WEBVTT
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Welcome to the lecture series Time value of
money-Concepts and Calculations. In this lecture
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will deal Discrete annually compounding. Discrete
compounding refers to the method by which
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interest is calculated and added to the principle
at specific interval of time. For example,
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interest may be compounded daily, weekly,
monthly, or even yearly. The frequency at
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which interest is compounded has its slight
effect on an inventorâ€™s effective annual
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yield.
With greater frequency of compounding that
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is has holding periods become smaller and
smaller. The effective rate gradually increases,
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but in small amounts. Discrete compounding
relates to measurable holding periods and
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a finite number of holding periods.
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For example Mr. XY deposits his saving Rupees
10,000 in a 2 year time deposit scheme of
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a bank which offers 8 percent interests compounded
quarterly. What is the maturity value at the
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end of second year?
Now, let us analyze this problem. Quarterly
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compounding means there are four compounding
periods 3 month each in a year. Instead of
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paying interest once in a year it is paid
in four equal installments after every 3 months.
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Using the above illustration there will be
eight 4 into 2 year that is equal to 8 compounding
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periods and the interest rate for each compounding
period will be 2 percent which we calculated
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as 8 percent divided by compounding period
in a year is equal to 8 percent divide by
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4 is equal to 2 percent.
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Now, if you see this. If I see here periods
it starts with 0 then 1 2 3 4 5 6 7 and 8,
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so eight compounding periods. And if I see
month wise the first compounding period hence
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after 3 months, second after 6 months, third
after 9 months, fourth after 12 months and
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fifth after 15 month, six after 18 month,
seventh after 21 month, and eighth after 24
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month. And year wise if we see its starting
year is 0 here is the first year and here
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is the second year. Now if we calculate what
will be the future value, the distance of
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between one period is 3 months the 0 to 1
is 3 month and so we are investing P amount
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at t equal to 0. And this amount grows to
P into 1 plus 0.8 divided by 4 to the power
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1 after the end of the first year. After the
end of the second year it is P into 1 plus
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0.8 divided by 4 square.
And in similar manner at the end of eighth
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period this is P into 1 plus 0.8 divided by
4 to the power 8. So, this is shown in the
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time line.
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So, Mr. XY deposits his saving of 10,000 in
a 2 year time period scheme when if the bank
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which offers 8 percent annual compounded quarterly
what is the matured value at the end second
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year, so we solve this. So, the interest period
is 8 divided by 4 is equal to 2 percent. So,
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interest rate per period is 2 percent which
is 0.02 and we have 2 years and principle
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is 10,000. Now if we calculate from the first
principle what will be the result?
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So, here a period end is 0 at 0 periods we
invest 10,000 which is the principal amount.
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Interest per period is 0.02, amount of interest
earn is 0, and beginning principal is 10,000,
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and period ending principal is 10,000. Because
at t equal to 0 obviously no interest will
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be learn, but at the end of that period this
is first period after 3 months we have beginning
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amount one 10,000 interest per period is 0.02.
So, interest earned is 10,000 into 0.02 comes
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out to be 10,000 plus 200.
So, period ending principal becomes 10,000
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200 now and this period ending principal will
become the period start principal for second
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period and again the 0.02 we earn 204 interest.
So, my beginning principal is 10,200 and ending
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principal is 10404. This will work as a starting
value for the starting up the period 3. So,
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beginning amount of period three is 10404
interest per period is 0.02 interest earn
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is 208.08. Now this beginning principal was
10404 so the ending principal is 10612.08.
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Similarly, for starting of the fourth period
this value which is 10612.08 becomes the principal
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amount here and interest rate is 0.02, so
interest earn is 212.24.
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Now the final value becomes 10,824.32 again
this becomes the principal of the start of
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the fifth period, and the value grows to 11040.81
at the end of the fifth period. So, it becomes
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the starting value at the start of the sixth
period and at the end of these six periods
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this becomes 11261.62. At the start of the
seventh period this is 11261.62 and the end
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of the seventh period it is 11486.86. So,
at the start of the eighth period this is
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the principal amount 11486.86. The interest
per period is 0.02, so the final value to
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become 11716.59.
