MAS1243 & MAS2243 Application of Mathematical methods to Finance Part I Semester 1, 2011/2012 Lecturer: Dr Alina Vdovina The aim of this course is to present an introduction to financial mathematics in order to support work on core modules. We will begin by introducing various forms of interest. Both theoretical and practical applications of several finan- cial topics will be covered. An introduction to financial assets, including bonds, shares, options and annuities will be given. Books 1. [LMW] D. Lovelock, M. Mendel and A.L. Wright, An Introduction to the Mathematics of Money, Springer, 2007. (In the university library as an ebook, available freely over the internet.) 2. [MS] A. Mizrahi and M. Sullivan, “Mathematics. An Applied Approach”, 7th Edition, Wiley, 2000. 3. [R] S. M. Ross, “An Elementary Introduction to Mathematical Finance”, 2nd Edition, Cambridge UP, 2003. Notes The printed notes consist of lecture notes, intended to supplement the notes you make during the lectures. There should be enough space in the printed notes for you to write down the notes you take in lectures. The notes, exercises and other course information can be found on the web at www.staff.ncl.ac.uk/alina.vdovina/teaching/mas1243/ from where they can be viewed or printed out. Alina Vdovina September 2011
Transcript
MAS1243 & MAS2243Application of Mathematical methods to Finance
Part I
Semester 1, 2011/2012Lecturer: Dr Alina Vdovina
The aim of this course is to present an introduction to financial mathematicsin order to support work on core modules. We will begin by introducing variousforms of interest. Both theoretical and practical applications of several finan-cial topics will be covered. An introduction to financial assets, including bonds,shares, options and annuities will be given.
Books
1. [LMW] D. Lovelock, M. Mendel and A.L. Wright, An Introduction to theMathematics of Money, Springer, 2007.(In the university library as an ebook, available freely over the internet.)
2. [MS] A. Mizrahi and M. Sullivan, “Mathematics. An Applied Approach”,7th Edition, Wiley, 2000.
3. [R] S. M. Ross, “An Elementary Introduction to Mathematical Finance”,2nd Edition, Cambridge UP, 2003.
Notes
The printed notes consist of lecture notes, intended to supplement the notesyou make during the lectures. There should be enough space in the printed notesfor you to write down the notes you take in lectures. The notes, exercises andother course information can be found on the web atwww.staff.ncl.ac.uk/alina.vdovina/teaching/mas1243/
Which would you prefer: to have £1000 today or £1000 this time next year.The answer may depend on several things but it’s worth bearing in mind that
• if you have the money now you can put it in a savings account and hope tohave more that £1000 by this time next year;
• if there is inflation (as there is now) then you won’t be able to buy as muchwith £1000 in a year’s time as you can now and
• there is always a risk that in a year’s time the money is no longer available.
Can you think of any reasons why it might be better to wait till next year.
• The bank you invest in may go bust;
• There may be deflation instead of inflation;
• By next year you may be responsible and mature
whereas now you may be inclined to spend it all in
nightclubs.
This means that the value of money depends on time: £1000 now is not thesame as £1000 in a years time (and very different from £1000 50 years ago) andso we refer to the
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 2
• present value of money; which is its worth at the present time and its
• future value which is its worth at some later time.
Because the present value of money is normally greater than its future value,when you borrow money you will usually be charged for its use. Such a chargeis called interest. Lenders commonly describe the interest as a percentage of theamount of the loan. There are many different ways in which these percentagesmay be used to calculate the actual charges, depending on how often interest iscalculated and whether or not interest is charged on interest. In this section andthe next we consider the main methods used.
1.1 Simple Interest
Example 1.1. Suppose that I put £1000 into a savings account which pays interestat a rate of 6% a year (per annum).
At the end of the first year I am paid 1000 × 6100 = 60
pounds, which I hide under my mattress. The same hap-
pens at the end of the second and third years. At the end
of the 3 years I have 1000 pounds in the savings account
and 3 × 60 = 180 pounds under the mattress. My cash
book looks like this.
Amount in account Interest TotalYear at year start paid at year end1 1000 60 10602 1000 60 11203 1000 60 1180
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 3
The future value of my £1000 is £1060, at the end of year
1, and £1180, at the end of year 3.
Before we derive a formula for calculating the balance on such accounts weintroduce some notation. We shall think of my investment in the savings accountfrom the point of view of the bank; which is borrowing the money from me andpaying me interest. The principal is the total amount of money borrowed, whetherby an individual from a bank in the form of a loan, or by a bank from an individualin the form of a savings account.
Definition 1.2. The annual rate of interest (also known as the nominal rate) isthe percentage of the principal that is paid as interest each year.
In Example 1.1 the annual interest rate is 6%. It is sometimes convenient toexpress the annual interest as a multiplier rather than a percentage: in which casewe write 0.06 instead of 6%.
Definition 1.3. Simple interest is interest computed on the principal, and onlythe principal, during the period it is borrowed.
The interest paid by the investment in Example 1.1 is simple interest, since thebalance in the account remained the same every year (we put the interest underthe mattress).
Suppose that instead of £1000 at 6% per annum, we invest a principal P0 at r%per annum. Again, we keep the interest payments separate from the investmentaccount. To simplify notation let i = r/100, the annual interest rate expressed asa multiplier. Then, at the end of year t, our cash book will look like this.
Year Amount in account at year start Interest paid Total at year end1 P0 P0i P1 = P0 + P0i2 P0 P0i P2 = P1 + P0i
where Pt is the future value of P0 after t years. This means that after t years wehave
Pt = Pt−1 + P0i
= (Pt−2 + P0i) + P0i = Pt−2 + 2P0i
...= P2 + (t− 2)P0i
= (P1 + P0i) + (t− 2)P0i = P1 + (t− 1)P0i
= (P0 + P0i) + (t− 1)P0i = P0 + tP0i.
This is the formula we are after.
Theorem 1.4 (The simple interest formula). If a principal P0 is borrowed at asimple interest rate of r% per annum, for a period of t years then, writing i =r/100, the interest It charged is
It = tP0i.
The amount Pt owed at the end of t years is
Pt = P0 + It = P0 + tP0i = P0(1 + ti). (1.1)
Remark 1.5. In this theorem t does not have to be an integer. For instance in thenext example we have t = 3/4.
Example 1.6. A loan of £2500 is made for 9 months at a simple interest rateof 17% per annum. What is the interest charged? What amount is due after 9months?
Solution. The period the money is borrowed for is 9
months, which is
t = 912 = 3
4
of a year. Thus the interest
It = P0r
100t = £2500× 0.17× 3
4= £318.75.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 5
The amount due after 9 months is
P3/4 = P0 + I3/4 = £2500 +£318.75 = £2818.75.
Example 1.7. Suppose that instead of the investment of Example 1.1 we are ableto put £1000 in a savings account with an annual interest rate of 6% paid every 6months. Is there any difference in the interest?
The interest per annum is still 6%. This means that the
interest for 6 months is 3%. After the first 6 month period
the interest is 1000 × 3100 = 30. We keep this separately
from the account. At the end of the first year another
3% interest on £1000 is paid; so we receive another £30.
Therefore at the end of the first year we have a total of
£1060, as in Example 1.1. The same applies to the sec-
ond and third years. Therefore the annual interest rate is
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 6
all that matters in the case of simple interest. We shall
see later that for other kinds of interest this is far from
true.
1.2 Discounted Loans
There are many different forms of loan and investment on the market, and asa result it can be very difficult to decide what gives the best value. We’ll considerso-called “discounted loans” as a first example.
Definition 1.8. If a lender deducts the interest from the amount of the loan at thetime the loan is made, the loan is said to be discounted. The interest deductedfrom the amount of the loan is the discount. The amount the borrower receives iscalled the proceeds.
Example 1.9. If I borrow £1000 at a simple interest rate of 6% per annum for 3years then, as in Example 1.1, I pay interest of £60 at the end of years 1, 2 and 3.The total interest paid is £180.
Suppose the lender offers an alternative discounted loan at 5.5%. If I need£1000 straight away, which loan should I take?
In the case of the discounted loan, if the total amount of
the loan is A and the total interest over 3 years is I then
the lender will give me an amount ofA−I straight away.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 7
I require
A− I = 1000.
The interest is calculated at an annual simple interest rate
of 5.5% per annum, which means that the interest I is
I = A× 5.5
100× 3 = A× 0.055× 3,
over the 3 years. So
A = 1000 + I = 1000 + A× 0.055× 3
and therefore
A =1000
1− 0.055× 3=
1000
0.835= 1197.61.
(Rounded up to the nearest penny.) The total interest I
pay is I = 1197.61 − 1000 = 197.61 pounds. This is
more than the original loan; and in addition I have to pay
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 8
all the interest immediately, rather than in instalments.
What happens if the interest rate is 5% on the discounted loan?
As in the above example it is often useful to be able to calculate the annualrate of simple interest that would give rise to a particular loan. To see how to dothis for a given loan suppose that
• I borrow an amount P0
• for n years
• and pay a total I in interest.
If I want to know what annual simple interest rate would give this result thensuppose that this (unknown) interest rate is r% and put i = r/100. Then, theinterest for one year is P0 × i; so the interest I over n years is I = P0 × i × n.Solving this to find i gives i = I
nP0, so
r =100I
nP0
,
as i = r/100.From the example we can also see how to obtain a formula relating the amount
of discounted loan, the interest, the proceeds (the amount the borrower actuallyreceives) and the interest rate.
Theorem 1.10 (The discounted loans formula). A borrower takes out a discountedloan of an amount A, over a period of n years. If r% is the annual rate of simpleinterest and i = r/100, then the total interest is I = Ain and the proceeds P0 is
P0 = A− I = A− Ain = A(1− in),
where i = r/100.
Example 1.11. You wish to borrow £10,000 for 3 months.There are 2 banks offering you loans and you want to choose between them.
• Bank 1 offers a discounted loan at 8% per annum.
• Bank 2 offers an ordinary loan at 9% per annum.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 9
In each case how much must you repay at the end of 3 months? What’s the bestdeal?
Solution. Bank 1.
We use Theorem 1.10. We have proceeds P0 = £10, 000,
i = r100 = 0.08, and the time the number of years n =
312 = 1
4 = 0.25.
By Theorem 1.10, P0 = A(1− in). Therefore
A =P0
(1− in)
=10000
(1− 0.08× 0.25)
=10000
(0.98
= 10, 204.08
Thus £10,204.08 must be repaid at the end and the total
interest in this case is
A− P0 = 204.08 pounds.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 10
Bank 2.
The amount to be repaid at the end is the future value
Pn of loan after n = 1/4 of a year. By Theorem 1.4, we
have Pn = P0(1 + in), so
P1/4 = 10000(1 + 0.09× 0.25) = 10225
Thus £10,225 must be repaid at the end and the total in-
terest in this case is
P1/4 − P0 = 225 pounds.
Less interest has to repaid with Bank 1, so this is a better
offer.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 11
1.3 Treasury Bills
United Kingdom treasury bills are one of the means by which the govern-ment borrows money. The bills have a face value or nominal value, (of at least£25,000) but do not specify a rate of interest. They are sold at public auction bythe United Kingdom “Debt Management Office” (DMO) with financial institu-tions making competitive bids, for amounts less than the nominal value.
For example, suppose that on 9th September 2011, a financial institution paid£998,841 for a 3 month treasury bill, with a nominal value of 1 million pounds.Then in 3 months time, on 5th December 2011, the government will pay the insti-tution 1 million pounds.
(In the week ending the 9th September 2011 the DMO sold 3 month bills withtotal nominal value of £1,000,000,000.)
What annual rate of simple interest is equivalent to the amount the financialinstitution above is making?
The amount of interest is I = 1, 000, 000 − 998, 841 =
1159 pounds. If the annual interest rate is i as a multi-
plier, then the 3 monthly interest rate is i/4. Thus P ×
(i/4) = I , where P is the amount paid. Therefore i =
4I/P = 1159 × 4/998841 = 4636/998841 = 0.0046; a
rate of 0.46% per annum.
Example 1.12. A bank wants to earn 7.65% per annum simple interest on a 6-month 1 million pound treasury bill. How much should it bid?
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 12
Solution. The amount to be repaid to the bank by the
government is A = £1, 000, 000.
The time is 6 months, that is, t = 0.5 year.
The proceeds P0 to the government are
P0 = A(1− r100t)
= £1, 000, 000[1− 0.0765× 0.5]
= £961, 750.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 13
2 Compound Interest
Example 2.1. As in Example 1.1, suppose I put £1000 into a savings accountwhich pays interest at a rate of 6% per annum.
At the end of the first year I am paid 1000 × 6100 = 60
pounds, and this time I keep the interest in the savings
account; which now has a balance of £1060. At the end
of the second year I am paid interest on the total balance:
that is 1060 × 6100 = 63.60 pounds. Again I keep the
interest in the account which now has a balance of 1060+
63.60 = 1123.60 pounds. At the end of the third year I
earn interest on this balance: that is 1123.60× 6100 = 67.42
pounds. The balance is then 1123.60 + 67.42 = 1191.02
pounds. My cash book looks like this.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 14
Amount at start Interest Total at endYear of year paid of year1 1000 60 10602 1060 63.60 1123.603 1123.60 67.47 1191.02
The future value of my £1000 is and £1191.02, at the end
of year 3. This way I earn 1191.02−1180 = 11.02 pounds
more interest than in Example 1.1.
Definition 2.2. Compound Interest is interest paid on the principal and previ-ously earned interest. In this case interest is calculated, and added to the principal,at various intervals called payment periods. Interest is said to be compoundedover the payment periods.
In Example 2.1 I earned compound interest on a loan to the bank, with pay-ment periods of one year: so the interest was compounded annually.
Example 2.3. Again, suppose I put £1000 into a savings account which paysinterest at a rate of 6% per annum. This time the interest is calculated and payedinto the account every 3 months. The interest over a 3 monthly period is onequarter of the annual interest; so the interest rate for 3 months is 6
4% which is
64×100 = 0.015 when expressed as a multiplier.
My cash book for a 3 year investment is now:
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 15
Payment Amount at start Interest Amount at endperiod of period for period of period1 1000 15 10152 1015 15.23 1030.233 1030.23 15.45 1045.684 1045.68 15.69 1061.375 1061.37 15.92 1077.296 1077.29 16.16 1093.457 1093.45 16.4 1109.858 1109.85 16.65 1126.59 1126.5 16.9 1143.410 1143.4 17.15 1160.5511 1160.55 17.41 1177.9612 1177.96 17.67 1195.63
The future value of my £1000 is and £1195.63, at the end of year 3. This way Iearn 1195.63 − 1191.02 = 4.61 pounds more interest than in Example 2.1. Thisshows that the amount of interest charged using compound interest depends onboth the annual rate of interest and the number of payment periods per year.
The payment period in the above account is 3 monthly: or in the jargon quar-terly. Other common payment periods are
annually : once per year,semi-annually : twice per year,quarterly : 4 times per year,monthly : 12 times per year,daily : 365 times per year,
but in principal any payment period could be used.
2.1 The Compound Interest Formula
To derive a formula for compound interest suppose that we invest
• a principal P0
• at a rate of compound interest of r% per annum,
• where the number of payment periods per year is k.
• Let i = r/100, the annual interest rate expressed as a multiplier; and
• let j = i/k, the interest rate per payment period, as a multiplier.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 16
• Assume that we invest for a total of n payment periods.
Our cash book will look like this.
Payment Amount at start Interest Amount at endperiod of period for period of period1 P0 P0j P1 = P0 + P0j2 P1 P1j P2 = P1 + P1j
where Pn is the future value of P0 after n payment periods. At least this seemsplausible; and later we’ll confirm it. For now assume it’s correct. This means thatafter n payment periods we have
Pn = Pn−1 + Pn−1j = Pn−1(1 + j)
= (Pn−2 + Pn−2j)(1 + j) = Pn−2(1 + j)2
...= P2(1 + j)n−2
= (P1 + P1j)(1 + j)n−2 = P1(1 + j)n−1
= (P0 + P0j)(1 + j)n−1 = P0(1 + j)n.
This is the formula we are after.
