Subject CT1 - Actuarial Education Company Upgrade/CT1-PU-16.pdf · 2015-09-24 · Subject CT1 CMP Upgrade 2015/16 CMP Upgrade This CMP Upgrade lists all significant changes to the

CT1: CMP Upgrade 2015/16 Page 1

The Actuarial Education Company © IFE: 2016 Examinations

Subject CT1

CMP Upgrade 2015/16

CMP Upgrade This CMP Upgrade lists all significant changes to the Core Reading and the ActEd material since last year so that you can manually amend your 2015 study material to make it suitable for study for the 2016 exams. It includes replacement pages and additional pages where appropriate. Alternatively, you can buy a full replacement set of up-to-date Course Notes at a significantly reduced price if you have previously bought the full price Course Notes in this subject. Please see our 2016 Student Brochure for more details.

This CMP Upgrade contains: all changes to the Syllabus objectives and Core Reading. changes to the ActEd Course Notes, Series X Assignments and Question and

Answer Bank that will make them suitable for study for the 2016 exams.

Page 2 CT1: CMP Upgrade 2015/16

© IFE: 2016 Examinations The Actuarial Education Company

1 Changes to the Syllabus objectives and Core Reading

1.1 Syllabus objectives

No changes have been made to the syllabus objectives.

1.2 Core Reading

Chapter 12 Page 29 The following sentence has been added to Core Reading: Conversely, if we know the real yield i ¢ which we have obtained from an equation of value using inflation-adjusted cashflows then we can calculate the money yield as follows: ( )1i i j i= + +¢ ¢

Replacement pages are provided. Chapter 14 Page 24 The following paragraph has been added to Core Reading: Note that another definition of duration exists: the modified duration. This is covered in more detail in Subjects ST5 and ST6 (Finance and Investment Specialist Technical A and B) but can be expressed in terms of the Macaulay Duration as:

( )1

pip

t+

where ( )pi and p are as defined in Chapter 3.

Replacement pages are provided.



2 Changes to the ActEd Course Notes

Chapter 5 A number of changes have been made to this chapter. In particular:

Section 2.2 on payment streams has been expanded.

The exam-style question and its solution have been replaced. As a result, some of the questions in the chapter have been reordered, and some solutions have been updated. A replacement chapter is provided. Chapter 8 Page 17 The summary has been updated to include the following: To find the yield on a transaction from an equation of value of the form ( ) 0f i , we

need:

to find 1i and 2i such that 1( )f i and 2( )f i are of opposite sign, and

the final value obtained for i to be greater than 1 . Replacement pages are provided. Chapter 10 Page 15 In the Example, Project C is now described as a “five-year” project, and Project E is now described as a “three-year” project.



Chapter 12 Page 18 Example 12.10 has been updated. Replacement pages are provided. Page 43 The summary has been updated. Replacement pages are provided. Chapter 15 Pages 21-23 Exam-style question 2 has been moved to be Question 15.12 in the chapter, and a new exam-style question 2 has been added. Replacement pages are provided. Page 36 The following paragraph has been added to Solution 15.11: Note that the distribution of nS obtained here is different to that derived in the chapter

of 2~ log ( , )nS N n nm s , which applies to the varying rate model, where the returns in

each year are independent.



3 Changes to the Q&A Bank

Q&A Bank 1

Question 1.26 The solution to this question has been expanded. Replacement pages are provided. Question 1.37 The solution to this question has been expanded. Replacement pages are provided. Question 1.39 This question has been replaced. Replacement pages are provided. Question 1.40 This question has been replaced. Replacement pages are provided. Q&A Bank 2

Question 2.10 An alternative solution has been added. Replacement pages are provided. Q&A Bank 3

Question 3.6 The words “immediately after the coupon payment” have been removed from the question. Question 3.20 This question has been replaced. Replacement pages are provided. Question 3.21 This question has been replaced. Replacement pages are provided.



Q&A Bank 4

No changes have been made to this Q&A Bank. Q&A Bank 5

Question 5.7 The solution to this question has been expanded. Replacement pages are provided.



4 Changes to the X Assignments

Assignment X1

Question X1.13 Parts (ii) and (iii) of this question have been changed. Replacement pages are included for those students who are submitting their assignments for marking as the most up-to-date version of the assignments needs to be used.



5 Other tuition services

In addition to this CMP Upgrade you might find the following services helpful with your study.

5.1 Study material

We offer the following study material in Subject CT1:

● Online Classroom

● Flashcards

● MyTest

● Revision Notes

● ASET (ActEd Solutions with Exam Technique) and Mini-ASET

● Mock Exam

● Additional Mock Pack. For further details on ActEd’s study materials, please refer to the 2016 Student Brochure, which is available from the ActEd website at www.ActEd.co.uk.

5.2 Tutorials

We offer the following tutorials in Subject CT1:

a set of Regular Tutorials (lasting two or three full days)

a Block Tutorial (lasting two or three full days)

a Revision Tutorial (lasting one full day)

Live Online Tutorials (lasting three full days). For further details on ActEd’s tutorials, please refer to our latest Tuition Bulletin, which is available from the ActEd website at www.ActEd.co.uk.



5.3 Marking

You can have your attempts at any of our assignments or mock exams marked by ActEd. When marking your scripts, we aim to provide specific advice to improve your chances of success in the exam and to return your scripts as quickly as possible. For further details on ActEd’s marking services, please refer to the 2016 Student Brochure, which is available from the ActEd website at www.ActEd.co.uk.

5.4 Feedback on the study material

ActEd is always pleased to get feedback from students about any aspect of our study programmes. Please let us know if you have any specific comments (eg about certain sections of the notes or particular questions) or general suggestions about how we can improve the study material. We will incorporate as many of your suggestions as we can when we update the course material each year. If you have any comments on this course please send them by email to [email protected] or by fax to 01235 550085.


All study material produced by ActEd is copyright and is sold for the exclusive use of the purchaser. The copyright is owned

by Institute and Faculty Education Limited, a subsidiary of the Institute and Faculty of Actuaries.

Unless prior authority is granted by ActEd, you may not hire out, lend, give out, sell, store or transmit electronically or

photocopy any part of the study material.

You must take care of your study material to ensure that it is not used or copied by anybody else.

Legal action will be taken if these terms are infringed. In addition, we may seek to take disciplinary action through the

profession or through your employer.

These conditions remain in force after you have finished using the course.

CT1-12: Elementary compound interest problems Page 29


Therefore 11

1

ii

j

ii j

j1.

This is sometimes approximated to i i j-¢ if only estimates are required. However,

you should only use this approximation if the question specifically asks for an estimate or an approximation since the “exact” method isn’t much more difficult. Conversely, if we know the real yield i ¢ which we have obtained from an equation of value using inflation-adjusted cashflows then we can calculate the money yield as follows, by rearranging the above equation for i¢ :

1 1i j i j i i j j i i i j j ( )1i i j ifi = + +¢ ¢

Example 12.14 Ten years ago, a saver invested £5,000 in an investment fund operated by an insurance company. Over this period the rate of return earned by the fund has averaged 12% per annum. If prices have increased by 80% over this period, calculate the average annual real rate of interest earned by the fund over this period. Solution The average annual rate of inflation j over the 10 year period can be found from:

( ) .1 1810 j j 18 1 0 06051 10. ./ ie 6.05%

The average annual rate of interest is i 012. . So the average annual real rate is:

i012 0 0605

106050 0561

. .

.. ie 5.61%

The graph on the following page shows the real growth over time of a single investment when the rate of inflation is (a) higher than, (b) equal to and (c) less than the rate of interest (taken to be 8% pa).

Page 30 CT1-12: Elementary compound interest problems


j=4%

j=8%

j=12%

Time

Real growth

If an investment is achieving a positive real rate of return ( i j ), then it is outstripping

inflation. If it is achieving a negative real rate of return ( i j ), then it is falling behind

inflation (which means that you would be better spending the money now, rather than “investing” it).

Question 12.15

You are the investment manager for a pension scheme. The trustees have paid over a sum of £250,000, instructing you to invest it so as to obtain the highest real rate of return over the next 5 years. Your researchers have advised you that suitable investments are available in the following countries: Expected rate of return Expected inflation rate Japan: 3% pa 1% pa United Kingdom: 7% pa 3% pa Malaysia: 15% pa 10% pa Mexico: 25% pa 40% pa Which country would you select? (Assume that the investment opportunities in each country are otherwise identical and ignore currency risks and all other factors that might influence your decision.)

In some cases a combination of known inflation index values and an assumed future inflation rate may be used to find the real rate of interest.

CT1-14: Term structure of interest rates Page 23


Question 14.14

Derive an expression for the DMT of a level 10-year annuity of 1 pa payable annually in arrears and sketch a graph of the DMT in this case as a function of the interest rate.

Another way of deriving the Macaulay duration is in terms of the force of interest, d :

1( )

1

( ) (1 ) ( )

d diA i

A d d

dii e e

d

e i i i

d d

d

t nd d

d

t n n

= - =

= - fi =

fi = = +

The equation for in terms of the cashflows Ctk

may be found by differentiating

A with respect to , recalling that k kt tiv e d-= .

The duration of an n year coupon paying bond, with coupons of D payable annually, redeemed at R, is:

D Ia Rnv

Da Rv

nn

nn

( )

We deduced this result in the last example. The duration of an n year zero coupon bond of nominal amount 100, say, is:

100

100

nv

vn

n

n

This last result should be intuitively obvious. The average term of a series of cashflows that has only one payment must be the time of that cashflow. Both the volatility and the discounted mean term provide a measure of the average “life” of an investment. This is important when considering the effect of changes in interest rates on investment portfolios since an investment with a longer term will in general be affected more drastically by a change in interest rates than an investment with a shorter term.

Page 24 CT1-14: Term structure of interest rates


Example How will the price of a conventional gilt that is redeemable at par with an annual coupon of 3% be affected if future rates of interest over all terms increase from 7% to 8%, if the term of the gilt is (a) 5 years and (b) 25 years? Comment on your results. Solution

Using the formula P a vnn 3 100| with n 5 and n 25, we find that:

The price of the 5 year stock would fall from £83.60 to £80.04 ie a fall of 4.3% The price of the 25 year stock would fall from £53.39 to £46.63 ie a fall of 12.7%. The change in interest rate has a greater effect on the longer 25 year stock, which has a DMT of 14.9 years (based on 7% interest), than it has on the shorter 5 year stock, which has a DMT of 4.7 years. Roughly speaking, a change in interest rates has the same effect on the present value of a cashflow series as it would have on a zero coupon bond with the same discounted mean term or volatility.

Note that another definition of duration exists: the modified duration. This is covered in more detail in Subjects ST5 and ST6 (Finance and Investment Specialist Technical A and B) but can be expressed in terms of the Macaulay Duration as:

( )1

pip

t+

where ( )pi and p are as defined in Chapter 3.

