WEBVTT
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Welcome to the lecture series on Time value
of money-Concepts and Calculations. Present
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lecture is on Amortization.
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What is Amortization? Amortization is the
paying of depth with a fixed repayment schedule
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in regular installments over a period of time.
Consumers are most likely to encounter Amortization
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with a mortgage or car loan.
In business Amortization refers to spreading
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payments over multiple periods. The term is
used for two separate processes; Amortization
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of loans and assets, it also refers to allocating
the cost of an in intangible asset over a
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period of time. Amortization happens when
you pay of a depth over time with regular,
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equal payments.
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With each payment generally, monthly payments
a portion of the money goes towards; one be
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interest costs what your lender gets paid
for the loan, and second reducing your loan
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balance as known as paying off the loan principal.
At the beginning of the loan your interest
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cost are at their highest especially with
long term loans. The majority of periodic
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payment is an interest expense and you only
pay of a small piece of balance. As time goes
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on more and more of each payment goes toward
your principal and you pay less in interest
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each month.
Amortized loans are designed. So, that after
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a certain amount of time your last loan payment
will completely pay off the loan balance.
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For example, after exactly 30 years or 360
months payment you will pay of a 30 year mortgage.
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Let us take an example through which will
explain this. Construct an Amortization schedule
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for a Rupees 1000, 10 percent annual rate
loan with 3 equal annual payments. So, P is
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here 1000, interest i is called 10 percent,
N is equal to 3 equal annual payments say
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A. That means, I am paying A amount each year.
From the above equation the value of A is
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1000 into 1.333 into 1 divided by in brackets
1.331 minus 1; this comes out to be Rupees
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402.12. Basically this is a equation which
was taken from Annuity and the principal is
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1000 and we are finding out annuity for 3
years and 1.33 to the power basically this
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is 1 plus i to the power N is equal to 1.331
and 0.1 is the i.
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So, here at P equal to 0 I am seeking 1000,
and at the end of the first year I am getting
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402. That means, if I take 402.11 at the end
of the first year, at the end of the second
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year, and at the end of the third year then
I can pay a 1000 Rupees loan. This is the
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equation which is been used P is equal to
A in brackets 1 plus i to the power N minus
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1 divided by i 1 plus i to the power N, and
this equation we have modified because we
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want the value of A not P. So, A is equal
to P into i 1 plus i to the power N divided
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by in brackets 1 plus i to the power N minus
1.
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So, the first step is to find out the annuity,
that we have already find out. Now in the
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step two find interest charge for year 1.
So, interest charge for year 1, If I say INT
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1 is equal to 1000 into interest rate that
is 10 percent is 0.1 comes out to be 100.
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Now step 3 find replacement of principal in
year 1. So, replacement principal is equal
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to A minus interest.
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So, A is my 402.11 minus 100 comes out to
be Rupees 302.11. Step 4 find ending balance
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after 1 year, ending balance is the beginning
balance minus repayment beginning balance
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is 1000 out of this 1000 the repayment Rupees
302.11 has already be done. So, the end balance
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becomes 697.89.
Now, we will have to repeat this steps for
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year 2 and 3 to complete the Amortization
table.
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This illustrates the Amortization payments,
where the money goes and the first year out
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of 402.11 which we have paid this much of
amount goes to the principal this much amount
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goes to the interest and the second payment
this much goes to the for the payment of principal
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this much amount goes to the return a interest
and in the third year this much amount goes
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for the payment of principal and this much
of amount goes for the payment of interest.
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So, this is how constant payments are divided
per year between the interest and the principal
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and the payment.
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Let us see the derivation of loan payment.
There is much type of loan repayments terms
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a very common type used for nearly all home
mortgages and many business loans call for
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constant periodic payments for a fixed period.
Each payment covers the current interest due
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and repays some of the remaining principal
balance. The total payment is constant, but
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the principal balance decreases. So, that
the interest portion of the payment is smaller
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than the previous one and the principal portion
of each payment is larger than the previous
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one.
Amortization tables are widely used for home
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mortgages, auto loans, business loans and
retirement plans.
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It is derived it let us L is equal to I j
plus P j I can write down, where L is the
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constant payment each period I j the j th
period interest payment and P j the j th period
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principal payment. The index j begins at 1
because payment is made at the end of the
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period. The symbol I in the equation represents
the effective annual interest rate for annual
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payments or the nominal rate per period r
by m for other periods.