Theorem 2.4 (The compound interest formula.). A principal P0 is invested at anannual rate of compound interest of r% per annum, with k payment periods peryear. Expressed as multipliers the annual rate of interest is i = r
100and the rate
of interest per payment period is j = i/k. The future value of P0 after n paymentperiods is then the amount Pn given by
Pn = P0(1 + j)n. (2.1)
2.2 Credit Cards
When you buy something using a credit card, the credit card provider paysfor your purchase. At the end of the month you can choose to pay off the totaloutstanding balance immediately, or pay off the balance in instalments over anextended period of time. If you do not pay off the total balance immediately, youwill be charged interest on the remaining balance until it has been paid.
There are more than sixty million credit cards in circulation in the UK at themoment, and billions of pounds are spent on these each month.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 17
(See http://www.monetos.co.uk/financing/credit-cards/ formore detail.) Interest rates on credit card balances are currently at around 16% to18% per annum compounded monthly.
Example 2.5. Suppose that my credit card account has a balance of £P0 =£1000. Assume that interest is charged at a rate of 18% per annum, compoundedmonthly. If I pay off nothing and spend nothing what will the balance be in oneyear’s time?
Solution. Note that the number k of payment periods
per year is 12. The rate of interest per payment period
is s = rk = 18
12 = 32 = 1.5%. Expressed as a multiplier
this is j = s/100 = 0.015. From the Compound Interest
Formula (2.1), with the number of payment periods n =
12, the amount P12 owed at the end of the year is
P12 = P0(1 + j)12
= P0[1 + 0.015]12 = 1.1956P0.
As P0 = 1000 the amount owing at the end of the year is
This means that the annual simple interest rate is
I
P0× 100 =
195.60
1000× 100 = 19.56.
That is a rate of = 19.56%: pretty high!
In this example we saw that the rate of simple interest paid for a year’s loanon a credit card is higher than the stated (compound) rate of interest. In the nextsection we consider some rates that borrowers or lenders are obliged to declare tomake it easier to understand their interest arrangements.
2.3 APR and EAR
Most lenders (banks, mortgage providers, credit card companies) have feeswhich are charged in addition to interest, on any money borrowed. These may takethe form of set-up charges, insurance, commission and so on. The Office of FairTrading, Consumer Credit Regulations, attempt to clarify the cost of borrowingwith an Annual Percentage Rate (APR), which all lenders must declare. TheAPR is a compound interest rate, calculated by treating fees as though they areadditional interest over the life of the loan. Although knowing the APR makes itmuch easier to compare different loans or investments, some types of charge maynot be included: for example charges for optional payment protection insurance,charges for missing or late payments or going over your credit limit, and balancetransfer fees. Thus APR must be treated with caution. The calculation of APRis not straightforward: we shall introduce it in stages, but begin by looking atanother common method of comparing rates of interest; the EAR.
Example 2.6. In Example 2.3 an investment of £1000 earned compound interestat a rate of 6%, compounded monthly, for 3 years. The balance in the accountafter 1 year was £1061.37 and after 3 years was £1191.02. What annual rate ofsimple interest would result in the same balances.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 19
If we invest P0 at an annual simple interest rate of r%
then, from Theorem 1.4, after n years the balance will be
P3 = P0(1 + ni), where i =r
100.
Thus, in this example, after 1 year the balance will be
P1 = 1061.37 = 1000(1 + i) so i = 0.06137.
Thus to achieve the same balance after 1 year we need an
annual simple interest rate of 6.137%.
After 3 years the balance will be
P3 = 1191.02 = 1000(1+3i) so 3i = 0.19102 and i = 0.06367.
Thus to achieve the same balance after 3 year we need an
annual simple interest rate of 6.3767%.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 20
This example illustrates one way of comparing 2 different rates of compoundinterest. That is, we calculate the annual rate of simple interest which gives thesame result as compound interest; after one year, as in the first part of Example2.6 above. To find a formula for this, suppose that we borrow a principal of P0,at a rate of interest of r% compounded k times per year. We should like to findthe annual simple interest rate s% that would result in the same payment after oneyear. From Theorem 2.4, the future value P1 after one year of compound interestis
P1 = P0
(1 +
r
100k
)k.
From Theorem 1.4, using an annual simple interest rate of s% would instead resultin a future value of
P1 = P0
(1 +
s
100
).
We are assuming that we know r and want to find an s that makes both theseversions of P1 the same. Therefore we solve the equation
P0
(1 +
r
100k
)k= P0
(1 +
s
100
),
for s. We have
P0
(1 +
r
100k
)k= P0
(1 +
s
100
)⇐⇒
(1 +
r
100k
)k=(
1 +s
100
)⇐⇒ s
100=(
1 +r
100k
)k− 1
⇐⇒ s = 100
[(1 +
r
100k
)k− 1
]This leads to the next definition and formula.
Definition 2.7. The effective annual rate (EAR) is the annual simple rate ofinterest that yields the same amount as compounding does after one year.
Theorem 2.8 (The effective annual rate). Assume that a principal of P0 is bor-rowed, at a rate of compound interest of r% compounded k times per year. Thecorresponding EAR is s% where
s = 100
[(1 +
r
100k
)k− 1
]. (2.2)
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 21
Remarks 2.9. 1. The EAR does not depend on the principal P0, but does de-pend on both the annual rate of compound interest, r%, and the number ofannual payment periods, k. Moreover (2.2) shows that s is not linear in r orin k.
2. If we express r/k and s as multipliers, say iear = s/100 and icom = r/100k,then we obtain another version of (2.2):
iear = (1 + icom)k − 1. (2.3)
3. In Example 2.5 we calculated an EAR. A credit card with an annual rate ofinterest of 18%, compounded monthly, was seen to have an EAR of 19.56%.
4. In Example 2.6 we found the EAR was 6.137%. However when we madethe same calculation for a 3 year period instead of a 1 year period, the resultwas different (6.376%). We shall return to this point when we discuss the“internal rate of return” later on.
5. The EAR is sometimes called the Annual Equivalent Rate (AER). Ap-parently one acronym is used for investments and the other for loans. Forinstance, fixed rate savings accounts declare their AER: as you can see, forexample, onmoneyfacts.co.uk/compare/savings/fixed-rate/short-term-bonds.
6. If there are no charges or fees, and the loan is for one year, then the APR isthe same as the EAR.
Example 2.10. A common form of investment in the US is a Certificate of Deposit(CD), which is issued by financial institutions. The institution pays a fixed interestrate on a loan, for a specified term.
A person has the choice between two CD’s, both of which mature in one year.One offers a nominal rate of 7% compounded semi-annually, and the other 6.85%compounded 365 times a year. Which is the better deal?
Solution. We use (2.2) to calculate the EARs s1% and
s2% for deals 1 and 2, respectively. For the first deal we
Another way of comparing two loans or investments is to calculate the presentvalue of the payments made and received. In the following example, knowing thefinal future value, we calculate the (unknown) present value of an investment.
Example 2.11. Suppose that I want to put money in a savings account (perhapsto pay my children’s future university fees) so that after 2 years I have an amountof £10,000. How much must I invest if the account pays interest of 5% per annumcompounded daily. (A year is assumed to have 365 days.)
Suppose that the amount I invest is P0. The annual inter-
est rate is 5/100, as a multiplier, and there are 365 pay-
ment periods each year, so the interest rate per payment
period is 5/36500. The compound interest formula (2.1)
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 23
gives the future value A after 2 years as
A = P2×365 = P0
(1 +
5
36500
)(365×2)
and I want A = 10, 000. Solving this for P0 gives
P0 = 10000
(1 +
5
36500
)−(365×2)
= 9048.44.
Thus, if I deposit £9048.44 now I’ll have £10,000 after 2
years.
In the general case, suppose that
• An amount At is required after t years and
• money can be invested at a compound interest rate of r% per annum,
• with k annual payment periods.
• This will mean that there are a total of n = tk payment periods and that thefuture value Pn should equal At.
To see what At should be the calculation of the previous example is repeated. Theinterest rate per payment period is j = r
100×k , expressed as a multiplier, and thecompound interest formula (2.1) tells us that the future value is
Pn = P0(1 + j)n,
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 24
after n payment periods. Solving for P0 we obtain
P0 = Pn(1 + j)−n.
This gives us a general formula.
Theorem 2.12 (The present value formula). Suppose an investment yields a totalamount of At after t years. If the compound interest rate is r% per annum, with kannual payment periods a year then the present value P0 of the investment is
P0 = At
(1 +
r
100× k
)−tk. (2.4)
Example 2.13. Supose a pub is to be renovated at a total cost of £100,000. Thereare two alternative methods of payment offered:
• pay £100,000 now, in full, or
• pay in four instalments; of £26,000 now, £26,000 in six months, £26,000 inone year, and £26,000 in 18 months time.
Which option is better if money can be invested at an annual rate of 7% com-pounded monthly?
Solution 1.
We calculate the present value of the second option
with four cash flows, using (2.4). To do this we calcu-
late the present values v1, v2, v3 and v4, of each of the
payments, and then add them. Working in pounds, the
present value v1 of the first payment is
v1 = 26000.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 25
The second payment is made after 6 months, so t = 1/2
and for this, and all the other present values, k = 12 and
j = r/100k = 7/1200. Therefore
v2 = 26000
(1 +
7
1200
)−12/2
= 25108.29.
The third payment is made after 1 year, so t = 1, and
v3 = 26000
(1 +
7
1200
)−12
= 24247.17.
The fourth payment is made after 18 months, so t = 3/2
and
v4 = 26000
(1 +
7
1200
)−12×3/2
= 23415.57.
Therefore the present value of this payment method is
As this present value is less than £100,000 the install-
ment plan is better. Moreover, we save £100, 000−£98, 771.03 =
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 26
£1, 228.97, which has a future value of
£1228.97×(1 +
7
1200
)18
= £1, 364.61,
in 18 months time.
Solution 2. Consider the installment plan. We first
pay £26,000 and invest £74,000 for 6 months at 7%. This
gives, at the end of 6 months, £76628.06. Then we pay a
second £26,000, leaving £50628.06. This is invested for
the next 6 months, giving £52426.09. After 1 year we
pay a third £26,000, leaving £26426.09. This is invested
for 6 months, giving £27364.60. After paying the final
£26,000 the amount left is £1364.60, which has a present
value of £1228.96. (This is the same as solution 1 apart
from a 1p rounding error.)
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 27
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 28
3 Annuities
3.1 Definition of an Annuity
In the previous sections we considered investments where a fixed amount ofmoney was deposited in an account that paid interest, compounded periodically.(From the borrowers perspective we considered loans that were made at the be-ginning of a fixed period and repaid in one instalment at the end.) Often, however,money is invested (or repaid) in small amounts at periodic intervals. Examples ofsuch investments are
– annual life insurance premiums,
– monthly deposits in a bank,
– repayments of loans in instalments.
Definition 3.1. An annuity is a sequence of deposits of a constant amount ofmoney at regular intervals, called payment periods. If the deposits are made atthe end of each period the annuity is called ordinary. If the deposits are made atthe beginning of each period the annuity is called due. The sum of all depositsmade plus all interest accumulated is called the amount of an annuity.
We will consider only ordinary annuities in this course. (Due annuities arediscussed in the recommended book [LMW].)
Example 3.2. Suppose that I invest £1000 at an annual rate of interest of 6%compounded annually, as an (ordinary) annuity.
At the end of the first year I make the first payment of
£1000, but receive no interest (because there has been
nothing in the account for the first year). At the end of
the second year the interest on £1000 for one year is £60
and I deposit a further £1000, so the balance in the ac-
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 29
count is £1000 + 60 + 1000 = 2060. At the end of the
third year interest is paid on £2060 at 5%: that is interest
of £103. I also deposit another £1000; so the balance at
the end of the third year is £2060 + 103 + 1000 = 3163.
Continuing this way my cash book at the end of the fifth year looks like this.
Balance Interest BalanceYear at year start paid Deposit at year end1 0.00 0.00 1000.00 1000.002 1000.00 60.00 1000.00 2060.003 2060.00 123.60 1000.00 3183.604 3183.60 191.02 1000.00 4374.625 4374.62 262.48 1000.00 5637.10
To derive a formula for the amount of an annuity we will need to be able tosum a finite geometric series.
Definition 3.3. The nth term of the geometric series with constant a and com-mon ratio q is the sum
Sn =n−1∑k=0
aqk = a+ aq + aq2 + · · ·+ aqn−1.
Theorem 3.4. If q 6= 1 then
Sn = a
(qn − 1
q − 1
).
For example
3(1 + 5 + 52 + · · ·+ 56) = 3
(57 − 1
5− 1
)= 3
78124
4= 234372.
Now, to derive a formula for an ordinary annuity, suppose
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 30
• an amount P is deposited at the end of every payment period;
• the rate of compound interest is r% per annum;
• there are k payment periods per year and
• a total of n payment periods.
• Then the interest rate per payment period, as a multiplier, is j = r100×k .
• The future value at the end of the nth payment period is Pn.
As in Example 3.2, at the end of period 1 the first deposit
of P is made and there is no interest, so the new balance
P1 is equal to P . At the end of the second period interest
of jP1 is paid on the balance P1; and a deposit of P is
made. Thus the balance P2 at the end of the second pe-
riod is P1 + jP1 + P .
The remaining entries are completed in the same way: if the balance at theend of period n− 1 is Pn−1 then the balance at the start of period n is also Pn−1.Therefore the interest for period n is jPn−1. A deposit of P is made at the end ofperiod n so the new balance Pn at the end of period n is
Pn = Pn−1 + jPn−1 + P = Pn−1(1 + j) + P. (3.1)
Then, as in Example 3.2, the cash book entries look like this.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 31
Payment At start of period At end of periodperiod Balance Interest Deposit Balance
1 0 0 P P1 = P2 P1 jP1 P P2 = P1 + jP1 + P3 P2 jP2 P P3 = P2 + jP2 + P4 P3 jP3 P P4 = P3 + jP3 + P5 P4 jP4 P P5 = P4 + jP4 + P...n− 1 Pn−2 jPn−2 P Pn−1 = Pn−2 + jPn−2 + Pn Pn−1 jPn−1 P Pn = Pn−1 + jPn−1 + P
As (3.1) holds for all n ≥ 1 we can use it to find a formula for Pn in terms ofP , j and n. We’ll try this for n = 1, 2 and 3 before attempting the general case.
When n = 1 we have
P1 = P
and there is nothing more to do. When n = 2 we have
P2 = P1(1 + j) + P = P (1 + j) + P,
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 32
since P1 = P . When n = 3 we have
P3 = P2(1 + j) + P
= [P1(1 + j) + P ](1 + j) + P
= P1(1 + j)2 + P (1 + j) + P
= P (1 + j)2 + P (1 + j) + P,
since P1 = P .
It will simplify the notation if we set q = (1 + j). Then we have
Ps = qPs−1 + P,
for all s ≥ 1, and the calculations we’ve just done show that P2 = Pq + P andP3 = Pq2 + Pq + P . To find Pn we repeat this process, as follows.
Pn = Pn−1q + P
= [Pn−2q + P ]q + P
= Pn−2q2 + Pq + P
= [Pn−3 + P ]q2 + Pq + P
= Pn−3q3 + Pq2 + Pq + P
...= P2q
n−2 + Pqn−3 + · · ·+ Pq3 + Pq2 + Pq + P
= [P1q + P ]qn−2 + Pqn−3 + · · ·+ Pq3 + Pq2 + Pq + P
since P1 = P . (At least this seems plausible and we shall verify it later.)Thus
Pn =n−1∑t=0
Pqt. (3.2)
That is, Pn is the nth term of a geometric series with constant P and commonratio q, as in Definition 3.3. From Theorem 3.4 we deduce the following formula.
Theorem 3.5 (The ordinary annuity formula). Suppose that
(i) an amount of P is deposited in an account at the end of each of n paymentperiods, where
(ii) the account pays interest at r% per annum, compounded over the paymentperiods, and
(iii) there are k equal payment periods each year.
Then the future value Pn of the investment after n payment periods is
Pn = Pn−1∑t=0
(1 + j)t = P
((1 + j)n − 1
j
), (3.3)
where j = r100×k .
Proof. As q = (1 + j) the left-hand equality follows from (3.2). From (3.2) andTheorem 3.4 we have also
Pn = P
(qn − 1
q − 1
).
Substitution of 1 + j for q now gives the second equality.
