Interest formulas for Uniform gradient series

Interest formulas for Uniform gradient series


Welcome to the lecture on interest formulas
for uniform gradient series So we have so far discussed about the single payment series
and equal annual payment series Now this is series where there is gradient
but this gradient is linear so this is known as Uniform Gradient Series or Linear Gradient
Series In this series basically end of year payment is increasing linearly means the first
year and if you pay F1 the second year end F1 will be increased by a constant value that
is suppose G So this G is the gradient this G is uniform every year So in the second year if it is F1 plus G the
third year it becomes F1 plus 2G and so on So in the nth year end we are basically paying
F1 plus n minus 1 into G Now in this case we have to find what is the equivalent annual
series So the aim is to find the equivalent annual end of yeur payment for the gradient
amount paid every year end So we can see by referring to the table about the gradient
series As we know the cash flow diagrams for such
series is like this We have 0 1 2 3 4 n minus 1 and n Now in this as we have seen we pay
F1 during the first year end and this increases to F1 plus G in the second year So this is
F1 and this is F1 plus G So this G is basically the gradient and this gradient will go on
increasing So in the third year this amount will be F1
plus 2G Hence in the n minus 1th this amount will be n minus 2 into G and in the nth year
it becomes F1 plus n minus 1 into G So these type of series are known as uniform gradient
series Now as we see in this series that this F1 is every time for every year it is constant
What is varying is G every year So here it is G here it is 2G and this way it is going So basically you can represent this cash flow
diagram as sum of two cash flow diagrams So in one cash flow diagram this can be represented
as two cash flow diagrams where in one of the cash flow diagram you have equal annual
payment series that is equal to F1 So this all amount is F1 plus another is its gradient
part So it has two part one is the fixed amount part another part is the gradient part So this part will have the cash flow diagram
like this so in this as we see the gradient value is
0 in the first year It starts only from the second year So in the second year its value
is G and then it goes on increasing So this value from first year it moves Here in the
second year it is G in the third year it is to G in the n minus 1th year it is n minus
2 into G and in the nth year it is n minus 1 into G We have to find an equivalent cash flow diagram
which should tell as the annual equivalent value for this gradient series so that is
what our aim is Now as we see that this is the end of year
0 1 2 to end This is the gradient amount as we have seen this gradient amount is in the
first year it is 0 In the second year this is G in the third year this is 2G so this
way in the nth year it is n minus 1 into G Now this has to be represented by one annual
payment series So basically this series should be equivalent to this series where a series
of such type will have the equal ear year end payment of amount A So if you try to find the equivalent basically
summation of all this should be summation of this and this way we can find an co relationship
between A and G So what we do for this For this so now let us see we will equate the
total amount Total amount at a future time will be summation of all this and its equivalent
value at this time and this will be nothing but A times the factor F by A i n So we have to calculate the future amount
at the end of year n Now what will be the amount at the end of year n Using this annual
payment series it will be we have already discussed that
when the year end equal amount is multiplied by F by A i n so this will be A times the
factor F by A i n This amount should be equal to the summation of all this G So now the value of this G if you can look
at this is nothing but this G will have a value and its component or its contribution
towards F will be G times F by A i 1 Basically what we see here the series goes like this
0 0 G G plus G G plus G plus G G plus G plus G plus G So basically we have seen this table
year end one 0 1 2 And in the nth period it will be G plus G plus G plus G plus G plus
G Now as we see here so we can see that this
is your future time in your future time one is this G in this this will be coming so this
way And here this component will come So like that you have to find the equivalent value
of all these quantities at the year end of n Now this G remains a G itself so this G
is multiplied by F by A i 1 F by A i 1 is 1 itself so this G remains as G plus this
2G G plus G is nothing but its component at the
future will be G times multiplied by F by A i 2 So this way it will move and it will
come upto here So it will go upto G times F by A i n minus 1 So what we have seen this is the expression
what we get A into F by A i n should be equal to G times F by A i 1 that is single G So
you can write here this is nothing but G times F by A i 1 This G will have component G times
F by A i 2 like that So this is moving and it is going up to G times F by A i n minus
1 Now you will find the expression and you have
to equate them So as we know F by A i 1 F by A i 1 is nothing but G times 1 plus i minus