CT1-05: Discounting and accumulating Page 1


Chapter 5

Discounting and accumulating

Syllabus objective (v) Calculate the present value and the accumulated value of a stream of equal or

unequal payments using specified rates of interest and the net present value at a real rate of interest, assuming a constant rate of inflation.

1. Discount and accumulate a sum of money or a series (possibly infinite) of

cashflows to any point in time where: the rate of interest or discount is constant

the rate of interest or discount varies with time but is not a continuous function of time

either or both the rate of cashflow and the force of interest are continuous functions of time

2. Calculate the present value and accumulated value of a series of equal or

unequal payments made at regular intervals under the operation of specified rates of interest where the first payment is:

deferred for a period of time

not deferred Real rates of interest are dealt with in Chapter 12.

0 Introduction

So far we have calculated present values and accumulated values of single payments. This chapter starts to look at present values and accumulations of a series of payments and continuous payments.

Page 2 CT1-05: Discounting and accumulating


1 Present values of cashflows

In many compound interest problems one must find the discounted present value of cashflows due in the future. It is important to distinguish between (a) discrete and (b) continuous payments. In Section 1.1 we will consider discrete payments before looking at continuous payments in Section 1.2.

1.1 Discrete cashflows

We have already seen that the present value of a cashflow, C, due at time t is tCv where 1/(1 )v i= + and i is the effective rate of interest per annum. Here we are assuming that

we are working in years and that the effective rate of interest is constant over the period. What if we have two payments, 1C due at time 1t , and 2C at time 2t ? The present

value of the these payments is the amount we would have to invest, say in a bank account, to be able to pay each of the payments at the times they are required. Rather than investing a single sum into a single bank account to provide for the payments, we could have set up a separate bank account to cater for each payment and invested the present value of each payment in the corresponding account. This alternative arrangement would have exactly the same result. So, we see that the present value of the two payments is just the sum of the individual present values. More generally, the present value of a series of payments of

1 2, , ...,

nt t tc c c due at times

1 2, , ..., nt t t is given by:

1

j

j

nt

tj

c v=Â



Example Under its current rent agreement, a company is obliged to make annual payments of £7,500 for the building it occupies. Payments are due on 1 January 2006, 1 January 2007 and 1 January 2008. If the company wishes to cover these payments by investing a single sum in its bank account that pays 7.5% pa compound, what sum must be invested on 1 January 2005?

1/1/2005 1/1/2006 1/1/2007 1/1/2008

£7,500 Payment

Time

£7,500 £7,500

Here: v i 1 1 1 1075 0 93023/ ( ) / . .

So: PV payment due on 1 January 2006 7,500 7,500 0.93023 6,976.74v

PV payment due on 1 January 2007 2 27,500 7,500 0.93023 6,489.99v

PV payment due on 1 January 2008 3 37,500 7,500 0.93023 6,037.20v

So: PV all payments 6,976.74 6, 489.99 6,037.20 £19,504

Question 5.1

Calculate the present value on 1 September 2002 of payments of £280 due on 1 September 2004 and £360 due on 1 March 2005. Interest is 15% pa effective.

If the effective interest rate is not constant then we could write the present value in terms of the function v t( ) , where ( )v t is the (discounted) present value of 1 due at time

t. The present value of the sums

1 2, , ...,

nt t tc c c due at times 1 2, , ..., nt t t (where

1 20 nt t t£ < < < ) is:

1 21 21

( ) ( ) ( ) ( )n j

n

t t t n t jj

c v t c v t c v t c v t=

+ + + = Â



If the number of payments is infinite, the present value is defined to be:

1

( )jt j

j

c v t•

=Â

provided that this series converges. It usually will in practical problems.

Question 5.2

Express the present value of the sums 1 2, , ...,

nt t tc c c due at times 1 2, , ..., nt t t in terms of

the force of interest at time t, denoted by ( )t .

Example Find the value at time 0t = of $250 due at time 6t = and $600 due at time 8t = if

( ) 3%t pad = for all t.

Solution If the force of interest, d , is constant then the present value of a payment of C due at time t is:

tCe d- Therefore the present value of the two payments in this example is:

6 0.03 8 0.03250 600 $680.79e e- ¥ - ¥+ =

1.2 Continuously payable cashflows (payment streams)

Suppose that 0T > and that between times 0 and T an investor will be paid money continuously, the rate of payment at time t being £ ( )t per unit time.

What is the present value of this cashflow?



In order to answer this question it is essential to understand what is meant by the “rate of payment” of the cashflow at time t. If M(t) denotes the total payment made between time 0 and time t, then by definition:

( ) ( )t M tr = ¢ for all t

Then, if 0 Ta b£ £ £ , the total payment received between time and time is:

- = ¢

=

Ú

Ú

( ) ( ) ( )

( )

M M M t dt

t dt

b

a

b

a

b a

r (1.1)

Thus the rate of payment at any time is simply the derivative of the total amount paid up to that time, and the total amount paid between any two times is the integral of the rate of payments over the appropriate time interval. You can think of Formula (1.1) as the sum of lots of small payments, each of amount

( )t dtr . It may help to consider this simple example:

If the rate of payment is a constant £24 pa then in any one year the total amount paid is £24, but this payment is spread evenly over the year. In half a year, the total paid is 24 ½¥ , ie £12.

In one month, the total paid is 11224¥ , ie £2.

So, in a small time period dt, the total paid is £24dt .



Example A life office starts issuing a new type of 10-year savings policy to young investors who pay weekly premiums of £10. Assuming that the life office sells 10,000 policies evenly over each year and that no policyholders stop paying premiums, what will the rate of premium income be for the life office during the first few years? Solution After t years the office will have sold 10 000, t policies. So the weekly premium income will be: 10 000 000, £10 £100,t t Since there are 52.18 (365.25 / 7 ) weeks in a year, this corresponds to an annual rate of income of:

5218 000 218 000. £100, £5, , t t

Question 5.3

Calculate the total premium income that would have been received during the first 3 years.

Between times t and t dt+ the total payment received is ( ) ( )M t dt M t+ - . If dt is

very small this is approximately ( )M t dt¢ or ( )t dtr . Theoretically, therefore, we

may consider the present value of the money received between times t and t dt+ as ( ) ( )v t t dtr . The present value of the entire cashflow is obtained by integration

as:

0

( ) ( )T

v t t dtrÚ

where T is the time of the last cashflow. For those who don’t like seeing integrals in formulae, don’t be put off. View this as summing, between times 0t = and t T= , the present values of an infinite number of payments. At each time point, t, the payment made (in the time interval of length dt) is

( )t dtr . We use ( )v t to discount the payment from time t to time 0, and the integral

sums the infinite number of payments.



If T is infinite we obtain, by a similar argument, the present value:

0

( ) ( )v t t dtr•

Ú

By combining the results for discrete and continuous cashflows, we obtain the formula:

0

( ) ( ) ( )tc v t v t t dtr•

+Â Ú (1.2)

for the present value of a general cashflow (the summation being over those values of t for which ct , the discrete cashflow at time t, is non-zero).

Assuming a constant interest rate this simplifies slightly to the important result for the present value of a series of discrete cashflows and a continuous cashflow:

Present value 0

( )t ttc v v t dtr

•

= +Â Ú

Example A company expects to receive for the next five years a continuous cashflow with a rate

of payment of 100 0.8t¥ at time t (years). Calculate the present value of this cashflow assuming a constant force of interest of 8% pa. Solution

5 5 50.08 0.08

0 0 0

50.08

0.080

0.08 5

( ) ( ) 100 0.8 100 ( 0.8)

100( 0.8)

log( 0.8)

100[( 0.8) 1]£257.42

0.08 log(0.8)

- -

-

-

-

= = ¥ ¥ = ¥

È ˘¥= Í ˙¥Í ˙Î ˚

È ˘¥ -= =Í ˙- +Í ˙Î ˚

Ú Ú Út t t

t

PV v t t dt e dt e dt

e

e

e

r



So far we have assumed that all payments, whether discrete or continuous, are positive. If one has a series of income payments (which may be regarded as positive) and a series of outgoings (which may be regarded as negative) their net present value is defined as the difference between the value of the positive cashflow and the value of the negative cashflow. The net present value will often be abbreviated to NPV. We will study net present values in Chapter 10.

Question 5.4

A company expects to receive for the next five years a continuous cashflow of £350 pa. It also expects to have to pay out £600 at the end of the first year and £400 at the end of the third year. Calculate the net present value of these cashflows if v t t( ) 1 100 for

0 5 t .



2 Valuing cashflows

Consider times 1t and 2t , where 2t is not necessarily greater than 1t . The value

at time 1t of the sum C due at time 2t is defined as:

(a) If 1 2t t≥ , the accumulation of C from time 2t until time 1t ; or

(b) If 1 2t t< , the discounted value at time 1t of C due at time 2t .

In both cases the value at time 1t of C due at time 2t is:

2

1

exp ( )t

tC t dtdÈ ˘-Í ˙Î ˚Ú (2.1)

This result was derived in the last chapter.

Question 5.5

Write down an expression for the value at time t1 of C due at time t2 if ( )t for all

t.

(Note the convention that, if 1 2t t> , 2 1

1 2

( ) ( )t t

t tt dt t dtd d= -Ú Ú )

Since:

2 2 1

1 0 0( ) ( ) ( )

t t t

tt dt t dt t dtd d d= -Ú Ú Ú

it follows immediately from Equation (2.1) that the value at time 1t of C due at

time 2t is:

2

1

( )

( )

v tC

v t (2.2)

This could also be written in the form CA t t

Cv t A t1

01 2

2 1( , )( ) ( , ) , remembering that

v t A t( ) ( , ) 1 0 .



Finding the value at time t1 of C due at time t2 involves discounting the payment C

from time t2 to time t1 . This could be performed directly using the expression

C A t t( , )1 2 .

Alternatively, we could first discount back to time 0 by multiplying by v t( )2 , and then

accumulate to time t1 by multiplying by A t( , )0 1 . The last step is equivalent to dividing

by v t( )1 and so this alternative uses Formula (2.2).

This is represented diagrammatically below: 0 t1 t2

1 1 2A t t( , )

v t( )2

1 1v t( )

Question 5.6 (easy)

Find the value at time 4 of a payment of 860 at time 10 if v( ) .10 0 76 and v( ) .4 0 91 .

The value at a general time 1t of a discrete cashflow of tc at time t (for various

values of t) and a continuous payment stream at rate ( )tr per time unit may now

be found, by the methods given in Section 1, as:

1 1

( ) ( )( )

( ) ( )tv t v t

c t dtv t v t

r•

-•+Â Ú (2.3)

where the summation is over those values of t for which 0tc π .