Example 3.6. In the annuity of Example 3.2 we invested £1000 at the end of everyyear, for 5 years, at a rate of 6% compounded annually. Here j = 6/100 = 0.06and if we apply the formula it gives a future value of
P5 = 1000
((1.06)5 − 1
0.06
)= 5637.09.
This is the same as the future value found in Example 3.2, almost. The differenceof 1p is due to rounding error: in Example 3.2 the balance was rounded after eachcalculation.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 34
Example 3.7. Building up a pension fund. A person, at age 35, decides to investin a pension fund. She will put aside £2,000 per year for the next 30 years. Howmuch will she have at age 65 if the rate of return is assumed to be 5% per annum,compounded annually?
Solution. This is an annuity with P = £2, 000, n = 30
and r = 5. The amount Pn of the annuity in her pension
fund after 30 years is
Pn = P ×(1 + r
100)n − 1
r100
= £2, 000×(1 + 5
100)30 − 1
5100
= £132, 877.695.
3.2 The Present Value of an Annuity
We saw, in Theorem 3.5, how to calculate the future value of an ordinary an-nuity: that is, it’s value after n payment periods. If we want to compare annuities,to each other and to other forms of investment, then it would be useful to knowthe present value. That is: the amount we would have to invest today, as a lumpsum, in order to make the amount as the annuity.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 35
Example 3.8. In Example 3.2 an annuity of 5 annual payments of £1000 resultedin a balance of £5637.10 at the end of 5 years. The annual interest rate was 6%,compounded annually.
Suppose that we invest a lump sum P0 in such an account,
at the beginning of year 1. From the compound interest
formula (2.1), after 5 years we have a balance of
P5 = P0
(1 +
6
100
)5
.
To arrange that P5 = £5, 637.10 we solve the equation
for P0 to give
P0 = P5
(1 +
6
100
)−5
=5637.10
1.3382255776= 4212.37 .
That is, if we invest £4,212.37 at an interest rate of 6%,
compounded annually, after 5 years we will have a bal-
ance of £5,637.10. The present value of the annuity is
therefore £4,212.37.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 36
In general suppose that an annuity yields an amount A after time t. The presentvalue of this annuity is the single amount of money that has to be deposited nowin order to result in a balance A at time t. (Here we assume that the rate, andfrequency of compounding, of interest is exactly the same for the annuity and forthe single deposit.)
We can use the argument of Example 3.8 to derive a for-
mula for the present value of an annuity. Suppose that
we have an ordinary annuity set up as in Theorem 3.5.
Then we have formula (3.3) for the future value Pn of the
annuity after n payment periods. Now suppose that a sin-
gle amount P0 is deposited in another account, which has
the same interest rate and payment periods as the annu-
ity. Then after n payment periods we see, from formula
(2.1), that this account has a balance of
A = P0(1 + j)n.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 37
To arrange that A = Pn we therefore need
P0 = Pn(1 + j)−n.
Combining these facts we obtain the following result.
Theorem 3.9 (The present value of an ordinary annuity). The annuity of Theorem3.5 has a present value P0 of
P0 = P
(1− (1 + j)−n
j
), (3.4)
where j = r100×k .
Proof. As above we require P0 = Pn(1 + j)−n and substituting for Pn, from (3.3)this becomes
P0 = P
((1 + j)n − 1
j
)1
(1 + j)n,
from which the result follows by simplification.
Example 3.10. Using formula (3.4) the present value of the annuity of Examples3.2 and 3.8, where P = 1000, j = 0.06 and n = 5, is
P0 = 1000
(1− 1.06−5
0.06
)= 4212.36.
This is what we found before, apart from a rounding error of 1p.
Example 3.11. To understand how the present value of this annuity is made upwe calculate the present value of Example 3.2 in another way; by finding thepresent value of each of the 5 deposits. The first deposit was £P = £1000 at theend of year 1. The present value of this deposit is the amount we would have toinvest now so that we would have £1000 in 1 years time. We can find this using
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 38
the present value formula for compound interest (2.4), with At = P , r = 6 andk = t = 1. This gives a present value for the first payment of £P , where
P(
1 +r
100
)−1=
1000
1.06= 943.40
(in pounds). The second deposit of £1000 was made at the end of the second year,and so (2.4) gives its present value (in pounds) as
P(
1 +r
100
)−2=
1000
(1.06)2= 890.00.
Similarly, the present values of the third, fourth and fifth deposits are
P(
1 +r
100
)−3=
1000
(1.06)3= 839.62,
P(
1 +r
100
)−4=
1000
(1.06)4= 792.09 and
P(
1 +r
100
)−5=
1000
(1.06)5= 747.26
The present value of the annuity is the sum of the present values of the 5 deposits,so is
as we found before. What is more interesting is that we have here a sum which isa term of a geometric series. If we simplify the notation by writing q = (1 + r
100)
then the present value of the kth deposit is Pq−k and so the present value of theannuity P0 is the sum
P0 = P[q−1 + q−2 + q−3 + q−4 + q−5
]= P
[q−1 + (q−1)2 + (q−1)3 + (q−1)4 + (q−1)5
]= Pq−1
[1 + q−1 + (q−1)2 + (q−1)3 + (q−1)4
]= Pq−1
[q−5 − 1
q−1 − 1
]= P
[q−5 − 1
1− q
],
using Theorem 3.4. Recalling that q = 1 + r100
= 1 + j we see that
P0 = P
[(1 + j)−5 − 1
−j
]= P
[1− (1 + j)−5
j
],
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 39
as in (3.4).
Example 3.12. Determining a Pension fund.Earlier, in Example 3.7, we calculated the amount built up in a pension fund,
with a given yearly payment. This raises an obvious question. Suppose we knowhow much we want the pension fund to deliver eventually. How do we decide howmuch to save each year so we’ll end up with this amount? In this example we’llsee how to do this, given some assumptions that simplify the calculations: namelythat we know the interest rates applicable and that they are fixed.
Suppose a person on retirement opens an account paying an annual interestrate of 6% compounded monthly. How much money does he need to invest, asa single lump sum, in his account in order to be able to withdraw £2, 000 eachmonth for 25 years?
Solution. This is the same as asking the present value
£P0 of an annuity of £P = £2, 000 per month at a rate
r% per annum, where r = 6, with k = 12 payment peri-
ods per year, for n = 25× 12 = 300 months. This means
that j = r/100k = 6/1200 and so he will need to invest
£P0 where
P0 = P
(1− (1 + j)−n
j
)=
= 2, 000
(1− (1 + 0.005)−300
0.005
)= 310, 413.728.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 40
That is £310, 413.73.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 41
4 Amortization
A loan with a fixed rate of interest is said to be amortized if both principaland interest are paid off fully by a sequence of equal payments made over equalperiods of time. (The clue is in the mort part of Amortize. The loan is killed offcompletely.)
In order to be able to pay off a given loan in this way we need to be ableto find out how much each of the regular payments should be: we call this anamortization problem.
Example 4.1. Suppose I wish to take out a loan of £1000 and pay it back in3 equal annual instalments, starting a year from now. If the interest rate is 5%compounded annually, how much should each payment be.
During the first year the balance is £1000, so the interest at the end of thefirst year is £1000× 5/100 = £50. Assuming that we make a yearly payment of£R the balance at the end of the first year is then £1000 + 500 − R. This is thebalance at the start of the second year. We cannot yet work out the interest in thesecond year numerically, as we don’t know R. To keep track we make a scheduleof balance and repayment over the length of the loan. As usual we express theinterest rate per payment period as a multiplier j = 5/100 = 0.05.
Balance at Interest paid Payment at Balance atYear year start for year end of year at year end1 P0 = 1000 jP0 = 50 R P1 = P0(1 + j)−R2 P1 jP1 R P2 = P1(1 + j)−R3 P2 jP2 R P3 = P2(1 + j)−R
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 42
We wish to fix R so that P3 = 0. We have
P3 = P2(1 + j)−R
= [P1(1 + j)−R](1 + j)−R
= P1(1 + j)2 −R(1 + j)−R
= [P0(1 + j)−R](1 + j)2 −R(1 + j)−R
= P0(1 + j)3 −R(1 + j)2 −R(1 + j)−R.
To achieve P3 = 0 we therefore need P0(1 + j)3−R(1 +
j)2 −R(1 + j)−R = 0; that is
P0(1 + j)3 = R(1 + j)2 +R(1 + j) +R,
or
P0(1+j)3 = R[(1+j)2+(1+j)+1] = R
((1 + j)3 − 1
j
).
(4.1)
In this example we have P0 = 1000 and j = 0.05 so we
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 43
have to solve
1000× 1.053 = R1.053 − 1
0.05,
so
R = 1000× 1.053 × 0.05
1.053 − 1= 367.21.
The table above, with this value of R filled in, is then (in pounds)
Balance at Interest paid Payment at Balance atYear year start for year end of year at year end1 P0 = 1000 jP0 = 50 367.21 1050− 367.21 = 682.792 P1 = 682.79 jP1 = 34.14 367.21 716.93− 367.21 = 349.723 P2 = 349.72 jP2 = 17.49 367.21 0
and the total amount paid is £3R = £1101.63.Notice that on the left-hand side of (4.1) we have the future value, after 3
years, of £P0 = £1000, while on the right-hand side we have the future value ofannuity, where 3 annual payments of £R are made. We can think of the paymentsas an annuity of equal yearly instalments of £R.
To derive a formula to find the repayment amount R for an arbitrary loan wereason as in the example. Suppose that
• we take out a loan of an amount P0,
• and repay in n equal instalments,
• made at the end of each of n equal payment periods,
• with k payment periods per year,
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 44
• and interest at a rate of r% compounded in each payment period.
• The interest for each payment period, expressed as a multiplier, is then j =r/(100k).
• We wish to find the amount R of each repayment.
As in Example 4.1, if the balance remaining to be paid at the end of paymentperiod t is Pt then this will also be the balance at the start of payment period t+1.The interest for payment period t+1 will then be jPt. This means that the balance(still to be repaid) at the end of period t+ 1 will be
Pt+1 = (Pt + jPt)−R = Pt(1 + j)−R.
As this holds for all t ≥ 1 we can derive a formula for Pt in terms of P0 and R.As in the example, where we derived a formula for P3, we have
Theorem 4.2. Suppose we take out the loan of P0 under the conditions outlinedabove and repay an amount R at the end of each payment period. Then at the endof the t-th payment period the balance Pt to be repaid is
Pt = P0(1 + j)t −R(
(1 + j)t − 1
j
)=
(P0 −
R
j
)(1 + j)t +
R
j, (4.2)
where j = r/(100k). To ensure that the loan is repaid after n repayment periodsthe repayment amount R should be
R = P0(1 + j)n(
(1 + j)n − 1
j
)−1= P0
(j
1− (1 + j)−n
). (4.3)
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 45
Proof. The first equality of (4.2) follows from the equations just above the the-orem combined with Theorem 3.4. The second equality of (4.2) is obtained byrearranging the middle expression to give the right-hand side. Equation (4.3) fol-lows directly from (4.2).
Example 4.3. Mortgage Payments. A mortgage is a loan used to buy a house.More precisely, it’s a loan which is secured on a property. That is, the bank orbuilding society which makes the loan will become the owner of the property ifthe borrower fails to make the repayments.
Suppose a person takes a mortgage loan for the amount £100, 000, for 30years, at an annual interest rate of 9%, compounded monthly. The mortgage is tobe repaid in equal monthly instalments over the 30 years.
(a) What are the monthly payments?
(b) What amounts of the monthly payment go towards capital repayment of theloan at the ends of the first 2 months and at the ends of the 2 last months?
(c) What amounts of the monthly payment go for interest at the ends of the first2 months and at the ends of the 2 last months?
Solution. (a) By the amortization formula (4.3), the
monthly payment £R
needed to pay off the loan of £P0 = £100, 000 at an
interest rate of r% per annum, where r = 9, compounded
over k = 12, monthly payment periods per year, over n =
(b) and (c) We have to find the amounts £Pt of the
monthly payment £R = £804.62 which go for capital
repayment at the ends of the first 2 months, that is, for
t = 1, 2; and at the ends of the 2 last months, that is, for
t = 359, 360. The remainder of £804.62 is interest paid
at the end of the corresponding month.
The amount of the capital balance paid off in payment
t is the difference between the amount £Pt−1 and £Pt.
That is, the amount of capital paid off in payment t is
£Pt−1 − Pt.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 47
By the amortization formulae (4.2),
Pt−1 − Pt ={(
P0 −R
j
)(1 + j)t−1 +
R
j
}−{(
P0 −R
j
)(1 + j)t +
R
j
}=
(P0 −
R
j
)(1 + j)t−1 (1− (1 + j))
= −j(P0 −
R
j
)(1 + j)t−1
= (R− P0j) (1 + j)t−1.
Let
M = (R− P0j) = 804.62− 9
1200× 100000 = 54.62.
Then the amount of capital paid off in payment t is £Ct,
where
Ct =M(1 + j)t−1 = 54.62× 1.0075t.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 48
We have
C1 =M(1 + j)0 = 54.62× 1.00750 = 54.62,
C2 =M(1 + j)1 = 54.62× 1.00751 = 55.02965,
C359 =M(1 + j)358 = 54.62× 1.0075358 = 792.6497 and
C360 =M(1 + j)359 = 54.62× 1.0075359 = 798.5946.
Therefore only £54.62 of the £804.62 paid during the
first month goes towards capital repayment and
I1 = R−C1
= £804.62−£54.62 = £750
is interest.
For the second month only £55.03 of the £804.62 paid
goes towards capital repayment and
I2 = R−C2
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 49
= £804.62−£55.03 = £749.59
is interest.
£792.65 of the £804.62 paid during the 359th month
goes towards capital repayment and only
I359 = R−C359
= £804.62−£792.65 = £11.97
is interest.
£798.59 of the £804.62 paid during the last (360th)
month goes towards capital repayment and only
I360 = R−C360
= £804.62−£798.59 = £6.03
is interest.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 50
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 51
5 Net Present Value and Internal Rate of Return
So far we’ve seen two methods of comparing loans. In Section 2.3 we usedthe effective annual rate (EAR) and is Section 2.4 the present value. Now weshall extend and generalise both these methods of comparison. We shall see howto calculate present values for accounts where there are repeated deposits andwithdrawals (as for example in the case of annuities). Then we shall calculate ancompound interest version of the EAR, which will also cope with accounts wheremany transactions are carried out over time.
5.1 Cash Flow
A cash flow is a list of amounts and times of money paid and received. For ex-ample, if I lend you £100 on 25/12/2011 and you pay me £5 interest on 25/12/2012,£6 interest on 25/12/2013 and then finally £104 on 25/12/2014, my cash flow list-ing these events is
Date 0 1 2 3Amount (£) -100 5 6 104
where I’m assuming that I’ve noted somewhere that date 0 is 25/12/2011, and thedates 1, 2 and 3 refer to the ends of year long payment periods.
The -100 indicates that I pay out £100 (an outflow). Positive figures representmoney I receive (inflows). Your cash flow looks the same except that the signs ofthe amounts in the second row are all reversed.
5.2 Net Present Value
If I have an account with cash flows at regular intervals then I can use thepresent value formula (2.4) to calculate the present value of each inflow and out-flow. This calculation requires an interest rate; which could, for example, be theinterest rate that money would earn elsewhere (the prevailing rate of interest).I can then use (2.4) for each item of the cash flow, sum all these present values,and obtain an overall present value for the cash flow. This leads to the followingdefinition.
Definition 5.1. The Net Present Value (NPV) of an investment or loan (withrespect to a given interest rate and frequency of compounding) is the sum of thepresent values of all the future cash flows (calculated using the given interest rate).
Example 5.2. For instance, with the cash flows of the
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 52
table above, suppose that the prevailing interest rate is 5%
compounded annually. Then the present value of each
cash flow can be calculated as follows. The −£100 at
time 0 has a present value of −£100. The amounts of
£5, £6 and £104 at times 1, 2 and 3, respectively have
That is: the investment with the given cash flows is £2.62
worse than a savings account with an interest rate of 6%.
A common use of NPV is to determine whether or not
a required target is met by a cash flow. For example, if an
investor requires a return of 5% per annum, compounded
annually, over 3 years, then an investment with the cash
flow of the table above would be ok. If the investor re-
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 54
quires 6% per annum: we should advise against such an
investment.