So, this is the amount which the bank will
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offer at the end of second year. So, what
calculation method we showed just now is from
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first principles.
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Now let see what is discrete compound interest
formula; in common industrial practice the
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length of the discrete interest period is
assume to be 1 year, and the fixed interest
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rate i is based on 1 year. However, there
are cases were other time units are employed.
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Even though the actual interest period is
not 1 year the interest rate is often expressed
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on annual basis this will always see that
8 percent compounded quarterly that 8 percent
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is 8 percent annually. And in such cases basically
we express nominal interest rate r. And nominal
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interest rates r is expressed always annually.
Consider the example in which interest rate
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is 2 percent per period and the interest is
compounded quarterly. So, a rate of this type
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would be refereed as 2 percent compounded
quarterly. Now i P r that is interest rate
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per period is equal to the nominal interest
rate divided by the number of periods. So,
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interest rate is given as 8 percent per annum
compounded quarterly that 8 percent comes
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from 2 percent into 4 which it becomes 8 percent.
So, in such cases will find that in variably
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normal interest rates are is given which reads
like 8 percent per annum compounded quarterly.
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The r is nominal rate compounded N times per
year the interest rate is r by m and the amount
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S n or FV at the end of 1 year or N years
has the case may be is given by S n is equal
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to P 0, S n is the final value of the sum
P 0, P 0 is the principal amount or present
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value in brackets 1 plus r by m brackets close
to the power m. This is when I am calculating
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the S n value which is the final value for
1 year, and this is equation 1. S n equal
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to P 0 1 plus r 2 divided by m whole to the
power m N is the value of S n which is the
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final value for N years when it is compounded
m times per year.
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Now take a problem we are solving it with
equations; first we are seen in example one
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which is Mr. X deposits saving 10,000 in 2
years time deposit scheme of a bank which
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offers 8 percent per annum compounded quarterly.
What is the maturity value at the end of 2
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years? This problem has solved earlier using
very fundamental concept here will solve the
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same problem using formula which we saw just
now.
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So, solution is r is 8 percent which is nominal
interest rate P is Rupees 10,000 N is 2 and
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m is 4 because it is quarterly compounded.
So, if we use this formula S n is equal to
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P 0 1 plus r divided by m to the power m N
done S n or FV is equal to 10,000 1 plus 0.08
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which is the nominal interest rate and m is
4 year because quarterly compounding to the
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power 4 into 2 which comes out to be 11716.59.
It should be noted that the result Rupees
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11716.59 often from the above problem using
first principle and that of the equation 2
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are the same. And enhance it can be concluded
that equation 2 is correct.
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Now, let us see what type of problems can
be created for the discrete compound interest.
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And we call it is a problem matrix. Now that
the problem matrix can be generated from the
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equation 2 is S equal to P into 1 plus r by
m to the power m N. Now if we analyze the
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unknowns and variables so we find that it
is one equation enhance one unknown can be
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solved using this one equation. Further, the
ever the equation which contains five variables,
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and what are those five variables S, P, r,
m, and N can be solved to find of the value
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of a single unknown variable only when other
four variables are known. Thus five types
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of problem can be generated out of this equation.
The problem matrix for the above equation
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is shown in the next slide. This is our problem
matrix.
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The first type of problem which I call problem
A; in the problem A, P, r, m and N are given
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and we have to find out the value of S, and
the formula used will be S is equal to P in
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brackets 1 plus r divided by m whole to the
power m N. And what is type of problem this
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is, that future value S has to be found out
when present value P, interest rate r, frequency
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of compounding per year m and number of years
N of compounding are known. If we come across
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type of problem will call it a problem A.
Now, for the second type of problem; the given
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is S, r, m, and N and we have to find out
what is the value of P. So, P is equal to
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S divided by in brackets 1 plus r divided
by m to the power N into m. In this problem
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we have to find out the present value of P
were S, r, m and N are known and we will call
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this type of problem; problem B.