We note that in the special case when 1 0t = (the present time), the value of the

cashflow is:

( ) ( ) ( )tc v t t v t dtr•

-•+Â Ú

where the summation is over those values of t for which 0tc π . This is a

generalisation of Formula (1.2) to cover the past as well as present or future payments. If there are incoming and outgoing payments, the corresponding net value may be defined, as in Section 1, as the difference between the value of the positive and the negative cashflows. If all the payments are due at or after time

1t , their value at time 1t may also be called their “discounted value”, and if they

are due at or before time 1t , their value may be referred to as their

“accumulation”. It follows that any value may be expressed as the sum of a discounted value and an accumulation. This fact is helpful in certain problems. Also, if 1 0t = and all

the payments are due at or after the present time, their value may also be described as their “(discounted) present value”, as defined by Formula (1.2).

Question 5.7

Consider the following four payments: £100 on 1 January 2005, £130 on 1 January 2006, £150 on 1 January 2008 and £160 on 1 January 2009.

If t 0 on 1 January 2004 and v tt

( ) .( )

0 922

100

3

, calculate the value of these

payments on 1 January 2007 and express the value as a sum of a discounted value and an accumulation.

It follows from Formula (2.3) that the value at any time 1t of a cashflow may be

obtained from its value at another time 2t by applying the factor 2 1( ) / ( )v t v t , ie:

1 2 2

1

Value at time Value at time ( )

of cash flow of cash flow ( )

t t v t =

v t

È ˘È ˘ È ˘Í ˙Í ˙ Í ˙

Î ˚ Î ˚ Î ˚

or:

[ ] [ ]1 21 2

Value at time Value at time ( ) ( )

of cash flow of cash flow

t t v t = v t

È ˘ È ˘Í ˙ Í ˙Î ˚ Î ˚

(2.4)



Each side of Equation (2.4) is the value of the cashflow at the present time (time 0). In particular, by choosing time 2t as the present time and letting 1t t= , we obtain

the result:

Value at time Value at the present 1

of cash flow time of cash flow ( )

t =

v t

È ˘ È ˘ È ˘Í ˙ Í ˙ Í ˙

Î ˚Î ˚ Î ˚

These results are useful in many practical examples. The time 0 and the unit of time may be chosen so as to simplify the calculations.

2.1 Constant interest rate

The special case when we assume that interest rates remain constant is of particular

importance. Using this assumption v t vt( ) for all t. Remember also that v i 1 1( )

and so 1 1v it t ( ) .

In the diagram below, each of the three payments P1 , P2 and P3 has the same present

value.

Payments P Xvn1 P X2 P X i n

3 1 ( )

Time –n 0 n

This shows that we can think of the factors ( )1 i n and vn as a way of adjusting

payments to a different point on the time line. If the present value of a series of definite payments at a particular date is X, then:

● the accumulated value at a date n years later is: X i n( )1

● the present value at a date n years earlier is: Xvn

Note that n does not have to be a whole number in these formulae.



Example Under its current rent agreement, a company is obliged to make annual payments of £7,500 for the building it occupies. Payments are due on 1 January 2004, 1 January 2005 and 1 January 2006. The nominal rate of interest is 8% per annum, convertible quarterly. (Remember this means that the effective quarterly rate of interest is 2%.) Working in quarters, the present value of these payments on 1 January 2003 is:

4 8 127,500( ) £19,243.72v v v+ + = where 1 1.02 0.980392v = =

Alternatively, you might prefer to first calculate the effective annual rate, i, as:

41.02 1 8.243216%- = Working in years, the present value of these payments on 1 January 2003 is:

2 37,500( ) £19,243.72v v v+ + = where 1 1.08243216 0.923845v = =

The accumulated value of the payments on 1 January 2007 is:

4 4 16(1 ) 19,244 1.08243216 19,244 1.02 £26,418PV i¥ + = ¥ = ¥ =

Question 5.8

What would the present value of the rent payments in the above example be as at (i) 1 January 2002, (ii) 1 January 2005 and (iii) 1 July 2017?



Example The force of interest takes the following values: ( ) 0.04td = 0 < t 10

2( ) 0.001( 10) 0.04t td = - + 10 < t

Calculate the accumulation of £150 from time t = 0 to time t = 20. Solution The accumulation is:

150 (0, 20) 150 (0,10) (10, 20)A A A= ¥

Calculating the two accumulation factors:

10 0.04(0,10) 1.4918A e ¥= =

20

2

10

203

10

(10,20) exp 0.001( 10) 0.04

0.001exp ( 10) 0.04

3

exp(0.7333) 2.0820

A t dt

t t

Thus the accumulation is equal to:

150 (0, 20) 150 1.4918 2.0820 £466A = ¥ ¥ =



2.2 Payment streams

We saw how to evaluate the present value of a continuous payment stream in Section 1.2. The present value of a continuous payment stream received from time 0 to time T , where the rate of payment at time t is ( )t , is given by:

0

( ) ( )T

v t t dt

In this section, we will consider a continuous payment stream paid at a rate of ( )t from

time a to time b, during which time the force of interest is ( )t .

The present value at time a of this payment stream is:

( )exp ( )b t

t aa a

PV t s ds dt

Question 5.9

Explain where this formula has come from.

The accumulated value at time b of this payment stream is:

( )exp ( )b b

t ba t

AV t s ds dt

Question 5.10

Explain where this formula has come from.

Notice that in the formulae for the present and accumulated values, the integral in the exponential term has t as one of the limits. It is easy to make a mistake here and substitute in the wrong limits. We will now look at some examples and questions to see how these formulae are used in practice.



Example The force of interest at time t , where 0 10t , is given by ( ) 0.07td = .

Calculate the accumulated value at time 10 of the payment stream 0.0510 te , which is received between time 5 and 10. Solution The payment stream starts at time 5 and finishes at time 10, so 5a and 10b . Using these values in the general formula for the accumulated value of a payment stream with

0.05( ) 10 tt e gives:

10 100.05

105

10100.05

5

100.05 0.7 0.07

5

100.7 0.02

5

100.020.7

5

0.2 0.10.7

10 exp 0.07

10 exp 0.07

10

10

100.02

100.02

86.699

tt

t

tt

t t

t

t

AV e ds dt

e s dt

e e dt

e e dt

ee

e ee

Question 5.11

A continuous payment stream is paid at rate 0.03te- from time 0t = to time 10t = . Calculate the present value of this payment stream at time 0t = , given that the force of interest over this time period is 0.04 pa.



You should recall the following result from A-Level Maths, Higher Level Maths or the equivalent:

( ) ( )( )b bf t f t

aa

f t e dt e

This is the chain rule for differentiation expressed in integral form.

Example The force of interest is:

( ) 0.01 0.04t td = + 0 5t

Calculate the present value at time 0 of the payment stream 0.5 2t , which is received between time 0 and time 5. Solution We can apply the general formula for the present value of a payment stream with 0a ,

5b and ( ) 0.5 2t t to give:

5

00 0

52

00

52

0

(0.5 2)exp 0.01 0.04

(0.5 2)exp 0.005 0.04

(0.5 2)exp 0.005 0.04

t

t

t

PV t s ds dt

t s s dt

t t t dt

Now, using the general result ( ) ( )( )b bf t f t

aa

f t e dt e , with 2( ) 0.005 0.04f t t t ,

so that ( ) (0.01 0.04)f t t , we know that:

5 5

2 2

00

(0.01 0.04)exp 0.005 0.04 exp 0.005 0.04t t t dt t t



So, since 0.5 2 50 (0.01 0.04)t t , we have:

52

00

52

0

52

0

(0.5 2)exp 0.005 0.04

50 (0.01 0.04)exp 0.005 0.04

50 exp 0.005 0.04

50 exp( 0.125 0.2) exp(0)

13.87

tPV t t t dt

t t t dt

t t

Alternatively, we can carry out the integral:

5

20

0

(0.5 2)exp 0.005 0.04tPV t t t dt

using the substitution 20.005 0.04u t t , so that:

0.01 0.04 ( 0.01 0.04) 50 (0.5 2)du

t du t dt du t dtdt

Also, when 0t , 0u and when 5t , 0.325u , so the integral becomes:

0.325 0.325 0.325

00

0

50 50 50 1 13.87u utPV e du e e

Question 5.12

The force of interest is: ( ) 0.01 0.05t td = + 0 10t

Calculate the accumulated value at time 10 of a payment stream of 0.3 1.5t that is received continuously from time 4 to time 8.



2.3 Sudden changes in interest rates

Where the force of interest is not a continuous function of time, it is necessary to break up the calculations at the points where the interest rate changes. Where payments do not start immediately, the appropriate discount factor must be included.

Example Calculate the present values as at 1 January 2005 of the following payments: (i) a single payment of £2,000 payable on 1 July 2009 (ii) a single payment of £5,000 payable on 31 December 2016. Assume effective rates of interest of 8% per annum until 31 December 2011 and 6% per annum thereafter. Solution (i) Here, the interest rate is constant throughout the relevant period, so the present

value is just:

4½@8%2,000 2,000 0.70728 £1,415v = ¥ =

(ii) Here, we need to break the calculation up at 31 December 2011 when the interest

rate changes:

7@8% 5@6%5,000 5,000 0.58349 0.74726 £2,180v v¥ = ¥ ¥ =

Question 5.13

An investment of £1,000 made at time 0 is accumulated at the following rates: 8% per annum simple for two years, followed by a rate of discount of 6% per annum convertible monthly for two years. Calculate the accumulated amount of the investment after 4 years.



3 Interest income

Consider now an investor who wishes not to accumulate money but to receive an income while keeping his capital fixed at C. If the rate of interest is fixed at i per time unit, and if the investor wishes to receive income at the end of each time unit, it is clear that the income will be iC per time unit, payable in arrear, until such time as the capital is withdrawn. This is because the effective rate of interest, i, is defined to be the amount of interest a single initial investment will earn at the end of the time period.

Simple example An investor who wishes to receive income deposits £1,000 in a bank account that pays an effective rate of interest of 8% per annum. The interest income is paid to the investor at the end of each year. The amount of each payment is £80, (8% of 1,000).

However, if interest is paid continuously with force of interest ( )td at time t then

the income received between times t and +t dt will be ( )C t dtd . So the total

interest income from time 0 to time T will be:

= Ú0( ) ( )T

I T C t dtd (4.1)

If interest is paid continuously to the investor then we are just considering a continuous cashflow with a rate of payment of C t ( ) . The total amount of interest received can

therefore be found by applying Formula (1.1), which gives Formula (4.1). The total amount of interest received is the sum, between 0 and T , of lots of small interest payments, each of amount ( )C t dtd .

If the investor withdraws the capital at time T, the present values of the income and capital at time 0 are:

0

( ) ( )T

C t v t dtdÚ (4.2)

and: ( )Cv T (4.3)

respectively.



