dewhurst ch12 week6-7slides

7/30/2019 Dewhurst Ch12 Week6-7Slides

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Chapter 12

Models for finance and accounting


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g

Simple Interest and growth

In practice we rarely encounter SIMPLE

interest as it only occurs when dealing

with a few institutions, such as the

government, when buying government

bonds. However understanding of

Simple Interest will help in

understanding Compound Interest.Simple interest occurs when the

investment reward cannot be left to

accumulate with the principle.


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Example

The alpha issue of 5 year Government

Bonds can be purchased in units of

1000 and offers 3% per annum simple

interest at the end of each year. If Mrs

Smith is considering purchasing 8 units

what is the interest that she receives

each year and in total?


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Simple interest formula

t is the number of periods

ris the rate of growth per period

F0 is the value of the initial investment

(i.e. principle) (at time zero) and is often

called the Present Value (PV)

Ft is the amount of money in the future

(after n periods) and is often called the

Future Value (FV) where:

Ft = F0+trF0 = F0(1+tr)


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Compound Interest and

growth Most financial institutions use

COMPOUND interest, in which the

interest received can be left toaccumulate with the principle.


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Example

Instead of purchasing government bonds,Mrs Smith is considering investing the 8000

in a high street bank savings account, which

offers 3% p.a. interest at the end of each

year. Mrs Smith does not intend withdrawing

any money from the account until the end of

the 5th year so that she can make a direct

comparison with the Alpha Government BondIssue. How much interest does Mrs Smith

receive on this account?


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Solution

After the first year, 8000 + 0.03* 8000 =8240 would be in the account.

At the start of the second year the accountholds 8240 and with a further 3% interest at

the end of the second year 8240 + 0.03*8240= 8487 would be in the account.

At the start of the third year the account holds

8487.2 and with a further 3% interest at the

end of the second year 8487 + 0.03* 8487 =8742 would be in the account.

etc


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Time scale of money

After 5 years Mrs Smith will receive 9274.19 of

which 8000 was the initial principle.

Mrs Smith will receive 1274.19 interest.


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BASIC COMPOUND GROWTH

FORMULA

F0 is the initial amount (principle) or

present value

t is the number of periods

r is the growth rate per period

Ft is the future value where:

Ft = F0 (1+r)t


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Using the formula

r = 0.03t = 5

F0 = 8000

F5 = 8000*(1+0.03)5 = 9274.19

NB 8000 was the principle amount and

1274.19 is the accumulated

interest, which itself has also

gained interest over 5 years


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Multiple compoundings

If we have c compoundings per year at

an annual rate, r, then the future value

after Y years will be:

FY = F0(1 + i )cY

c


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Effective and Annual Percentage

Rates

The annual effective interest rate (AER) is

the annual rate of growth if we only

compounded once at the end of each year:

AER = (1 + i )c - 1

c


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Example

Consider investing just 1in an investment offeringa Nominal Annual Rate of interest of 24% p.a. with

different compounding periods throughout the year

(e.g. annually, monthly, weekly and daily).

In this case we have:

i = 0.24, F0 = 1 and Y =1

and we considerc =1, 12, 52 and 365.

This example will also show how the AER

changes according to the number of multiple

compoundings


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If annual compounding is undertaken at the

end of the year (i.e. c =1) then:

F1 = 1(1 + 0.24 )1 = 1.24In this case the AER is the same as the

nominal annual rate(i.e. 24% p.a.).

If monthly compounding is undertaken at theend of each month (i.e. c = 12) then:

F1 = 1(1 + 0.24 )1*12 = 1.268

12A nominal annual rate of 24% is equivalent to

a nominal monthly rate of 2% (i.e. 24/12 =2)

and the AER is 26.8%


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If weekly compounding is undertaken at the endof each week (i.e. c = 52) then:

F1 = 1(1 + 0.24 )1*52

= 1.27

52The nominal annual rate of 24% is equivalent to anominal weekly rate of 0.46%, (i.e. 24/52 = 0.46) and the

AER = 27%

If daily compounding is undertaken at the end

of each day (i.e. c = 365) then:

F1 = 1(1 + 0.24 )1*365 = 1.271

365

The nominal annual rate of 24% is equivalent to a

nominal daily rate of 0.065%, the AER = 27.1%


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Continuous compounding

When we have long periods of investment anda large number of compounding periods per

year (i.e. as c ) then we find another

formula being used, i.e.

FY = F0eiY where e = 2.718

This is the CONTINUOUS COMPOUNDING

FORMULA. This only applies for large valuesof c and Y and will only provide approximate

results in most practical circumstances as it

implies that interest is being calculated and

added continuously.