In general, assume that there are k payment periods per year, and that the cashflow is C0 at the beginning of payment period 1 (time 0), C1 at the end of paymentperiod 1, . . ., Cn at the end of payment period n. If the given interest rate is r% perannum, compounded over the k payment periods per year, then let j = r/100k,as usual. From (2.4) the present value of the cash flow Cp at the end of paymentperiod p is
Vp = Cp(1 + j)−p.
This gives us the following theorem.
Theorem 5.3. Suppose an account has k equal payment periods per year, and acash flow sequence of amounts C0, C1, . . . , Cn, where C0 is to be received at thebeginning of payment period 1 and Cp is to be received at the end of paymentperiod p, p = 1, . . . , n.
Given an interest rate of r%, compounded over the k payment periods, the NetPresent Value of this sequence of cash flows is
NPV =n∑
p=0
Cp × (1 + j)−p , (5.1)
where j = r/100k.
Example 5.4. Suppose that you have option of saving money in two differentinvestments. Both have payment periods of 6 months and you wish to invest£2000 for 2 years. The cash flows for both the investments are as shown below.On the basis of these cash flows, decide which is the better investment assumingthe interest rate is
We use (5.1) to calculate the NPV of each investment
in each of cases (i) and (ii).
(i) The interest rate is 7%, with k = 2, so j = 7/200 =
0.035 and 1 + j = 1.035. The total number of pay-
ment periods is n = 4.
For investment 1 we have a NPV of £V1 where
V1 = −2000 + 100× 1.035−1 + 200× 1.035−2
+ 200× 1.035−3 + 2200× 1.035−4
= 380.88.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 56
For investment 2 we have a NPV of £V2 where
V2 = −2000 + 350× 1.035−1 + 200× 1.035−2
+ 130× 1.035−3 + 2000× 1.035−4
= 385.00.
Therefore, if the interest rate is 7% per annum, in-
vestment 2 is better.
(ii) The interest rate is 5%, with k = 2, so j = 5/200 =
0.025 and 1 + j = 1.025. The total number of pay-
ment periods is n = 4.
For investment 1 we have a NPV of £V ′1 where
V ′1 = −2000 + 100× 1.025−1 + 200× 1.025−2
+ 200× 1.025−3 + 2200× 1.025−4
= 466.74.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 57
For investment 2 we have a NPV of £V ′2 where
V ′2 = −2000 + 350× 1.025−1 + 200× 1.025−2
+ 130× 1.025−3 + 2000× 1.025−4
= 464.44.
Therefore, if the interest rate is 7% per annum, in-
vestment 1 is better.
5.3 Rate of Return
Another way to compare cash flows is to use rates of return. At their simplest,these are just another way of describing simple interest rates. Suppose an amountS is invested and after a certain time yields an amount T . Then we say that therate of return on this investment is
s =T − SS
.
(Or, as a percentage, 100 × (T − S)/S = 100s%.) Rearranging this expressiongives
T = S + sS = S(1 + s).
If the investment lasted a period of t years and we set i = s/t then
T = S(1 + ti),
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 58
so, comparing to Theorem 1.4, we see that s/t is the annual rate of simple interestthat results in the amount T , if S is invested for t years.
Next we’ll consider the rate of return for a more general cash flow. First wemake the following restriction, which is unlikely to make any difference to ourcalculations and avoids the possibility that we divide by zero in some situations.
For the rest of this section we consider only interest rates i% such that
i
100> −1.
5.4 Internal Rate of Return
The rate of return above is a simple interest rate. In the next example we finda compound interest rate which corresponds to a simple cash flow.
Example 5.5. Suppose £1000 is invested in an account, and after 10 years (withno further deposits or withdrawals) the account has a balance of £2000. Assumingthat interest is compounded semi-annually, what annual rate of compound interestwould result in this balance?
Suppose the (unknown) rate is r%. We shall use the
compound interest formula (2.1) to find r%. There are
two payment periods a year, so let k = 2 and set j =
r/100k = r/200. There are a total of n = 10k = 20 pay-
ment periods, so from the compound interest formula, we
have Pn = P0(1 + j)n, where P0 = 1000 and Pn = 2000.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 59
That is
2000 = 1000(1 + j)20.
Solving for j we find
j = 2120 − 1.
Therefore
r = 200j = 200(2120 − 1) = 7.052.
That is: if the annual rate of interest is 7.052%, com-
pounded semi-annually, then the balance will be £2000
after 10 years.
In the previous example, if we assume that the account is closed after 10 yearsand the balance withdrawn, then the cash flow over the 20 payment periods is
We calculated the value r% such that, the future value of −1000, at the end of thetwentieth payment period, using a percentage rate r%, was −2000: or
−1000(1 + j)20 + 2000 = 0,
where j = r/200. Thus the sum of the future values of the cash flow is 0 (and thenthe account is closed). Put another way, we could say that we found the interestrate r% so that, compounding semi-annually, the NPV of the cash flow is 0. Infact this follows from the equation above, which implies
−1000 + 2000(1 + j)−20 = 0,
and the net present value formula (5.1).In general suppose we have a cash flow sequence for an account, over n pay-
ment periods, of C0, . . . , Cn. We assume that after the nth payment period theaccount is closed and has a balance of 0. We wish to find the annual interest rater%, such that if we compound over the payment periods then we obtain the samecash flows. This means that, when we use this rate, the sum of the future valuesof the cash flows should be 0. Equivalently, the sum of the present values, whichis the NPV, should be 0. This is the motivation for the following definition.
Definition 5.6. Let C0, . . . , Cn be a cash flow, over n regular payment periods.The internal rate of return (IRR) of this cash flow is the annual rate of interestr%, such that the NPV of the cash flow (with respect to this interest rate com-pounded over the payment periods) is 0.
Remark 5.7. It is not necessarily the case that such a number r exists, as we shallsee below. For more discussion of the existence and uniqueness of internal rate ofreturn see the book [LMW].
We can repeat the argument of Example 5.5 to derive the following expressionfor the IRR.
Theorem 5.8. Let C0, . . . , Cn be a cash flow, over n regular payment periods,with k payment periods a year. Then the IRR (if it exists) is r%, where
Moreover, r% is the interest rate such that the sum of the future values of the cashflows after n payment periods is 0.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 61
Proof. The first part of the theorem is just the definition
of NPV, Definition 5.1. If we have an interest rate of
r% then the future value, at the end of the nth payment
period, of the cash flow Cp, which occurs at the end of
the p-th payment period can be calculated using (2.1) and
is
Cp(1 + j)n−p,
where j = r/100k.
Multiplying both sides of (5.2) (or (5.1)) by (1 + j)n
we have
n∑p=0
Cp(1 + j)n−p = 0.
That is, the sum of future values of the cash flows is 0, as
required.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 62
Example 5.9. Returning to Example 5.5 we have k = 2, n = 20 and only twonon-zero cash flows, namely C0 = −1000 and C20 = 2000. Using Theorem 5.8,the IRR is r%, where r = 200j and
C0 + C20(1 + j)−20 = −1000 + 2000× (1 + j)−20 = 0
which implies that j satisfies
1000× (1 + j)20 = 2000
exactly as in Example 5.5.
Example 5.10. Suppose that you wish to save by investing money every month.You are going to pay £250 into an investment account at the beginning of everymonth, for two years. The manager of the investment account tells you that at theend of the two years you will have a balance of £9000. Assuming this to be true(and that interest is compounded monthly) what is the IRR on this investment?
(The 1st payment is at the beginning of the 1st month, which is time 0. The 24thpayment is at the beginning of the 24th month, which is the end of the 23rd.)
We have k = 12 and n = 2k = 24. Also C0 = C1 =
· · ·C22 = C23 = −250 and C24 = 9000. From Theorem
5.8, the IRR is r%, where r = 200j and j satisfies
−250− 250(1 + j)−1 − 250(1 + j)−2 − · · ·
· · · − 250(1 + j)−23 + 9000(1 + j)−24 = 0.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 63
Multiplying both sides of this equation by (1 + j)24 we
see that j satisfies
−250(1 + j)24 − 250(1+j)23 − 250(1 + j)22 − · · ·
· · · − 250(1 + j) + 9000 = 0,
which can be written as
250
24∑p=1
(1 + j)p = 9000,
or
24∑p=1
(1 + j)p =9000
250= 36,
Using Theorem 3.4 we have
24∑p=1
(1 + j)p = (1 + j)
23∑p=0
(1 + j)p
= (1 + j)
((1 + j)24 − 1
(j + 1)− 1
)= (1 + j)
((1 + j)24 − 1
j
).
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 64
Therefore j satisfies
(1 + j)
((1 + j)24 − 1
j
)= 36.
This is not something that we can solve analytically, but
we can use numerical approximations to find solutions
which are quite accurate enough for these purposes. For
instance we could use the bisection method or the Newton-
Raphson method. This can be done in Maple and we find
a solution
j = 0.03111 so r = 200j = 6.222.
Note that what we have found r = 6.222, such that if
we open a savings account and
• deposit £250 every month,
• for 24 consecutive months,
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 65
• then, if the interest rate is r% per annum, compounded
monthly,
• we shall finish up with a balance of £9000.
The next two examples illustrate some of the difficulties that may arise whentrying to find the IRR of a cash flow.
Example 5.11. Calculate the IRR of the following cash flow, which has paymentperiod of one year.
Date 0 1 2Amount (£) -1000 2000 -1100
Here k = 1 and n = 2 and, according to Theorem 5.8,
we want to find j which satisfies
−1000 + 2000(1 + j)−1 − 1100(1 + j)−2 = 0.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 66
We have
−1000 + 2000(1 + j)−1 − 1100(1 + j)−2 = 0
⇐⇒ −1000(1 + j)2 + 2000(1 + j)− 1100 = 0
⇐⇒ −1000(1 + 2j + j2) + 2000j + 900 = 0
⇐⇒ −100− 1000j2 = 0
⇐⇒ 1 + 10j2 = 0
⇐⇒ j2 = − 1
10.
There is no real number j satisfying this equation, so
the cash flow has no IRR.
Example 5.12. Calculate the IRR of the following cash flow, which has paymentperiod of one year.
Date 0 1 2Amount (£) -1000 2205 -1215
Here k = 1 and n = 2 and we find j by solving the
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 67
equation given by Theorem 5.8:
−1000 + 2205(1 + j)−1 − 1215(1 + j)−2 = 0
⇐⇒ −1000(1 + j)2 + 2205(1 + j)− 1215 = 0
⇐⇒ −1000(1 + 2j + j2) + 2205j + 990 = 0
⇐⇒ −10 + 205j − 1000j2 = 0
⇐⇒ 2− 41j + 200j2 = 0
⇐⇒ j =41±
√412 − 4× 200× 2
2× 200
⇐⇒ j =41±
√81
400
⇐⇒ j =50
400or
32
400
⇐⇒ j =1
8or
2
25.
As k = 1 we have r = 100/8 = 12.5 or r = 200/25 = 8 and both values of rcan be regarded as the IRR of this cash flow.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 68
An examination of the balances of accounts with these interest rates may helpto shed light on this apparently strange result and on IRR in general.
If the interest rate is 12.5% a year then the balance sheet for the account is thefollowing. (All amounts are in pounds.)
Payment Balance at Interest Payment at Balance atPeriod start of period for period end of period end of period
The UK Consumer Credit Regulations 2010 oblige anyone advertising creditto state the cost to the borrower in the form of an interest rate called the AnnualPercentage Rate (APR). The advertiser must include in the calculation of thisrate any non-interest charges, such as commission, set-up fees and so on. If thepayment periods for the credit agreement are regular and all the cash flows (thatis all payments and repayments) are known, then the APR will be the IRR.
In practise, as interest rates are constantly changing, a loan agreement doesnot usually specify the precise amounts of repayment to be made over the life ofthe loan. These will vary with the prevailing interest rates. Therefore, what theadvertiser is obliged to do is to provide an example of what the IRR would be,under certain assumptions about interest rates and the behaviour of the borrower.The Office of Fair Trading documentation says
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 69
“... if an advertisement includes an interest rate ... this triggers a ’represen-tative example’ including a ’representative APR’ ... and at least 51 per cent ofborrowers must be expected to get the advertised APR or better. ... The repre-sentative example must include ... any non-interest charges required to be paidfor the credit, and the amount of credit ... the duration of credit, the total amountpayable, the periodic instalments, and the cash price of goods or services ... .”
In any event, the calculation of the APR, once suitable assumptions have beenmade, is the calculation of the IRR of the expected cash flows of the agreement... except that, in most situations there are some charges that do not have to beincluded! (For example, the APR for a mortgage on a house purchase does nothave to include survey fees or mortgage insurance fees.) Of course the phrase“must be expected to get” could be used as a convenient escape clause by anunscrupulous advertiser.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 70
6 Financial Bonds
One method of raising money, used by governments and corporations, is toissue bonds. A bond is a loan, from an investor to a borrower, who is either agovernment or a corporation. The borrower pays interest to the investor at regularintervals. The interest payments are called coupons. At some fixed time in thefuture, called the redemption or maturity date the borrower agrees to repay theloan to the investor. Bonds play an important role in the global economy. “Quan-titative Easing”, a term which has become well-known over the last 3 or 4 years,is the process by which governments attempt to restore health to their economiesby buying up bonds. We’ll return to this later.
Background material on bonds is covered in videos “The Basics of Bonds”and “Corporate Bonds”, both by Tim Bennett, an editor of the magazine MoneyWeek. The videos can be found on the Money Week websitewww.moneyweek.com/investments/bondsor on the webpage for this modulewww.staff.ncl.ac.uk/alina.vdovina/teaching/mas1243
The main points of these videos are summarised in the next two sub-sectionsof these notes.
6.0.1 Government Bonds
The government has income, from taxes for example, which it spends on schools,universities, hospitals, roads and so on. It may be that these costs are greaterthan the government’s income, so the government has to borrow money. Oneway it does this is to issue IOUs; that is bonds. Government bonds in the UKare known as “gilts” (apparently because they used to be sold as certificates withgilded edges). For example, here is a line (adapted) from the Financial Times (FT)on 24th September 2011, under the heading “Bonds”:
Treasury 3.75 2021 112.25
• “Treasury” indicates that the bond was issued by the UK government. Infact it is the Central Bank that issues bonds, not the government itself. Inthe UK that means the Bank of England. In the US the central bank is theFederal Reserve.
• “2021” indicates that the bond has a maturity date on some fixed day in2021. On this day the government will repay the loan to the investor (or tothe owner of the bond on that date).
• Each bond has a fixed face value (or nominal value) which is the amountwhich the holder of the bond will be paid on the maturity date. In the news-paper line above the face value is £100, by convention.
• “3.75” indicates that the interest rate is 3.75%. Interest is paid on the facevalue, so is the same every year. In this case an amount of £3.75, called thecoupon will be paid to the investor every year (for each bond with a facevalue of £100). The coupon will be paid in two instalments; six monthsapart.
• “112.25” is the market price, in pounds, of the bond (assuming the facevalue is £100). The bond may be be sold by the investor at any time be-fore maturity. If the current interest rate (on other investments or bonds) ismore that 3.75% then you would probably not want to buy this bond fromme for £100, because you could get a better return by investing your £100elsewhere. Therefore the market price would probably fall below £100. Onthe other hand if 3.75% is greater than the current interest rate on similarinvestments then the market price is likely to be more than £100.
Government bonds are similar to Treasury Bills, which we touched on in Section1.3: except that bonds pay interest and Treasury bills do not. (In fact TreasuryBills are called “zero-coupon bonds”.)
Government bonds attract investors for two main reasons.
– Firstly they are secure. If you lend money to a person or institution whichgoes bust, then you’ll probably lose all your money. Investors buying agovernment bond hope and expect that the government will not go bust.Lending money to, say a new company, is clearly more risky. As bonds areseen as safe they are commonly used by pension funds which are approach-ing retirement date, since at this stage security is much more important thangrowth.
– Secondly, for a secure investment, they can offer a reasonable rate of return:at least better than putting the money a bank account (which may also besecure but is likely to have a worse rate of return).