Since:

0 0 0

00

( ) ( ) ( )exp ( )

exp ( )

1 ( )

T T t

Tt

t v t dt t s ds dt

s ds

v T

d d d

d

È ˘= -Í ˙Î ˚

È ˘Ê ˆ= - -Í ˙Á ˜Ë ¯Î ˚

= -

Ú Ú Ú

Ú

we obtain:

0

( ) ( ) ( )T

C C t v t dt Cv Td= +Ú (4.4)

as one would expect by general reasoning. If we invest an amount of capital C, then the present value of the proceeds we receive from this investment should equal our original amount of capital.

Example An investor deposits £2,000 in a bank account and receives income at the end of each of the next three years. The rate of interest is 4% pa effective. The investor withdraws the capital after three years. At the end of each year the investor receives 0.04 2,000 £80¥ = .

The present value of the interest received is:

2 380( ) £222.01v v v+ + =

The present value of the capital received after three years is:

32,000 £1,777.99v =

The present value of the capital plus the present value of the interest equals the initial investment, ie: 1,777.99 222.01 £2,000+ =



Question 5.14

A woman deposits £200 in a special bank account. Interest is paid to the woman every year on her birthday for five years. The capital is returned after exactly five years, along with any interest accrued since her last birthday. Interest is calculated at an effective rate of 6% pa. Calculate the present value of the interest received by the woman.

So far we have described the difference between money returned at the end of the term and the cash originally invested as “interest”. In practice, however, this quantity may be divided into interest income and capital gains, the term capital loss being used for a negative capital gain. If you invest some capital then not only might you receive income but the value of your capital may also increase (or decrease). Equities or shares are a good example of this. These were introduced earlier in the course. If you buy some shares in a company then you should receive dividends or interest from the company. However the capital value that you receive back will depend upon the market price of the shares when you decide to sell. We will consider this in more detail later in the course.

Question 5.15

A rich woman pays £2m and in return expects to receive a continuous cashflow for the next six years with a constant rate of payment. Calculate the annual payment from this cashflow and the accumulated amount of the cashflow after six years if the interest rate is 9% pa effective.

Question 5.16

True or false:

(i) 21 2

1

(0, )( , )

(0, )

A tA t t

A t=

(ii) 2 1 1 2(0, ) (0, ) ( , )A t A t A t t= +

(iii) 2 1 2 1( ) ( ) ( )v t v t v t t= -

(iv) 1 1 22

1(0, ) ( , )

( )A t A t t

v t=



4 Exam-style question

This is a typical exam-style question on this chapter. Have a go at it before turning over and having a look at the solution.

Question The force of interest at any time t (measured in years) is given by:

0.04 0 1

( ) 0.05 0.01 1 5

0.24 5

t

t t t

t

d< £Ï

Ô= - < £ÌÔ >Ó

(i) Derive and simplify as far as possible expressions for ( )A t , where ( )A t is the

total accumulated value at time t ( 0 ) of an investment of 1 at time 0. [5]

(ii) A continuous payment stream is received at a rate of 0.0225 te units per annum between time 5 and time 10. Calculate the present value (at time 0) of this payment stream. [4]

[Total 9]



Solution (i) We have to break down the expression for A t( ) at times when the force of

interest changes. Since A t( ) represents the accumulated value at time t of an

investment of 1 at time 0, we have: 0 1 t :

A t e t( ) . 0 04

1 5t< £ :

2

2

0.04

1

0.04 2

1

0.04 0.025 0.01 0.025 0.01

0.025 0.01 0.025

( ) exp 0.05 0.01

exp 0.025 0.01

t

t

t t

t t

A t e s ds

e s s

e e

e

- - +

- +

È ˘Í ˙= ¥ -Í ˙Î ˚

È ˘= ¥ -Î ˚

= ¥

=

Ú

5 t< :

20.025 5 0.01 5 0.025

5

0.6 0.24 1.2

0.24 0.6

( ) exp 0.24t

t

t

A t e ds

e e

e

¥ - ¥ +

-

-

È ˘Í ˙= ¥Í ˙Î ˚

=

=

Ú



(ii) It is easiest to calculate the present value at time 5 first and then discount it back to time 0. The present value at time 5 is:

[ ]( )

100.02

55 5

100.02

55

100.02 0.24 1.2

5

101.2 0.26

5

100.261.2

5

1.22.6 1.3

25 exp 0.24

25 exp 0.24

25

25

250.26

25

0.26

63.2924

tt

t

tt

t t

t

t

PV e ds dt

e s dt

e e dt

e e dt

ee

ee e

-=

-

- - +

-

-

- -

È ˘Í ˙= -Í ˙Î ˚

= -

=

=

È ˘= Í ˙-Í ˙Î ˚

È ˘= -Î ˚-

=

Ú Ú

Ú

Ú

Ú

We then need to discount this back to time 0. We know from part (i) that the

accumulation factor from time 0 to time 5 is:

20.025 5 0.01 5 0.025 0.6(5)A e e¥ - ¥ += =

So the discount factor from time 5 to time 0 is 0.6e . The present value at time 0 of the payment stream is therefore:

0.60 63.2924 34.7356tPV e



This page has been left blank so that you can keep the chapter summaries

together for revision purposes.



Chapter 5 Summary In many compound interest problems you may need to find the discounted present value of cashflows due in the future. It is important to distinguish between discrete and continuous payments. The present value of a series of discrete payments is the sum of the individual present values. The present value of continuous payments is found by integrating the rate of payment multiplied by a discount factor. So the formula for present value is:

0

( ) ( ) ( )tc v t v t t dtr•

+Â Ú

The net present value is defined as the difference between the value of the positive cashflow and the value of the negative cashflow. The value of payments that are due after the time of valuation is called a discounted value. The value of payments that are due before the time of valuation is called an accumulated value. The value of a cashflow at one particular time can easily be found from the value of the cashflow at a different time. The formula for moving along the timeline is:

1 2 2

1

Value at time Value at time ( )

( ) of cash flow of cash flow

t t v t =

v t

È ˘ È ˘ È ˘Í ˙ Í ˙ Í ˙Í ˙ Í ˙ Î ˚Î ˚ Î ˚

An investor may wish to receive an income while keeping the amount of capital fixed. The present value of the income plus the present value of the returned capital equals the initial capital invested ie:

0( ) ( ) ( )

TC C t v t dt Cv Td= +Ú







Chapter 5 Solutions Solution 5.1

PV 2 2.5 2 2.5280 360 280 1.15 360 1.15 211.72 253.84 £465.56v v - -= + = ¥ + ¥ = + = Solution 5.2

01

exp ( )

j

j

n tt

j

PV c s ds

Solution 5.3

Total premium income:

33 2

0 0

5,218,0005,218,000 2,609,000 9 £23,481,000

2

ttdt

È ˘= = = ¥ =Í ˙

Í ˙Î ˚Ú

Solution 5.4

Present value of the income:

5 5

0 0

52

0

350 ( ) 350(1 100)

350 1.75 £1,706.25

v t dt t dt

t t

= -

È ˘= - =Î ˚

Ú Ú

Present value of the outgo:

600 (1) 400 (3) 600 0.99 400 0.97 £982v v+ = ¥ + ¥ =

So NPV 1,706.25 982 £724.25= - = .



Solution 5.5

2 1

1 2

( )

( )

Value t t

t t

Ce

Ce

d

d

- -

-

=

=

Solution 5.6

(10) 0.76860 860 718.24

(4) 0.91

v

v¥ = ¥ =

Solution 5.7

From the formula 3( 2)

( ) 0.92100

tv t

-= - , we can calculate (1) 0.93v = , (2) 0.92v = ,

(3) 0.91v = , (4) 0.84v = , (5) 0.65v = .

The accumulation is the value on 1 January 2007 of the first two payments which is equal to:

1

(100 (1) 130 (2)) £233.63(3)

v vv

+ =

The discounted value is the value on 1 January 2007 of the last two payments which is equal to:

1

(150 (4) 160 (5)) £252.75(3)

v vv

+ =

So the net value 233.63 252.75 £486.38= + = . Solution 5.8

(i) To find the PV as at 1 January 2002, we must move our reference point back in time 1 year (from 1 January 2003 to 1 January 2002) ie we need to multiply

19,243.72 by a factor of 4 4@ 2% 1/1.02 0.923845v = = .

So PV as at 1 January 2002 419,244 £17,778v= = .



(Note that the PV will be less, since the payments are now more distant from the reference point.)

(ii) PV as at 1 January 2005 819,244 1.02 £22,547= ¥ = .

(iii) PV as at 1 July 2017 5819,244 1.02 £60,688= ¥ = .

Solution 5.9

The formula is ( )exp ( )

b t

a a

t s ds dt . Consider the payment at time t, which is at a

rate of ( )t . This payment needs to be discounted back to time a and the discount

factor will be exp ( ) t

a

s ds .

Finally, we need to add together all the present values of the payments at the different times. Since we are receiving payments continuously, we integrate these present values between the limits a and b, ie the times between which the payment comes in. Solution 5.10

The formula is ( ) exp ( )

b b

a t

t s ds dt . Consider the payment at time t, which is at a

rate of ( )t . This payment needs to be accumulated to time b and the accumulation

factor will be exp ( ) b

t

s ds .

Finally, we need to add together all the accumulated values of the payments at the different times. Since we are receiving payments continuously, we integrate these present values between the limits a and b, ie the times between which the payment comes in.



