Lesson One level more complicated non-standard problems on ordinary annuity plans

One level more complicated non-standard problems on ordinary annuity plans

Problem 1

It is a non-traditional accumulative saving plan.

The complication is that the deposits are made annually, while compounding are made quarterly.


So, we should construct an equivalent model, which will smoothly combine/treat these features.


An account with 8% annual interest rate compounded quarterly works as the effective quarterly rate r = 0.08/4;
then the effective annual growth coefficient is

    t =  =  = 1.08243216,  or an effective annual rate q = 0.08243216.


Now this given saving plan as an equivalent to the ordinary annuity 

with annual deposits of $10,000 and with effective annual rate q = 0.071859031 compounding yearly.


Therefore, we can apply the standard formula for future value of such ordinary annuity

    FV =  = 146549.56.  (rounded).


ANSWER.  The future value is $146549.56.

Problem 2

Again, it is a non-traditional accumulative saving plan with $950 deposited semi-annually and compounded 
monthly at 3% per annum. 


It means that the monthly effective growth factor is  (1+0.03/12) = 1.0025.


Then the semi-annual effective growth factor is  =  = 1.01509406308652.


So, we can now write the formula for the future value of the ordinary annuity in 15 years 
with the semi-annual deposits of $950 at the end of every 6 months with the found effective 
rate

    FV =  = 35713.39  (rounded).


ANSWER.  In 15 years, the accumulated amount will be about 35713.39 dollars.

Problem 3

It is a non-traditional accumulative saving plan with $10,000 deposited semi-annually and compounded 
quarterly at 4% per annum. 

It means that the quarterly effective rate is  0.04/4 = 0.01 and the equivalent
semi-annual effective rate is  =  = 1.0201.


    +-------------------------------------------------------------------+
    |  So, this non-traditional accumulative saving plan is equivalent  |
    |   to the ordinary annuity with semi-annual deposits of $10,000    |
    |   and the semi-annual effective rate of compounding r = 1.0201.   |
    +-------------------------------------------------------------------+


Now use the general formula for a classic Ordinary Annuity saving plan


    FV = ,    (1)


where  FV is the future value of the annuity;  P is the semi-annual deposit; r is the semi-annual 
effective percentage yield presented as a decimal; n is the number of deposits.


Under the given conditions, P = 10000;  r = 0.0201.  So, according to (1), the formula for
the future value is


    FV = .


So, we should find n, the number of deposits (or the number of semi-annual periods)  from this equation

    50000 = .


Simplify it by dividing both sides by 10000

     = ,

or

    5 = .


Simplify it further, step by step

    5*0.0201 = ,

    0.1005 = ,

    0.1005 + 1 = ,

    1.1005 = .


Take logarithm base 10 of both sides

    log(1.1005) = n*log(1.0201)

and find n

    n =  = 4.81  (approximately).


Finally, round the decimal value of 4.81 to the closest GREATER integer value of 5 
in order for the bank be in position to complete the last semi-annual compounding.


At this point, the solution is complete.


The ANSWER is: 5 semi-annual periods are needed.

Problem 4

As it is given, this annuity is not standard: the payments are made at the end of each year,
while compounding is made at the end of each month.


Analytic formulas exist only for coinciding schedules of payments and compounding.


But we can transform to an equivalent standard scheme, considering payments at the end of each year 
and compounding at the end of each year with the effective annual multiplicative growth rate  

    1+r =  = 1.042818007.    (1)


Now we can use a well known formula for such ordinary annuity 

    FV = .    (2)


In this formula, FV is the future value in 18 years; P is the annual payment, the unknown value
which we should find.


We calculate the factor in the formula (2) first

     =  = 26.31908947.


Then from formula (2) we find

    P =  =  = 607.93 dollars.


Thus we found out the annual payment value. It is $607.94.    ANSWER