6.0.2 Corporate Bonds
Companies may also issue bonds as a means of raising money. If a company issuccessful it may be able to build up as much money as it needs just by investingits profits. However, if the company wants to expand it may need to raise moneymore quickly than its profits come in. There are 4 main ways a company can raisemoney.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 72
1. Making profit, as discussed. This is called “organic growth”.
2. Taking out a bank loan. This may be expensive.
3. Issuing a bond. For example, suppose Tesco wishes to raise 1 million poundsto build new stores. As an alternative to borrowing from the bank it couldissue 10,000 bonds each with a face value of 100 pounds. Essentially it wouldborrow 100 pounds from each of 10,000 investors. This may be a cheaper wayof raising the money than taking out a bank loan.
4. Issuing shares, which means taking in more owners. We’ll go into this optionin a later Section. This is likely to be the most expensive way of raising money.
Here is more information from the FT, in the section headed “Retail Bonds” on25/08/2011.
Tesco 5.2 24/08/18 105.80 6mth
This is a corporate bond issued by Tesco. The bond has a coupon rate of 5.2%,a maturity date of 24/08/18, and a payment period of 6 months. Its face value is£100 and its price on 25/08/2011 was £105.80.
As with government bonds, various factors affect the prices of corporate bonds.
– If interest rates elsewhere are rising then the bond, with its fixed couponrate, becomes less attractive.
– If the maturity date is well in the future it does not have a big influence onthe price. On the other hand, when the maturity date is close then the priceof the bond will approach the face value. This is because, on the maturitydate, the bond holder will be paid exactly the face value of the bond, by thebond issuer.
– If the company is seen to be secure the bond will be attractive and this willkeep the price up. The risk of the company going bust is called the defaultrisk. There are firms called “ratings agencies” who give ratings to all bondson the market: from AAA to D, where AAA is as good as possible, andD means likely to go bust (default). Bonds with ratings between BBB andAAA are regarded as of good enough quality to invest in. For example Stan-dard & Poor’s, Moodies, and Fitch are three well known ratings agencies.
6.1 Bond Basics
To see, in more detail, how bonds work let’s suppose we have a bond with thefollowing characteristics.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 73
• The bond has a face value of F . This is the amount that the issuer will paythe holder of the bond on the maturity date.
• The bond has k payment periods per year. Government bonds have 6 monthpayment periods, so k = 2. Corporate bonds may have other paymentperiods.
• The bond has n payment periods until redemption.
• The bond has an annual coupon rate of c%. We write d = c/100k; so d isthe coupon rate per payment period, expressed as a multiplier.
• The bond has a price P0, which is the current market value of the bond: thatis how much it is being traded for today (or at close of trading yesterday).Usually this price is quoted as a percentage of the face value, as in thenewspaper cuttings reproduced above. For simplicity however we take P0
to be the actual amount the bond is fetching today.
Remarks 6.1. 1. We are assuming that the bond has n payment periods left torun. It may be that the bond has just been issued, and so n is the total numberof payment periods of the bond, or that the bond is already part way throughit’s life. Our analysis covers both possibilities.
2. The price of the bond depends on the coupon rate and the number of paymentperiods n to run. It also depends on how much of the current payment periodis left to run. To make life easier we’ll always assume that we make our calcu-lations on the first day of the current payment period. This means that the nextcoupon will be paid in exactly 12/k months time. What happens to the pricepart way through a payment period is discussed in Chapter 8 of [LMW].
6.2 The Present Value of a Bond
A key notion when considering investing in a bond is that the coupon rate isfixed for the life of the bond, but the market interest rate is variable. This meansthat investment in bonds involves an element of betting against future prevailinginterest rates.
The price that a buyer is willing to pay for a bond is based on its presentvalue, as well as other considerations such as security. The present value of abond is the NPV of the cash flow consisting of all its future coupon payments,plus the final face value payment at maturity. These present values are calculatedusing the current prevailing interest rate: that is the interest rate that the buyercould get by investing the money elsewhere. Of course this is not a fixed andeasily identifiable quantity; interest rates vary alot between different investments,
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 74
even on a particular day, let alone over the life of a bond, which may be severalyears. Thus we certainly have to make some assumptions about the prevailinginterest rate. Here, to make our calculations reasonably straightforward, we shallassume a fixed annual prevailing interest rate of r%, compounded each paymentperiod. As usual we write j = r/100k, the interest rate per payment period, as amultiplier.
We next derive a formula for the present value of a bond, with respect to agiven interest rate.
Theorem 6.2. Assume that a bond has a face value of F , k payment periods peryear, a total of n payment periods to maturity and a coupon rate of c%. Writed = c/100k.
Given an annual interest rate of r%, compounded over the payment periods,set j = r/100k. Then the present value of the bond (with respect to the interestrate r%) is
V0 = F
[d
j
(1− 1
(1 + j)n
)+
1
(1 + j)n
]. (6.1)
Proof. We calculate the present value of the coupons first.
Each coupon payment is for an amountC = d×F , which
is fixed for the life of the bond. Therefore the coupon
payments can be regarded as an ordinary annuity, of a
fixed amount C, over n payment periods, with k pay-
ment periods per year, and an interest rate of r% com-
pounded over the payment periods. From Theorem 3.9
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 75
the net present value of the coupon payments is therefore
U1 = C
(1− (1 + j)−n
j
)= d× F
(1− (1 + j)−n
j
).
The final payment is a single sum of F paid at the end of
the nth payment period, so its present value is given by
(2.4) and is
U2 = F (1 + j)−n.
Therefore the present value of the bond is
V0 = U1 + U2
= d× F(1− (1 + j)−n
j
)+ F (1 + j)−n
= F
[d
(1− (1 + j)−n
j
)+ (1 + j)−n
]= F
[d
j
(1− (1 + j)−n
)+ (1 + j)−n
],
as claimed.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 76
Example 6.3. Suppose a bond has a face value of £1000, a coupon rate of 5%and a 6 monthly payment period. What is the coupon payment at the end of eachpayment period? Suppose there are exactly 7 years until maturity and that theprevailing interest rate is 6%. What is the present value of the bond?
Solution The annual coupon rate is 5% and there are 2
payment periods per year, so k = 2. Therefore d = 5/200
and the coupon payment is d × F = 5 × 1000/200 = 25
pounds.
The prevailing interest rate is 6%, so j = 6/200. There
are 7 years to maturity, so there are a total of n = 14
payment periods to run. Hence, using (6.1), the present
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 77
value is
V0 = 1000
[(5/200)
(6/200)
(1− 1
(1 + (6/200))14
)+
1
(1 + (6/200))14
]= 943.52
The present value of this bond is thus £943.52. If the in-
terest rates stay at 6% then probably the price of the bond
will be below £943.52.
Example 6.4. To calculate the present value of the Tesco bond above we make asimilar calculation. The face value is £100. The coupon rate is 5.2% and againk = 2. Hence d = 0.026 and the coupon payment is d × F = 2.6 pounds, every6 months. The bond has exactly 7 years to run, so again n = 14. As a roughestimate let’s say the prevailing interest rate is 3%. Substituting these values into(6.1) gives a present value of £V0, where
V0 = 113.80.
The price in this case is £105.80; so less than the present value, so possibly agood buy. But is it? What is the rate of return realised by buying this bond at thisprice?
6.3 The Yield to Maturity of a Bond
A first attempt to measure the return from a bond is the flat yield (FY) ofthe bond. This is the annual coupon as a percentage of the current price. In thenotation we have established above, the annual coupon is C = cF/100 and so theFY is (C/P0)× 100, that is
FY =c× FP0
.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 78
The flat rate takes no account of the face value payment, or of the times at whichpayments are made. If the maturity date is well in the future, then it’s a usefulguide, but as maturity date approaches the face value payment becomes a moreimportant issue, and other measures are needed.
A more accurate idea of return is given by the IRR of all the cash flows of thebond, from the time it’s bought to the maturity date. This is called the Yield toMaturity (YTM) of the bond.
Example 6.5. Let’s consider the Tesco bond again, and assume that we bought itfor £105.80 on 25/08/2011. What is the FY and what is the YTM of the bond?
The FY iscF/P0 = 5.2× 100/× 105.8 = 4.91%.
The YTM is the IRR so is, from Theorem 5.8, the rate of interest which makesthe net present value of the cash flow 0. The cash flows of this bond consist ofthe price we paid, the coupon payments and the final payment, of the face value,on the redemption date of 24/08/2018. There are 14 payment periods to go, each6 months long, and the coupon payments are £2.60 at the end of each one. Thecash flows are therefore
(The -105.80 is the initial payment and the 102.60 is the final payment of 100 plusthe final coupon of 2.60.) Let’s write rirr for the IRR and let jirr = rirr/100k. Inthis example jirr = rirr/200. Now we can use (6.1) to write down the equation forthe IRR. In (6.1) we have a formula for V0, the NPV of all the future cash flows,with respect to a given interest rate r%, where j = r/200. To obtain the NPVof all cash flows we must add the present value of the initial payment, which is−£P0, with P0 = 105.80, in this example. Therefore jirr is the solution to thefollowing equation.
− P0 + F
[d
jirr
(1− 1
(1 + jirr)n
)+
1
(1 + jirr)n
]= 0. (6.2)
We have d = 5.2/200 = 0.026 and F = 100, so the equation is:
−105.80 + 100
[0.026
jirr
(1− 1
(1 + jirr)14
)+
1
(1 + jirr)14
]= 0.
This equation is not so easy to solve analytically, but using approximation, suchas Newton-Raphson, we find a solution
jirr = 0.02117.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 79
Therefore the yield to maturity is rirr% = 200jirr% = 4.234% per annum. Thiscompares favourably with other interest rates in August 2011, so the bond doesseem attractive.
In Tim Bennett’s video an approximation to the yield to maturity was de-scribed. This was calculated by first finding the difference between the currentprice and the face value: namely P0 − F = 5.80. (Normally the price is higherthan the face value, so we expect this to be positive, but for the buyer it will bea loss.) This loss is spread over the remaining years of the bond: if there are kpayment periods per year and n payment periods to go, there are n/k years left.Therefore the loss per year will be
L =P0 − F(n/k)
=k(P0 − F )
n=
2× 5.8
14= 0.83.
The yearly return is then estimated as the the annual coupon payment minus theannual loss. The annual coupon payment is C = cF/100 = 5.2, so the yearlyreturn is C−L = 5.2−0.83 = 4.37. Finally the approximate YTM is the yearlyreturn divided by the current price, as a percentage. That is the approximate YTMis
C − LP0
× 100 =4.37
105.80× 100 = 4.13.
The approximate YTM, unlike the FY, accounts for the final value, but makesno consideration of the timing of the cash flows over the life of the bond. How-ever, as can be seen in this example it is a good enough approximation for manypurposes. One use for it would be as an initial value for the Newton-Raphsonmethod, when finding the actual YTM.
We record the underlying findings of the example above in a theorem.
Theorem 6.6 (Yield to maturity of a bond). The YTM of the bond of Section 6.1is rirr%, where rirr = jirr × k × 100, for jirr satisfying the equation (6.2) above.The approximate yield to maturity is a%, where a is given by the formula
a =100
P0
[c× F100
− k(P0 − F )
n
].
Proof. The proof of the first statement is contained in Example 6.5. For the secondstatement; from the example we had a = 100(C −L)/P0. Here C = cF/100 andL = k(P0 − F )/n, so the result follows.
From this Theorem and Theorem 6.2 we see that the YTM is the percentagerate r% that makes the price P0 equal to the present value of the bond V0. As men-tioned after the proof of Theorem 6.2, as the interest rate increases the presentvalue decreases, and vice-versa. Therefore, as the YTM increases the price de-creases, and vice-versa. This is what we should expect: if we pay more for thebond, this decreases the YTM, whereas if we pay less, this increases the YTM.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 80
6.4 Quantitative Easing
Quantitative Easing is one of the methods being used by governments aroundthe world to try to restore health to the global economy. When banks, companiesand individuals become nervous of borrowing and lending money, the economyslows down and in extreme situations, like the present, this may endanger theviability of banks and of the economy of entire countries.
To help restore confidence and encourage banks to lend and companies toborrow, governments try to spend alot of money: so this money is injected intothe economy, to start things moving again. One way governments can do this, isby buying government and corporate bonds: and this is Quantitative Easing (QE).The idea is that this will increase the price of bonds, so depress the yield. It ishoped that the effect of this will be that the banks will stop investing in securegovernment bonds and will lend money to companies and individuals. Whetheror not this works is a matter of debate. A more detailed discussion of QE is givenon the video “Quantitative Easing”, by Tim Bennett of Money Week, shown inthe lecture. (See the webpage for this modulewww.staff.ncl.ac.uk/alina.vdovina/teaching/mas1243.)
7 Variable, Continuous and Continuously Variable, Interest Rates
7.1 The Variable Interest Rates
In practise interest rates on most loans and investments vary from one paymentperiod to another. In this case we say the interest rate is variable.
Example 7.1. Suppose that I invest £1000 over 3 years and that interest is com-pounded annually, at rates of 5, 6 and 3 percent, in years 1, 2 and 3, respectively.What is the balance at the end of year 3?
In the first year interest is earned at 5% over the payment period; so the balanceis £1000 + 0.05× 1000 = 1050 at the end of year 1.
In the second year the interest rate is 6% so the balance at the end of the secondyear is £1050 + 0.60× 1050 = 1050 + 63 = 1113.
In the third year the interest rate is 3% so the balance at the end of the thirdyear is £1113 + 0.03× 1113 = 1146.39.
The balance after 3 years is £1146.39. In tabular form:
Balance at start Interest Balance at endPeriod of period over period of period1 P0 = 1000 1000× 0.05 = 50 P1 = 10502 P1 = 1050 1050× 0.06 = 63 P2 = 11133 P2 = 1113 1113× 0.02 = 33.39 P3 = 1146.39
To find a formula for the general case of this exam-
ple note that, if Pn is the balance at the end of the nth
payment period then the balance Pn+1 at the end of the
(n + 1)st payment period is
Pn +rn+1
100Pn = Pn
(1 +
rn+1
100
).
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 82
Therefore, if we set jt = rt100, for t = 1, . . . , n, then
(If j1 = j2 = · · · = jn then we recover Theorem 2.1.) As well as variableinterest rates it is also common for investors to deposit different amounts at theend of each payment period.
Example 7.2. Suppose that I make yearly deposits over 3 years into an accountthat earns interest, compounded annually, at rates of 5, 6 and 3 percent, in years
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 83
1, 2 and 3, respectively. I deposit amounts of £1000, £900 and £1200 at thebeginning of years 1, 2 and 3. What is the balance at the end of year 3?
The investment can be seen as 3 separate investments.
1. The first is £1000 over 3 years at the given interest rates, as in Example 7.1.This has final balance of £1146.39.
2. The second is an investment of £900 for 2 years at rates of 6% and 3%. From(7.1) the balance on this at the end of year 3 (when it has been invested for 2years) is 900× 1.06× 1.03 = 982.62.
3. The third is an investment of £1200 for the last of the 3 years, at a rate of 3%.This part has a final balance of 1200× 1.03 = 1236.
The balance of the total investment after 3 years is the sum of these three finalbalances; that is £1146.39 + 982.62 + 1236 = 3365.01.
If we write this out algebraically using (7.1), for each
of the three parts we have the following.
1. A deposit of D1 = 1000, and rates of j1 = 0.05, j2 =
0.06 and j3 = 0.03, so a final balance B1 of
B1 = D1(1 + j1)(1 + j2)(1 + j3) = 1146.39,
as before.
2. A deposit of D2 = 900, and rates of j2 = 0.06 and
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 84
j3 = 0.03, so a balance B2 of
B2 = D2(1 + j2)(1 + j3) = 932.62.
3. A deposit of D3 = 1200 and j3 = 0.03, giving a bal-
ance
B3 = D3(1 + j3) = 1236.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 85
Then the final balance is the sum
B = B1 +B2 +B3
= D1(1 + j1)(1 + j2)(1 + j3)
+D2(1 + j2)(1 + j3)
+D3(1 + j3)
= D1
(3∏s=1
(1 + js)
)+D2
(3∏s=2
(1 + js)
)
+D1
(3∏s=3
(1 + js)
)
=
3∑t=1
Dt
(3∏s=t
(1 + js)
).
Extending this to the general case of variable interest rate investments, we havethe formula of the following theorem.
Theorem 7.3. Suppose that
• money is deposited in a savings account at the beginning of each of n equalpayment periods;
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 86
• the amount deposited at the beginning of period t is Dt and
• the interest rate for period t is rt%.