Solution 5.11

0 010 10 10 100.04 [0.04 ]0.03 0.03 0.03 0.04 0.07

0 0 0 0

100.07

0

0.7

1

0.07

1(1 ) 7.192

0.07

ttds st t t t t

t

PV e e dt e e dt e e dt e dt

e

e

- -- - - - -

-

-

Ú= = = =

È ˘= -Í ˙Î ˚

= - =

Ú Ú Ú Ú

Solution 5.12

We can calculate the accumulated value at time 8 first and then accumulate that value to time 10. Using the general formula for the accumulated value of a payment stream with

4a , 8b and ( ) 0.3 1.5t t :

8 8

84

8 82

4

82

4

(0.3 1.5 )exp 0.01 0.05

(0.3 1.5 )exp 0.01 0.025

(0.3 1.5 )exp 1.68 0.01 0.025

tt

t

AV t s ds dt

t s s dt

t t t dt

Now, using the general result that ( ) ( )( )b bf t f t

aa

f t e dt e , we know:

8 8

2 2

44

(0.01 0.05 )exp 1.68 0.01 0.025 exp 1.68 0.01 0.025t t t dt t t

So, since 0.3 1.5 30 (0.01 0.05 )t t , we have:

82

84

82

4

30 (0.01 0.05 )exp 1.68 0.01 0.025

30 exp 1.68 0.01 0.025

tAV t t t dt

t t



Evaluating this gives:

8

0 1.24

30 exp(1.68 0.08 1.6) exp(1.68 0.04 0.4)

30

73.6684

tAV

e e

We now need to accumulate this value at time 8 forward to time 10:

10 8

10

8

102

8

0.92

(8,10)

73.6684 exp (0.01 0.05 )

73.6684 exp 0.01 0.025

73.6684 exp 0.1 2.5 0.08 1.6

73.6684

184.855

t tAV AV A

t dt

t t

e

Alternatively, we can carry out the integral:

8

28

4

(0.3 1.5 )exp 1.68 0.01 0.025tAV t t t dt

using the substitution 21.68 0.01 0.025u t t , so that:

0.01 0.05 0.01 0.05 30 0.3 1.5du

t du t dt du t dtdt

Also, when 4t , 1.24u and when 8t , 0u , so the integral becomes:

0 0 1.24

81.24

1.24

30 30 30 1 73.6684u utAV e du e e



Solution 5.13

The accumulated value of the investment is:

( ) 240.06121,000(1 2 0.08) 1 £1,308.29

-+ ¥ - =

Solution 5.14

The present value of the interest received plus the present value of the capital returned will equal the initial deposit. Therefore the present value of the interest received equals the initial deposit less the present value of the capital returned:

55

200200 200 200 £50.55

1.06v- = - =

Solution 5.15

If C is the constant rate of payment per annum then we need to solve the equation:

66 6

0 0

12,000,000

log log(1 )

2,000,000£0.427m

4.68489

tt v v

Cv dt C Cv i

C

È ˘ -= = =Í ˙ +Í ˙Î ˚

fi = =

Ú

Because we know the present value of the cashflow is £2m we can easily work out the

accumulated value after six years by multiplying this by 6(1 )i+ :

Accumulated value 62,000,000 1.09 £3.354m= ¥ =



Solution 5.16

(i) True. (ii) False. A correct relationship is 2 1 1 2(0, ) (0, ) ( , )A t A t A t t= ¥ .

(iii) False. The function ( )v t is a discount factor from time t to time 0, ie it gives

the present value at time 0 of a payment of 1 at time t . So, the LHS, 2( )v t , discounts a payment from time 2t to time 0.

On the RHS, 1( )v t discounts a payment from time 1t to time 0 and 2 1( )v t t

discounts a payment from time 2 1t t to time 0. This is not the same as the

LHS.

A correct relationship is 2 11 2

1( ) ( )

( , )v t v t

A t t , where

1 2

1

( , )A t t discounts a

payment from time 2t to time 1t and 1( )v t discounts a payment from time 1t to

time 0. (iv) True.










CT1-08: Equations of value Page 17


Chapter 8 Summary An equation of value equates the present value of money received to the present value of money paid out. You will need to use the following formulae for equations of value:

PV income – PV outgo = 0

01

( ) 0d dr•- -

=+ =Â Úr

r

nt t

tr

c e t e dt

01

(1 ) ( )(1 ) 0r•- -

=+ + + =Â Úr

r

nt t

tr

c i t i dt

To find the yield on a transaction from an equation of value of the form ( ) 0f i , we

need:

to find 1i and 2i such that 1( )f i and 2( )f i are of opposite sign, and

the final value obtained for i to be greater than 1 . Linear interpolation is often needed to find the yield from an equation of value. The formula is:

i iP P

P Pi i

11

2 12 1( )

If there is uncertainty about the payment or receipt of a cashflow at a particular time, allowance can be made in one of two ways:

apply a probability of payment/receipt to the cashflow at each time

use a higher rate of discount such as a new force of interest , where .

Page 18 CT1-08: Equations of value






It follows from equations (1.3) and (1.4) that if > -( )1(1 )p D

i tR

then the purchaser

will receive a capital gain when the security is redeemed. From the investor’s viewpoint, the sooner a capital gain is received the better. The investor will therefore obtain a greater yield on a security which is redeemed first.

Question 12.7

An investor purchases a £100 zero-coupon bond for £80. Calculate the yield obtained if the bond is redeemed after (a) five years, and (b) ten years.

So to ensure the investor receives a net annual yield of at least i then they should assume the worst case result: that the redemption money is paid as late as possible, ie = 2n n .

Similarly if < -( )1(1 )p D

i tR

then there will be a capital loss when the security is

redeemed. The investor will wish to defer this loss as long as possible, and will therefore obtain a greater yield on a security which is redeemed later.

Question 12.8


So to ensure the investor receives a net annual yield of at least i then they should assume the worst case result: that the redemption money is paid as soon as possible, ie = 1n n .

Finally, if = -( )1(1 )p D

i tR

then there is neither a capital gain nor a capital loss. So

it will make no difference to the investor when the security is redeemed. The net annual yield will be i irrespective of the actual redemption date chosen.

Question 12.9




Example 12.10 A fixed interest stock with a coupon of 8% per annum payable half-yearly in arrears can be redeemed at the option of the borrower at any time between 10 and 15 years from the date of issue. The stock is redeemable at par. What price should an investor subject to tax at 25% on income only, who wishes to obtain a net yield of at least 7% per annum, pay for £100 nominal of this stock? Solution Carrying out the capital gains test:

(2) 0.52(1.07 1) 0.0688i = - = 18

(1 ) 0.75 0.06100

Dt

R- = ¥ =

Since (2)1(1 ) D

Ri t> - there is a gain on redemption. The worst case scenario is the

latest possible redemption date (ie time 15 years). This will give the minimum yield. So:

P a v 0 75 8 100 7% 5559 36 24 8315

2 15. @ . . £91.|( )

Suppose, alternatively, that the price of the loan is given. The minimum net annual yield is obtained by again assuming the worst case result for the investor. So if: (a) <P R , then the investor receives a capital gain when the security is

redeemed. The worst case is that the redemption money is repaid at the latest possible date. If this does in fact occur, the net annual yield will be that calculated. If redemption takes place at an earlier date, the net annual yield will be greater than that calculated.

(b) P R> , then the investor receives a capital loss when the security is

redeemed. The worst case is that the redemption money is repaid at the earliest possible date. The actual yield obtained will be at least the value calculated on this basis.

(c) P R= , then the investor receives neither a capital gain nor a capital loss.

The net annual yield is i , where = -( )1(1 )p D

i tR

, irrespective of the actual

redemption date chosen.



Chapter 12 Summary The price for a fixed interest security can be calculated as the present value of the interest and redemption payments. When the proceeds are subject to income tax or capital gains tax, the net payments must be used. The formulae used for calculating the price are:

Ignoring tax: ( )p nn

P Da Rv= +

Allowing for income tax: ( )1(1 ) p n

nP D t a Rv= - +¢

Allowing for income and capital gains tax: ( )1 2(1 ) ( )p n n

nP D t a Rv t R P v= - + - -¢¢ ¢¢

Based on a given purchase price, the running yield and the redemption yield can be calculated, on either a gross or a net basis. The running yield is the coupon divided by the price. The redemption yield is the rate of interest that equates the price with the present value of the interest and redemption payments. For a security with a redemption date that is selected at the option of the borrower, an investor who wishes to achieve a yield of at least i should value the security on the assumption that will give the lowest yield, ie the worst case. The worst case is the latest possible date for a gain and the earliest possible date for a loss. Equities are usually valued by assuming that dividends increase at a constant rate and continue in perpetuity. If A is the value of an equity paying annual dividends just after a dividend payment, then:

AD g

i g

( )1 where D is the amount of the dividend just paid

Property is valued in a similar way although rents are generally fixed for a number of years at a time and some contracts are for a fixed term. The equation of value for property is:

/1/

1

k m

k mmk

A D v where 1/k mm D is the rental income at time t .



The real rate of interest of a transaction is the rate of interest after allowing for the effect of inflation on a payment series:

1

-= -¢+

i ji i j

j

Index linked bonds can be valued by allowing for the increases in the coupons and the redemption payment. The equation of value for an index linked bond is:

AD Q k

Qv R

Q n

Qvi

kin

k

n

2

2

0 02

1

2 ( )

( )

( )

( )/ where ( )Q t is the index value at time t

Capital gains tax is a tax levied on the difference between the sale or redemption price of a stock (or other asset) and the purchase price, if lower. Sometimes it is possible to offset capital losses against capital gains or to reduce the capital gain by the inflation over the period of the gain. This latter practice is generally referred to as the “indexation of gains”.

CT1-15: Stochastic interest rate models Page 21


3 Exam-style questions

Before you finish, try the following two exam-style questions on stochastic interest rates. The solutions are over the page.

Question 1 Interest rates over the next five years are fixed at either 4% pa with probability 0.2 or 5% pa with probability 0.8. What is the standard deviation of the present value of a payment of £25,000 in 5 years’ time? [3]

Question 2 The annual returns, i , on a fund are independent and identically distributed, with a mean of 6% and a standard deviation of 3%. Each year, the distribution of 1 i is

log-normal with parameters and 2 .

(i) Calculate the values of and 2 . [4]

(ii) Calculate the probability that the accumulation of a single investment of £1 will

be greater than 110% of its expected value after 10 years. [4] [Total 8]

Page 22 CT1-15: Stochastic interest rate models


Solution 1

The present value will either be 5

25,00020548.18

1.04= with probability 0.2, or

5

25,00019588.15

1.05= , with probability 0.8.

The mean of these values is 20,548.18 0.2 19,588.15 0.8 19,780.16¥ + ¥ = .

The standard deviation of these values is:

2 2 220,548.18 0.2 19,588.15 0.8 19780.16 £384¥ + ¥ - =

Solution 2 (i) Parameters of the log-normal distribution

We know that 21 log ,i N , and that 0.06E i and 2var 0.03i . Using

the formulae for the mean and variance of the log-normal distribution from the Tables:

21

21 1 1.06E i E i e (Equation 1)

and: 2 22 2var 1 var 0.03 1i i e e (Equation 2)

Squaring Equation 1, we have:

21 2

2

22 21.06e e

Substituting this into Equation 2 gives:

2 22 2 2

2

0.031.06 1 0.03 ln 1 0.00080068

1.06e

CT1-15: Stochastic interest rate models Page 23


So, using Equation 1:

212ln 1.06 0.057869

(ii) Probability Let 10S denote the accumulated value at time 10 of an investment of £1 at time 0. We

require the probability 10 101.1P S E S .

As the interest rate in each year is independent of that in other years, using the varying rate model, we have:

10 1010 1 1.06E S j since 0.06j E i

The distribution of 10S under this model is:

210 log 10 ,10 log 0.57869,0.0080068S N N

So:

10 1010 10 10 101.1 1.1 1.06 ln ln 1.1 1.06P S E S P S P S

Using 0,1Z N , we have:

1010

10

ln 1.1 1.06 0.57869ln ln 1.1 1.06

0.0080068

1.110

1 1.110

1 0.86650

0.1335

P S P Z

P Z

ie a probability of 13.35%.

Page 24 CT1-15: Stochastic interest rate models


4 Appendix – basic statistical results

This appendix gives a brief summary of the statistical results that are used in this chapter. All of these can be found in the CT3 course notes, the StatsPack or any A level (or equivalent) statistics textbook.