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Example

Consider investing 10 at

%

per day (i.e. 12%nominal annual rate) for 5 years.

If we apply the discrete multiple compounding

formula we get:F5 = 10(1 + 0.12 )5*365 = 18.2194

365

If we apply the continuous compoundingformula we get:

F5 = 10e0.12*5 = 18.2212


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Discounting

Discounting occurs when we want tomake a one-off investment for a

particular future purpose and may need

to know how much to invest now (i.e.the present value) to realise a particular

future value. Discounting also occurs or

when a business wants to know thepresent value of its investments or

assets for accounting purposes.


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Example

A grandparent wants to invest a lumpsum on the birth of a grandchild to givethe child 5000 towards their educationat the age of 18. If an investment trust

offers a guaranteed rate of growth of 4% p.a. how much should be investednow?

In effect we know the future value F18 = 5000,the rate of growth, r = 0.04 and the

investment horizon, t = 18 years but we do

not know the present value F0.


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Solution

If we take our basic compound growthformula Ft = F0(1+r)t then we can simply

rearrange it to find F0 in terms ofFt , rand t,

i.e.F0 = Ft

(1+r)t

We can now answer the question byapplying the above formula giving:

F0 = 5000 = 2468.14

(1+0.04)18


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Discounting in business

When calculating the present value of a futurevalue using the above formula we usually

refer to DISCOUNTING future values to their

present day equivalents and call rthe

discounting rate. When compiling end of year accounts many

companies include the current or present

value of investments or assets that are held.

Often the monetary equivalent of these

investments or assets cannot be realised (i.e.

they cannot be cashed in) but they need to be

included in the company accounts.


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Calculating t and r

We can also rearrange the compound

growth formula to find t and r, i.e.

t = log (Ft/ F

0)

log (1 + r)

r = (Ft / F0 )1/t

1


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Examples

How long does it take 50 to double if theannual growth rate operating is 6% p.a?

Ft =100, F0 = 50 , r = 0.06 so t = 12 years

What is the annual growth rate if an

investment has increased by 50% over 5

years?

Ft =1.5, F0 = 1 , t = 5 so r = 24.5%


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When calculating their end of year accounts,businesses include the current book value of their

assets. These assets (e.g. equipment) may have been

purchased several years ago and due to wear and

tear would not realise their original purchase price if

sold. Accountants use several mechanisms for

calculating the reduction in the book value of assets

(e.g. straight line depreciation) but one popular

method, called Reducing balance depreciation,

operates in a similar fashion to compound growth

except that present values (i.e. original price or value)

are reduced (i.e. marked down), rather than increased

(i.e. marked up), by a particular percentage each

period.

Reducing balance depreciation


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In effect all we need to do is to replace the+ sign by a - sign in the compound growth

formula to reflect reducing rather than

increasing value over time, i.e.

Current book value = Ft = F0(1 - r)t

Where F0 is the original purchase price or value, r

is now called the rate of depreciation and t is the

number of periods over which the asset is

depreciated.

Reducing balance formula


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A computer system was purchased 3

years ago for 1500 and for accounting

and tax purposes, is subject to 30%

depreciation each year. What is thecurrent book value of the computer

system?

Applying the equation gives the current

book value = 1500(1 0.3)3

= 514.5

Example


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Savings, Endowments and

Sinking fundsSo far we have considered problems

involving a single amount of money (e.g.

the future value resulting from a singleinvestment or the present book value of a

single asset). Frequently we come across

problems involving investing a stream ofmoney or cash (i.e. cash flow) to save up

for a future purchase


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Example

A manufacturing company plans toreplace its current production line in 5years time and expects that it can setaside 5,000 from its end of year profits

to create a Sinking Fund to put towardsthe purchase of the new equipment.The company can obtain a guaranteed

8% p.a. on such a fund from its currentbank. What is the total amount that thecompany will have available in 5 yearstime to put towards its new production

line?


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Solution

End of year 1 2 3 4 5 Futurevalues

Payments 5000 5000 5000 5000 5000 = 5000

Previous 5000(1+0.08) = 5400

paymentswith their 5000(1+0.08)

2= 5832

compound

interest 5000(1+0.08)3

= 6299

added

5000(1+0.08)4

= 6802

Total = 29333


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In general

End of year 1 2 etc t-2 t-1 t Futurevalues

Payments A A . . . A A A = A

Previous A(1 + r) = A(1 + r)

payments

with their A(1 + r)2

= A(1 + r)2

compoundinterest A(1 + r)

3= A(1 + r)

3

added

.

etc . etc

.

