Financial Maths
Mathematics of Finance
This topic is concerned with the time value of money. The important point is that the same sum of money due at different points in time is worth different amounts; $10 is only worth $10 if it is in your hands now – if you have to wait a year for it, it is worth less than $10.
A mathematical background useful (but not necessary) for this topic is an understanding of geometric progressions. While answers which are in money terms are presented rounded to the nearest cent, it is worthwhile at least for intermediate values to carry a substantial number of decimal places, especially for interest rates. This is because premature rounding can have severe effects on the accuracy when a rounded number is raised to a large power. Note that several examples in these notes might have slightly different answers if interest rates are calculated to a larger number of decimal places. The usual approach taken is to express these to six decimal places; clearly an eight figure value will give a more accurate result.
Simple Interest
Compound InterestTime Lines and Equations of Equivalence
Solving for unknowns in accumulation and present value formulae
This is calculated by multiplying the interest rate per period by the principal at the beginning of the period, and multiplying the result by the number of periods over which interest is to be calculated. The normal formula is I = PRT, where R is expressed as a decimal.
Examples
(a) $100 is banked for three years at a simple interest rate of 6 % p.a.
The interest earned is I = P*R*T = 100*0.06*3 = $18
(b) $150 is banked for 4 years 9 months at a simple interest rate of 4.5% p.a.
Interest earned is I = 150*0.045*4.75= $32.06
- Let the principal, or present value, be represented by P or PV.
- The compounded, or future, value may be denoted by F, or FV.
- The number of interest periods over which compounding occurs is n.
- Interest rates are normally given as annual rates and are qualified by the number of times they are to be compounded, or converted, per year, and are denoted by jm, where m indicates the number of times interest is to be converted per year.
- We use i to represent the interest rate per interest conversion period. It is good practice to indicate the period over which interest is to apply.
A rate j4 = 12% p.a. represents an annual rate of 12%, converted quarterly, i.e. at a rate i = 3% per quarter.
In general, i = jm/m
Suppose P is to be invested at a compounding rate of i per period for n periods.
| Example if P=1000, i=10% | ||
| During the first period, it earns interest: | iP | 100 |
| At the end of the first period, it is worth: | P + iP = P(1 + i) | 1100 |
| During the second period, it earns interest: | iP(1 + i) | 110 |
| At the end of the second period, it is worth: | P(1 + i) + iP(1 + i) = P(1 + i)2 | 1210 |
| During the third period, it earns interest: | iP(1 + i)2 | 121 |
| At the end of the third period it is worth: | P(1 + i)2 + iP(1 + i)2 = P(1 + i)3 | 1331 |
FV = 24(1.0225)1500 = $ 7.5*1015
(NB: 7.5*1015 is known as scientific notation, and is used to simplify the communication of extremely large or small numbers. $ 7.5*1015 = $7500 000 000 000 000. On most calculators this will be displayed as 7.5E15. If you have forgotten about this form of notation a quick revisit of a grade 9 or 10 maths text might be appropriate. It is very important that you are comfortable with the way that your calculator handles very large and very small numbers.)
PV = 12 000 + 18 000(1.03)-8 = 12 000 + 14 209.37 = $ 26 209.37.
Consider FV = PV(1 + i)n
By taking logs (to the base e) of each side, ln FV = ln PV + n*ln (1 + i)
Hence, 
Similarly, FV/PV = (1+ i)n, and taking the n th root of each side,
Examples
(a) Find the time for a bank deposit of $ 400 to double in value if the bank compounds monthly at j12 = 7.75 % p.a.
We have PV = $ 400, FV = $ 800, i = 0.006458 per month
Then 
= 107.67 months = 8.97 years. => 9 years
(round up to next month as interest is paid each month.)
(b) Compare this result with the time required if interest is compounded daily.
The daily rate is j365 = 0.0775 / 365 = 0.00021233.
Then 
= 3264.62 days => 3265 days (approx 8.94 years)
(round up to next day as interest is paid daily)
This is an example of the impact of continuous compounding to be explored in an exercise later.
(c) At what rate of interest will a bank deposit of $ 400 double in value in 10 years if interest is converted quarterly ?
We have PV = $ 400, FV = $ 800, n = 40 quarters.
Then 
= (800/400)1/40 – 1 = 21/40 – 1 = 0.017480 per quarter, or j4 = 6.9919 % p.a.
