WEBVTT
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Welcome to the course Depreciation Alternate
Investment and Profitability Analysis In this
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lecture we will be covering module two which
is alternative investment and in this module
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we will cover annual cost method part two
In this part two we will take some difficult
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problems and will solve it using annual cost
method Now in the first slide I have shown
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the formula in the first slide I have shown
how to derive the formula for the annual cost
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method and based on this formulae we will
solve our numericals
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Now this is the first numerical and this objective
of this numerical is given different capital
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investments different salvage values attractive
rate of return unequal estimated life span
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annual operating cost compare two different
investment based on annual cost method Now
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here again we have machine A and machine B
now capital investment is 45000 this is 30000
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estimated useful life that is N value is 15
and 10 in earlier cases we have taken N to
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be same salvage value
this is 5000 and this is 2500 this is also
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different Operating cost
this is 8000 this is 15000
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Now we see that though the machine B requires
less capital investment that is 30000 it needs
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more operating cost that is 15000 and here
it takes more capital investment 45000 but
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less operating costs and hence in such type
of problems basically we need annual cost
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method to decide upon whether we should go
for machine A or machine B Now the investment
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for machine A
the initial investment
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is equal to rupees 45000 the salvage value
is 5000
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Now if I take the timeline at t equal to 0
I am investing 45000 t equal to 0 and at t
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equal to 15 years I am getting rupees 5000
So in this case the depreciable amount is
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not 45000 minus 5000 because they are at different
time periods So what what we have to do we
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have to bring this 5000 to this timeline t
equal to 0 and then we deduct from 45000 to
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find out the value of the initial initial
investment basically
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Now if I want to do this so I have to find
out the present worth of this present worth
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of rupees 5000 is equal to 5000 divided by
1 plus the interest rate is 10 percent I am
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sorry I have left it this is 10 percent this
is 10 percent is equal to 01 due to 10 percent
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to the power 15 and this converts into rupees
1196.96 that means when I take this 5000 to
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t equal to 0 to present value is becomes 1196.96
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So the capital expenditure at the start of
first year this is a start of first year is
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equal to 45000 minus 1196.96 that comes out
to be rupees 43803.04 Now I have to find out
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the annual cost of capital recovery annual
cost of capital recovery is equal to capital
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investment this is rupees 43803.04 into 0.1
divided by 1 minus 1 by 0.1 to the power minus
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15 This is my capital recovery factor and
this gives me rupees 5758.95 so here I can
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write down my capital recovery factor that
is capital recovery per year is 5758.95
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Similarly I should calculate for machine B
Now for the machine B this is rupees 30000
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and salvage value is 2500 and this 2500 I
am going to get at the end of tenth year So
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in the timeline at t equal to 0 this is t
equal to 10 I am getting 2500 here and t equal
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to 0 I am investing 30000 So I am sinking
this 30000 here and I am getting 2500 here
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so this has to be brought to my zero timeline
that means I have to find out the present
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value of 2500
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So present value of rupees 2500 is equal to
2500 divided by 1 by 0.1 to the power 10 and
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this comes out to be rupees 963.86 So when
I ported from here to here this becomes 963.86
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So the capital expenditure at the start of
first year is equal to 30000 minus 963.86
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It comes out to be rupees 29036.14 Now annual
cost of capital recovery annual cost of capital
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recovery equal to this 29036.14 into this
factor i that is 0.1 from minus 1 plus 0.1
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minus 10 and this comes out to be rupees 4725.5
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So here this becomes 4725.5 now to calculate
the total annual cost I have to add the operating
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cost
so I have to add this operating cost with
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this and I have to add this with this So if
I add this is 8000 plus 5758.95 equal to 13758.95
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and if I add this 15000 plus 4725.5 this is
19725.5 As the annual cost of machine A is
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less than machine B so my selection will be
machine A Now let us take another example
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Now the objective of the example two is that
given the capital investment at different
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timeline attractive rate of return unequal
estimated life span annual operating cost
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Compare two different investments based on
annual cost method Now the example two tells
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two plans A and B exist for building construction
For plan A an total initial investment of
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6000000 is required of which 4000000 will
be spent on permanent construction now this
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is a trick Permanent construction means life
span N is equal to infinite
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Now how to handle the cases which has got
N equal to infinite and how to calculate the
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capital recovery factor will pose some problems