Now if we see the third type of problem. In
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the third type of problem S, N, P and m are
given and we have to find out r which is the
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nominal interest rate. And we can calculate
these by the formula r are equal to m in brackets
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S by P to the power 1 by m N minus 1. In this
case we have to find out the interest rate
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r where S, N, P and m are known. And such
type of problems will be called problem C.
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The fourth type of problem is when S, P, m
and r are given and we have to find out N.
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So, we will use the formula N is equal to
ln S by P divided by m into ln 1 plus r. In
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this case we have to find out the number of
years N when S P m and r unknown. Such type
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of problem will be called problem D.
Now the fifth type of problem is where S,
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P, r, and N are given and we have to find
out the value of m. The formula here will
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be S by P to the power 1 by m N minus 1 is
equal to r by m. Here we have to determine
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the value of m the frequency of compounding
per year when S, P, N, and r are known. The
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value of m can be determined using trial and
error method, because we find that in the
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left end side there is m and in the right
end side there is m so we have to use trial
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and error method to solve this problem and
to calculate m. And such type of will be designated
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has problem E.
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What conclusion we derive out of this? The
conclusion number 1; under an annual interest
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r the annual compounding, money in a bank
account will grow with a multiplier 1 plus
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r each year. Number 2; if compounding occurs
m times per year the multiplier for each compounding
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at the end of the period will be 1 plus r
by m. And third conclusion is to find out
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the final value of amount compounding m times
per year for N years following equation should
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be used S n and FV is equal to P 0 in brackets
1 plus r divided by m whole to the power m
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into N.
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Fourth conclusion; to convert involves compounding
to discrete compounding change r to r by m
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and N to m into N has shown below. So, the
first formula which is simple compounding
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or annually compounding is FV is equal to
1 plus r to the power N, but if I go for discrete
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compounding the FV will be 1 plus r divided
by m and whole power m into N. So, what I
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am doing r is changed to r by m and m is changed
to m into N.
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Now let us see, what is the value of m; that
is compounding frequency when different time
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periods are decided? In a discrete compounding
problem the value of m may vary widely depending
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on frequency of compounding.
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So, m is equal to 1 when yearly compounding
is used. When m is equal to 2 semi-annually
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compounding is used. When m is equal to 4
quarterly compounding is used. When m is equal
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to 12 monthly compounding is used or other
way we can say for monthly compounding m is
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equal to 12 or quarterly compounding m is
equal to 4.
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For daily compounding m assumes a value of
365. For hourly compounding m assumes a value
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of 8736. And for compounding per minute the
aim as a value of 524160.
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Now, see the effect of discrete compounding
factor that is 1 by r by m to the power m.
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On different values of m and nominal interest
rates. Now suppose r is 5 which is nominal
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interest rate per year and for this value
r equal to 5 percent if I am doing annually
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compounding then my discrete compound factor
which is given by 1 plus r divided by m to
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the power m becomes 1.05. For semi-annually
when m value is 2 it is 1.050625. When it
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is quarterly that is m equals to 4 it is 1.0509454.
And what is monthly 12 m is equal to 12 it
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is 1.051162. When it is daily that m is 365
it is 1.051267. And when it is hourly 8760
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it is 1.051271. And what is minutely that
means each minute it is compounding an m is
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equal to 524160 this is 1.051271.
What indication it gives? That when frequency
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is increasing the factor is not changing much.
For example, when m is equal to 8760 and m
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is equal to 524160. The value of discrete
compounding factor which is 1 plus r by m
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into the power m becomes almost same up to
say 6 digits. So, what we find that after
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certain amount of frequency this factor does
not take much. The similar conclusions can
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be found out for r equal to 8 percent; r is
equal to 10 percent, and r equal to 20 percent.
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We see that for hourly and minutely compounding
the values are almost same for all nominal
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interest values whether it is 5 percent, whether
it is 8 percent, whether it is 10 percent
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or whether it is 20 percent. Let us take problems
now.
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Problem type A is example 2. If Rupees 10,000
where loan for a total time of 1 year. At
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nominal interest rate of 18 percent per year
compounded monthly then find the future value
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of amount after 1 year. Now let us analyze
this problem. In this problem P is given that
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is Rupees 10,000, r nominal interest rate
is given 0.18 that is 18 percent, N is equal
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to 1 and m is equal to 12 has the compounding
is monthly and there are 12 months in a year.