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So, the interest payment I can write down
I j is equal to I into P j minus 1, where
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I is the interest rate and P j minus 1 the
principal balance after payment j minus 1.
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The remaining principal balance after j minus
1 period is P j minus 1 is equal to P0 Minus
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summation m equal to 1 P j minus 1 Pm. Where
Pm is the m th principal payment and P0 is
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the initial amount of the loan. Then based
on the equation 1,2 and 3 we can write down
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this P j is equal to l minus I j equal to
Ll minus I j can replace by this quantity
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1 into this quantity P0, this is the quantity
because this I j into P j minus 1.
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Now P j 1 minus can be replaced by this quantity.
So, this is replaced here. So, this is equation
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four. Now P j equal to L minus I j equal to
L minus I j into this factor this is equation
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number four I am rewriting it.
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From equation number 4, P1is equal to L minus
I P0. If I put this j is equal to 1, this
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is L minus I1 equal to L minus I and this
m is equal to 1 to j minus 1. So, it become
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0. So, it is only P0. So, I into P0 now for
P2, this comes out to be L minus I in brackets
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P0 minus P1 is equal to L minus I then P0
minus P1 is replaced by this quantity, P1
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is replaced by this quantity.
So, finally it comes out to be L1 plus I minus
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I into P0 in brackets 1 plus I , and similarly
for P3, it becomes L1 plus I whole square
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minus I P0 1 plus I whole square and for general
we can write down P j is equal to L into 1
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plus I to the power j minus 1 minus I P0 1
plus I to the power j minus 1, because if
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I see P3 here, the power of 1 plus I here
is 2 that means, 3 minus 1. So, when I take
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j here, it would be j minus 1. Similarly here
also to 2 that is 3 minus 1. So, this would
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be j minus 1. The some of the all principal
payments must equal to the original loan principal
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that is pay off the original loan.
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Accordingly summation P j j 1 to N is equal
to P0. So, I can write down for this, the
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summation of L 1 plus I to the power j minus
1 from 1 to n minus summation 1 to N, I P0
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into 1 plus I to the power j minus 1 . So,
this comes out be this, L, I can take out
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because it is a constant values similarly
I and P0 I can take out. So, this becomes
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this. I solve this for L. So, L becomes this,
P0 in brackets 1 plus I summation 1 to N 1
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plus I to the power j minus 1 divided by summation
1 to N 1 plus I j minus 1 .So, this gives
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me the yearly or monthly payment what so ever
be for the loan.
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Now, let us take an example, construct a Amortization
schedule for Rupees 10000, 10 percent, annual
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interest loan with 3 equal payments in 3 years.
So, the component of interest as well as principal
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for each payment draw you conclusions from
the above data. Solution principal P or PV
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is equal to 10000, N is equal to 3 years,
i equal to 10 percent. So, i is equal to I
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by 100 is 0.1. So, R or the annuity this is
the annuity R is equal to P into this formula,
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basically is a annuity formula and we are
finding out the annuity value which is represented
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here by R and many exercise we are represent
this by A also. So, equal to 10000 into 0.1
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into this factor comes out to be 4021.15.
So, 3 equal installment of 4021.15 is to be
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paid off to get 10000 loan.
In the first year interest component at the
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end of the first year is 10000 into 0.1 which
comes out be 1000, please note the money was
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taken and start of the year.
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Hence the amount of principal paid at the
end of the first year, in the first installment
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is Rupees 4021.15 minus 1000, which is the
interest. So, comes out to be Rupees 3021.15.
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So, left out principal amount at the end of
the first year is 10000 minus 3021.15 it comes
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out to be 6978.85; that means, at the start
of the first year the principal was 10000,
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but at the end of the first year, when we
pay the annuity it comes out to be 6978.85.
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Now, for the second year interest component
at the end of the second year is 6978.85 into
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01 is comes out to be 6978.85, principal paid
at the end of the second year is Rupees 4021.15
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minus the interest value that is Rupees 6978.85,
it comes out to be Rupees 3323.265. So, the
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remaining principal at the end of the second
year is, the principal was left out at the
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end of the first year which is 6978.85 minus
this value 3323.265. So, it comes out to be
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3655.585; that means, this is the principal
which is left out at the end of second year.