(d) A business loan of $ 6000 has to be repaid by a single payment of $ 12 500 in 8 years, 3 month’s time. What monthly interest rate and what annual rate is implied by this contract ?
We have PV = $ 6000, PV = $12 500, n = 99 months.
Then i = (12500/6000)1/99 – 1 = 0.007441 per month => j12 = 8.9297 % p.a.
Equivalent Interest Rates
, from which: 
(b) Money is invested at j12 = 4 % p.a. for 6 months. What rate j4 will accumulate the same amount in the same time ?
- (i) from first principles:
Suppose $ 1 is invested, at a rate i = j12/12 = 0.003333 per month.
In 6 months it will be worth 1.0033336 = $ 1.020167
If the applicable rate is compounded quarterly, there will be two conversions over the six-month period.
Hence (1 + i)2 = 1.020167, from which i = 0.010033 per quarter, and j4 = 4.013 % p.a.
(i.e. (1 + i) = Ö 1.020167 => i = Ö 1.020167 – 1 = 0.010033 => j4 = 0.010033 * 4 = 4.013%)
- (ii) using equivalent rates formula:
The general approach to solving this problem is to use the equation
(1 + j4/4)2 = (1 + j12/12)6, from which j4 = (1 + j12/12)3 – 1.
If j12/12 = 0.003333, then j4/4 = (1.003333)3 – 1 = 0.010033 per quarter. => j4 = 4.013% p.a.
This is the rate compounded once per year which is equivalent to a rate converted over some different period, usually more than once per year. The virtue of such a rate is that it allows unequivocal comparisons to be made between financial deals which may be offered with differing interest periods, where it is not clear whether, say, a j12 rate is lower than a j4 rate. Commercial lenders are required to indicate in financial contracts what the effective rate is.
The conversion follows from the expression above, where we denote the annual effective rate by r. (NB r is the same as j1)
Replacing p with 1, in the equivalent rates formula we get 1 + r = (1+ jm/m)m, from which
r = (1+ jm/m)m – 1
The equation of equivalence is 1 + r = (1 + j4/4)4, from which j4/4 = (1.06)1/4 – 1 = 0.014674
Then j4 = 5.8695 % p.a.
(i.e. using (1 + r)1/4 = 1 + j4/4 => (1 + r)1/4 – 1 = j4/4 )
Time Lines and Equations of Equivalence
These are useful devices to depict cash flows, especially when annuities are being analysed. They emphasise the notion that dated amounts may only be validly compared when replaced by equivalent amounts due at the same time.
Examples
(a) Suppose $ 1000 is due in three months, $ 500 is due in seven months, and $ 800 is due in 15 months. What is the PV of the debt?
It should be clear by now that we may not add these to claim a debt of $ 2300 is due, either now or at the end of three months. The amounts due may be depicted on a time line as below :
……….1000………….500……………………….800
|__________|________________|______________________________|
0……….3…………….7…………………………15 months
This set can be replaced by a single payment due at any one date, given an interest rate of j12 = 15% p.a., or i = 1.25 % = 0.0125 per month.
The present value (dated today) can be stated in an equation of equivalence as follows:
PV = 1000(1.0125)-3 + 500(1.0125)-7 + 800(1.0125)-15
= 963.42 + 458.36 + 663.99 = $ 2085.77.
Similarly, the value of the stream can be determined at any other date by manipulating the equation of equivalence, e.g.
Value at the end of month 10 = 1000(1.0125)7 + 500(1.0125)3 + 800(1.0125)-5
= 1090.85 + 518.99 + 751.82 = $ 2361.66.
(b) Although this example is somewhat “contrived” it is a useful example of how payment (or debt) streams can be manipulated when refinancing is required.
A firm has two debts:
- A bond it issued which has a face value of $10 000 to be paid in seven years time,
- A loan of present value $5000, to be retired with a single payment (of principal and interest) in four years, where the contacted interest rate is j2 = 7.5% p.a.
A forecast of cashflows indicates that money is likely to “tight” at these times, but that the firm should have excess cash in three years time and then again in six years time. The firm also has excess cash now, and could afford a current payment of $2000.
The firm is therefore looking for somebody to take over the debt. It is offering to pay $2000 now and two equal payments in 3 and 6 years time. If the interest rate that the company taking over the debt requires is j4 = 6.5% p.a., what is the size of the two equal payments?
This problem is best viewed in a time diagram, where the equal payments are X.
