Financial Mathematics of Housing and Personal Loans

Mortgage and other reducible rate loans

In this article I will discuss two methods of repaying loans that ar commonly used to individuals. The first method discussed is the compound interest or reducible rate method and used for mortgage and credit union loans. The second method discussed is the flat rate method and is used by banks and finance companies for personal loans.

Under the compound interest method, a series of periodic payments is made with each payment paying the interest due on the outstanding loan and, with the portion of the payment remaining, also apart of the outstanding principal. As time goes on, the outstanding principal is gradually reduced so the the interest portion of successive payments also reduces. It is for the reason that these loans may be known as reducible rate loans enen though the interest rate charged on the loan is normally unchanged.

When a loan is repaid by level installment of principal and interest at equal time intervals, as is usually the case, then the loan (which is the discounted value of future installments) becomes the present value of an annuity. Hence, the size of the level of installment may be determined as a unknown variable in the present value equation.

Example

A loan of 10000 is to be repaid by equal annual installments of principal and interest over 5 years at 10% per year effective. Find (a) the annual installment, and (b) the interest and principal portion of the first two payments.

Solution

(a) First find the installment by using the annuity equation Where A = 10000, i= 0.10 and n = 5.

There fore, annual installment R = 10000/ a10/5 = $2637.08. Where “a10/5” is the annuity factor for 10 periods and at a interest rate of 10% per year effective.

(b) The interest due at the end of 1st year is 10% of 10000 0r 1000. This leaves 1637.98 of the first installment at the end of the first year. Thus the laon reduces by 1637.98 from 10000 to 8632.02.

At the end of second year, the interest due will be 836.20 so that the remaining 1801.78 can be used to repay part of the loan. Thus the loan outstanding at the end of 2nd year is 6560.24.

Loan Outstanding

It is important to be able to find the amount of principal remaining to be paid in respect of a loan at any time. For instance, the borrower may wish to pay off some or all of the loan with a lump sum payment or the loan may need to be re-arrranged due to a change in interest rates.

We could find the loan outstanding by drawing up a loan repayment schedule but this becomes rather tedious and long-winded when a large number of payments are involved. In this instant one shall calculate the outstanding principal directly from an appropriate equation of value. There are two methods for finding the value at any time k .

The retrospective method

This method uses the past history of the loan – the payments that have been made already. The outstanding principal, P at k, is calculated as the difference between the accumulated value of the loan and the accumulated value of the payment already made. Thus, P = A (1+i)k – R (s 1/k ). In this equation (s 1/k) is the future accumulated value of $1.00 for a period of k times at an effective interest rate i.

The Prospective method

This method looks to the future – that the installment that are going to be made. The outstanding principal P, at k, is calculated as the present value of (n-k) payments yet to be made. If the payment are equal, we obtain:

P = R (a i/n-k) where (a 1/n-k) is the annuity factor for n-k period for dollar 1.00 at the interest rate for the period i.

In the case there is a reduced payment then P = R (a i/n-k) + X (1+i)-(n-k+1)

These equations are very useful if one does not know the original value of the loan or if ther have been several interest rate changes in the past. We can remember that the loan outstanding is the always the present value of future installments.

Example

Five years ago a man borrowed $40000 from their building society with 25 year housing loan repayable by equal monthly installments at (5 payable monthly. Constrcut the loan repayment schedule for the next 4 months of the loan.

Solution

Monthly Repayments = 40000/ a.0075/300. That is the loan is for 25 years and there fore the the period n = 25*12 = 300 and interest rate per month is 9/12 per cent that equals to 0.0075. The monthly repayments = 335.68.