Then the balance in the account Pn at the end of the nth payment period is
Pn =n∑
t=1
Dt
(n∏
s=t
(1 + js)
), (7.2)
where jt = rt/100, for t = 1, . . . , n.
Proof. The proof is the same as the argument following Example 7.2, but with 3replaced by n. Regard the investment as n separate investments. Investment t runsfrom the beginning of the t-th payment period to the end of the n-th, and consistsof one deposit of Dt at the beginning of period t. From (7.1) the final balance Bt
of this investment is
Bt = Dt
n∏s=t
(1 + js).
The final balance of the entire investment is therefore
Pn = B1 + · · ·Bn =n∑
t=1
Bt,
and the result follows.
Example 7.4. Suppose that 4 deposits of £1,000,000, £500,000, £500,000, and£700,000 were made, at a rate of one a month, in a savings account with variableinterest rates compounded monthly. For the first 2 months the annual interest ratewas 4%, then the bank reduced the rate by 0.5%.
– What was the balance at the end of 3 months?
– What was the interest paid?
Solution. By the Variable Interest Rates Formula 4.1, the
balance in the account B3 at the end of the 3rd month is
Limits of sequences are covered in MAS1041 and MAS1242, and later in moredepth at Stage 2. Here we shall consider a very particular limit which underliesthe definition of the power of one real number by another.
Definition 7.5. For every real number x,
ex = limn→∞
(1 +
x
n
)n.
Note that this definition includes a definition of the number e: this is the casewhere x = 1. (There are other possible definitions of ex, one of the most commonbeing that
ex =∞∑n=0
xn
n!.
All involve a limit at some point (and all result in the same function). A propertreatment of ex, including a proof that this limit really does exist, will be given inStage 2.)
One helpful consequence, which we shall state but not attempt to justify, isthat
limn→∞
(1 +
x
n
)nt= ext. (7.3)
(This can be proved using the theory of continuous functions, which is coveredin Stage 2 modules.) The usual rules, that we are used to applying to powers ofnumbers, now follow. Namely, for x, y ∈ R,
1. ex+y = exey;
2. exy = (ex)y and
3. e−x = 1/ex.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 89
Once the exponential ex has been defined for all real numbers x we can de-fine real powers of numbers other than e. Recall that, for all real numbers a weunderstand how to evaluate the following “powers” of a.
(i) For positive integers n ≥ 2, we have an = a×· · ·×a (where there are n−1of the × symbol).
(ii) For the special cases n = 0 and n = 1 we have a0 = 1 and a1 = a (for alla ∈ R).
(iii) For integers n > 0, we have a−n = 1/an, if a 6= 0 (but this is undefinedotherwise).
(iv) For integers n > 0, we have a1/n = n√a, if a ≥ 0. (This may also be
understood for negative a, using complex numbers when n is even.)
(v) For integers m,n > 0, we have am/n = n√am, if a ≥ 0.
However, if r is a real number which is not rational (like√
2 or π) then none of theabove give any meaning to ar. To do so we shall use the exponential function, andit’s inverse the natural logarithm. The latter is the function ln defined as follows:
ln(x) = y, where ey = x, for all x ∈ R, x > 0.
Note: this function is defined only for positive x.Now we can define arbitrary powers of positive numbers. First notice that if
a > 0 then, for m,n ∈ Z (with n 6= 0), we have
emn
ln(a) = (eln(a))mn = a
mn .
We use this as the starting point of the next definition.
Definition 7.6. Let a, t ∈ R and let a > 0. Then
at = e[t ln(a)].
The rules of exponentiation for ex now imply the analagous rules for ax.Namely for x, y ∈ R,
1. ax+y = axay;
2. axy = (ax)y and
3. a−x = 1/ax.
There are two functions associated to the definition of at above.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 90
• The power function p, given by fixing t and defining
p(x) = xt,
for all x ∈ R (for which the function makes sense according to one of thedefinitions above).
• The base a exponential function q, given by fixing a and defining
q(x) = ax,
for x ∈ R.
7.3 Continuously Compounded Interest
Example 7.7. Suppose that we borrow £1000 for one year at an annual interestrate r = 5%. If the loan is compounded at k equal intervals in the year the amountowed at the end of the year is
Pk = 1000
[1 +
(5
100× k
)]k= 1000
[1 +
(0.05
k
)]k.
As the number k, of intervals in the year, tends to infinity and hence the com-pounding interval tends to zero, then the future value after one year tends to £P ,where
P = limk→∞
Pk
= limk→∞
(1000
[1 +
(0.05
k
)]k)
= 1000 limk→∞
([1 +
(0.05
k
)]k)= 1000e0.05,
using Theorem 7.5.Thus the future value in one year’s time is
£P = £1000× er
100 = £1000× e5
100 = 1051.27.
Note that this is quite general and does not depend on r being 5: it works for allvalues of r.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 91
From this we can calculate the EAR, which is the annual simple interest ratewhich gives the same future value after one year. If the EAR is s% per annumthen, from Theorem 1.4, we require that s satisfies
1000(
1 +s
100
)= 1000× e
r100 .
That is, the EAR is s%, wheres
100= e
r100 − 1.
In this example r = 5, so the EAR is
s = 100(e
5100 − 1
)= 100(1.0512711− 1) = 5.127%.
Definition 7.8. As the number k, of intervals in the year, tends to infinity andhence the compounding interval tends to zero, then we say that compounding iscontinuous.
The example above suggests the following theorem.
Theorem 7.9 (The continuous compound interest formula). An amount A0 is bor-rowed at an annual rate of interest r% compounded continuously. The futurevalue At after t years (i.e. the amount owed) is
At = A0 ert100 . (7.4)
The EAR of the loan is s%, wheres
100= e
r100 − 1. (7.5)
Proof. If interest had been compounded k times during
the year, then there would have been kt payment periods
by time t; and, from Theorem 2.4, the amount owed at
time t would be
Pkt = A0
[1 +
r
100× k
]kt.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 92
If interest is compounded continuously the amount owed
at time t is
At = limk→∞
Pkt
= limk→∞
A0
[1 +
r
100× k
]kt= A0 lim
k→∞
[1 +
r
100× k
]kt,
= A0er
100×t,
using (7.3).
The argument above Definition 7.8 shows that the EAR
is given by (7.5).
As a result of this theorem continuously compounded interest rates are easy tomanipulate, even when rates are varying over periods of different lengths. Ofcourse, for a given annual interest rate, continuous compounding results in ahigher EAR than compounding over a finite number of payment periods per year.A combination of these facts makes continuously compounded interest very com-mon.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 93
Example 7.10. Suppose that £1000 is invested for 2 years. For the first 18 monthsthe interest rate is 4% per annum, and for the last 6 months it is 3% per annum,compounded continuously in both cases. What fixed annual rate of interest, com-pounded continuously would result in the same balance after 2 years?
First we calculate the balance after 18 months: that
is 1.5 years. For this period the interest rate is 4% per
annum, so from Theorem 7.4, with A0 = 1000,
A1.5 = A0e0.04×1.5 = 1000e0.04×1.5.
Now we calculate the balance A2 after the following 6
months; that is at the end of 2 years. This time we start
with A1 which is invested for 0.5 of a year at a continu-
ously compounded interest rate of 3% per annum. There-
fore, using Theorem 7.9 again,
A2 = A1e0.03×0.5 = 1000e0.04×1.5e0.03×0.5
= 1000e0.04×1.5+0.03×0.5 = 1000e0.075.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 94
Now we want to find the rate r% so that compounding
continuously at this rate for 2 years gives the same result.
The result of compounding continuously for 2 years is,
A0e2r100 = 1000e
2r100 ,
so we need r such that
1000e2r100 = 1000e0.075.
Therefore (after cancelling 1000 and taking logs of both
sides) we see that r% = 7.5/2% = 3.75%. What we have
found here is the IRR for continuous compounding, of
the cash flow with −A0 at time 0 and A2 at time 2. The
calculation is much simpler than the analogous calcula-
tion for interest compounded over fixed intervals, as in
Section 5.4.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 95
The previous example leads to the following theorem.
Theorem 7.11. An amount A0 is borrowed and interest is compounded continu-ously at annual rates of,
– r1%, between time t0 = 0 and time t1;
– r2% between time t1 and time t1 + t2, and so on: till,
– rn% between times t1 + · · ·+ tn−1 and t1 + · · ·+ tn.
That is at rp% between times t1 + · · · + tp−1 and t1 + · · · + tp, for p = 1, . . . , n.Then the future value at time T = t1 + · · ·+ tn is
AT = A0 e(∑n
p=1 rptp)/100. (7.6)
Proof. We use induction on p. If p = 1 then we have
A1 = A0er1t1100 ,
from Theorem 7.9, and this shows that the theorem holds in the case p = 1.Suppose it holds when p = s− 1, for some s > 1, so
At1+···+ts−1 = A0 e(∑s−1
p=1 rptp)/100.
Then, from Theorem 7.9,
At1+···+ts = At1+···+ts−1ersts100 = A0 e
(∑s−1
p=1 rptp)/100ersts100 = A0 e
(∑s
p=1 rptp)/100.
Therefore, by induction, the result holds for all p ≥ 1, and this proves the theorem.
Example 7.12. As in Example 7.10, Suppose that £1000 is invested for 3 years.Interest is compounded continuously. As in Example 7.10, for the first 18 monthsthe interest rate is 4% per annum, and for the last 6 months it is 3% per annum.Then for the last year the interest rate is 5% per annum. What fixed annual rate ofinterest, compounded continuously would result in the same balance after 3 years?
From Theorem 7.11 the balance after 3 years is
A3 = 1000e0.04×1.5+0.03×0.5+0.05×1 = 1000e0.125.
Therefore we want to find r such that e3r/100 = e0.125; and we arrive at r = 12.5/3.The required rate of interest is r% = 4.17%.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 96
Example 7.13. Suppose, as in Example 5.5, that £1000 is invested for 10 years,resulting in a final balance of 2000. What annual rate of continuously compoundedinterest gives this result.
Suppose the required rate is r%. Then, from Theorem 7.9
2000 = 1000e2r/100,
soe2r/100 = 2,
and2r/100 = ln(2).
Thus r = 50 ln(2).
Theorem 7.14. The continuously compounded IRR of a cash flow with payments−A0 at time 0 and At after t years is r%, where
r
100=
ln(At)− ln(A0)
t. (7.7)
The proof of this is left to the exercises. Note that in the example above wehave r
100= ln(2)/2 = (ln(2000) − ln(1000))/2, with t = 2, A2 = 2000 and
A0 = 1000.
7.4 Continuously Varying Interest Rates
In this section we suppose that interest is continuously compounded with arate which is varying. Let the present time be time 0, and let r(s) denote theinterest rate per unit, at time s units, s ≥ 0. The quantity r(s) is called the spot orthe instantaneous interest rate at time s.
As we shall use the composition of the exponential function with some inte-grals it’s convenient to use the notation
exp(x) = ex.
Theorem 7.15 (The continuously varying interest rate formula). Suppose an amountP0 is deposited in an account with instantaneous interest rate r(s). Then the futurevalue D(t) of the account at time t is
D(t) = P0 · exp
(∫ t
0
r(s)ds
). (7.8)
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 97
Proof. In order to determine D(t) in terms of the interest rates r(s), 0 ≤ s ≤ t,note that by the Simple Interest Formula, for small h, we have
D(s+ h) ≈ D(s)(1 + r(s) · h).
(≈ means “is approximately equal to”.) That is
D(s+ h) ≈ D(s) +D(s) · r(s) · h,
soD(s+ h)−D(s) ≈ D(s) · r(s) · h
andD(s+ h)−D(s)
h≈ D(s) · r(s).
By the definition of the derivative, taking the limit as h→ 0, we have
D′(s) = D(s) · r(s)
orD′(s)
D(s)= r(s).
To solve this differential equation, integrate both sides:∫ t
0
D′(s)
D(s)ds =
∫ t
0
r(s)ds.
Thus
[lnD(s)]t0 =
∫ t
0
r(s)ds
or
lnD(t)− lnD(0) =
∫ t
0
r(s)ds.
Since D(0) = P0, we obtain from the preceding equation that
lnD(t)− lnP0 =
∫ t
0
r(s)ds
so
exp (lnD(t)− lnP0) = exp
(∫ t
0
r(s)ds
).
We have
exp (lnD(t)− lnP0) = exp (lnD(t)) · exp(lnP−10
)= D(t) · P−10 .
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 98
Therefore
D(t) = P0 · exp
(∫ t
0
r(s)ds
).
Example 7.16. Suppose that interest is continuously compounded with a ratewhich is changing in time. Let the present time be time 0, and let
r(s) = 0.10− 0.051
1 + s
be the interest rate per unit at time s units, s ≥ 0.
(i) Find the amount D(t) that you will have in your account at time t, t ≥ 0, ifyou deposit £1, 000 at time 0.
(ii) Suppose the unit is equal to one year. How much money do you have inyour account after 10 years?
Solution.
(i) By the continuous varying interest rate formula
(7.8),
D(t) = £1, 000 exp
(∫ t
0
r(s)ds
).
Note that ∫ t
0
r(s)ds =
∫ t
0
0.10− 0.051
1 + sds
(by the linearity of the integrals)
=
∫ t
0
0.10ds− 0.05
∫ t
0
1
1 + sds
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 99
= 0.10 · t− 0.05 · [ln |1 + s|]t0
= 0.10 · t− 0.05 · (ln |1 + t| − ln |1|)
= 0.10 · t− 0.05 · ln |1 + t|.
Hence
D(t) = £1, 000 · exp (0.10t− 0.05 · ln |1 + t|)
D(t) = £1, 000 · e(0.10t+ln |1+t|−0.05)
= £1, 000 · e0.10t · eln |1+t|−0.05
= £1, 000 · e0.10t · |1 + t|−0.05.
(ii) For t = 10,
D(10) = £1, 000 · e0.10·10 · (1 + 10)−0.05
= £1, 000 · e · 11−0.05
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 100
(Recall that e ≈ 2.71828182. )
= £1, 000 · 2.71828182 · 0.88701378 = £2, 411.15.
7.5 The Continuous Compound Interest Formula
The continuous compound interest formula 7.4 is a special case of the contin-uous varying interest rate formula 7.8, arising when the interest rate r(s), s ≥ 0,is always equal to r
100. Let a principal P0 be borrowed at an annual rate of interest
r% compounded continuously for t years.Then, according to 7.15, for D(0) = P0 and r(s) = r
100, s ≥ 0, the amount At
owed on the principal P0 at time t is
At = D(t) = P0 · exp
(∫ t
0
r(s)ds
)
= P0 · exp
(∫ t
0
r
100ds
)= P0 · exp
( r
100· t)
= P0 er
100×t.
7.6 The Present Value.
The continuous varying interest rate formula 7.8 states that a principal P0 earn-ing continuously compounded interest r(s) per unit at time s units will be worth
D(t) = P0 · exp
(∫ t
0
r(s)ds
)at time t. Therefore
P0 = D(t) ·[exp
(∫ t
0
r(s)ds
)]−1AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 101
= D(t) · exp
[−(∫ t
0
r(s)ds
)].
We record this as a theorem.
Theorem 7.17. The present value P0(t) of an amount At at time t, with respect tothe variable interest rate r(s), s ≥ 0, is given by
P0(t) = At · exp
[−(∫ t
0
r(s)ds
)], t ≥ 0,
and is equal to the amount to be invested at time 0 to accumulate At at time t bycontinuous compounding at the interest rate r(s), s ≥ 0.
Definition 7.18. The function
P0(t) = exp
[−(∫ t
0
r(s)ds
)], t ≥ 0,
which is at each point t equal to the present value P0(t) of the amount 1, at timet, with respect to a variable interest rate r(s), s ≥ 0, is called the present valuefunction.
In other words, the present value function at t is equal to the amount to beinvested at time 0 to accumulate 1 at time t by continuous compounding at theinterest rate r(s), s ≥ 0.
Example 7.19. How much money should be invested at time 0 at a continuouslycompounded interest rate
r(s) = 0.10− 0.051
1 + s
per unit at time s units, s ≥ 0, in order to accumulate £5, 000, 000 at time t?