and we will tackle this and the rest 20000
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will spent immediately on a temporary structure
which will require renewal every 15 years
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The annual operating cost of plan A will be
50000 every year however for plan B an initial
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investment of 4000000 is required on a permanent
structure again permanent structure in addition
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to it an additional investment of 3000000
on a permanent structure is required after
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10 years from now
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So after 10 years again a investment of 3000000
will be done on permanent structure is required
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after 10 years from now The annual operating
cost of plan B is 20000 per year if attractive
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return attractive rate of return is 10 percent
land is free of cost and is not to be included
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in computation then using annual cost method
find out whether you will prefer plan A or
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plan B So complication in his plan has come
due to this permanent structure in this problem
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investments are done in different timeline
and these are to be brought into one timeline
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before the application of annual cost method
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While finding annual capital recovery of permanent
structure is assumed in a permanent structure
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it is assumed that the N will tend to infinite
and if this N tends to infinite than i 1 plus
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i to the power n divided by 1 plus i to the
power N minus 1 that will tend to i Thus the
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case becomes a case of perpetuity if you remember
in the perpetuity to find out the perpetuity
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we multiply the investment with i so it becomes
a perpetuity So for plan A A if we calculate
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the annual capital recovery cost for rupees
4000000 and at 10 percent interest rates will
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be 4000000 into 0.1 why 0.1 because this is
the value of i and here we have seen that
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while to convert this factor which is used
for capital recovery the whole factor moves
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towards i when N moves towards infinite
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So the annual capital recovery cost for a
permanent factor can be calculated by multiplying
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its cost with the value of i which comes out
to be rupees 400000 Now another capital recovery
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another 2000000 is being done life of the
structure is 20 years so annual capital recovery
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cost for rupees 2000000 is easy to calculate
this is 2000000 into 0.1 divided by 1 minus
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1 plus 0.1 to the power minus (20 years) this
is 15
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Here in the question the renewal will be required
every 15 years so it is 15 Now this comes
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out to be rupees 262947.55 now after 15 years
this structure has to be rebuild and when
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it will be rebuild at that time also the annual
capital recovery will remain this 262947.55
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provided the cost of the structure does not
change Now annual operating cost
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is rupees 50000 so to
find out the annual capital cost sorry annual
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cost this is total annual cost we will add
all these three that is rupees 400000 plus
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rupees 262947.55 plus rupees 50000 this comes
out to be rupees 712947.55
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So annual cost for A is this total annual
cost of project A is equal to rupees 712947.55
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Now let us do it for now for plan B If we
see the problem plan B plan B is also investing
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4000000 rupees on a permanent structure at
t equal to 0 So annual recovery cost of rupees
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4000000 on a permanent structure is equal
to 4000.000 into 0.1 which is the value of
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i Now this becomes 400000 now the present
worth now if you see the timeline for B at
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t equal to 0 when we investing 4000000 after
10 years when investing 3000000 on a permanent
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structure
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So first we have to do bring this to this
timeline and then multiply with the i So
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the present worth
rupees 3000000 is equal to 3000000 divided
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by 1 plus i to the power 10 and this comes
out to be rupees 1156629.87 so the present
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worth is this 1156629.87 Now taking this structure
to be permanent this is a permanent structure
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so the annual capital recovery recovery is
equal to rupees 1156629.87 into the value
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of i which is 1 because this is a permanent
structure so annual capital recovery will
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be equal to the value at t equal to zero into
0.1 that comes out to be rupees 115662.99
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Now we have annual operating cost
cost is equal to rupees 20000 So the total
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annual cost for plan B is equal to rupees
400000 plus rupees 115662.99 plus rupees 20000
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and when we add this up then it becomes rupees
535662.99
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So here total annual cost of project B is
equal to 535662.99 as the annual cost of project
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B is less than project A This project is selected
Let us summarize this common problem confronted
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in engineering economics are those where alternative
comparison between two or more mutually exclusive
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alternative investment compete involving different
series of capital disbursement that means
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the disbursement are at different time In
the present lecture I have taken one of the
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methods which is called annual cost method
to tackle such problems and to select one
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of the alternatives from mutually exclusive
alternatives and this annual cost part two
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we have taken some difficult problems and
we have shown how to tackle this using annual
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cost method Thank you