As 5 years have since passed the start of the loan, we know that 60 payments have been paid leaving 240 payments to go. Hence using the prospective equation P = 335.68* a 0.0075/240= 37309.14. There fore the end of 60th payment the loan outstanding is 37309.14. There fore beginning of the 61st month loan outstanding is 37309.14. The schedule is as follows:

Month loan outstanding interest due Principal Loan outstanding

start at 0.75% Repaid end of month

61 37309.14 279.82 55.86 37253.28

62 37253.28 279.40 56.28 37179.00

63 37179.00 278.98 56.70 37140.30

64 37140.30 278.55 57.15 37083.17

Changes in the loan interest rate

In the previous examples I have assumed the interest rate is constant over the loan period. However, in practice, the lender has the right to vary the interest rate consonant with the market conditions. A rate of interest change means that the installment may change or the period of the loan changes. For example if interest rate increases, then the loan repayment amount will increase or kepping the loan repayments constant the period may increase. If the interest rate falls then the installment will be reduced or the term of the loan.

Example

A housing loan for $20000 is repayable over 20 years by monthly installment at J12= 9%. After 5 years the interest rate increases to j12 = 12%. The installment remains unchanged so the term of the loan increases and there is a reduced final installment. (a) Draw up the loan repayment schedule for the month before and after the interest rate changes. (b) Calculate the the new term of the loan.

Solution

(a) As the interest rate changes at the end of the 60th month, we need to draw up the loan repayment schedule for the 60th month (when the interest rate is 0.75% per month) and the 61st month (when the interest rate is 1% per month). However, first we have to find the original installment using retrospective equation.

Monthly installment = 20000/ (a 0.0075/240) where n= 12^20 and i=0.0075 for the annuity factor (a 0.0075/240). There fore Monthly installment = 179.95

To set up the loan repayment schedule for the 60th month we need the loan outstanding after 59 months. Using R = 179.95, n- k = 181 where k = 59 and i=0.0075, we we can calculate the loan outstanding at the beginning of the 60th period.

Loan outstanding at 59th month = 179.95* (a 0.0075/181) = 17788.42.

The schedule is as follows:

Month Loan outstanding interest due principal Loan outstanding

start of month repaid end of month

60 17788.42 133.42 46.54 17741.88

61 17741.88

but the interest rate now changes to 1% per month so the next line is

61 17741.88 177.42 2.53 17739.35

(b) According to the repayment schedule, the interest rate due in the 61st month at the new interest rate is less than the monthly installment so the new term of the loan will be finite and we will not have the problem of infinite period. To find the new term, we set up an equation of value equating the loan outstanding with the present value of the future installments at the new interest rate.

17741.88 = 179.95* a0.01/n

There fore, a 0.01/n = 98.5934

1- (1.01)-n/0.01 = 98.5934

(1.01)-n = 0.014066

-n log(1.01) = log 0.014066

-n = log (0.014066)/log (1.01)

n = 428.53

This shows that the there need to be 428 complete installments and a further with a reduced installment in 429 month time. This result mean that the term of the loan has been extended by 249 months (429-180) or 20.75 years. The total term of the loan is now 40.75 years.

Flat rate loans

Repayments for personal (or hire purchase) loans by banks, finance companies and retail stores are normally not calculated using compound interest. Instead a flat rate of interest is quoted and this rate is used on the total loan to the whole term of the loan. Hence the total interest payable under a flat rate loan may be calculated as follows:

loan interest = loan * flat rate per year * term of the loan (in years)

We can now see that here is a difference between a compound interest (or reducible rate) loan where the interest is paid on the loan outstanding, which reduces as the loan goes on, and a flat rate loan where the interest is paid on the original lain for the whole term.

After the total interest under flat rate loan has been worked out, we are then able to calculate the required installment as the loan plus the interest must be repaid by the installments.

Required installment = original loan + loan interest/ number of installments

As the flat rate of interest applies to the original loan for the whole term for flate rate loan we cannot directly compare the rate of interest quoted for these loans with the rate used by a compound interest rate loan. To carryout this comparison, we have to calculate the effective interest rate of compound interest that applies in respect of the flat rate loan. To do this, we set up the appropriate equation of value and find the effective rate compound interest per per payment period. This rate then be expressed as n equivalent effective rate of compound interest per year, using the efffective interest equation.