4.1 Expectation

The expectation (or mean) of a discrete random variable which takes values x with probabilities ( )P X x= is given by:

( ) ( )x

E X xP X x= =Â

We can also find the expectations of functions of X :

2 2( ) ( )x

E X x P X x= =Â

In general:

[ ( )] ( ) ( )x

E g X g x P X x= =Â

The following relationships hold: ( ) ( )E aX b aE X b+ = +

1 2 1 2( ) ( ) ( )E X X E X E X+ = +

1 2 1 2( ) ( ) ( )E X X E X E X= if 1X , 2X are independent

CT1: Q&A Bank Part 1 – Questions Page 13


Question 1.37

(i) Calculate the present value of a continuously payable annuity that is initially paid at a rate of £200 pa but decreases linearly to £100 pa after 10 years, assuming a force of interest of ( ) 0.2 0.01t td = - , where t is the time from

commencement (measured in years). [6] (ii) Calculate the accumulated value of the annuity after 10 years at the same force

of interest. [2] [Total 8] Question 1.38

(i) Assuming a rate of interest of 6% pa, calculate the present value as at 1 January 2008 of the following annuities, each with a term of 25 years:

(a) an annuity payable annually in advance from 1 January 2009, initially of

£3,000 pa, and increasing by £500 pa on each subsequent 1 January (b) an annuity as in (i), but only 10 increases are to be made, the annuity then

remaining level for the remainder of the term. [6] (ii) An investor is to receive a special annual annuity for a term of 10 years in which

payments are increased by 5% compound each year to allow for inflation. The first payment is to be £1,000 on 1 November 2009. Calculate the accumulated value of the annuity payments as at 31 October 2026 if the investor achieves an effective rate of return of 4% per half year. [4]

[Total 10] Question 1.39

The force of interest at time t , where t is measured in years, is given by:

20.002 0.01 0.0004 0 6

( ) 0.01 0.003 6 10

0.04 10

t t t

t t t

t

Calculate the present value at time 0 of a continuous payment stream of £120 per annum payable from time 10 to time 15. [6]

Page 14 CT1: Q&A Bank Part 1 – Questions


Question 1.40

The force of interest, t , is a function of time and at any time t , measured in years, is

given by the formula:

0.06 0 4

0.10 0.01 4 7

0.01 0.04 7

t

t t t

t t

(i) Calculate the value at time 5t of £1,000 due for payment at time 10t . [5]

(ii) Calculate the constant rate of interest per annum convertible monthly which

leads to the same result as in (i) being obtained. [2]

(iii) Calculate the accumulated amount at time 12t of a payment stream, paid continuously from time 0t to 4t , under which the rate of payment at time t

is 0.02100 tt e . [6]

[Total 13]

CT1: Q&A Bank Part 1 – Solutions Page 11


Solution 1.26

The discount factor for a payment at time t is:

20.005

0

( ) exp 0.01t

tv t rdr e [1]

This continuously-increasing, continuously-payable annuity has a rate of payment of t at time t , so its present value can be expressed as an integral as:

2

100.005

100

( ) tIa te dt [1]

Now, using the general result:

( ) ( )( )b bf t f t

aa

f t e dt e

we know that:

2 2

10 100.005 0.005

00

( 0.01 ) t tt e dt e- -È ˘- = Í ˙Î ˚Ú

So, we have:

2 2

2

10 100.005 0.005

100 0

100.005 0.5

0

( ) 100 ( 0.01 )

100 100 1 39.35

t t

t

Ia te dt t e dt

e e

[2]

Alternatively, we can use the substitution 20.005u t , so that:

0.01 0.01 100du

t du t dt du t dtdt

Page 12 CT1: Q&A Bank Part 1 – Solutions


Noting that when 0t , 0u , and when 10t , 0.5u , we have:

( )210 0.5

0.50.005 0.510 0

0 0

( ) 100 100 100 1 39.35t u uIa te dt e du e e- -- -È ˘= = - = - = - - =Î ˚Ú Ú

Solution 1.27

(i) Expressed in terms of v’s, this function is:

( ) 1 2 3 ( 1)|

1 1 1 1 1 p p p p np p nn

a v v v v vp p p p p

[1]

Multiplying by p, then multiplying through by 1(1 ) pi and subtracting gives:

( ) 1 2 3 ( 1)|

( )1 1 2 3 ( 1)|

( )1|

(1 ) 1

[(1 ) 1] 1

p p p p np p nn

pp p p p np pn

pp nn

pa v v v v v

i pa v v v v

i pa v [2]

So:

( )| 1 ( ) ( )

1 1 1 (1 )

[ 1 1]

n n np

p p pn

v v ia

i ip i [1]

(ii) We have a rate of payment of t pa. So considering a period of t at time t, then

the present value will be tt t v .

t 0 n time

t v

t

discounting to

payment

t = 0

Summing for all the times from 0 to n gives:

|0

( ) n

tnIa t v dt [1]



So:

12

5,487 £94070.0484

X [1]

(iii) The accumulated value of the first 7 years’ payments in (i) is:

84 72| |72 12

1000 400(1 )

12 12

1000 4002.3067 51.1504 0.48850 11.2551 £10,255

12 12

AV i a v a

[2]

The accumulated value for (ii) is:

|84940 940

131.324 £10,28712 12

s [1]

Solution 1.36

(i) The formula for 1 2( , )A t t in terms of the force of interest ( )td is:

2

1

1 2( , ) exp ( )t

t

A t t t dt [1]

Note that this formula is given on page 31 of the Tables. (ii)(a) The nominal rate of interest over a one year time period is the same as the

equivalent constant effective annual rate of interest, which can be found as follows:

2 2

1 1

22

1

0.055

1 exp ( ) exp (0.01 0.04)

exp 0.005 0.04

exp[(0.02 0.08) (0.005 0.04)] 1.05654

i t dt t dt

t t

e [3]

So the annual rate of interest is 5.65%.



(ii)(b) The accumulation factor from time ½ to time 6 is:

6212 1/ 2

0.39875

( ,6) exp 0.005 0.04

exp[ 0.18 0.24 (0.00125 0.02)] 1.4900

A t t

e [3]

(iii) The accumulated values are:

(a) 1

2 0.05100 100 1.025315 £102.53 e [1]

(b) 6

0.05100 1 100 1.025262 £102.53

12

[1]

(c) 1

2100 1.05 100 1.024695 £102.47 [1] Solution 1.37

(i) Present value The rate of payment starts at £200 pa, but decreases linearly to £100 pa after 10 years. So the rate of payment at time t , ( )t is 200 10t .

The present value of this payment stream at time 0 is given by:

10

0 0

(200 10 )exp 0.2 0.01t

PV t s ds dt

[2]

This can be evaluated as:

102

00

102

0

(200 10 )exp 0.2 0.005

(200 10 )exp 0.2 0.005

tPV t s s dt

t t t dt

[1]




( ) ( )( )b bf t f t

aa

f t e dt e

we know that:

( ) ( )10 10

2 2

00

( 0.2 0.01 )exp 0.2 0.005 exp 0.2 0.005t t t dt t tÈ ˘- + - + = - +Î ˚Ú

Since 200 10 1,000( 0.2 0.01 )t t , we have:

102

0

102

0

102

0

(200 10 )exp 0.2 0.005

1,000 ( 0.2 0.01 )exp 0.2 0.005

1,000 exp 0.2 0.005

PV t t t dt

t t t dt

t t

[2]


( ) ( )102 1.5 0

01,000 exp 0.2 0.005 1,000 £776.87PV t t e e-È ˘= - - + = - - =Î ˚ [1]

Alternatively, we could carry out the integration:

10

2

0

(200 10 )exp 0.2 0.005t t t dt

using the substitution 20.2 0.005u t t , so that:

0.2 0.01 ( 0.2 0.01 ) 1,000 (200 10 )du

t du t dt du t dtdt

Also, when 0t , 0u , and when 10t , 1.5u , so we have, as before:

( ) ( )10 1.5

2 1.5 0

0 0

(200 10 )exp 0.2 0.005 1,000 1,000ut t t dt e du e e-

-- - + = - = - -Ú Ú



(ii) Accumulated value The accumulated value at time 10 can be found by accumulating the present value at time zero for 10 years: 776.87 (0,10)A

where:

2

10

0

102

0

0.2 10 0.005 10 1.5

(0,10) exp 0.2 0.01

exp 0.2 0.005

A t dt

t t

e e

[1] So, the accumulated value at time 10 is:

1.5776.87 (0,10) 776.87 £3,482A e [1]

Solution 1.38

(i)(a) The present value of the payments can be expressed as: | |25 252,500 500( ) PV a Ia [1]

Using annuity values from the Tables: 2,500 12.7834 500 128.7565 £96,337 PV [2]

Alternatively: | |25 252,500 500 ( ) PV va v Ia

(i)(b) The present value of the payments is:

11| | |25 11 142,500 500( ) 5,500 PV a Ia v a [1]



So, using annuity values from the Tables:

112,500 12.7834 500 42.7571 5,500 1.06 9.2950 £80, 268 PV [2]

Alternatively:

12| | |25 11 142,500 500 ( ) 5,500 PV va v Ia v a

(ii) The accumulated value of the payments is:

2 9

342 4 18

1.05 1.05 1.051,000 1.04 1

1.04 1.04 1.04

AV [1]

@3.0095%34|10

1,000 1.04 a [1]

1,000 3.7943 8.7827 £33,324 [2]

Solution 1.39

The present value of the payment stream at time 10 (when the payment stream begins) is:

15

1010 10

120exp 0.04t

tPV ds dt

[1]

15

1010

15

10

150.4 0.04

10

120exp 0.04

120exp 0.04 10

120

t

t

s dt

t dt

e e dt

[1]

Carrying out the integration, we have:

150.04 0.04 15 0.04 100.4 0.4

10

10

120 120 £543.810.04 0.04

t

te e e

PV e e

[1]



Alternatively, since there is a constant force of interest over the time period when the payment stream is being received, and the rate of payment is constant, we can express the present value of the payment stream at time 10 as:

5 5

10 5

1 1120 120 120t

v ePV a

Using 0.04 , we have:

5 0.04

101

120 £543.810.04te

PV

We then need to discount the present value at time 10 back to time 0 using the appropriate values of ( ) t :

10 6

20

6 0

543.81exp (0.01 0.003 ) exp (0.002 0.01 0.0004 )tPV t dt t t dt

[½] Now:

10 102

66

0.136

exp (0.01 0.003 ) exp 0.01 0.0015t dt t t

e

[1] and:

6 62 2 3

00

0.2208

0.0004exp (0.002 0.01 0.0004 ) exp 0.002 0.005

3t t dt t t t

e

[1] So:

0.136 0.22080 543.81 £380.62tPV e e

[½]



Solution 1.40

This question is taken from Subject CT1, April 2008, Question 9. (i) Value at time 5 of £1,000 due at time 10 The value at time 5 of £1,000 at time 10 is given by the expression:

7 10

5 7

1,000exp 0.1 0.01 exp 0.01 0.04t dt t dt [1]

Now:

7 72

55

exp 0.1 0.01 exp 0.1 0.005

exp( (0.455 0.375))

exp( 0.08)

t dt t t

[1½]

and:

10 102

77

exp 0.01 0.04 exp 0.005 0.04

exp( (0.1 ( 0.035)))

exp( 0.135)

t dt t t

[1½]

Therefore, the value at time 5 is:

0.08 0.135 0.2151,000 1,000 £806.54e e e [1]



(ii) Equivalent rate of interest convertible monthly

Let (12)i be the equivalent constant rate of interest convertible monthly over the 5 years (ie 60 months). Therefore:

60(12)

806.54 1 1,00012

i

[1]

Hence:

160(12) 1,000

12 1 4.3077%806.54

i

[1]

(iii) Accumulated value of payment stream The accumulated value of the payment stream at time 4 is:

4 4

0.02

0

100 exp 0.06t

t

e ds dt

[1]

Carrying out the integration, we have:

4 440.02 0.02 0.24 0.06

0 0

40.24 0.04

0

40.040.24

0

0.240.04 4

100 exp 0.06 100

100

1000.04

1001

0.04

469.9052

t t tt

t

t

e s dt e e dt

e e dt

ee

ee

[1½] We now need to accumulate this value at time 4 forward to time 12.