The appropriate interest rates are as follows :
If j2 = 7.5 % p.a., i = 0.0375 per half-year; (contract rate on loan)
if j4 = 6.5 % p.a., i = 0.01625 per quarter. (market rate)
Payments 2000 X X
Owing 5000(1.0375)8 10 000
|___________________|_________|______________|________|
0……………….3………4……………6…….7 years
0……………….6………8……………12……14 half-years
0……………….12……..16…………..24……28 quarters
PV amount owing = 5000(1.0375)^8(1.01625)^-16 + 10 000(1.01625)^-28
= 5186.40 + 6367.73 = $ 11 554.13.
PV of new scheduled payments = 2000 + X(1.01625)^-12 + X(1.01625)^-24
Solving, PV(debt) = PV(new payment scheme)
=> 11 554.13 = 2000 + X(1.01625)^-12 + X(1.01625)^-24
=> 9 554.13 = X(0.824125) + X(0.679183)
=> 9 554.13 = X(1.503308)
=> X = 9554.13 / 1.503308 = $ 6355.41
which is the amount to be paid at the end of three and six years.
Discussion:
Why are there two different interest rates?
- Presumably the loan (with present value of $5000) was taken out sometime in the past when prevailing interest rates were higher. If j2 = 7.5% was written into the contract as a fixed interest rate, it still applies to the loan, even though market rates may have fallen since. The j4 = 6.5 % rate reflects the current market rate (for whatever the assumed level of risk is.)
- Where do these figures come from? 5000(1.0375)8(1.01625)-16 , 10 000(1.01625)-28. ? The second is merely the present value of the $10 000 to be paid in seven years. The former has two components, the 5000(1.0375)8 is how much the contracted debt will have grown to by the time it is due, and the (1.01625)-16 merely discounts this back to a present value.
A simple annuity is a series of uniform payments over a set term, made at uniform intervals of time, with payments made at the end of the payment period, where the interest conversion period is the same as the payment period.
An annuity due differs from the above only by having payments made at the beginning of the respective payment period, e.g. rents on a property.
A general annuity differs from a simple annuity if the interest conversion period does not coincide with the payment period, e.g. quarterly payments and monthly interest rests.
In this Unit, we will be concerned only with simple annuities. The treatment of annuities due can be handled quite simply within the framework of simple annuities. General annuities can be converted to equivalent simple annuities by converting the interest rate to an equivalent one with the same conversion period as the payment period, as discussed above in Section 5.
Formulae can be developed for finding the present value and future value of simple annuities. Suppose R is the annuity payment. An annuity of n payments can be depicted on a time line.
Remember, the formulae have been developed on the basis of payments being made at the end of the respective periods. (On the time line, “1″ represents “end of period 1″, i.e. in one period’s time. Start of period 2 and end of period 1 are coincident.)
PV………………………………………………..FV
R………R………R………R………R…….R………R
|_________|_________|_________|_________|_ _____|_________|
0………1………2………3………4…….n-1…….n periods
Derivation of PV of Simple Annuity
Suppose the interest rate is i per payment period. The annuity’s PV can be determined by discounting each of the n payments as follows :
PV = R(1 + i)^-1 + R(1 + i)^-2 + R(1 + i)^-3 + . . . + R(1 + i)^-n
= R{(1 + i)^-1 + (1 + i)^-2 + (1 + i)^-3 + . . . + (1 + i)^-n}
The bracketed coefficient of R is a geometric series with a first term of (1 + i)^-1 and a common ratio of (1 + i)^-1.
Recall that for the general geometric series of n terms with a as its first term and common ratio of r, the sum of the n terms is given by S = a(1 – rn) / (1 – r)
Substituting, PV = R{(1 + i)-1[1 - (1 + i)-n]} / (1 – (1 + i)-1)
Simplifying:
The annuity’s FV is the PV compounded for n periods, so that
which simplifies to:
0……….1………2………3……..41……..42 months
Accumulated amount = FV = 200{(1.0075)42 – 1} / 0.0075 = $ 9830.66
Present value = PV = 200{1 – (1.0075)-42} / 0.0075 = $ 7182.74
(b) A parent reckons that $ 10 000 per year will be needed for each of four years’ tertiary education for a child, beginning at the end of 15 years from now. To provide the necessary funds, $100 per month will be deposited in an account earning j12 = 8 % p.a. beginning at the end of this month, for the next 15 years.
Assuming the $ 10 000 is needed at the beginning of each of years 16, 17, 18 and 19 from now, by how much will the accumulated fund exceed, or fall short of, that required to fund the studies, where both amounts are valued at the beginning of 16 years from now ?