Example

What is the monthly installment required to repay aloan for 400 at 10 per year flat over 2 years? What is the effective rate of compound interest rate per year in respect of the loan 400?

Solution

Interest = 400* 0.10* 2 = 80

Installment = 400 +80/2812 = 20 per month.

The equation for the value of the flat rate loan is as follows:

20* (a i/24) = 400 =20

By linear interpolation (a 0.015/24) = 20.0304

( a1/24) = 20.0000

(a 0.02/240 = 18.9139

1- 0.015/0.02 – 0.015 = 20.000 – 20.0304/18.9139 – 20.0304

i- 0.015 = 0.005* 0.02723

i= 0.01514 per month

There fore the effective rate per year say I will be calculated as follows:

(1+I) = (1+0.01514)12

I = (1.01514)12 – 1 = 0.1976 or 19.76% per year.

Rule of 78

As shown above it is easy to calculate loan outstanding for a compound interest loan using appropriate equations depending on the nature of installment payments and terms. However, with flat rate loans different method is normally used to find the loan outstanding part-way through the term. The method is known as the Rule 78. The rule 78 states that “ the interest column in the loan repayment schedule is a decreasing arithmetic progression through out the term of the loan”. As the Rule 78 determines the interest column of the loan repayment schedule it is the column that is first developed in the construction of loan repayment schedule for the flat rate loans.

If t = number of repayments remaining

R = the repayment per period

I = total loan interest

Then for a flat rate loan

Loan outstanding = t*r – (1=2+5———t)/(1=2+3————n)* I

Example

A leading finance company offers 4 year loans at 11% flat repayable with monthly installments. For a loan of 600 find (a) the level of monthly installments (b) The effective interest rate per month c) the loan outstanding according to the compound interest method (d) the loan outstanding according to rule 78 (e) The effective rate of compound interest per month paid by the borrower, if the laon is repaid after 1 year with a lump sum payment based on the Rule 78.

Solution

(a) Loan interest = 6008 0.115* 4 = 276.00

Installment = 600+ 276/48 = 18.25 per month

(b) The equation of value to determine effective interest rate is:

1.  

   1. * (a i/48) = 600.00

(a i/48) = 32.8767

For the linear interpolation, the following figures are used:

(a 0.015/48) = 34.0426

(a i/480) = 32.8767

(a 0.0175)/480 = 32.2938

That is i-0.015/0.0175 – 0.015 = 32.8967 – 34.0426/ 32. 2938 – 34.0426

i-0.015 = 0.0025* 0.6666

i= 0.01667 or 1.667% per month

(c Using the compound interest technique using the effective rate loan outstanding after 1 year is as follows:

18.25* (a 0.01667/360 = 491.05

1. Using the Rule 78 and using t= 36, n=48, R=18.25 and I = 276

Loan outstanding = 36* 18.25 – (1=2=3………..36)/(1=2+3……….480*276

= 657.00 – 156.11 = 500.69

(e) If the loan is repaid after 1 year with final payment of 500.69 (based on the rule 78) then the equation of value for the loan becomes:

600 = 18.25* (a i/12) + 500.69* (1+i)-12

Value of the right hand side at 1.75% = 602.59

value of right had side at i% = 600.00

Value of right hand side at 2% = 587.59

That is i-0.0175/0.02 – 0.0175 = 600.00 – 602.59/ 587.79 – 602.59

i- 0.0175 = 0.0025* 0.175

i= 0.01794 or 1.794%

The use of rule 78 figure to calculate loan repayments figure after 1 year has raised the overall effective rate of loan interest from 1.667% per month to 1.794% per month or from 21.9% per year to 23.8% per year effective.

Comparing the cost of borrowing long-term funds

When a firm borrows long-term funds it can borrow a compound interest (or reducible rate) loan with level installment of principal and interest payable through out the term of the loan. Alternatively, the company or firm can borrow a fixed interest loan with interest only payments during the loan and a full repayment at the end of the term (possibly accumulated in a sinking fund).