The accumulated value of the payment stream at time 7 is: 469.9052 (4,7)A

where:

7 72

44

0.135

(4,7) exp (0.1 0.01 ) exp 0.1 0.005

exp(0.455 0.32)

A t dt t t

e

[1½] The accumulated value of the payment stream at time 12 is:

0.135469.9052 (7,12)e A

where:

12 122

77

0.275

(7,12) exp (0.01 0.04) exp 0.005 0.04

exp(0.24 ( 0.035))

A t dt t t

e

[1½] So the final accumulated value of the payment stream at time 12 is:

0.135 0.275469.9052 708.0615e e [½]












Solution 2.9

We have: Total interest paid = flat rate term amount of the loan¥ ¥

So, in this case, the amount of interest payable, assuming that the loan is for an amount 1, will be 0.075. The monthly repayment is the total amount paid out divided by the number of payments:

1.075

6 [1]

since the loan is for one year. Working in years, the equation of value is:

(12)0.5|0.5

1.07512 1

6v a¥ ¥ = [1]

We need to solve this using trial and error. Try 9.5%:

(12)0.5|0.5

1.07512 1.0006

6v a¥ ¥ = [1]

Try 9.6%:

(12)0.5|0.5

1.07512 0.9998

6v a¥ ¥ = [1]

Since the 9.6% value is closer to 1 than the 9.5% value, the APR is 9.6%. [1] Note that here, the flat rate is 7.5% and the APR turns out to be 9.6%. So, the APR is therefore quite a lot less than 15% (twice the flat rate), which is our usual first guess. The APR 2 Flat Rate approximation relies on an assumption that loan repayments start to be made straight away. So here, where the repayments on the loan are deferred for 6 months (the first repayment is in July 2008), this approximation is not very good.



Solution 2.10

This question is taken from Subject 102, April 2003, Question 3. The following timeline illustrates the cashflows involved in the project:

Time (years)0 8/12 2 6 10 14 18 22

-60,000 -25,0009,000 13,000 17,0005,000 21,000

Cashflows

50,000

p ap ap a p ap a

The value at time 0 of the investment required is:

8

129%60,000 25,000 60,000 25,000 0.94417 83,604.20v [2]

The total income received between time 2 and time 6 is 20,000; that received between time 6 and time 10 is 36,000 etc. So, working in a time period of four years, the income can be divided into two parts:

● a level payment of 4,000 per period for 5 periods,

● a payment of 16,000 in the first period, 32,000 in the second period, increasing by 16,000 in each period with a final payment of 80,000 in the 5th period.

The first part can be valued using a level annuity, and the second part using an increasing annuity. Each annuity will have a term of 5. Using our time period equal to 4 years, the value at time 2 of the income from the project is:

55 5

4,000 16,000 50,000 kk ka I a v where 41.09 1 0.41158 k

So, the value of the income at time 2 is: 4,000 2.3834 16,000 5.5860 50,000 0.17843 107,830 [3]

The value of the income at time 0 is:

29%107,830 107,830 0.84168 90,759v [1]

So the net present value of the project is 83,604 90,759 £7,155 . [1]



Alternatively, the whole solution could be approached by working in years, using an annual effective interest rate of 9%:

812 2 6 10

4 4 4

14 18 224 4

60,000 25,000 5,000 9,000 13,000

17,000 21,000 50,000

NPV v v a v a v a

v a v a v

This simplifies to:

8

12 2 6 10 14 184

22

60,000 25,000 1,000 5 9 13 17 21

50,000

NPV v a v v v v v

v

Evaluating this gives the same answer as before. Solution 2.11

This question is taken from Subject A1, September 1998, Question 9. First calculate the amount of each monthly repayment, M, from an appropriate equation of value:

(12)|5 @15.4%

4,000 12 M a [1]

5 5

(12) 1/12

1 1 1.15412 12

12(1.154 1)

v

M Mi

42.5877 M

£93.92 M [1]

Flat rate Total interest paid 5 12 93.92 4,000

8.2%Term Loan 5 4,000

pa

[2]



Solution 2.12

This question is taken from Subject 102, September 2003, Question 13 (i) and (ii). (i)(a) The discounted payback period is the smallest time t for which the present (or

accumulated) value of the returns up to time t exceeds the present (or accumulated) value of the costs up to time t . [2]

(i)(b) The payback period is the same as the discounted payback period, except that

the present value calculation (or accumulation) is carried out using an interest rate of 0. In other words, it is the earliest time for which the monetary value of the returns exceeds the monetary value of the costs. [1]

(ii) Unlike the NPV, neither the DPP nor the PP give any indication of how

profitable a project is, as they ignore cashflows after the accumulated value of zero is reached. [1]

There may not be one unique time when the balance in the investor’s account changes from negative to positive. However, the NPV can always be calculated. [1] The PP can give misleading results, as it does not take into account the time value of money. [1]

Solution 2.13

This question is taken from Subject 102, September 2003, Question 2. The time-weighted rate of return is the solution, i , of the equation:

( )380 200 2001

120 110 210i¥ ¥ = + [2]

Solving this gives:

1/31.1544 1 4.90%i = - = [1]

CT1: Q&A Bank Part 3 – Questions Page 7


Question 3.19

Outline the differences between government bonds, unsecured loan stock and Eurobonds. [6] Question 3.20

(i) State what is meant by a “forward contract”. Your answer should include reference to the terms “short forward position” and “long forward position”. [3]

(ii) Six months ago, an investor entered into an 18-month forward contract to

purchase a share that pays dividends continuously with a constant dividend yield of 4% pa. The forward price was calculated assuming a risk-free force of interest of 6% pa over the 18 months. The value of the share is now 95% of its original value.

Calculate the minimum value of the risk-free force of interest over the remaining

year such that the original forward contract still has a positive value to the investor. [6]

[Total 9] Question 3.21

An investor purchases £100 nominal of a fixed interest stock, which pays coupons of 7% pa half-yearly in arrears. The stock is redeemable at par and can be redeemed at the option of the borrower at any time between 5 and 10 years from the date of issue. The investor is subject to tax at the rate of 40% on income and 25% on capital gains. (i) Calculate the price that the investor should pay in order to obtain a net yield of at

least 6% pa. [5] (ii) Given that this was the price paid by the investor, calculate his net annual

running yield. [1] [Total 6]

Page 8 CT1: Q&A Bank Part 3 – Questions


Question 3.22

A woman purchased a government bond on 1 January 2000. The bond pays coupons of 6% pa six monthly in arrears on 30 June and 31 December. The bond is due to be redeemed at 105% at the end of the year 2010. The woman expects to achieve a net yield of 5% pa effective interest on her investment. She pays income tax at the rate of 23% on 1 April for any coupon payments received in the previous year (1 April to 31 March). She also pays capital gains tax on that date at the rate of 40% on any capital gains she realised in the previous year. Calculate the price she paid for the bond. [7] Question 3.23

A man bought a 5-year forward contract on 1 May 2006 to buy £400 nominal of a stock that pays coupons of 4% pa payable quarterly on 31 March, 30 June, 30 September and 31 December. The stock is also expected to pay out a lump sum of £50% on 1 August 2010. The stock is expected to yield 4.5% pa effective if purchased on 1 May 2006 and held forever. (i) Calculate the forward price for the contract, given that the risk free rate of

interest is 5%. [5] (ii) Determine the value of the forward contract on 1 September 2008, when the

stock price is £140%. [4] [Total 9]



Solution 3.20

(i) A forward contract is an agreement made at some time 0t = , say, between two parties under which one agrees to buy from the other a specified amount of an asset (denoted S) at a specified price on a specified future date. [1]

The investor agreeing to sell the asset is said to hold a short forward position in

the asset, and the buyer is said to hold a long forward position. [2] (ii) Let 0S denote the share price at time 0 (when the forward contract is entered)

and let 0K denote the forward price agreed at time 0, to be paid at time 18

months (ie 1.5 years) based on a risk-free force of interest of 6% pa. Then:

(0.06 0.04) 1.5 0.030 0 0K S e S e [1½]

Now, letting 0.5K denote the forward price agreed at time 6 months, to be paid

at time 18 months for the same share, and using for the unknown risk-free force of interest over the year from time 6 months to time 18 months, we have:

( 0.04) 1 0.040.5 0.5 00.95K S e S e [1½]

The value of the forward contract at time 6 months is:

0.5 0

0.04 0.030 0

0.04 0.030 0

( )

0.95

0.95

V K K e

S e S e e

S e S e

[1]

We require the value of for which this is positive. Hence:

0.04 0.030 0

0.04 0.030 0

0.95 0

0.95

S e S e

S e S e

[1]

Cancelling the factor of 0S and rearranging, we find that:

0.07 0.07

ln 0.12130.95 0.95

e ee

So the minimum value for the new risk-free force of interest is 12.13% pa. [1]



Solution 3.21

(i) We first check whether there is a capital gain on redemption. To do this, we compare:

71 100(1 ) 0.6 0.042D

Rt- = ¥ =

with:

(2) @6% 0.059126i =

As (2)1(1 ) D

Ri t> - , there is a capital gain. [1]

We need to look at the worst case scenario to ensure that the investor obtains a minimum yield of 6%. The worst case under a capital gain is to have the latest possible redemption date (ie the stock will be redeemed after 10 years). [1] Let P denote the price that the investor pays in order to achieve a net yield of at least 6% pa. The equation of value (allowing for tax) is:

