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Loan repayments

Loan repayments involve repaying the amount of money borrowed.

When we borrow an amount (usually called a loan or a mortgage) from a bank or financial institution, we have to repay not only the amount borrowed (called principal), but also the interest on the principal. These loan repayments are made periodically at regular intervals.

The interest payable is calculated on the balance owing after each repayment until the balance becomes zero. The interest charged is called reducible interest.

 

Let us look at an example of calculating loan repayments:

Example: Samantha borrows $100,000 from Citibank at 12% p.a. reducible interest. She agrees to repay the bank on a monthly basis for 5 years. What is her

  • monthly repayment?
  • total amount she repaid after 5 years (principal + interest) ?

Let’s assume Samantha’s monthly repayment to be M.

r = 12% p.a. or 1% p.m.  =  0.01 p.m

n = 5 years = 5 x 12  =  60 months

Total amount including interest at the end of first month  =  100000 (1 + r)1

=  100000 (1.01)

The amount owing to the bank at the end of first month after making the first repayment

A1  = 100000 (1.01)  – M

This amount A1 now becomes the Principal to calculate the compound interest for second month

Interest  =  A(1.01)1  =  (100000 (1.01)  – M) x 1.01

= 100000 (1.01)2 – M (1.01)

A2, the amount owing at the end of the 2nd month after the second repayment is

A2  =  100000 (1.01)2 – M (1.01) – M

Now, the amount owing at the end of the 3rd month after making the third repayment is

A3  =  (A2 x 1.01) – M

=  (100000 (1.01)2 – M (1.01) – M) 1.01  –  M

=  (100000 (1.01)3 – M (1.01)2 – M (1.01) – M

Similarly, we get the other amounts A4, A5, A6, . . . . .  and so on till A60

After 60 months, the amount owing is zero, since all the amount is repaid. So A60 = 0. 

A60  =  (100000 (1.01)60 – M (1.01)59 – M (1.01)58 – . . . . . . . . . . . – M (1.01)– M (1.01) – M

        =  (100000 (1.01)60 – M [(1.01)59 – (1.01)58 – (1.01)57. . . . . . . . . . . – (1.01)– 1.01 – 1]

The numbers in the square bracket form a geometric series with first term = 1 = a, and common difference r = 1.01. Using sum to n in geometric series

Sn  =   {a (r^n - 1)}/{r - 1} ; |r| >1

So first we need to get the value of this geometric series.

Substituting these values of a and r in the geometric series formula, we get

S60  =   {1 (1.01^60~ - ~1)}/{1.01~-~1}

=   {(1.01^60~ - ~1)}/{0.01}

=  81.6697

A60   =  (100000 (1.01)60 – M [(1.01)59 – (1.01)58 – (1.01)57. . . . . . . . . . .- (1.01)2 – 1.01 – 1]

=  (100000 (1.01)60 – M x 81.6697

We know A60 = 0, so

0      =  (100000 (1.01)60 – M x 81.6697

M    =   {100000~*~{1.01}^60}/81.6697

=  $2224.44   (to the nearest cent)

So Samantha makes a monthly repayment of $2224.44 for 5 years (60 months).

Total amount repaid by Samantha = $2224.44 x 60  =  $133466.64

Interest paid by Samantha on her borrowing I  =  Total amount repaid – Amount borrowed

= $133466.64 – 100000  =   $33466.64.

 

The above example can be consolidated to find the repayments using the formula:

A   =   { {M ~((1+r)^n - 1)}/ {r (1+r)^n } }, or

M =    {A~*~(1+r)^n * r } / {(1+r)^n~-~1} where

  • A is the amount borrowed,
  • M is monthly repayment made,
  • r is the rate of interest per time period, and
  • n is the total time period (in years, or months) for repaying the borrowing.

M  =   {100000~*~(1+0.01)^60~*~0.01 } / {(1+0.01)^60~-~1}

=  $2224.44