Chapter 01 - Engineering Economics 04-10-2010

8/7/2019 Chapter 01 - Engineering Economics 04-10-2010

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Interest Formula Derivations

I Makalanda

MSc(Fin. Maths), BSc (Eng.), DPM(UK), ACIM

Chapter 01


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Simple & Compound Interest

Simple interest

The interest owed upon repayment of a loan isproportional to the length of time the principal

sum has been borrowed. Let I represent the interest earned, P the

principal amount, n the interest period, and Ithe interest rate.

PniI !


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Example

Suppose that LKR 1,000 is borrowed at a

simple interest rate of 18% per annum. At the

end of one (01) year, the interest owed would

be,

180

18.011000

!

vv!

!

I

I

niI


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Describing cash flows over time

To aid in identifying & recording the economic

effects of investment alternatives, a graphical

description of each alternatives cash

transactions may be used.

This graphical descriptor, referred to as a cash

flow diagram, will provide the information

necessary for analyzing an investment

proposal.


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Describing cash flows over time

Borrower Lender

1000

0

1 2 3 4

160 160 160

1160

0

1 2 3 4

1000

160 160 160

1160

Net cash flow is the arithmetic sum of the receipts (+) and the

disbursements (-) that occur at the same point in time.


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Interest formulas (discrete compounding,

discrete payments)

i the annual interest rate

n the number of annual interest periods

P a present principal sum A a single payment, in a series of n equal

payments, made at the end of each annual

interest period. F a future sum, n annual interest period

hence.


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Interest formulas (discrete compounding,

discrete payments)

The end ofone (01) year is the beginning of

the next year.

Pis at the beginning of a year at a timeregarded as being the present.

Fis at the end of the nth year from a time

regarded as being present.

An Aoccurs at the end of each of the period

under consideration.


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Single - payment compound amount

factors

If an amount P is invested now and earns at

the rate i per year, how much principal and

interest are accumulated after n years?

1 2 3 .. n-1 n

P

F

0


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factors

Year Amount

beginning of year

Interest earned

during year

Compounded amount

at the end of year

1 P Pi P(1+i)1

2 P(1+i) P(1+i)i P(1+i)2

3 P(1+i)2 P(1+i)2i P(1+i)3

n P(1+i)n-1 P(1+i)n-1i P(1+i)n = F

(1+i) is known as the single-paymentcompounded amount

factor and is designated

),,/( niPF

),,/()( niPFPiiPF n !!


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factors

The equations for solving problems may be setup prior to looking up values of the factorsfrom the tables and inserting them in theparenthesis.

The source and the identity of values takenfrom the tables are maintained during thesolution.

),,/( niPF


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Single-payment present worth factor

1 2 3 .. n-1 n

P ?

F

0

!n

i

FP

1

1

),,/( niFP


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Equal-payment-series compound amount

factor

0 1 2 3 .. n-1 n

F??

A A A A A A

!

i

iAF

n1)1(

),,/( niAF

The resulting factor, [(1+i)n-1]/i, is known as the equal-payment-series

compound-amount factor & is designated


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Equal-payment-series sinking-fund

factor

0 1 2 3 .. n-1 n

F

A? A? A? A? A? A?

!

1)1(n

i

iFA

),,/( niFA

The resulting factor, [(1+i)n-1]/i, is known as the equal-payment-series

sinking-fund factor & is designated


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Equal-payment-series capital recovery

factor

1 2 3 .. n-1 n

P

A?

0

A?A?A?A?A?

!

1)1(

)1(n

n

i

iiPA

),,/( niPA

The resulting factor, i(1+i)n/[(1+i)n-1], is known as the equal-payment-

series sinking-fund factor & is designated


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Equal-payment-series present worth factor

1 2 3 .. n-1 n

P?

A

0

AAAAA

!

n

n

ii

iiAP

)1(

1)1(

),,/( ni

The resulting factor, [(1+i)n-1]/i(1+i)n, is known as the equal-payment-series sinking-fund factor & is designated


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Example 01

What interest rate is required to triple 1000/

in 10 years?


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Answer for Example 01

%6.11

11612.1)10986.0exp(%)1(

%)1ln(10986.0

%)1ln(100986.1

%)1ln(10)3ln(

%)1.(10003000

%)1(

10

!

!!

!

!

!

!

!

i

i

i

i

i

i

iPFn


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Or

11.6%i

ion,interpolatlinearBy

3220.0)10%,12,/(

rate,interest12%atSimilarly

3855.0)10%,10,/(

10%atthatfindcanwetablesUsing

10%iAssume,

3

1)10,,/(

),,/(

!

!

!

!

!

!

FP

FP

iFP

NiFPPF


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What annual year-end payment must be

made each year to have 20000 available 5

years from now? The compound annualinterest rate is 6%.


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A A A A A

F=20000

1 2 3 4 5

DR 6%

Year 0 1 2 3 4

DR 1 1.06 1.1236 1.1910161.262477

Payments A A A A A

FV 20000

3548

)63709.5(20000

262477.1191016.11236.106.1

!

!

vvvv!

A

A

AAAAAFV

.3548

)1774.0(20000

)5%,6,/(

!

!

!

A

A

FAFA


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Example

A machine needed for 3 years can be

purchased for 77662 and sold at the end of

the period for about 25,000. A comparable

machine can be leased for 30000 per year. If

the firm expect a return of 20% on their

investment, should it buy or lease the

machine?


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A A A

S S S

77662

25000


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For a New Product Development Project you are required capital of

LKR 5,000,000.00. Prepare the loan amortization plan showing the

yearly loan commitment using the following criterion.

1 Repayment period - 8 years

2 Installments per year 2

3 Interest charge 9% per 6 months

Clearly state assumptions you made as foot notes. ( 10 marks)


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Chapter 01 - Engineering Economics 04-10-2010

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