A(1 + r)t-2 = A(1 + r)t-2

A(1 + r)t-1 = A(1 + r)t-1


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A formula for the future value

FV = A + A(1 + r)1 + A(1 + r)2 + A(1 + r)3 + A(1 + r)t-

1

This type of mathematical expression is called a

series


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Geometric seriesIn general a series of the form:

F + FC1 + FC2 + FC3+ ..

is known as a GEOMETRIC SERIES.

and F is the FIRST term and C is the COMMONRATIO

The SUM to n terms is:Sn= F+FC1

+FC2

+FC3

++FCn-1

Sn = F(1 - Cn)

(1 - C)


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The future value

In this case the geometric series for future value has

First term F = A

Common ratio C = (1 + r)

Therefore: FV = A[1- (1+r)t ] = -A[1- (1+r)t ]

(1 - 1 - r) r


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Example

A = 5000, r = 0.08 and t = 5

Substituting this data into our formula

gives

FV = -5000*[1- (1+0.08)5]/0.08

FV = 29333


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In the previous case we considered makingpayments at the end of each year. We can

also derive a formula for schemes where

payments are made at the start of the year.

Payments at start of year


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Cash flowEnd of year 0 1 2 3 4 5 Future

values

Payments 5000 5000 5000 5000 5000 0 = 0

Previous 5000(1+0.08) = 5400

payments

with their 5000(1+0.08)2

= 5832

compoundinterest 5000(1+0.08)

3= 6299

added

5000(1+0.08)4

= 6802

5000(1+0.08)5

= 7347

Total = 31680

In general


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In generalEnd of year 0 1 2 etc t-1 t Future

values

Payments A A A . . . . A 0 = 0

Previous A(1 + r) = A(1 + r)

payments .

with their etc . etc

compound .

interest A(1 + r)t-2

= A(1 + r)t-2

added

A(1 + r)t-1 = A(1 + r)t-1

A(1 + r)t

= A(1 + r)t

FV = A(1 + r)1 + A(1 + r)2 + A(1 + r)3 + + A(1 + r)tUsing the sum of a geometric series with:first term F = A(1+r) and common ratio C = (1+r)

FV = A[1- (1+r)t ] = -A[1- (1+r)t ]

(1 - 1 - r) r


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Example

A = 5000, r = 0.08 and t = 5 but nowwith payments at the start of each

period.

FV = 5000*(1+0.08)*[1 - (1+0.08)5 ]/0.08

FV = 31679.64


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Loans and mortgages

From a mathematical point of view amortgage is essentially just a long-termloan, usually taken out to purchase a

house or business. Several types ofloan or mortgage are sold by mostfinancial institutions but essentially thereare only three basic types: interest

only, repayment, and endowment. Allother types of loan or mortgage arespecial variants of these.


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Interest only loans and

mortgages

An interest only loan or mortgage is

exactly that, i.e. an amount is borrowedfor a specified period of time and the

borrower agrees to pay the interest on a

regular basis and at the end of theagreed period repays the amount

borrowed.


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Example

Ms Jones, a speculative builder, has bought aplot of land on which to build a house to sell.

She has insufficient cash to pay for the

building work and wants an interest only loan

of 10000 over 3 years. Her bank canprovide such a loan at 12% p.a. with interest

to be paid at the end of each month. The

bank also charge an arrangement fee of 5%

which is paid on completion. What is theAER, what are Ms Joness monthly payments

to the bank and what does she pay in total?


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AnswerAlthough the bank have stated the interest rate onan annual basis (i.e. 12% p.a. nominal) it is actuallybeing calculated and charged on a monthly basis(i.e. 1% per month nominal) with an AER of 12.68%p.a.

Her annual interest is: 10000*0.1268 = 1268

Her monthly interest is: 1268/12 = 105.67

In total she pays:

Interest over 3 years = 36* 105.67= 3804The original loan = 10000

The arrangement fee = 0.05*10000 = 500

Total = 3804 + 1000 + 500 = 14304


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Repayment loans and mortgages

The most common form of loan or mortgage inwhich the loan and interest are paid off by equal

regular installments. In a repayment loan

compound interest is calculated by the lender

and added to the loan at the end of eachcompounding period before any installment by

the borrower is deducted. At the end of each

compounding period it is possible to calculate the

outstanding debt owed by the borrower and atthe end of the agreed loan period the entire loan

and interest will have been paid off.

E l


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Example

To answer this question we need to calculate the

outstanding debt at the end of each year andinclude a variable to represent the, as yetunknown, instalments that John will be making.