When comparing these two types of loans the company must calculate cost per period of these two alternatives. For the first loan the cost of borrowing is the cost of the size of installments in the plus the interest payments. For the second loan, the cost of borrowing per period is the cost of interest per period plus the installment in the sinking fund.

If say loan = A, interest = i per period, the rate of sinking fund interest rate is = r, then the cost of borrowing per period sat C1 = A i + A/ (s r/n)

For the compound interest loan the cost of borrowing applying the same notations, the cost of borrowing say C2 = A* i+ A/(a i/n). These equations are only valied when the interest rate of the two loan types are the same. When they are different the cost of borrowing has to be calculated separately.

Example 1

A company wishes to borrow 100000 for 5 years. One source will lend the money at j2= 20% if it is repayable with level half-yearly installments. A second source will lend the money at j2=19% if only the installment is paid each half-year and the principal is returned ina lump sum at the end of 5 years. If the second source is used, a sinking fund will be established by half-yearly deposits that accumulate at j12=14%. How much can the company save each half-yearly by using the cheaper plan?

Solution

When the first source is used, the half yearly cost is:

Ci= 10000/ (a 0.10/10) = 16274.54

When the second source is used, the interst on the loan paid each year on a half-yearly basis at 9.5% of 10000 =9500.00.

To calculate the half-yearly deposit into the sinking fund we must first calculate the half-yearly rate i equivalent to j12=14%.

That is (1+i)2 = (1+0.14/2)12

i=(1+0.14/2)12-1= 0.0720737

There fore half-yearly deposit to sinking fund R = 10000/(s 0.0720737/10= 7167.16

There fore the total cost per period or half yearly = 9500+ 7167.16= 16667.16

Thus the first loan is cheaper and the company can save 16667.16 – 16274.54 = 392.62 each half yearly.

Example 2.

A firm wants to borrow 500000. One source will lend the money at j4=18% if interest is paid quarterly and principal returned in a lump sum at the end of 10 years. The firm can set up a sinking fund at j4=16%. At what rate j4 would be less expensive to repay the debt with a compound interest loan over 10 years.

Solution

First calculate the the quarterly expense of the debt

Interest payment = 500000* 0.045 = 22500.00

Sinking fund deposit = 500000/ (s 0.04//40) = 5261.74

Total cost per period = 22500.00 + 5261.74 = 27761.74

The compound interest loan will be expensive if the quarterly repayment is equal to 27761.74. Thus we want to find the interest rate I per quarter (and then j4) given A = 500000 R = 27761.74 and

n=40.

There fore 500000 = 27761.74 * (a i/40)

(a i/400) = 18.010398

By linear interpolation the following values are as follows:

(a 0.047/40) = 18.401584

(a i/40) = 18.010398

(a 0.05/40) = 17.159086

That is, 1-0.045/0.05 – 0.045= 18.010398 – 18.401584/17.159086 – 18.401584

i-0.045 = 0.005* 0.314838

i= 0.04657 or 4.657 per quarter.

Or j4 = 18.63% where (1+j4) = (1+0.04657)4

Hence if the firm is able to borrow at a rate less than 18.63% for a compound interest laon repayable with level installments, then it should do so.

Summary

As mentioned above, to compare a flat rate loan with a compound interest is not straight forward and one must calculate the effective interest rate to compare a reducible compound interest rate loan. In addition, if interest rates vary, one must use the appropriate equations depending on the values known and use the retrospective or prospective method yo calculate loan outstanding for a compound interest loan. In addition for a flat rate loan one can use the Rule 78 equation to calculate loan outstanding part-way of the term of the loan. Above all, to compare alternative long-term funds one must calculate the cost per period and must choose the cheapest one, particualrly when the interest rate differs for the alternatives and a sinking fund is used for a fixed rate loan. In addition effective interest rate must be used to calculate the borrowing cost per period to decide at what rate a compound interest rate become more feasibke to that of a fixed interest rate loan.