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And what is demanded is the future value.
So I can use my future value equation, future
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value is equal to P into 1 plus r by m to
the power m N; that is 10,000 into 1 plus
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0.18 divided by 12 in brackets to the power
12 into 1 which becomes Rupees 11956.18. That
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means, if I invest Rupees 1000 with 18 percent
interest rate compounded monthly then it will
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convert into 11956.18. However, you will find
that this value will be always more if compounding
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is used for yearly.
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Now, let us take a mix problem which is example
3. For the case of nominal annual interest
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rate of 20 percent per year determine. Part
a; the total amount to which Rupees 1 of initial
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principal would accumulate after 1 year with
annual compounding and the effective interest
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rate. Part b; the total amount to which Rupees
1 of initial principal will accumulate after
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1 year with monthly compounding and the effective
interest rate. Part c; the total amount to
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which Rupees 1 of initial principal would
accumulate after 1 year with daily compounding
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and the effective annual interest rate. Number
d; the total amount to which Rupees 1 of initial
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principal would accumulate after 1 year with
hourly compounding and the effective annual
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interest rate.
Now, here if you see what we doing that the
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time period is been changed in part a time
period is 1 year, in part b it is the compounding
26:05.290 --> 26:15.350
is monthly, in c it is daily and in d it is
hourly. And then we are also asking to calculate
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the effective interest rate which will give
us a feeling that how this effective interest
26:23.340 --> 26:32.600
rate increases when frequency of compounding
per year is increase, but it at aesthetically
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remains constant.
26:34.190 --> 26:44.250
Now for answer part a; P 0 is the principal
which is Rupees 1, N is equal to 1 r is equal
26:44.250 --> 26:53.750
to i and i effective is 0.2 or 20 percent,
because for annual compounding r is equal
26:53.750 --> 27:00.910
to i and i is equal to i effective and that
is why r is equal to i is equal to i effective
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is equal to 0.2 or 20 percent. But S n that
is the final value is P 0 1 plus i effective
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to the power N is equal to 1 into 0.02 to
the power 1 is equal to 1.2. Now, this S n
27:18.590 --> 27:24.810
can be found out in different way I can replace
i is equal to r is equal to i effective value
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for annual compounding, but part b the things
will change.
27:30.300 --> 27:38.640
For part b; P 0 is equal to 1, r is equal
to 0.20, m is equal to 12 and N is equal to
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1. So, i have use this formula S n is equal
to P 0 in brackets 1 plus r divided by m to
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the power m. So, S n is equal to P 0 into
1 plus r divided by m in the whole bracket
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to the power m is equal to 1.2194. So, what
we saw that if it is annually compounding
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the S n value is only 1.2, when it is monthly
compounding the value rises to 1.2194. So,
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slight rise in the value. So, any guy who
is doing this will get more money.
28:16.750 --> 28:25.620
Now in part c; P 0 is equal to 1, r equal
to 0.20, N equal to 1 and m is equal to 365.
28:25.620 --> 28:34.010
Here compounding is daily, so S n is equal
to P into 1 plus r by m whole to the power
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m comes out to be Rupees 1.2213. Now this
is 1.2213 is more than Rupees 12194; and hence,
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when I increase the frequency of compounding
by compounding it daily, then my money S n
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value that is final value increases slightly.
And i effective is equal to this is formula
28:55.812 --> 29:04.440
of i effective 1 plus r by m whole to the
power m minus 1 so comes out to be 0.2213.
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That means that I effective have also increase
then the r value if frequency of compounding
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is increasing.
For part d; P 0 is equal to 1, r 0 is equal
29:18.940 --> 29:26.560
to 0.20, N equal to 1 and m is 8760 because
it is hourly compounding and i is equal to
29:26.560 --> 29:35.090
0.2213999. So, here we see that up to four
digits it is matching with the daily compounding.
29:35.090 --> 29:43.910
So, i effective have not increase much when
the frequency has gone beyond daily to hourly
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and S n is equal to 0.2213999.
