OF ILS Mortgages
Kindly Contributed by W J Waghorn

I am indebted to Bill Waghorn for his guidance on terminology and the choice of mathematical symbols used throughout the mathematics of Mortgage Finance.

It should be noted that Mr. Waghorn uses the suffix ‘0’ to represent the values at the start of the first year whereas I have used the suffix ‘1’ in my mathematics.


'ILS Mortgages' are those which were originated by Edward C D Ingram and are referred to as Ingram Lending and Savings (ILS) Mortgages.

'Entry Cost' is the cost of the first year's annual payment 'P1'

It is assumed that readers have already read 'MATHS STRAIGHT IN'

For those who may not have read that:

Most students of mortgage finance are told that the repayments are fixed unless or until the interest rate changes. The monthly or annual repayments are then calculated over the repayment period of N years for the given rate of interest r% p.a.

When the interest rate r% p.a. changes the new repayment calculation once more assumes that the interest rate will never change and then the new cost of the repayments of capital and interest is calculated over the remaining period.

With ILS Mortgages it is assumed that the rate of repayments, or more accurately, annual payments 'P' will rise (escalate) at e% p.a. over the whole period, and that everything else like the rate of Average Earnings Growth, AEG% p.a. and the rate of interest r% will remain fixed.

For more information on the definitions please read the definitions at the start of  'MATHS STRAIGHT IN'

And read the whole of that (later) for more on how to manage the repayments and the changing level of payments as conditions change.

This page finds the equations for:
(a) the total repayment period and 
(b) the entry cost for a given repayment period
Assume a loan of L0 at a fixed interest rate of r%, with annual repayments starting at P1 and escalating at e%. 
To work out the repayment period N consider the repayments being invested, at the same interest rate, until the end of the loan (i.e. for N years), and only then used to pay it off.  To keep the algebra simple, let R=1+r/100 and let E=1+e/100.

Then on completion the loan will have accumulated N years of compound interest. In other words, L= L0 * RN 

At the same time, each payment Pn (= P1*En-1) will have accumulated (N-n) years of compound interest, and thus be worth P1*En-1*RN-n. The total value of these repayments, from the first in year 1 to the last in year N, will thus be

T  = P1*E0*RN-1 + P1*E1*RN-2 + P1*E2*RN-3 + … + P1*EN-2*R1 + P1*EN-1*R0

To calculate this total, note that
T*E  = P1*E1*RN-1 + P1*E2*RN-2 + P1*E3*RN-3 + … + P1*EN-1*R1 + P1*EN*R0 
T*R  = P1*E0*RN + P1*E1*RN-1 + P1*E2*RN-2 + … + P1*EN-2*R2 + P1*EN-1*R1 

Subtracting these two (thus cancelling the shaded areas), we get: -
T*R - T*E = P1*E0*RN - P1*EN*R0
whence T = P1*(RN - EN) / (R – E)

The loan is paid off after N years if this accumulated value equals the loan value, which implies that P1*(RN - EN) / (R – E) = L = L0 * RN

Multiplying the two sides by (R - E) and dividing by P1*RN we get
(1 – (E/R)N) = (R – E)*L0/P1  

That equation enables us to calculate the initial payment required to achieve completion after a given period of N years: -

P1  = (R – E) * L0 / (1 – (E/R)N) ………………………… W1.

It also enables us to calculate the termination period for a given initial payment. It gives us (E/R)N = 1 – (R – E)*L0/P1, whence: -

N = log(1 – (R – E)*L0/P1) / log(E/R) ……………………… W2.

Note that the above algebra does not work if R = E. 

But in that case the initial equation for T simply reduces to T = N*P1*RN-1, and the termination condition is thus

N*P1*RN-1 = L0*RN. This reduces to: -

P1 = L0*R/N …………………W3a


N = L0*R/P1, ………………..W3b

These are the characteristic equations of the original ‘Ingram Sliced Mortgage’[1], which entails payment escalation at the interest rate (i.e. E = R).

E&OE, of course. 

Edward’s Comment: We now know NOT to escalate at the same rate as nominal interest as that would normally be far above AEG% p.a!


Readers may like to check that these general equations reduce to the old familiar ones when e% = 0%

This means that E = 1

Substitute that into the above equations - and you do indeed get the usual equations for an annuity (Level Payments) Mortgage.

[1] A reference to the early exploratory mathematics done in the 1980s or earlier.

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