The accumulation problem has an interest rate of i = 0.6667 % = 0.006667 per month
The time line for the accumulation problem is
PV……………………………………….FV
……….100…….100…….100….100……..100
|__________|_________|_________|___ __|__________|
0……….1………2………3……179…….180 months
The accumulated amount is FV = 100{(1.006667)180 – 1}/ 0.006667 = $ 34 605.05.
The time line for the spending side of the equation is
……………..10000……….10000………10000……….10000
_____|______________|_____________|______________|_____________|
….15………….16…………17………….18………….19 years
This part of the problem is an example of a general annuity, because the payment period is one year, while we have a monthly interest rate. The solution is straightforward – convert the rate to an equivalent effective rate.
The annual rate equivalent to j12 = 8 % p.a. is given by j1 = (1 + j12/12)12 – 1 = 8.3004 % p.a.
Then the present value (at the beginning of year 16 = end of year 15) of education costs for the four years is
PV = 10 000 + 10 000{1 – (1.083004)-3}/ 0.083004 = 10 000 + 25 631.93 = $ 35 631.93
Hence the accumulated amount falls slightly short of the required funds.
Readers may wish to determine the required amount to be deposited each month to accumulate the amount needed at the start of 16 years from now.
Solving for unknowns in accumulation and present value formulae
Payment, R
and (i) 20 000 = R{1 – (1.055)-12}/0.055 = 8.6185R, so that R = 20000/8.6185 = $ 2320.58
(ii) 20 000 = R{(1.055)12 – 1}/0.055 = 16.3856R, so that R = $ 1220.58
(b) A home-buyer borrows $ 70 000 from a bank at a rate j12 = 8.25 % p.a. with a requirement that the loan be paid off in 12 years by uniform monthly instalments.
What payments are necessary ?
Given j12 = 8.25 % p.a., i = 0.006875 per month, n = 144
The equation of equivalence is 70 000 = R{1 – (1.006875)-144} / 0.006875
Hence 70 000 = R * 91.2236, that is R = $ 767.35 per month
II Time, or number of payments, n
It is often necessary to know for how long payments need to be made to retire a debt, or to accumulate a given sum. Begin with the annuity formulae, do a little algebra and take logs of both sides:
Similarly we find for a FV:
0……….1………2………3………..n…….n+1
n = – ln{1 – 15 000*0.00875/300}/ ln (1.00875) = – ln 0.5625 / ln 1.00875 = 66.04
Hence 66 complete payments are required, which have a present value of
300{1 – (1.00875)-66}/ 0.0875 = $ 14 992.75
The PV of the partial payment must be 15 000 – 14 992.75 = $ 7.25.
Its value when due (in 67 periods) is 7.25(1.00875)67 = $ 13.00
(b) A municipal council decides to establish a sinking fund to replace a sewer system. The council’s engineers estimate the present cost of the replacement to be $ 3.2 m., and that this cost will remain stable into the foreseeable future. The council plans to put $ 300 000 into the fund at the end of each quarter, with the fund guaranteeing to pay interest of 7.5 % p.a., converted quarterly. How long will it take to accumulate sufficient funds to proceed with the works ?
The fund’s quarterly accumulation interest rate is 7.5% / 4 = 0.01875.
A time diagram, in $ ’000, showing the accumulating fund’s deposits, is as follows :
3200
……….300…….300…….300………300……300
|__________|_________|_________|_____ _____|________|
0……….1………2………3………..n-1……n quarters
- By frist principles:
The equation of equivalence is 3200 = 300{(1.01875)n – 1}/ 0.01875,
Simplifying, 0.2 = 1.01875n – 1, from which ln 1.2 = n ln 1.01875
Thus n = 0.182322 / 0.018576 = 9.8 quarters.
(ii) Using formula: 
Hence it will take 10 quarterly deposits of $ 300 000 to accumulate the required $ 3.2 m to undertake the works.
III Interest rate
Inspection of the formulae indicate that it is not possible to directly solve for i when the other three values are given. However, successive calculator approximations will produce a solution as close as is required.
| Approximation | Value of expression |
| Try i = 3.5 % per quarter | 21.355072 |
| Try i = 4 % per quarter | 19.792774 |
Begin with a monthly interest rate of 1 %, whereby LHS = 26.9735 – too little.
If i = 0.015, LHS = 28.6335 – too much.