29:47.400 --> 29:58.990
Now takes the problem type A, this is example
4. If Rupees 1000 were loaned for a total
29:58.990 --> 30:07.070
time of 1 year at nominal interest rate of
24 percent per year compounded monthly. Then
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find the future value amount after 1 year.
Or if 1000 were loan for a total time of 1
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year at a monthly interest rate of 2 percent
then find the future value of amount after
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1 year.
Now, the second part of the problem it is
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monthly interest rate is given and in the
first part of the problem 24 percent per year
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compounded monthly is a nominal interest rate
is given. So, we should not confuse here.
30:39.580 --> 30:46.540
The nominal interest rate is i P r that is
per period interest rate into number of period
30:46.540 --> 30:55.090
that is 12, so r it comes out to be 24 percent
per year. S n is equal to P 0 1 plus r divided
30:55.090 --> 31:02.500
by m 2 to the power m because N here is 1,
so m into 1 so that is why we have written
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m because the compounding is for 1 year, but
compounding monthly.
31:07.050 --> 31:13.490
So, m is equal to 12, S n is equal to 1000,
1 plus 0.24 divided by 12 whole to the power
31:13.490 --> 31:25.810
12 is equal to 1268.5. That means, if I invest
Rupees 1000 for 1 year with a nominal interest
31:25.810 --> 31:35.320
rate of 24 percent then after 1 year it will
convert into 1268.5 when it is compounded
31:35.320 --> 31:43.430
monthly. And these values will be more than
the value of S n if it is only compounded
31:43.430 --> 31:44.600
yearly.
31:44.600 --> 31:49.740
Let us take a mix type of a problem which
is example 5. For a nominal interest rate
31:49.740 --> 31:56.720
of 18 percent per year compute for a principal
of 1000 what will be the final amount after
31:56.720 --> 32:03.300
1 year of annual compounding. Also find the
effective annual interest rate. Part b; for
32:03.300 --> 32:09.240
a principal of 1000 what will be the final
amount after 1 year with monthly compounding.
32:09.240 --> 32:16.700
And also effective annual interest rate. Part
c; for the same principle of 1000, find out
32:16.700 --> 32:22.990
the final amount after 1 year when daily compounding
is used. And also find out the effective interest
32:22.990 --> 32:27.450
rate.
Here also we are finding the final value when
32:27.450 --> 32:33.950
frequency of compounding is changing and as
a same time is also finding out the effective
32:33.950 --> 32:41.380
annual interest rate which gives you a feel
that your final value will increase with number
32:41.380 --> 32:47.670
of compound when number of the frequency of
compounding is increasing.
32:47.670 --> 32:57.470
Now the solution is for part a; P is equal
to Rupees 1000, r is equal to 0.18, N is equal
32:57.470 --> 33:05.210
to 1. In this case r is equal to i i effective
is equal to 0.18 because for yearly compounding
33:05.210 --> 33:10.640
r is equal to i effective is equal to i. So,
final amount FV is equal to P into 1 plus
33:10.640 --> 33:16.930
i to the power of 1 or P equal to 1 plus i
effective to the power 1 is comes out to be
33:16.930 --> 33:23.470
1180.
For part b; P is equal to 1000, r is equal
33:23.470 --> 33:32.900
to 0.18, m is equal to 12. So, FV is equal
to 1 plus r by divided by m whole to the power
33:32.900 --> 33:40.300
m. This is 1000 1 plus 0.18 divided by 12
whole to the power 12 comes out to be 1195.62.
33:40.300 --> 33:51.000
Here we see very clearly that when compounding
is yearly my final value is 1180.
33:51.000 --> 34:00.280
And when compounding is monthly my final value
is 1195.62 that means, it has increased when
34:00.280 --> 34:08.389
my compounding period has increase. So, i
effective will also increase because effective
34:08.389 --> 34:14.129
is a parameter this shows whether be compounded
value will be more or less. So, i effective
34:14.129 --> 34:21.240
is equal to 1 plus r by m to the power m minus
1 is comes out to be 19.56 percent. Now, this
34:21.240 --> 34:31.119
percent is more than the r value of 18 percent
nominal interest rate. It shows clearly that
34:31.119 --> 34:38.250
if I am increasing the frequency of compounding
the every value will increase.
