00:34:58.749 --> 00:35:05.299
now as an illustrative example let us try
to see this a municipality wishes to install
00:35:05.299 --> 00:35:10.460
a new electricity distribution net worth and
it is estimated that this will cost fifty
00:35:10.460 --> 00:35:15.670
lakhs and the project last for twenty years
and the municipality estimates that its discount
00:35:15.670 --> 00:35:21.240
rate is seven percent how much must the municipality
charge the customers for the capital cost
00:35:21.240 --> 00:35:27.039
of the distribution net worth now if you want
to use this formula how do we go about doing
00:35:27.039 --> 00:35:28.089
it
00:35:28.089 --> 00:35:36.470
the equivalent annual charge which is eac
is fifty times a by p given seven percent
00:35:36.470 --> 00:35:42.359
of interest in twenty years of use and a fact
that turns out to be point zero nine four
00:35:42.359 --> 00:35:50.509
four and therefore the equivalent annual charge
is four point seven two lakhs for an investment
00:35:50.509 --> 00:36:00.849
which is being made for fifty lakhs to last
for twenty years effectively the customers
00:36:00.849 --> 00:36:06.740
are buying the distribution net worth and
paying the cost for twenty years at an interest
00:36:06.740 --> 00:36:15.279
rate of seven percent in equal instalments
so that is the interpretation of the eac
00:36:15.279 --> 00:36:22.170
now going back to this table that we started
we have completed our discussion up to the
00:36:22.170 --> 00:36:29.360
equivalent annual charge and these two methods
which is the cost benefit ratio or the benefit
00:36:29.360 --> 00:36:33.880
cost ratio and the incremental rate of return
methods these would be covered in a lecture
00:36:33.880 --> 00:36:39.829
subsequently possibly by professor jha iit
delhi i look forward to seeing you once again
00:36:39.829 --> 00:36:43.539
in another lecture but before that let me
give you some references
00:36:43.539 --> 00:36:48.400
these are the books which you could probably
refer to understand the subject more clearly
00:36:48.400 --> 00:36:49.900
thank you