Solution. By Theorem 7.17, the present value P0(t) of
As discussed in Section 6.0.2 if a company needs money, perhaps to build newfactories or stores, there are four sources it can call on. It can use its profit, it cantake a bank loan, it can issue bonds, or it can sell shares. Taking a loan or issuingbonds are both called raising money on the debt market. Here we consider thelast of these options, selling shares, which is called raising money on the equitymarket. (An equity is another name for a share.)
To raise money in this way the company calculates the value of all its assets:that is the total value of all it owns. It divides this value into equal shares. Then itoffers some of these shares for sale. For example, if the company has total assetsworth £10 million then it may divide this into 10 million shares each worth £1. Itcan then offer say 300,000 of these shares for sale to buyers of its choosing. Thebuyer and the company have to agree on the price of each share (and this pricemight not be £1.) The buyers of shares are called shareholders and become partowners of the company.
The shareholders have a right to vote on various matters regarding the runningof the company. If you hold 10% of the shares in the company then you have 10%of the vote on such decisions. For example, the shareholders elect the board ofdirectors of the company, and the board of directors then appoints the people whomanage the company on a day to day basis.
A share of the profits of the company is divided between the shareholders, andpaid out to them, periodically, usually two or four times a year. These paymentsare called dividends. A shareholder with p% of the total number of shares willreceive p% of the total dividend. The directors decide, how much the dividendswill be at each payment period.
If the company wishes to make it’s shares more widely available, so anyonecan buy them, then it must register on a stock market. These are markets whereshares are bought and sold. (The company must satisfy some conditions, about itssize, credibility and future prospects, before it can be registered.) Some famousstock markets are the London Stock Exchange (LSE), the New York Stock Ex-change (NYSE), the National Association of Securities Dealers Automatic Quo-tation System (NASDAQ) and the Tokyo Stock Exchange (TSE). On some daysover a billion shares are traded on each of these markets.
A short introduction to the working of stock markets in given in the animation“Stock Markets” on the module web page:www.staff.ncl.ac.uk/alina.vdovina/teaching/mas1243
8.1 Buying and Selling Shares
The value of a share to an investor depends on the market price and the amountof dividend that it will pay in the future. Both of these are liable change, the price
daily and the dividends from payment period to another.Shares must be bought and sold through investment firms, called brokers,
that are registered with a stock market. Brokers charge a commission on eachtransaction, which may be a flat rate or a percentage of the price of the contract.Many people now buy and sell shares over the internet, because the commissionis usually much less this way, and it’s quick and convenient.
Example 8.1. Suppose that two people, Anne and Bert, buy shares of the samecompany every month. Anne buys 10 shares every month, while Bert buys £100worth of shares. What Bert is doing is called unit cost averaging.
Let’s calculate the average cost of a share for each of
Anne and Bert, after the first two purchases. Denote the
cost of a share by S(t) at the end of the t-th month, and
suppose they both buy at the ends of months 1 and 2.
Anne buys a total of 2×10 shares for a cost of £10S(1)+
10S(2). The average cost of a share to Anne is therefore
MA =10S(1) + 10S(2)
2× 10=S(1) + S(2)
2.
Bert buys 100/S(1) shares, at the end of month 1, and
100/S(2) shares, at the end of month 2; for a total cost of
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 106
2× 100. The average cost of a share to Bert is therefore
MB =2× 100
100S(1) +
100S(2)
=2
1S(1) +
1S(2)
=2S(1)S(2)
S(1) + S(2).
To see that MA ≥MB we consider the difference
MA −MB =
(S(1) + S(2)
2
)−(
2S(1)S(2)
S(1) + S(2)
)=S(1)2 − 2S(1)S(2) + S(2)2
2(S(1) + S(2))
=(S(1)− S(2))2
2(S(1) + S(2)).
We are assuming S(1) and S(2) are positive, so the de-
nominator of this expression is positive, and the numer-
ator is greater than or equal to zero. That is, MA ≥ MB,
which is to say that Anne pays an equal or higher average
price than Bert.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 107
In the previous example we saw that the unit cost averaging method of buyingshares, over two months, results in cheaper average costs per share. In fact thisconclusion holds for any number of purchases. If we consider the point of viewof the seller of shares then the benefits are reversed: selling shares using unit costaveraging results in a lower average sale price per share. (If Anne always buysfrom Andrew and Bert always buys from Belinda, then the price per share thatAndrew receives is greater than or equal to the price paid to Belinda: so Andrewshould be happier.)
Theorem 8.2. When buying or selling shares on a regular basis the unit costaveraging method results in a lower average price per share than the method ofbuying or selling a fixed number of shares every time (except when both methodsgive the same).
Both methods give the same average price if and only if prices are the samefor all transactions.
Proof. We generalise the argument of Example 8.1 to n,
rather than 2, transactions. Suppose A buys a fixed num-
ber F of shares at regular intervals 1, . . . ,T . AlsoB buys
shares to a value of £P at the same times. (SoB is using
unit cost averaging.) Assume the cost of a share at time
t is S(t).
In totalA buysF×T shares for a total cost of∑T
t=1 F×
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 108
S(t). The average cost per share for A is
MA =
∑Tt=1 F × S(t)F × T
=
∑Tt=1 S(t)
T.
B buys P/S(t) shares at time t so buys a total of∑Tt=1 P/S(t) shares. B spends P × T on these shares
so the average cost per share for B is
MB =P × T∑Tt=1 P/S(t)
=T∑T
t=1 1/S(t).
Therefore
MA −MB =
(∑Tt=1 S(t)
T
)−
(T∑T
t=1 1/S(t)
)
=1
T∑T
t=1 1/S(t)
[(T∑t=1
S(t)
)(T∑t=1
1/S(t)
)− T 2
].
As T and S(t) are positive this means thatMA−MB ≥ 0
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 109
if and only if(T∑t=1
S(t)
)(T∑t=1
1/S(t)
)− T 2 ≥ 0. (8.1)
For this we use the Cauchy-Schwarz inequality (see equa-
tion (10.1) in the Appendix). This states that(T∑t=1
atbt
)2
≤
(T∑t=1
a2t
)(T∑t=1
b2t
)
with equality if and only if at = λbt, for some constant λ,
or bt = 0; for all t. With at =√S(t) and bt = 1/
√S(t),
we obtain(T∑t=1
1
)2
≤
(T∑t=1
S(t)
)(T∑t=1
1/S(t)
),
that is
T 2 ≤
(T∑t=1
S(t)
)(T∑t=1
1/S(t)
),
from which (8.1) follows; so MA ≥MB.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 110
Moreover MA = MB if and only if the left-hand side
of (8.1) is equal to 0, which occurs if and only if there is
a constant λ such that√S(t) = λ/
√S(t), for all t; since
1/√S(t) 6= 0. We deduce from
√S(t) = λ/
√S(t) that
S(t) = λ, for all t. That is MA = MB if and only if
S(1) = · · · = S(T ) = λ.
8.2 Leverage
Leverage means buying (houses, cars, shares or anything else) using borrowedmoney rather than your own money. A mortgage taken out to buy a house is anexample of leverage. Leverage can be used to boost the profits made by investingmoney, but comes at a price: it may result in big profits, but can also result incatastrophic losses. As a simple example, suppose that I invest £10 in someshares. Let’s say that after a year the price of the shares has gone up and I sellthem for £15. Then I have made a profit of £5 and as a proportion of my originaloutlay this is 5/10 = 1/2; a rate of return of 1/2.
Now suppose that you also have £10 but that you borrow £90 and then buy£100 of the same shares (at the same time as I did). After one year you can sellthem for £150. You must then pay back the loan of £90, plus interest of say £10.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 111
This leaves you with £50 so you have made a profit of £40: a rate of return of40/10 = 4, which is 8 times higher than mine.
However, let’s now see what happens if the price of the shares is halved overthe year. Then, at the end of the year, I sell my shares for £5, which means I’velost £5 and have a rate of return of −5/10 = −1/2. I also still have £5.
You, on the other hand, sell your shares for £50. Now you must pay backthe loan, plus interest, which is £100. Therefore you owe £50. Your loss is theoriginal £10 plus this £50, so you have a rate of return of −60/10 = −6, and are£50 in the red.
We shall not go into leverage here. An overview of how it works can be foundin the videos “Leverage” and “Spread betting” on the module webpage:www.staff.ncl.ac.uk/alina.vdovina/teaching/mas1243and more details in the book [LMW].
8.3 Stock Market Indices
Stock market indices track the performance of certain selected stock from dayto day and publish this data. This allows investors to compare the performanceof individual companies to the market as a whole, and to make decisions aboutbuying and selling particular stocks. For example the most important index in theUK is the FTSE 100 Index, which tracks the performance of the 100 largest com-panies on the London Stock exchange. (FTSE because it was originally owned bythe Financial Times and the Stock Exchange.) Companies in this list are worth onaverage £1.3 trillion (1.3× 1012 pounds).
The FTSE 100 is a capitalisation weighted index: that is the larger the com-pany the more it contributes to the index. To be more precise, the market valueof a company is the total number of all the shares it has issued multiplied by thecurrent market price (on the London Stock Exchange). The index is the sum ofthe market values of the 100 companies listed, divided by a constant which wasfixed when the index was first set up in 1984. The value is recalculated every 15seconds.
For example suppose we have a list of three companies, Builder, Baker andCandlestick Maker, with the following market values.
Company Share Price No. of Shares Market ValueBuilder £100 1,000,000 £100,000,000Baker £50 1,500,000 £75,000,000
Candlestick maker £100 500,000 £50,000,000
The sum of these market values is (100+75+50)×106 = 225×106 pounds. Ifthe constant chosen was 3×106 then the value of the index would be 225/3 = 75.The capitalisation weighting means that a change in the price of Builder shares has
twice as much effect on the index as the same change in the price of Candlestickmaker shares.
Globally the most quoted index is the Dow Jones Industrial Average (DJIA)started in 1896, which tracks 30 stocks. Roughly speaking the Dow Jones iscalculated by taking the average of the share prices of each of the 30 companieslisted. (A more precise description can be found in [LMW].)
The index for list of three companies above, calculated this way, would thenbe £250/3 = 83.33. Here a change in the price of Builder shares has the sameeffect as a change in the price of either of the other 2 shares.
Other well-known indices are Standard & Poor’s 500 (S&P 500) and the NAS-DAQ Composite index, both of which are capitalisation weighted.
One drawback with all these indices, which an investor must bear in mind, isthat the index may be dominated by a change in price of very few stocks. Forexample if both Builder and Baker shares decrease in value by £1 and Candle-stick Maker shares increase by £6 per share, then the value of a capitalisation-weighted index will increase, although most shares are losing value. In fact thecapitalisation-weighted index reduces the effect of such bias, over straight averageindices such as the DJIA.
8.4 Share Prices
The investors return on a share is determined by
• future dividend payments and
• the share price.
Both of these are unknown so investors have to make predictions on which to basetheir decisions. Here we shall make some basic observations about the pricing ofshares. In MAS3312 “Stochastic and Financial Modelling” the topic is pursuedfurther, but this requires some knowledge of probability theory.
Let’s consider an investor who wishes to buy a partic-
ular share, and wants to decide how much to pay. We
shall assume that the investor wishes to earn a return of
r% per annum on the investment in this share. In fact we
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 113
shall assume that other investors also want to earn the
same rate of interest, and that this is fixed indefinitely.
Suppose that there are k dividend payments per year,
at regular intervals. Dividends are payed at the end of
each of payment period, and we denote byDt the amount
dividend expected at the end of payment period t. Let
V (t) be the future value of a share at the end of payment
period t, calculated after dividend Dt has been paid. The
investor is willing to pay an amount P0 equal to V (0),
the present value of all future dividend payments. (Of
course we set the rate of return at r% precisely so that
this is true.)
First we assume that the investor keeps the share in-
definitely. If the firm does not go bust then this is in prin-
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 114
ciple possible (assuming the investor has an heir). In this
case the present value V (0) depends only on the future
dividend payments. The (expected) cash flow for these
payments is:
Date 1 2 . . . t . . .Amount D1 D2 . . . Dt . . .
Therefore, the present value V (0) is the limit as t tends to
infinity of the net present values of the first t payments.
That is the limit of the sequence S1, S2, . . ., where
Sn = D1(1 + j)−1 + · · ·Dt(1 + j)−n =
n∑t=1
Dt(1 + j)−t,
with j = r/100k, the rate of return per payment period.
Assuming this limit exists we write it, as usual, as
limt→∞
St =
∞∑t=1
Dt(1 + j)−t.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 115
Therefore we have
V (0) =
∞∑t=1
Dt(1 + j)−t. (8.2)
The price the investor would be willing to pay, under
these assumptions, is P0 = V (0) =∑∞
t=1Dt(1 + j)−t.
Now suppose that the investor intends to sell the share
at some time in the future, say after T payment periods.
Using the same assumptions the price that an (other) in-
vestor will be willing to pay at time T is the value at time
T of all dividend payments from time T+1 onwards: that
is V (T ) which, from (8.2), is
V (T ) = DT+1(1 + j)−1 +DT+2(1 + j)−2
+DT+3(1 + j)−3 + · · ·
=
∞∑t=1
DT+t(1 + j)−t.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 116
Therefore our investor will sell the share at time T for
a price
PT = V (T ) =
∞∑t=1
DT+t(1 + j)−t.
The future cash flow of the investment is now
Date 1 2 . . . T − 1 T
Amount D1 D2 . . . DT−1 DT + PT
and the net present value of this cash flow is
V ′(0) =
(T−1∑t=1
Dt(1 + j)−t
)+ (DT + PT )(1 + j)−T
=
(T∑t=1
Dt(1 + j)−t
)+ PT (1 + j)−T . (8.3)
As
PT (1 + j)−T = (1 + j)−T∞∑t=1
DT+t(1 + j)−t,
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 117
we have
V ′(0) =
(T∑t=1
Dt(1 + j)−t
)+
((1 + j)−T
∞∑t=1
DT+t(1 + j)−t
)
=
(T∑t=1
Dt(1 + j)−t
)+
( ∞∑t=T+1
Dt(1 + j)−t
)
= V (0).
We record all this as the following theorem.
Theorem 8.3. Suppose that a particular share
• pays dividends k times per year,
• the expected dividend is Dt at the end of payment period t,
• the investors required rate of return is r% per annum,
• the price the investor pays for a share at time 0 is P0
• and the investor will sell the share for PT at the end of payment period T .
Then, setting j = r/100k,
P0 =∞∑t=1
Dt(1 + j)−t (8.4)
and
P0 =
(T∑t=1
Dt(1 + j)−t
)+ PT (1 + j)−T . (8.5)
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 118
To see how this might be useful we consider an example.
Example 8.4. Assume that you have bought 100 shares (at time 0), and that div-idends are paid quarterly. You expect to receive dividends of £1.50 at the end ofthe first and second quarters, dividends of £2 at the end of the third and fourthquarters and dividends of £2.50 at the end of the fifth and sixth quarters. Youexpect to be able to sell the shares for £125 at the end of the sixth quarter. Howmuch should you pay for these shares in order to realise a rate of return of 5% perannum?
From Theorem 8.3 you should pay P0, where P0 is
given by (8.5). We have quarterly dividend payments, so
Derivatives are contracts to buy or sell some underlying asset at some time inthe future. For example a farmer may agree in February to sell 400 tonnes of bar-ley to a grain merchant on 1st November, at £120 per tonne. The benefit to bothparties is that they can plan for the future on the basis of this fixed price and quan-tity. If the price of barley falls below £120 by November then the grain merchantis overpaying, and similarly if the price rises above £120 the farmer is losingsome profit. However the stability involved may well offset these disadvantages.
Derivatives are by no means a new idea. Such contracts have been found writ-ten on clay tablets from Mesopotamia that date to 1750 B.C. Aristotle mentionedan option on the use of olive oil presses in his “Politics” written around 2,400 b.c..He tells the story of a philosopher, Thales, who wanted to demonstrate that phi-losophy was not a useless occupation, as some critics had suggested. Thales haddevised some basic methods of weather prediction, and reckoned that the comingolive harvest would be good. He made contracts with all the olive-press owners inhis area to give him exclusive use of their presses at harvest time. He paid a smalldeposit up front and agreed to pay the rest of the rent at harvest time. The olivepress owners were happy to have the deposit, as they had no idea whether or notthere would be a good harvest; and Thales was able to negotiate very low rentalprices. Aristotle relates that
“When the harvest-time came, and many [presses] were wanted all at once andof a sudden, he let them out at any rate which he pleased, and made a quantity ofmoney. Thus he showed the world that philosophers can easily be rich if they like,but that their ambition is of another sort. ”
In this story Thales had no obligation to rent the presses; and if the harvesthad been bad could have decided to waive his right and lose his deposit. Thiskind of derivative, involving a right to something but not an obligation to take it,is nowadays called an “option”.