(2) 10 1010

0.6 7 100 0.25(100 )P a v P v= ¥ + - - [1]

10 10

0.25 751 4.2 7.46888

1.06 1.06P

[1]

£85.13Pfi = [1]

(ii) The net running yield is defined to be the net coupon per £100 nominal divided

by the price per £100 nominal. So, the investor’s net annual running yield is:

7

0.6 0.0493 4.93%85.13

¥ = = [1]

CT1: Q&A Bank Part 5 – Revision Solutions Page 3


(ii) The APR on the transaction is the solution of the equation of value, expressed as an annual rate and rounded to the nearest 0.1%. So, working in years:

(12)|5

10,000 3,000 a=

ie (12)|5

3.3333a = . The APR is roughly twice the flat rate, so we can use trial

and error:

(12)|5

0.19 3.3155i a= fi =

(12)|5

0.187 3.3338i a= fi =

(12)|5

0.188 3.3277i a= fi =

Since the 18.7% value is closer to 3.3333 than the 18.8% value, the APR is

18.7%. [2] Solution 5.6

Since the bond is ex-dividend, the purchaser does not receive the coupon due in 8 days’ time. The present value of the future payments on the day that the coupon is due is:

(2)77

100 8v a [1]

To find the price paid ( P ), we need to discount this back by 8 days:

8

(2)7 3657

100 8 @ 6%P v a v [1]

So we have:

8

365(66.5057 45.3192) 1.06 111.68P

[2]

Page 4 CT1: Q&A Bank Part 5 – Revision Solutions


Solution 5.7

The accumulated value of the payment stream is:

( )10 10

0

(10 2 )exp 0.05 0.01t

AV t s ds dtÊ ˆ

= + +Á ˜Ë ¯

Ú Ú [2]

This can be evaluated as follows:

( )

10102

0

102

0

(10 2 )exp 0.05 0.005

(10 2 )exp 1 0.05 0.005

tAV t s s dt

t t t dt

È ˘= + +Î ˚

= + - -

Ú

Ú [1]


( ) ( )( )b bf t f t

aa

f t e dt e

we know that:

( ) ( )10 10

2 2

00

( 0.05 0.01 )exp 1 0.05 0.005 exp 1 0.05 0.005t t t dt t tÈ ˘- - - - = - -Î ˚Ú

Since 10 2 200( 0.05 0.01 )t t , we have:

( )

( )

( )

102

0

102

0

102

0

(10 2 )exp 1 0.05 0.005

200 ( 0.05 0.01 )exp 1 0.05 0.005

200 exp 1 0.05 0.005

AV t t t dt

t t t dt

t t

= + - -

= - - - - -

È ˘= - - -Î ˚

Ú

Ú

[2]


0 1200( ) 343.66AV e e= - - = [1]

CT1: Q&A Bank Part 5 – Revision Solutions Page 5


Alternatively, we could evaluate the integral:

( )10

2

0

(10 2 )exp 1 0.05 0.005t t t dt+ - -Ú

using the substitution 21 0.05 0.005u t t , so that:

0.05 0.01 ( 0.05 0.01 ) 200 (10 2 )du

t du t dt du t dtdt

Also, when 0t , 1u , and when 10t , 0u , so we have:

( ) ( )10 0

2 0 1

0 1

(10 2 )exp 1 0.05 0.005 200 200ut t t dt e du e e+ - - = - = - -Ú Ú

as before. Solution 5.8

(i) Increasing annuity

|( )nIa represents the present value of payments of 1 at the end of the first time period, 2

at the end of the second time period, …, n at the end of the nth time period. Thus:

2 3|( ) 2 3 n

nIa v v v nv= + + + + [1]

Hence:

2 1|(1 )( ) 1 2 3 n

ni Ia v v nv -+ = + + + + [1]

By subtraction, we obtain:

2 3 1| |( ) 1 n n n

n ni Ia v v v v nv a nv -= + + + + + - = - [1]

so:

|

|( )

nn

n

a nvIa

i

-= [1]

Page 6 CT1: Q&A Bank Part 5 – Revision Solutions


(ii) Present value

0

20

1

30

2

40

3

100

9

95

10

90

11

25

24

20

25

amount

time

2 8 9 10 24 2520 30 90 100 95 25 20PV v v v v v v v = + + + + + + + + [1]

8 8 2 17 2 17

8 8 2 17 2 17

88 8 17 17

20 90 (105 105 105 5 10 85 )

20 90 (105 105 105 5 2 5 17 5 )

10 10( ) (105 5( ) )

61.48815 257.20376 0.6156991(1,080.59912 389.54155)

= + + + + + + - - - -

= + + + + + + - - ¥ - - ¥

= + + -

= + + -

v v v v v v v v v

v v v v v v v v v

a Ia v a Ia

[2] £744.18= [1]

CT1: Assignment X1 Questions Page 3


Question X1.10

A man receives payments half-yearly in arrears for 20 years. The first payment is £250, and each payment is higher than the previous one. The payments increase by 5% pa effective for 7 years (ie the last increase of this amount is at time 7.5 years). The remaining payments increase by 4.5% pa effective. The interest rate is 6% pa convertible quarterly for the first 10 years and 5.5% pa effective for the last 10 years. Calculate the present value of the payments. [9] Question X1.11

(i) An annuity payable annually in arrears has a first payment of £300, with subsequent payments decreasing by £10 each year to £110 in the final year.

Find an expression for the present value of this annuity. Hence, calculate the

present value of the annuity payable at an effective rate of interest of 6% pa. [3] (ii) A 15-year annuity-due provides annual payments starting at £50 in year 1, £70

in year 2, £90 in year 3, and so on, until the payments have increased to £150. Payments then continue at £150 pa until the 15th payment has been made.

Calculate the present value of this annuity at an effective rate of interest of

5.2% pa. [4] (iii) An annuity certain provides payments annually in arrear for 8 years. The first

payment is £500, with subsequent payments increasing by 5% pa compound. Calculate the present value of this annuity at an effective rate of interest of

8% pa. [3] [Total 10]

Page 4 CT1: Assignment X1 Questions


Question X1.12

(i) Prove from first principles that:

|

|( )n

nn

a nvIa

d-

= [5]

(ii) An annuity certain payable continuously for a term of 10 years provides

payments at the rate of:

1 pa for the first 3 months 2 pa for the next 3 months 3 pa for the next 3 months, and so on.

Calculate the accumulated amount of the annuity payments at the end of the term

if interest is 10% pa convertible half yearly. [6] [Total 11] Question X1.13

The force of interest ( )td is a function of time, and at any time t , measured in years, is

given by the formula:

0.04 0.005 0 6

( ) 0.16 0.015 6 8

0.04 8

+ £ <ÏÔ= - £ <ÌÔ £Ó

t t

t t t

t

d

(i) Derive expressions in terms of t for the accumulated amount at time t of an

investment of 1 at time 0. [6] (ii) (a) Calculate the value at time 0 of £100 due at time 9. (b) Calculate the annual effective rate of discount implied by the transaction

in (a). [4]

(iii) A continuous payment stream is received at a rate of 0.0145 te units per annum between time 10 and time 15. Calculate the present value at time 4 of this payment stream. [5]

[Total 15]

END OF ASSIGNMENT

CT1: Assignment X1 Solutions Page 19


For 8 t£ , the integral of the force of interest is:

8

0.04 0.04( 8)t

ds t= -Ú [1]

So:

0.44 0.04( 8) 0.12 0.04(0, ) t tA t e e e- += = (8 t£ ) [1]

In summary:

2

2

0.04 0.0025

0.36 0.16 0.0075

0.12 0.04

0 6

(0, ) 6 8

8

t t

t t

t

e t

A t e t

e t

+

- + -

+

Ï £ <ÔÔ= £ <ÌÔ £ÔÓ

(ii)(a) Present value The present value here is:

0.480.12 0.04 9 0.48

100 100 100100 £61.88

(0,9)e

A e e

[2]

Alternatively, we could calculate this from first principles:

6 8 9

0 6 8

6 82 2 90 6 8

0.04 0.005 0.16 0.015 0.04

0.04 0.0025 0.16 0.0075 0.04

0.33 (0.8 0.69) (0.36 0.32)

0.48

100

100

100

100 £61.88

t dt t dt dt

t t t t t

PV e e e

e e e

e e e

e

(ii)(b) Equivalent annual effective rate of discount Letting d denote the equivalent annual effective rate of discount:

0.48 9 0.48 9100 100(1 ) (1 )e d e d [1]

Page 20 CT1: Assignment X1 Solutions


Therefore:

1

90.481 0.051936 5.1936%d e ie [1]

Alternatively, we could calculate the equivalent annual effective interest rate first:

1

90.48 9 0.48100 100 1 0.054781e v i e

We can then use this to calculate the annual effective rate of discount:

0.054781

0.051936 5.1936%1 1.054781

id ie

i

(iii) Present value of payment stream at time 4 The present value at time 10 is:

10

15 0.040.01

1010

45

t

dst

tPV e e dt-

=Ú

= Ú [1]

Carrying out the integration:

[ ]

( )

10

150.040.01

1010

150.01 0.04 0.4

10

150.4 0.03

10

150.03 0.40.4 0.03 15 0.03 10

10

45

45

45

4545 230.912

0.03 0.03

tstt

t t

t

t

PV e e dt

e e dt

e e dt

e ee e e

-=

- +

-

-- ¥ - ¥

=

=

=

È ˘= = - =Í ˙- -Í ˙Î ˚

Ú

Ú

Ú

[2]

We now need to discount this back to time 4.

CT1: Assignment X1 Solutions Page 21


The present value at time 4 is:

4230.912

(4,10)tPVA

where:

2

0.12 0.04 10 0.520.32

0.20.04 4 0.0025 4

(0,10)(4,10)

(0,4)

A e eA e

A ee

[1]

So the present value of this payment stream at time 4 is:

0.324

230.912230.912 167.677

(4,10)tPV eA

[1]

Alternatively, the discount factor from time 10 to time 4 could be calculated from first principles:

8 106

6 84

86 22 106 84

0.16 0.015 0.040.04 0.005

0.16 0.00750.04 0.0025 0.04

(0.33 0.2) (0.8 0.69) (0.4 0.32)

0.32

t dt dtt dt

t tt t t

e e e

e e e

e e e

e

Alternatively, the answer could be calculated by first calculating the present value at time 0, using the integral:

15 15

0.01 0.01 0.12 0.040

10 10

145 45 137.282

(0, )t t t

tPV e dt e e dtA t

- -= = = =Ú Ú

and then accumulating this forward to time 4:

20.04 4 0.0025 4 0.2

4 0 (0,4) 137.282 137.282 167.677t tPV PV A e e

where the expressions for (0, )A t are taken from part (i).










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