161_Lectures_2011-10-04

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GE161 Fall2011Printed: 10/4/201

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1

Time Value of Money - 1

Basic Principles and Formulas

2Examples

Matsushita and Toray (Japan) will invest 180 billion yen ($1.46billion) to build a plasma display panel (PDP) plant, increasingproduction by 6 million panels/yr.

Cargill (USA) announced a $20 million investment to increasecapacity of liquid chocolate and a production line for chocolateflakes, drops and chunks in Mouscron, Belgium.

Good decisions???

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3Time Value

Easy decisions if value of money over time = constant

(Interest rate = 0)

Interest > 0 typically

Interest : Accumulation of value (paid or accrued) as apercentage of the principal at the end of a time period

Where does interest come from??

4Interest

Contractual agreement between parties

Simple interest paid on the principal only

Compound interest accumulation interest, as theinterest is added to principal, forming an exponentialrelationship

Einstein the most powerful force in the universe is

compound interest.

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5Cash Flow diagrams

Cash receipts or disbursements

(Cash in or cash out) At a specific time

From a specific standpoint

Receipt: positive

Disbursement: negative

Time line

Arrows = Cash flowsUp: ReceiptsDown: Disbursements

8

Time

timeperiod

6Cash Flow Diagram Example

Receiving a loan

Making a loan

Savings and withdrawal

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7Cash Flow Terms

P = Present sum of money (relative)

F = Future sum of money (relative)

A = a uniform series of sums of money at the end of a timeperiod

i = interest rate per period

n = number of periods

Convention With the exception of P all cash flows aremade at the end of a time period

8Simple Interest

Interest is paid on the principal only No interest is paid on previously unpaid interest

Example: You borrow $1000 at 10% simple annual interest and payback a lump sum after 5 years. What is your lump-sum payment?

P =1000

F =??

5

( )( )( ) 1 ( )( )

1000 1 (0.5)(5)

1500

F P P i n P i n

F

F

Total interest paid = F P = 1500 1000 = 500

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9MARR

Engineering alternatives are evaluated based on a reasonablerate of return(ROR).

The minimum ROR acceptable ROR is referred to as theminimum acceptable rate of return (MARR)

The MARR is higher than the interest that must be paid to borrowthe money.

ROR MARR CostofCapital

10Equivalence

Sums of money or Cash Flows at different times whichhave equal economic value

Cash flow

Time

Interest rate

Example: At 10% interest annual interest,

$100 today = $110 in one year = $121 in two years

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11Equivalence Formulas

Single-payment compound-amount factor

(1 )F P i 1P

F=?

One Year

(1 )(1 )F P i i 2P

F=?

Two Years

nP

F=?

n Years(1 )(1 )...(1 )

(1 )n

F P i i i

F P i

F=P(F/P,i,n)

(F/P,i,n)= f(i,n)Single-payment compound-amount factor

12

Example

Put $100 in the bank for 8 years at 10% per year,interest compounded annually. What is it worth in 8years?

P=100

F=??

8

Equivalence Formulas

Single-payment compound-amount factor

8

(1 )

100(1 0.1)

100(2.1436) 214.36

nF P i

F

F

F=100(F/P,i,n)F=100(F/P,0.1,8)

F=100(2.1436) = 214.36

Compound Interest FactorPage 174 in Text

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Single-payment present-worth factor

nP=?

Fn Years

(1 )

1

(1 )

n

n

F P i

P Fi

Solve for P

P=F(P/F,i,n)

(P/F,i,n) = f(i,n)Single-payment present-worth factor

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Single-payment present-worth factor

5P=?

F=100

5

1

(1 )

1100

(1 0.1)

100(.6209)

62.09

nP F

i

P

P

P

P= F(P/F,0.1,5)

P= 100(.6209)

P= 62.09

Example: What is the present value of a future payment of $100 fiveyears from today, if the interest rate is 10% compounded annually?



Uniform-series compound-amount factor

n

F=??

1 2

1

(1 ) (1 ) ... (1 )

(1 ) (1 ) (1 ) ... (1 )

(1 ) (1 ) 1

(1 ) 1

n n

n n

n n

n

F A i A i A i A

F i A i A i A i

Fi A i A A i

iF A

i

A is given

0

Treat each cash flow A as a

single cash flow P and

find the future value for each

using (1 )nF P i

Multiply both sides by (1+i)

(1)

(2)

Subtract equation (1) from (2)

F= A(F/A,i,n)

(F/A,i,n) = f(i,n)Uniform-seriescompound-amount factor

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Example: What is the future value of 5 end-of-period payments of$100 each, at the end of the 5 th period at 10% per period?