In the U.S., derivatives contracts have been formally traded on the ChicagoBoard of Trade since 1849. Today the size of derivatives markets is estimatedby the Bank of International Settlements to exceed $109 trillion in outstandingcontracts and over $400 trillion a year in trading volume on derivatives exchanges.
Derivatives are a useful and widespread method of managing financial risk,and are in everyday use by most large companies. Unfortunately they can also beused for less worthy purposes, such as tax-avoidance, manipulation of accountsand outflanking of financial regulations. Moreover, speculation in derivatives hasled to several well known financial disasters in the last 10 years, for examplethe collapse of Barings Bank and of Enron, were both the results of speculativederivatives trading.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 120
There are many different kinds of derivatives and alot of jargon surroundingtheir use. Here is a list of some of the most common.
• Forwards. The original and most basic form of a derivative contract, a for-ward transaction is an agreement to buy or sell a certain quantity of an assetor commodity in the future at a specified price, time and place. For exam-ple, A agrees to sell to B, 1 million Euro, in six months time, at $0.9402 pereuro.
• Futures. Futures contracts are like forwards, but they are highly standard-ised and publicly traded. Futures are traded on organised exchanges knownas “pits”, traditionally by shouting out bids from the floor (but now elec-tronically).
• Options. An option contract gives the buyer or holder the right to buy(or sell) the underlying item at a specific price at (or sometimes before) aspecific time in the future. We go into more detail below.
• Swaps. Swap contracts, in comparison to forwards, futures and options, areone of the more recent innovations in derivatives contract design. The firstcurrency swap contract, between the World Bank and IBM, dates to Augustof 1981.
A swap is an agreement between two parties to swap the net value of twoseries of payments, one of which is usually based on a fixed interest rate,and the other linked to a variable interest rate of some sort.
• Structured note. A hybrid that combines a bond or loan with a derivative.
9.2 Call Options
The simplest financial option, a European call option, is a contract with thefollowing conditions:
• At a prescribed time in the future, known as the expiry date or expirationdate, the holder of the option may
• purchase a prescribed asset, known as the underlying asset for
• a prescribed amount, known as the exercise price or strike price.
Remarks 9.1. 1. The word “may” in this description implies that for the holderof the option, this contract is a right and not an obligation.The other party to the contract, who is known as the writer, does have a po-tential obligation: he must sell the asset if the holder chooses to buy it.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 121
2. Since the option confers on its holder a right with no obligation it has somevalue. Conversely, the writer of the option must be compensated for the obli-gation he has assumed.
3. The word “European” in the name of the option does not mean anything ge-ographical: these options are traded globally. There are also American calloptions which allow the holder (or the buyer) to exercise the option at anytime up to the expiry date; whereas the European option specifies a particulardate on which the option may be used, and it can only be used on this date.
We shall discuss two main concerns in connection with options:
• How much should one pay for this right, that is, what is the value of anoption?
• How can the writer minimise the risk associated with his obligation?
There are sophisticated mathematical models designed to deal with these ques-tions. Here we shall take some first steps in this direction.
In the following example a situation arises in which an investor can makean immediate profit without any risk of loss. Such a sure-win betting scheme isknown in the financial world as an arbitrage. An arbitrage can be obtained, forexample, if there are two prices for the same shares on different exchanges, bybuying the shares at the cheaper price and immediately selling them at the moreexpensive one.
Example 9.2. A European Call Option. Assume today’s date is 5th December2011. Suppose one share of X costs 250p now. Consider an option which entitlesthe holder to
• purchase one X share for 250p,
• on 19 July 2012.
(The holder can also choose not to buy this share.) How much is the above optionworth now?
In order to gain an intuitive feel for the price of this option let us imagine twopossible situations that might occur on the expiry date, 19 July 2012, nearly eightmonths in the future.
If the X share price is 270p on 19 July 2012, then the holder of the optionwould be able to purchase the asset for 250p. This action, which is called exer-cising the option, yields an immediate profit of 20p. That is, he can buy the sharefor 250p and immediately sell it in the market for 270p:
270p− 250p = 20p profit.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 122
This is an arbitrage opportunity. (Here we are trying to find an appropriate pricefor the option, so we consider only the profit made on buying and selling theshare.)
On the other hand, if the X share price is only 230p on 19 July 2012, then itwould not be sensible to exercise the option. Why buy something for 250p whenit can be bought for 230p elsewhere?
If we assume that the X share only takes the values 230p or 270p on 19 July2012, with equal probability, then the expected profit to be made is
1
2× 0 +
1
2× 20 = 10p.
Ignoring interest rates for the moment, it seems reasonable that the order of mag-nitude for the value of the option is 10p.
Of course, valuing an option is not as simple as this, but let us suppose that theholder did indeed pay 10p for this option. Now if the share price rises to 270p atexpiry he has made a net profit calculated as follows:
profit on exercise = 20pcost of option = -10p
net profit = 10p
This net profit of 10p is 100% of the up-front premium.The downside of this speculation is that if the share price is less than 250p at
expiry, then the holder has lost all of the 10p invested in the option, giving a lossof 100%.
If the investor had not bothered with options, but instead purchased the sharefor 250p on 5 December 2011, the corresponding profit or loss of 20p would havebeen only ±8% of the original investment.
9.3 Put Options
The option to buy an asset, as above is a call option. The right to sell an assetis known as a put option and has payoff properties which are opposite to those ofa call option.
A European put option allows its holder to sell the asset on a certain specifieddate for a prescribed amount. The writer is then obliged to buy the asset at thisprice.
An American put option allows the holder to exercise the option to sell at anytime up to the expiry date; whereas a European put option can only be exercisedat the expiry date.
To arrive at some conclusion about the price that should be paid for an optionwe must make some assumption about the amount that it costs to borrow money.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 123
We therefore assume that we know a prevailing interest rate for borrowing. Weshall also assume that interest is compounded continuously, as this is the mostcommon method of compounding in use, and the easiest to work with.
Example 9.3. Consider a European put option that allows its holder to sell oneshare of a stock for £3 in six months. Let the price of one share of the stock be£2.80 at time 0. Given an annual interest rate of 4% compounded continuously,find an upper bound for the price £P of the put option, so that a profit can beexpected from buying the option.
Solution. Initially we buy one share of the stock and
one put option. This initial payout of £(2.80 + P ) is
borrowed from a bank to be repaid in six months. By
the Continuous Compound Interest Formula (7.4), the
amount owed on the principal
£(2.80 + P )
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 124
at time t = 6/12 = 1/2 is £A, where
A = (2.80 + P ) er
100×t
= (2.80 + P )× e4100×
12
= (2.80 + P )× e0.02
= (2.80 + P )× 1.02020134.
Let us now consider the value of our holdings in six
months. There are two cases that depend on £F , the mar-
ket price of the share in six months.
Case 1. If F ≤ 3, then we can exercise our put option to
sell the share for the amount £3. In this case to pay off
our bank loan and realise a positive profit we need to find
P such that the following inequality holds
A < 3
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 125
⇐⇒
(2.80 + P )× 1.02020134 < 3
⇐⇒
2.80 + P < 2.94059602
⇐⇒
P < £0.14.
Thus is we pay £P , where P < 0.14 we shall, in this
case, make a profit.
Case 2. If F > 3, then it would not be sensible to exer-
cise our put option; we can sell our share for the market
price F elsewhere. In this case we shall make a profit if
A < F,
and since F > 3, this will hold, as in Case 1, if P < 0.14.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 126
Therefore if we pay £P where
P < 0.14,
we shall in both cases be able to pay off our bank loan
and realise a positive profit.
Theorem 9.4 (The European put option inequality). Suppose that shares in acompany AAA are for sale.
• Let S be the price of one AAA share at time 0.
• Let P be the price of a European put option that enables its holder to sellone AAA share for an amount F at time t.
• Suppose that interest is compounded continuously at a rate of r(s)%, 0 ≤s ≤ t.
Then the holder of the AAA put option will realise a positive profit at time t if
P < F × exp
(−∫ t
0
r(s)ds
)− S. (9.1)
Proof. Initially we buy one share of the stock and one put option. This initial pay-out of (S + P ) is borrowed from a bank to be repaid in time t. By the ContinuousVarying Interest Formula (7.8), the amount A owed on the principal
(S + P )
at time t is
A = (S + P ) exp
(∫ t
0
r(s)ds
).
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 127
Let us now consider the value of our holdings at time t. There are two casesthat depend on S(t), the market price of the share at time t. (Note that S(0) = S.)Case 1. If S(t) ≤ F , then we can exercise our put option to sell the share for theamount F . In this case to pay off our bank loan and realise a positive profit weneed to find P such that
A < F.
That is, P such that
(S + P ) exp
(∫ t
0
r(s)ds
)< F
⇐⇒
S + P < F × exp
(−∫ t
0
r(s)ds
)⇐⇒
P < F × exp
(−∫ t
0
r(s)ds
)− S.
Therefore, in this case, we realise a profit if and only if (9.1) holds.Case 2. If S(t) > F , then it would not be sensible to exercise our put option;we can sell our share for the market price elsewhere; for the amount S(t). If Psatisfies (9.1) then, as in Case 1 above,
A = (S + P ) exp
(∫ t
0
r(s)ds
)< F.
Since F < S(t) we shall then have A < S(t), so we can again pay off our bankloan and realise a positive profit.
Therefore in both cases, if (9.1) holds a profit is realised.
Theorem 9.5 (The put-call option parity formula).
• Let S be the price of one share of a stock at time 0.
• Let P be the price of a European put option that enables its holder to sellone share of the stock for the amount F at time t.
• Let C be the price of a European call option that enables its holder to buyone share of the stock for the amount F at time t.
• Suppose that the interest rate is compounded continuously with a rate r(s),0 ≤ s ≤ t.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 128
Then either
S + P − C = F × exp
(−∫ t
0
r(s)ds
)(9.2)
or there is an arbitrage opportunity.
Proof. We shall show first that if the left hand side of (9.2) is less than the righthand side, then there is an arbitrage opportunity. Then we shall show that if theleft hand side is greater than the right hand side there is also and arbitrage oppor-tunity. This is sufficient to prove the theorem.
Case 1. Suppose that
S + P − C < F × exp
(−∫ t
0
r(s)ds
).
That is
(S + P − C)× exp
(∫ t
0
r(s)ds
)< F.
Then we shall show that we can effect a sure win by initially
• buying one share of the stock,
• buying one put option and
• selling one call option.
The initial payout ofS + P − C
is borrowed from a bank to be repaid at time t. By the Continuous Varying InterestFormula (7.8), the amount A owed on the principal
S + P − C
at time t is
A = (S + P − C)× exp
(∫ t
0
r(s)ds
).
Let us now consider the value of our holdings at time t. There are two casesthat depend on S(t), the market price of the share at time t.
Case 1.1. If S(t) ≤ F , then we can exercise our put option to sell the share forthe amount F . Note that the call option we sold is worthless in this case.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 129
Case 1.2. If S(t) > F , then the call option we sold will be exercised, forcing usto sell our share for the price F . Note that our put option is worthless in this case.
Thus, in either case we will realise the amount F at time t. Since
A = (S + P − C)× exp
(∫ t
0
r(s)ds
)< F,
we can pay off our bank loan and realise a positive profit in both cases.
Case 2. Suppose now that
S + P − C > F × exp
(−∫ t
0
r(s)ds
)⇐⇒
(S + P − C)× exp
(∫ t
0
r(s)ds
)> F.
We shall show that we can effect a sure win by
• selling one share of the stock,
• selling one put option and
• buying one call option.
The initial amount ofS + P − C
can be invested at the varying interest rate r(s), 0 ≤ s ≤ t, compounded contin-uously for time t. By the Continuous Varying Interest Formula (7.8), the futurevalue A of the principal
S + P − Cat time t is
A = (S + P − C)× exp
(∫ t
0
r(s)ds
).
Let us now consider the value of our holdings at time t. There are two casesthat depend on S(t), the market price of the share at time t.
Case 2.1. If S(t) ≤ F , then the put option we sold will be exercised, forcing usto buy our share for the price F . Note that our call option is worthless in this case.
Case 2.2. If S(t) > F , we can exercise our call option to buy the share for theamount F . Note that the put option we sold is worthless in this case.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 130
Thus, in either case we will get one share back and will spend at most F attime t. Since the amount A accrued on the principal
S + P − C
at time t is greater than F :
A = (S + P − C)× exp
(∫ t
0
r(s)ds
)> F
we can realise a positive profit in both cases.
9.4 Hedging
We introduce hedging by a simple example. What happens to the value of aholding containing both shares and put options when a share price falls?
The answer depends on the ratio of shares and options in the holding. Whenthe share price falls the price of put options increases, as the put option enablesthe holder to sell shares at fixed price, no matter how far the market price falls.Conversely, when the share price rises the price of put options falls.
The value of a holding that contains only shares falls when the share pricefalls, while one that is all put options increases in value. Somewhere in betweenthese two extremes is a ratio at which a small unpredictable movement in pricesdoes not result in any unpredictable movement in the value of the holding.
The reduction of risk by taking advantage of such correlations between theshare and option price movements is called hedging. This idea is central to thetheory and practise of option pricing.
Remark 9.6. We have given here an exposition of the main ideas of optionpricing based on simple examples; a fuller picture can only be presented at a moreadvanced level. One will need a good knowledge of Real Analysis, ProbabilityTheory and Differential Equations to follow those arguments. An account of theanalysis of option pricing can be found in
P. Wilmott, S. Howison and J. Dewynne, “The Mathematics of Financial Deriva-tives. A Student Introduction”, CUP, 2002.
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 131
10 Appendix
10.1 Percentages
The word percent means “per hundred”. The familiar symbol % thus meansto divide by one hundred.
Example 10.1. 1% = 1100
= 0.01,17.8% = 17.8
100= 0.178.
Example 10.2. What percent of 40 is 18?Solution. Let x be the unknown percentage. Thenx% of 40 is 18x100· 40 = 18
Therefore x% = 18×10040
% = 45%.
10.2 The Cauchy-Schwarz inequality
This inequality comes up in almost all branches of mathematics at some point.From the book “The Cauchy-Schwarz Master Class” by J. Michael Steele
“Cauchy’s inequality for real numbers tells us that
and there is no doubt that this is one of the most widely used and most importantinequalities in all of mathematics.”
This is the usual statement.
Theorem 10.3. (n∑
i=1
aibi
)2
≤
(n∑
i=1
a2i
)(n∑
i=1
b2i
), (10.1)
with equality if and only if ai = λbi, for some constant λ, or bi = 0; for all i.
Proof. If b1 = · · · = bn = 0 then both right and left sides of (10.1) equal 0, andthe theorem holds. Assume then that bi 6= 0, for at least one i. Then (10.1) holdsif and only if
0 ≤
(n∑
i=1
a2i
)(n∑
i=1
b2i
)−
(n∑
i=1
aibi
)2
,
AV December 7, 2011
MAS1243/MAS2243 Notes Lecturer’s copy 132
if and only if
0 ≤
(n∑
i=1
a2i
)− (∑n
i=1 aibi)2
(∑n
i=1 b2i )
=
(n∑
i=1
a2i
)− 2
(∑n
i=1 aibi)2
(∑n
i=1 b2i )
+(∑n
i=1 aibi)2
(∑n
i=1 b2i )
=
(n∑
i=1
a2i
)− 2λ
(n∑
i=1
aibi
)+ λ2
(n∑
i=1
b2i
),
where
λ =(∑n
i=1 aibi)
(∑n
i=1 b2i ). (10.2)
Thus, if some bi 6= 0, (10.1) holds if and only if
0 ≤n∑
i=1
(a2i − 2λaibi + λ2b2i
)=
n∑i=1
(ai − λbi)2 , (10.3)
and both sides of (10.1) are equal if and only if ai = λbi, for all i, where λ is givenby (10.2).
