Thread: Financial Maths - Future Value Calculation

The problem:

Zanele plans to save R50 000 towards buying a new car in three years' time. She makes three equal deposits at the beginning of each year into a savings account, starting immediately. The interest paid on the money in the savings account is 11% p.a. compounded quarterly. Calculate how much money she will need to pay on each occasion.

My attempt:



Which gets me x = R16216.62
According to the textbook, the answer should be R13362.60
What am I doing wrong?

The formula you are using assumes a deposit at end of each year;
but with your problem, the deposit is made at beginning of each year.

Also, the "i" used in formula must be the annual "i", to coincide with the deposit frequency.
This means i = (1 + .11/4)^4 - 1 =~ .11462, or ~11.462%;
makes sense that an annual rate paid more frequently than annual ends up higher, right?

The formula for Future Value of deposits made at beginning of period is:
F = D * {[(1 + i)^(n + 1) - 1] / i - 1}

So, to calculate D:
D = F / {[(1 + i)^(n + 1) - 1] / i - 1} : see NOTE below

D = deposit required (?)
F = future value (50000)
n = number of periods (3)
i = interest per period [(1 + .11/4)^4 - 1] = .11462

D = 50000 / [(1.11462^4 - 1) / .11462 - 1] = 13,362.60

And your "account" will look like:

Code:

DATE         DEPOSIT        INTEREST      BALANCE
Jan.01/1    13,362.60            .00    13,362.60
Jan.01/2    13,362.60       1,531.64    28,256.84
Jan.01/3    13,362.60       3,238.84    44,858.28
Dec.31/3                    5,141.72    50,000.00

NOTE:
Can be simplified to:
D = F * i / [(1 + i) * ((1 + i)^n - 1)]

Usually a good idea to make (1 + i) a variable, like let r = 1 + i, then equation easier to handle:
D = F * i / [r * (r^n - 1)] where r = 1 + i

To avoid searching fior the right factor
50000= R(1.0275)^12 +R(1.0275)^8 + R 91.02750^4 which gives R =13362.60

Originally Posted by bjhopper

To avoid searching fior the right factor
50000= R(1.0275)^12 +R(1.0275)^8 + R 91.02750^4 which gives R =13362.60

True BJ...but lotsa typing if for 25 years!
But it's a good way to show a student how the formula is arrived at:
F = Future value (50000)
D = Deposit per period (?)
n = number of periods (3)
r = 1 + rate per period (1.0275^4)

Code:

  -F   = -Dr  -  Dr^2  -  Dr^3  - ....... -  Dr^(n-1)  -  Dr^n
  Fr   =         Dr^2  +  Dr^3  + ....... +  Dr^(n-1)  +  Dr^n  + Dr^(n+1)
--------------------------------------------------------------------------
Fr - F = -Dr  +   0    +   0    + ....... +     0      +    0   + Dr^(n+1)

F(r - 1) = Dr^(n + 1) - Dr

F(r - 1) = D[r^(n + 1) - r]

D = F(r - 1) / [r^(n + 1) - r]

D = F(r - 1) / [r(r^n - 1)]

SO: D = 50000(1.0275^4 - 1) / [1.0275^4(1.0275^(4n) - 1] = 13362.60