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For the third year interest component at the
end of the third year is 3655.585 into 0.1
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is comes out to be 365.559.
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So, amount of principal paid at the end of
the third year is 4021.15 minus 365.59. So,
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amount paid for again principal is 3655.591.
So, this is the left out principal end of
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the third year and hence it is paid. Now if
I draw the Amortization table then i divide
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into 3 columns, end of the first year, end
of the second year, end of the third year.
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So, installment is the same for all the 3
years, that is 4021.15, 4021.15, 4021.15 and
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4021.15 interest paid is Rupees 10000 at the
end of the first year at the end of the second
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year the interest draws down to 6978.85 and
the end of the third year drops down further
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to 365.559.
Principal paid on the first year is 3021.15
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in the end of the second year rises to 3323.265
and it further rises to 3655.591 at the end
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of third year and remaining principal at the
end of first year is 6978.85 at the end of
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second year is this becomes 3655.591 and end
of the third year the remaining principal
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becomes 0; that means, loan is paid totally.
From the above data is clear that, the interest
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charge is maximum at the end of the first
year and minimum at the end of third year.
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Where as principal as amount paid at the end
of the first year is minimum but at the end
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of the third year is maximum.
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Now, take an example where we calculate the
loan a loan of Rupees 50000 at a nominal interest
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of rate of 15 percent per year is made, for
a repayment period of 2 years determine the
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constant payment per period, the interest
and principal paid each period and the remaining
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unpaid principal at the end of each period
by using constant end of the month payments
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assume 12 equal length months per year. Now
you will be using the equation which we have
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derived for the payment.
Now constant end of the period that is month
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payment is equal to L the principal amount
is 50000 I equal to R by m is 0.115 divided
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by 12N is 12 into 2 is 24. This is a discrete
compounding problem now to compute L first
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compute summation 1 to m 1 plus I to the power
j minus 1. This comes out to be 27.78808.
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So, we will see that this is here I have computed
this factor.
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So, first compute this factor and do the summation
and here in the summation it is 27.78808.
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So, this factor can be very easily computed
without the information of others parameters
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only takes into a count of value of I and
other fixed parameters. So, I can calculate
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this very easily 27.78808. Now where P0 is
the principal amount 50000 plus I am using
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this formula this 50000 in brackets 1 plus
i this is 0.15 into summation this. This already
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I have calculated. So, it is 27.78808 divided
by 12 because this i will be divided by 12.
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So, the 12 is here I divided by 12 because
here I equal to r by m this is 1.5, 0.15 divided
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by 12 and then whole divided by this values
which is 27.78808 divided by 27.78808.
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So, the value of L the monthly payment comes
out to be Rupees 2424.33, at the end of the
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first month the interest would be 50000 into
0.15 divided by 12 which comes out to be 625.
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So, principal paid at the end of the first
month this, this L value subtracted by this
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interest is comes out to 1799.33. So, this
is the principal amount which would be paid
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at the end of first month.
So, remaining amount of principal at the end
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of first month is 50000 minus this value which
comes out to be 48200.67. Similarly interest
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charge at the end of the second month will
be interest will be charged on this amount
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because this is the left out principal. So,
48200.67 into this is the rate 0.15 divided
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by 12 because this is the month period of
interest comes out to be 602.51, principal
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paid at the end of the second month is the
value of L that is payment this is each month
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minus this interest comes out be 1821.82.
So, remaining amount of principal at the end
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of second month is this is the remaining amount
at the end of the first month minus this 1821.82
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comes out to 46378.85.
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Now, we see that the we fill up this table,
this is the parameter we have computed here
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this is 1 and this is 1.0125 this is the constant
payment which is being made 2424.33 the interest
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paid when the first month this we have already
calculated principal paid for the month this
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we have already calculated and remaining principal
we have already calculated. For the month
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2 this values we have already calculated.
So, once this is calculated we can fill up
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this values in this table in this table and
then we can sum up this is 27.788 and this
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is value is this value is 58183.92 that is
through constant payment for a principal of
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50000. I will be paying 58183.92 this is the
interest paid total interest paid is 80183.99
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and the total principal is 50000 Rupees and
this was due to the remaining principal.
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Thank you.