If i = 0.014, LHS = 28.2916 – too much.
If i = 0.013, LHS = 27.9547 – too little.
If i = 0.0135, LHS = 28.1225 – too much.
We could settle on i = j12/12 = 1.3 % per month, whereby j12 = 15.6 % p.a.
Alternatively, interpolation between 1.3% and 1.35 % per month would yield a more precise figure for the required interest rate, about 1.31 % per month. This is left for the reader.
Amortisation refers to paying off a debt, e.g. relating to a property purchase, usually with uniform payments at equal intervals of time.
The outstanding principal is the present value of the remaining payments at any point, and is sometimes referred to as seller’s equity.
By subtracting seller’s equity from the original debt, buyer’s equity can be found – this represents the degree of ownership attained in an asset after making a given number of payments.
An amortisation table shows the composition of periodic payments between interest charges and principal repayments.
Examples
(a) A person buys a block of land for $ 48 000 and agrees to make monthly payments at j12 = 12 % p.a. for four years, i.e. i = 1 % per month. We may complete several investigations into this debt.
(i) The value of monthly payments required is solved for by
48 000 = R{1 – (1.01)-48}/ 0.01, hence 48 000 = R*37.973959
=> R = $ 1264.02, say $ 1264.
(ii) Buyer’s equity after 36 payments = 48 000 – PV of remaining 12 payments
BE = 48 000 – 1264{1 – (1.01)-12}/0.01 = 48 000 – 14 226.42 = $ 33 773.58.
(iii) The amortisation table for the final four payments can be constructed as follows, noting that the outstanding principal after the 44 th payment is given by
OP = 1264{1 – (1.01)-4}/ 0.01 = $ 4932.08
| End | Monthly | Interest | Decrease in | Outstanding |
| Month | Payment | Accrued | O.P. | Principal |
| 44 | 4932.08 | |||
| 45 | 1264 | 49.32 | 1214.68 | 3717.40 |
| 46 | 1264 | 37.17 | 1226.83 | 2490.57 |
| 47 | 1264 | 24.91 | 1239.09 | 1251.48 |
| 48 | 1264 | 12.51 | 1251.49 | - |
If the buyer wishes to retire the loan with monthly payments of $ 500, how many payments are necessary, and what partial payment is necessary one month later to pay it out ?
Prepare an amortisation table for the last four months.
The interest rate per month is i = 0.009583.
The initial monthly repayment is obtained from 45 000 = R{1 – (1.009583)-60}/ 0.009583
Then R = $ 989.6673 per month.
Outstanding principal after 18 th payment
= 989.6673{1 – (1.009583)-42}/0.009583 = $ 34 086.3136
New interest rate = 0.00875 per month.
New monthly payment commencing with 19 th is obtained from
34 086.3136 = R{1 – (1.00875)-42}/0.00875, from which R = $ 973.3261 per month.
Outstanding principal after 33 rd payment = 973.3261{1 – (1.00875)-27}/0.00875 = $ 23 315.88
Amount owing after paying lump sum $ 18 000 = $ 5315.88
New repayment rate commencing with 34 th payment from
5315.88 = R{1 – (1.000875)-27}0.00875 , from which R = $ 221.9126 per month.
Outstanding principal after 43 rd payment = 221.9126{1 – (1.00875)-17}/0.00875 = $ 3491.1972
New interest rate = 0.0075 per month.
Now: 
Hence there are to be seven complete payments of $ 500, made at the end of months 44 to 50, whose PV as at end month 43 is
500{1 – (1.0075)-7}/0.0075 = $ 3397.3189,
which means the PV at end month 43 of the partial payment is 3491.1972 – 3397.3189 = $ 93.8783
Final partial payment made at end of month 51 = 93.8783(1.0075)8 = $ 99.6611
To complete the amortisation table, we require the outstanding principal after the 47 th payment.
This is found from 500{1 – (1.0075)-3}/0.0075 + 99.6611(1.0075)-4 = $ 1574.5046
The amortisation table is
| End month | Payment | Interest accrued | Decr O.P. | Outst Principal |
| 47 | 1574.5046 | |||
| 48 | 500 | 11.8088 | 488.1912 | 1086.3133 |
| 49 | 500 | 8.1474 | 491.8526 | 594.4606 |
| 50 | 500 | 4.4585 | 495.5415 | 98.9191 |
| 51 | 99.6611 | 0.7419 | 98.9192 | - |
$180,000
……….R………R………R……..R………R
|_________|_________|_________|_____ __|_________|
0………1………2………3……..35……..36 months
The monthly deposit into the sinking fund is determined from 180 000 = R{(1.0075)36 – 1}/0.0075,
from which R = $ 4373.95, say $ 4374.