34:38.250 --> 34:47.149
Now for part c; P is equal to Rupees 1000,
r is equal to 0.18, and m is equal to 365.
34:47.149 --> 34:54.440
So, now in frequency has increased to daily
free compounding. So, FV is equal to 1 plus
34:54.440 --> 35:03.210
r by m to the power m which comes out to be
Rupees 1197.16. So, we see that further view
35:03.210 --> 35:09.329
FV value has increase and so the i effective
will also increase and i effective have become
35:09.329 --> 35:14.819
19.72.
Now if we see here the monthly compounding
35:14.819 --> 35:22.819
and daily compounding the i effective value
has not changed much only of the tune of say
35:22.819 --> 35:31.299
point 0.2 percent also. But you see from yearly
compounding to monthly compounding it has
35:31.299 --> 35:41.710
changed by 1.56 percent. So, when we increase
number that is compounding frequency the difference
35:41.710 --> 35:49.529
had increased i effective becomes lower and
lower and assumes at an aesthetic value.
35:49.529 --> 35:58.299
Let us take another example which is example
6. If Mr. XX pays 5.12 to nominal interest
35:58.299 --> 36:03.630
compounded quarterly, what is the effective
annual interest rate? So, the problem here
36:03.630 --> 36:08.640
is very clear if somebody is cleaver, he can
pick up the result very quickly. The nominal
36:08.640 --> 36:15.529
interest rate is 5.12 and compounded quarterly.
So, the effective interest rate will be slightly
36:15.529 --> 36:25.160
more than 5.12 and that we will observe here.
So, results are a 5.219 percent, b 7.250 percent,
36:25.160 --> 36:32.700
c 9.893 percent, and then d is 12.556 percent.
Now solution i effective is the formula of
36:32.700 --> 36:41.589
i effective 1 plus r by m whole to the power
m minus 1 comes out to be 5.2191 percent.
36:41.589 --> 36:46.970
So, we see that this is little bit higher
than the nominal interest rate. So, any answer
36:46.970 --> 36:52.380
which is very close to nominal interest rate
or it is not very different than the nominal
36:52.380 --> 36:53.990
interest will be the correct answer.
36:53.990 --> 37:03.180
Now let us see the problem type C, example
7. Ten years ago Rupees 10,000 was deposited
37:03.180 --> 37:15.480
in a bank account, and today it is worth 22595.
The bank pays interest semi-annually. What
37:15.480 --> 37:23.009
was the nominal annual interest rate paid
on this account? So, what is available what
37:23.009 --> 37:33.609
is demanded is P is given 10,000, S is given
22595, m is given 2 because it is semi-annually
37:33.609 --> 37:41.319
that means, in a year 2 times N is equal to
10. So, what is r? So, here we see that the
37:41.319 --> 37:52.109
formula S is equal to P 1 plus r divided by
m to the power m N. So, here 22595 is equal
37:52.109 --> 38:00.390
to 10,000 1 plus r by 2 whole to the power
2 into 10. So, we have to find out the value
38:00.390 --> 38:07.950
of r. So, if we solve this equation then it
is 1 plus r by 2 is equal to 2.2595 to the
38:07.950 --> 38:20.480
power 0.05, and r comes out to be by solving
comes out to be 8.32 percent. So obviously,
38:20.480 --> 38:22.880
your answer d is correct.
38:22.880 --> 38:31.880
Now, let us take another problem which is
problem type B example number 8. Five years
38:31.880 --> 38:44.030
ago a certain sum of money was deposited in
a bank account, and today it is worth 10775.
38:44.030 --> 38:53.289
The bank pays 6 percent nominal interest compounded
quarterly. What was the amount of money deposited?
38:53.289 --> 39:09.980
Solution; yes that is FV is given as 10775,
N is given as for because it is quarterly,
39:09.980 --> 39:18.089
so quarterly compounded m is 4, number of
years N is given 5, r is given 6 percent.