1 by i plus G times 1 plus i raised to the power 2 minus 1 by i So this will go on G
times 1 plus i raised to the power n minus 1 minus 1 by i This amount basically the expression
which we get that will be equated to this that we will do later Now let us find the expression for this so
this will be G by i we can take out 1 plus i plus 1 plus i 2 plus 1 plus i upto n minus
1 We have n minus 1 terms at n minus 1 terms this 1 comes so this comes as minus n minus
1 After this you can write this as G by i this 1 will come here 1 plus 1 plus i plus
1 plus i 2 up to 1 plus i raised to the power n This will be one term minus G n by i So we have separated now this term with n
outside and the term which is inside the bracket this is a geometric progression term and summation
of the GP series term will be used So this will be written as G by i 1 plus i you have
n terms n minus 1 upon i That is in the denominator we have 1 plus i minus 1 so it is i minus
G n by i Now we have to find the expression for A so this term will be equated to this
term So what we will calculate now A into this
factor F by A i n will be equal to G by i 1 plus i n minus 1 upon i minus G n by i As
we know we know F by A i n this factor is nothing but 1 plus i raised to the power n
minus 1 upon i So we will put this factor here in that case A will be equal to 1 by
1 plus i n minus 1 In place of this 1 we will have i the reciprocal of this term multiplied
by G by i into 1 plus i n minus 1 by i minus G n by i Once you get this now you have to again multiply
these terms one by one so you get G by i because this and this cancels minus G n by i times
i by 1 plus i raised to the power n minus 1 So the expression is if you look at you
can write this as G multiplied by 1 by i minus n by 1 plus i raised to the power n minus
1 So fine finally we are getting A equal to G into 1 by i minus n by 1 plus i raised to
the power n minus 1 This is the factor which when multiplied with
G the constant gradient amount will give you the ah equal year end amount equivalent value
So this is a factor which is known as linear gradient series factor
Now this amount is basically corresponding to the gradient amount Now this gradient value
can be either positive or negative So once you get this gradient value you have another
fixed component that is F1 So ultimately your net equal annual equivalent
value will be F1 plus this A which you obtained through this value because these are the known
components with you you know the G you know i and n So from here you can calculate the
net value of annual equivalent payment annual payment either it will be F1 plus A or minus
A depending upon the sign of A you get The expression can further be written in other
form you can also write this expression what you have received G 1 by i minus n by i into
i by 1 plus 1 n minus 1 And if we recall this is basically a factor this factor is nothing
but A by F i n So you can write it as G 1 by i minus n by i A by F i n So what we see we have basically derived this
expression G A equal to G 1 by i minus n by i A by F i n and this is known as the linear
gradient series factor Now let us discuss a problem based on these type of gradient
series factors So what we have seen that you have A as a function of G 1 by i minus n by
i A by F i n or you can also write it as G 1 by i minus n by 1 plus i n minus 1 Now let us see you have a question an example
tells that a person is paying in the first year end a sum of Rs 20000 and for the next 14 years that is up to 15 years end the
year end payment decreases by Rs 1000 What equivalent annual payment will represent the
same cash flow diagram as that of the gradient series diagram Now let us see because many a times we deal
with certain situations where the person has paid certain amount in the first year end
and there is a change in that amount In this case it is decreasing so for this case if
you look at and also you will be given i as 15percent compounded annually So if your pro problem is like this it means
you are given G as Rs 1000 i you know as 15 percent n i 15 so your job is to find so your
basically cash flow diagram the first cash flow diagram where in the first year he is
paying 20000 second year he pays Rs 1000 less so he will be paying 19000 like that it will
come and in the 15th year he will be paying 5000 Basically your job is to find equivalent cash
flow with equal annual payment and that will be like this 0 1 2 14 and 15
and basically you have to find what is this A So basically whatever value you get here
that will be deducted from Rs 20000 This is sorry this is only 1000 So what you do is you use this series so what
you will do you will find A as G is given as 1000 multiplied by one by i 1 by 0 point
15 minus 15 by 1 point 15 to the power 15 minus 1 Whatever comes here suppose it comes
as X then your final answer is because this is the equivalent to this gradient
series this gradient series is X This X amount will be basically deducted from this 20000
so your final answer will be 20000 minus X So this way you solve such problems Here the
gradient amount is a negative amount that is why we have subtracted it If it is a positive
amount we will add it here it will be 20000 plus all X because this line will go like
this So this way we calculate the equivalent annual amount for a linear gradient series
factor Thank you

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