The interest payment to the lender is at a rate of 1 % per month on the whole $180,000, i.e. a constant monthly outlay of $1,800.
The overall monthly cost to the borrower is 1800 + 4374 = $ 6174.
Immediately after the 33 rd deposit has been made, the sinking fund’s balance is
4374{(1.0075)33 – 1}/ 0.0075 = $ 163 084.33
The final months of the accumulation process can be tabulated :
| End | Monthly | Interest | Increase in | Sinking |
| Month | Deposit | Accrued | S.F. | Fund |
| 33 | 163 084.33 | |||
| 34 | 4374 | 1223.13 | 5597.13 | 168 681.46 |
| 35 | 4374 | 1265.11 | 5639.11 | 174 320.57 |
| 36 | 4374 | 1307.40 | 5681.40 | 180 001.97 |
……….300…….300……..300……..300……300
|__________|_________|_________|_____ _____|________|
0……….1………2………3……….n-1…….n quarters
To find the number of deposits required, we need to solve
3200(1.006192)n = 300{(1.01875)n – 1} /0.01875 for n.
This simplifies to 0.2(1.006192)n = (1.01875)n – 1, which cannot be directly solved for n.
A simple approach is to use trial and error, knowing that in the absence of inflation, the sinking fund required 10 deposits, which is a lower bound for n.
If n = 10, LHS = 0.2127, RHS = 0.2041 – not enough.
If n = 15, LHS = 0.2194, RHS = 0.3213 – too much.
If n = 11, LHS = 0.2141, RHS = 0.2267
Hence, it appears that 11 deposits of $ 300 000 are required to accumulate the sum required, given a 2.5 % p.a. inflation rate.
To check this, the accumulated amount (in $ ’000) will be:
300{(1.01875)11 – 1}/ 0.01875 = 3627.45,
while the capital cost will have been inflated to (in $ ’000) 3200(1.006192)11 = 3424.83.
The amortisation schedules above indicate that the portion of successive payments which comprise interest decreases. In a flat rate contract the amount of interest paid remains constant in absolute terms, so that the interest rate payable on a decreasing outstanding principal rises the closer the completion of the deal becomes. This means that the effective interest rate payable over the life of a flat rate contract, which is the typical hire purchase deal, is much higher than the stated rate.
Example
A car yard has a vehicle with a cash price of $ 18 000. If it is bought on terms (hire purchase) over 30 months, there are terms charges of 9 % p.a. and monthly payments are calculated by dividing the amount owing by 30.
The terms charges are calculated using the simple interest procedure, i.e. I = P*R*T.
They amount to 18 000*0.09*2.5 = $ 4050, making the amount owing $ 22 050, which is to be repaid in 30 uniform monthly payments of $ 735.
The interest rates j12 and r (annual effective) implied by this contract are given by
18 000 = 735{1 – (1 + i)-30}/i, i.e. 1 – (1 + i)-30 /i = 24.489796
Substituting i = 0.01 and i = 0.015 into this expression gives left side values of 25.8077 and 24.0158 which means 1 % < i < 1.5 % per month.
A substitution of i = 0.013628 per month gives a solution close to the left side value, 24.4895, for which we settle.
Hence, j12 = 16.35 % p.a. and it can be verified that the effective rate is j1 = r = 17.64 % p.a.
Single amount problems
(a) Simple interest : I = PRT,
where I is interest earned over the period, P is principal at beginning of period, R is rate per year, expressed as a decimal, T is number of years.
(b) Compound interest, where i is interest rate per period, n is number of periods over which compounding or discounting occurs.
Future value : FV = PV(1 + i)n, Present value : PV = FV(1 + i)-n
(c) Time required to compound to a certain FV :
(d) Interest rate required to compound to a certain FV :
(e) Interest rate compounded p times p.a. equivalent to interest rate compounded m times p.a. :
(f) Effective interest rate, r = (1+ jm/m)m – 1
Annuities
All these should be treated as simple annuities, i.e. where the interest conversion period coincides with the payment period.
Notation : R is end of period payment, i is interest rate per interest conversion period, n is number of payments.
(a) Present value
(b) Future value :
(c) Number of payments (as appropriate) :, 
, 