Uniform-series compound-amount factor

n

F=??

F= A(F/A,0.1,5)F= 100(6.1051)

F= 610.51

A is given

0

5

(1 ) 1

(1 0.1) 1100

0.1

100(6.1051)

610.51

niF A

i

F

F

F



Sinking-fund deposit factor

(1 ) 1

(1 ) 1

n

n

iF A

i

iA F

i

Solve for A

A= F(A/F,i,n)

(A/F,i,n) = f(i,n)Sinking-fund deposit factor

n

F is given

A =??

0

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Sinking-fund deposit factor

8

(1 ) 1

0.11000

(1 0.1) 1

1000(0.08744)

87.44

n

iA F

i

A

A

A

Example: What amount must be deposited annually for 8 years toyield $1000 when withdrawn in 8 years, with 10% interestcompounded annually?

A= F(A/F,0.1,8)

A= 1000(0.08744)

A= 87.44


8

F =1000

A =??

0


Capital-recovery factor

nP is given

A =??

(1 ) 1

(1 )

(1 )

(1 ) 1

n

n

n

n

iA F

i

F P i

i iA P

i

Knowing that

and

Substituting for F..

A= P(A/P,i,n)

(A/P,i,n) = f(i,n)Capital-recovery factor

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Example: What is the required annual revenue amount needed for fiveyears, to equal an initial investment of $1000 if the interest rate is10% per year compounded annually?


Capital-recovery factor

5P = 1000

A =??

5

5

(1 )

(1 ) 1

0.1(1 0.1)1000

(1 0.1) 1

1000(0.26380)

263.80

n

n

i iA P

i

A

A

A

A= P(A/P,0.1,5)

A= 1000(0.26380)

A= 263.80



Uniform-series present-worth factor

nP =??

A is given

P= A(P/A,i,n)

(P/A,i,n) = f(i,n)Uniform-series present-worth factor

(1 )

(1 ) 1

(1 ) 1

(1 )

n

n

n

n

i iA P

i

iP A

i i

Solve for P

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Example: You can afford to make loan payments of $1000/year for 5years. How much money can you borrow now at an interest rate of10% per year?

5

P = ??

A =10005

5

(1 ) 1

(1 )

(1 0.1) 11000

0.1(1 0.1)

1000(3.7908)

3790.80

n

n

iP A

i i

P

P

P

P= P(P/A,0.1,5)

P= 1000(3.7908)

P= 3790.80



Uniform-series present-worth factor

24Sample ProblemsSimple vs. Compound Interest

You borrow $5000 at 6% per yearsimple interest and repay in alump sum after 10 years. What isyour payment?

1 ( )( )

5000 1 (.06)(10)

5000(1.6)

8000.00

F P i n

F

F

F

You borrow $5000 at 6% per yearcompound interest and repay in a lumpsum after 10 years. What is yourpayment?

F= P(F/P,i,n)F= 5000(F/P,0.06,10)

F= 5000(1.7908)

F= 8954.00

5

P = 5000

F =??

10

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To build a new manufacturing facility you must pay the architectural

fee of $70,000 in one year, the building contractor installment of$1,400,000 in 2 years, and the final contractor payment of$2,600,000 in three years. What is the present cost of these futurepayments at an annual interest rate of 7%?

Sample ProblemsMultiple Cash Flows

3

P = ??

70K1,400K

2,600K

P = 70,000(P/F,.07,1) + 1,400,000(P/F,.07,2) + 2,600,000(P/F,.07,3)

P = 70,000(0.9346) + 1,400,000(0.8734) + 2,600,000(.8163)

P = $ 3,410,562

26Sample ProblemsFinding a Factor Value

How long will it take to at least triple your money if the rate or returnis 15% per year compounded annually?

P

F = 3P

n = ?i = 15%

Find n in table

n = 8 years

( / , , )

3 ( / ,0.15, )

( / , 0.15, ) 3

F P F P i n

P P F P n

F P n

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27Shifted Cash Flows

10

P=??

A =100

1 4

i =10%

1. Find equivalent [P] at yr. 3

2. Find equivalent P at yr. 1from [F] at year 3

[P]

[F]

Example: What is the present value of seven annual end of period

payments of $100 which begin in 4 years at 10% interest?

( / , , )

100( / ,0.1,7)

( / ,0.1,3)100( / ,0.1,7)( / ,0.1,3)

100(4.8684)(0.7513)

365.76

P A P A i n

P P A

P F P F

P P A P F

P

P

161_Lectures_2011-10-04

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