Let us call the equal annual instalment A

John wants to purchase a car costing 5000. The

car dealer offers him a 3 year repayment loanwith an interest rate of 15% p.a. compounded

annually to be paid by equal end of year

installments. What are the three installments that

John would pay?


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Year 1 2 3

Initial debt 5000 5750 - A 6612.5 - 2.15A+ + + +

Interest 0.15*500 =750 0.15*(5750A) = 862.5 - 0.15A 0.15*(6612.5 - 2.15A) =991.88-0.32A-A

- - - -payment A A A= = = =

Outstanding 5000 + 750 - A 5750 - A + 862.5 - 0.15A - A 6612.5 - 2.15A + 991.88 - 0.32A - A

Debt. at end = = =of year 5750 - A 6612.5 - 2.15A 7604.375 - 3.4725A

The time flow of money for Johns loan


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Johns instalments

At the end of these three years John will havepaid off ALL the outstanding debt andtherefore we must have:

7604.375 3.4725A = 0

Hence: A = 7604.375/3.4725 = 2189.885

Therefore John will pay 2189.89 each yearfor three years.


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General formula

L is the amount borrowedr is the compound rate of interest charged on

the outstanding debt at any time

t is the number of periods over which the loanis paid off

A are the equal regular instalments


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Debt at end of year =

L(1 + r)n-A(1 + r)n-1- . . . -A(1 + r)3-A(1 + r)2-A(1 + r)1-A

Period 1 2 3

Initial debt L L(1 + r)A L(1+r)2-A(1+r)-A+ + + +

interest rL r[L(1r) A] r[L(1+r)2-A(1+r)-A]- - - -

payments A A A= = = =

Outstanding L + rLA L(1 + r)A + r[L(1 + r)A]A L(1+r)2-A(1+r)-A+r[L(1+r)

2-A(1+r)-A]-A

debt. at end = = =of period L(1 + r)A L(1 + r)2A(1 + r)A L(1 + r)3 - A(1 + r)2A(1 + r)A

General repayment loan


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As before, after t years the outstanding debt is zero so:

L(1 + r)t-A(1 + r)t-1-. . -A(1 + r)3-A(1 + r)2-A(1 + r)1-A = 0

L(1 + r)t

= A(1 + r)t-1

+ . . . . +A(1 + r)3

+A(1 + r)2

+A(1 + r)1

+A

Taking out the common factor A on the right hand side gives:

L(1 + r)t = A [(1 + r)t-1 + . . . . +(1 + r)3 + (1 + r)2 + (1 + r)1 + 1]

hence: A = L(1 + r)t

[(1 + r)t-1 + . . . . +(1 + r)3 + (1 + r)2 + (1 + r)1 + 1]

and: L = A [(1 + r)t-1

+ . . . . +(1 + r)3

+ (1 + r)2

+ (1 + r)1

+ 1](1 + r)t


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Johns repayment loan

L = 5000

t = 3 years

r = 0.15

A = 5000*(1 + 0.15)3 =2190

(approx)

[(1 + 0.15)2 + (1 + 0.15)1 + 1]


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Endowment loans

Endowment loans and mortgages operate ina completely different way to Interest only andRepayment loans. These types of loans andmortgages were once quite popular but have

since become less so as they depend upongrowth rates as well as on interest rates.Essentially an Endowment loan, or morefrequently a mortgage, is based on theborrower paying interest on the loan (as withan Interest only loan) and also taking out anEndowment or other savings scheme to payoff the loan at the end of the agreed period.

E l


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ExampleMr and Mrs Smith approach their building society for a

mortgage of 100,000 to put towards the purchase oftheir new home. The building society, offer them twochoices:

(a) A repayment mortgage with an interest rate of

12% p.a. compounded annually.(b) Endowment mortgage with an interest rate of12% p.a. compounded annually on the loan and anEndowment scheme with a guaranteed maturity valueof at least 100,000 with a growth rate of 10% p.a.

In both cases, annual compounding is undertaken bythe building society over 20 years. Which option hasthe lowest cost to Mr and Mrs Smith?


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Solution

(a) For a repayment mortgage they would pay:

A = 100000*(1.12)20 = 13388

72.05

(b) For an Endowment mortgage:

Annual interest = 100000*0.12 = 12000

Instalments on endowment= 100000 = 1746

57.275

Total annual payments = 12000+1746=13746


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Annuities

Annuities are usually used to providepensions. An investor will purchase an

annuity for a particular price and in return

receive an equal regular income for a

specified period of time. From a mathematicalpoint of view an Annuity is the same as a

Repayment mortgage in which the purchase

price of the annuity is the same as the

amount borrowed (L) and the income is the

same as the instalments (A) made to pay off

the loan.


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