39:18.089 --> 39:24.109
Now question is what is the value of P?
So, we have this equation S is equal to P
39:24.109 --> 39:30.590
in brackets 1 plus r by m whole to the power
m into n. So, if we put these values into
39:30.590 --> 39:41.700
this 10775 is equal to P into 1 plus 0.05
divided by 4 to the power 4 into 5. So, it
39:41.700 --> 39:57.340
comes out to be P into 1.346855 or P is equal
to 8000.12. So, if I spend 8000 Rupees it
39:57.340 --> 40:05.920
will grow to 10775 at the interest rate nominal
interest rate is 6 percent and if the compounding
40:05.920 --> 40:07.509
is quarterly.
40:07.509 --> 40:20.160
Now, let us the problem type D and example
number 9. Some years ago Rupees 7000 was deposited
40:20.160 --> 40:30.019
in a bank account, and today it is worth 11259.
The bank pays 8 percent nominal interest compounded
40:30.019 --> 40:34.069
quarterly. For how many years the money was
invested?
40:34.069 --> 40:43.529
Now, if we analyze this problem the given
is P is equal to 7000, S is equal to 11259,
40:43.529 --> 40:53.279
r is equal to 8 percent, m is equal to 4 and
what is the value of N m is N is unknown,
40:53.279 --> 40:59.759
so four variables known four variables unknown.
So, we can solve because we have one equation.
40:59.759 --> 41:13.770
So, this is 11259 is 7000 into 1 plus 0.08
divided by 4 to the power 4 into N. So 4 into
41:13.770 --> 41:22.670
N, if I take the log 4 into N is equal to
ln 11259 divided by 7000 now whole divided
41:22.670 --> 41:37.099
by ln 1.02. So, if we solve this then 4 N
is equal to 23.9997 and N is equal to 23.99969
41:37.099 --> 41:45.140
divided by 4 this comes out to be 5.9999 and
which is approximately equal to 6 years.
41:45.140 --> 41:54.980
Now problem type E, example 10. Five years
ago Rupees 9000 was deposited in a bank account,
41:54.980 --> 42:03.749
and today it is worth 14091.13. The bank pays
9 percent nominal interest compounded for
42:03.749 --> 42:11.819
an unknown frequency per year. Find the value
of the frequency of compounding per year.
42:11.819 --> 42:20.849
Now the same equation S equal to P in brackets
1 plus r by m to the power m N. Now the solution
42:20.849 --> 42:29.339
what is given and what is required let us
see P is given 9000, S is given 14091.13,
42:29.339 --> 42:40.950
r is given 9 percent, m is not given, and
N is given 5. So, P, S, r and N is available
42:40.950 --> 42:46.860
and we have to find out what is the value
of N. Now if you remember in the first part
42:46.860 --> 42:53.980
of this lecture I have told you that then
we find out m it is trial and error crosses,
42:53.980 --> 42:57.710
because left hand side contents same and right
hand side also contents same. So, you have
42:57.710 --> 43:03.770
to go through a trial and error crosses. So,
we can right down using this equation 14091.13
43:03.770 --> 43:12.930
equal to 9000 in brackets 1 plus 0.09 divided
by m to the power m into 5.
43:12.930 --> 43:21.450
So, we converted into two parts that is 1
plus 0.09 by m to the power 5 m is equal to
43:21.450 --> 43:36.730
1.565981. So, let m is equal to 4 then LHS
is equal to 1.5605. And RHS is equal to 1.565981.
43:36.730 --> 43:46.021
When m is equal to 10 LHS is equal to 1.565158,
and RHS is equal to 1.565981. And for m equal
43:46.021 --> 43:52.430
to 12 LHS becomes equal to 1.565981 because
RHS remains constant so whenever changing
43:52.430 --> 43:58.339
m LHS is changing. Now for m equal to 12 the
LHS has become equal to RHS that is right
43:58.339 --> 44:03.799
hand side LHS is left hand side becomes is
equal to right hand side. So, as at m equal
44:03.799 --> 44:12.440
to 12 LHS equal to RHS the assumed value is
correct.
44:12.440 --> 44:23.390
Thank you.