(c) 2002 contemporary engineering economics 1 chapter 4 time is money interest: the cost of money...

(c) 2002 Contemporary Engineering Economics

1

Chapter 4Time Is Money

• Interest: The Cost of Money

• Economic Equivalence• Development of

Interest Formulas• Unconventional

Equivalence Calculations


2

Key Concepts of Last Lecture

1. Interest

2. Principal?

3. Interest rate?

4. Interest period?- number of interest periods.

5. Plan for receipts or disbursements

6. Future amount of money

7. Simple vs Compound


3

Time Value of Money

• Money has a time value because it can earn more money over time (earning power).

• Time value of money is measured in terms of interest rate.

• Interest is the cost of money—a cost to the borrower and an earning to the lender


4

Elements of Transactions involve Interest

1. Initial amount of money in transactions involving debt or investments is called the principal.

2. The interest rate measures the cost or price of money and is expressed as a percentage per period of time.

3. A period of time, called the interest period, determines how frequently interest is calculated.

4. A specified length of time marks the duration of the transactions and thereby establishes a certain number of interest periods.

5. A plan for receipts or disbursements that yields a particular cash flow pattern over a specified length of time.

6. A future amount of money results from the cumulative effects of the interest rate over a number of interest periods.


5

• An = A discrete payment or receipt occurring at the end of some interest period.

• i = The interest rate per interest period.

• N = The total number of interest periods.

• P = a sum of money at a time chosen for purposes of analysis as time zero, sometimes referred to as the present value or present worth.

• F = A future sum of money at the end of the analysis period. This sum may be specified as Fn.

• A = An end of period payment or receipt in a uniform series that continues for N periods. This is a special situation where A1 = A2 = …= AN.

• V n = An equivalent sum of money at the end of a specified period n that considers the effect of time value of money. Note that V0 = P and VN = F.


6

Example of an Interest TransactionRepayment Plans (Table 4.1)

End of Year Receipts Payments

Plan 1 Plan 2

Year 0 $20,000.00 $200.00 $200.00

Year 1 5,141.85 0

Year 2 5,141.85 0

Year 3 5,141.85 0

Year 4 5,141.85 0

Year 5 5,141.85 30,772.48

P = $20,000, A = $5,141.85, F = $30,772.48


7

Cash Flow DiagramRepresent time by a horizontal line marked off with the number of interest periods specified. Cash flow diagrams give a convenient summary of all

the important elements of a problem.


8

Methods of Calculating Interest

• Simple interest: the practice of charging an interest rate only to an initial sum (principal amount).

• Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.


9

Simple Interest

• P = Principal amount• i = Interest rate• N = Number of

interest periods• Example:

– P = $1,000

– i = 8%

– N = 3 years

– I=(iP)N

End of Year

Beginning Balance

Interest earned

Ending Balance

0 $1,000

1 $1,000 $80 $1,080

2 $1,080 $80 $1,160

3 $1,160 $80 $1,240


10

Compound Interest (Example 4.1)

• P = Principal amount• i = Interest rate• N = Number of

interest periods• Example:

– P = $1,000

– i = 8%

– N = 3 years

– F=P(1+i)^N

End of Year

Beginning Balance

Interest earned

Ending Balance

0 $1,000

1 $1,000 $80 $1,080

2 $1,080 $86.40 $1,166.40

3 $1,166.40 $93.31 $1,259.71


11

Comparing Simple to Compound Interest


12

Simple Vs Compound Interest

0.05000.0

10000.015000.020000.025000.030000.035000.040000.045000.0

1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46

P=1000i=8%

Simple

Compound


13

Economic Equivalence

• What do we mean by “economic equivalence?”

• Why do we need to establish an economic equivalence?

• How do we establish an economic equivalence?


14

Economic Equivalence (EE)

• Economic equivalence exists between cash flows that have the same economic effect and could therefore be traded for one another.

• EE refers to the fact that a cash flow-whether a single payment or a series of payments-can be converted to an equivalent cash flow at any point in time.

• Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal.


15

Typical Repayment Plans for a Bank Loan of $20,000

Repayments

Plan 1 Plan 2 Plan 3

Year 1 $5,141.85 0 $1,800.00

Year 2 5,141.85 0 1,800.00

Year 3 5,141.85 0 1,800.00

Year 4 5,141.85 0 1,800.00

Year 5 5,141.85 $30,772.48 21,800.00

Total of payments

$25,709.25 $30,772.48 $29,000.00

Total interest paid

$5,709.25 $10,772.48 $9,000.00


16

• If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N.

• F dollars at the end of period N is equal to a single sum P dollars now, if your earning power is measured in terms of interest rate i.

N

F

P

0

NiPF )1(

NiFP )1(


17

Equivalence Between Two Cash Flows

• Step 1: Determine the base period, say, year 5.

• Step 2: Identify the interest rate to use.

• Step 3: Calculate equivalence value.

$3,000$2,042

50

i F

i F

i F

6%, 042 1 0 06 733

8%, 042 1 0 08 000

10%, 042 1 0 10

5

5

5

$2, ( . ) $2,

$2, ( . ) $3,

$2, ( . ) $3,289


18

Example 4.3 Equivalence

Various dollar amounts that will be economically equivalent to $3,000 in 5 years, given an interest rate of 8%


19

Example 4.4 (Prnpl:1) Equivalent Cash Flows are Equivalent at Any Common Point In Time


20

The Five Types of Cash Flows

(a) Single cash flow

(b) Equal (uniform) payment series

(c) Linear gradient series

(d) Geometric gradient series

(e) Irregular payment series


21

Single Cash Flow Formula (Example 4.7)

• Single payment -compound amount factor (growth factor)

• Given:

• Find: P

F

N

0

F P i

F P F P i N

N

( )

( / , , )

1

i

N

P

10%

8

000

years

$2,

F

F P

$2, ( . )

$2, ( / , )

$4, .

000 1 010

000 10%,8

28718

8


22

Single Cash Flow Formula (Example 4.8)

• Single payment present worth factor (discount factor)

• Given:

• Find: P

F

N

0

P F i

P F P F i N

N

( )

( / , , )

1

i

N

F

12%

5

000

years

$1,

P

P F

$1, ( . )

$1, ( / , )

$567.40

000 1 0 12

000 12%,5

5


23

Uneven Payment Series (example 4.11)

P P F

P P F

P P F

P P P P

1

2

4

1 2 4

000 10%,1

000 10%,2

000 10%,4

622

$25, ( / , )

$3, ( / , )

$5, ( / , )

$28,


24

Example 4.12 Calculating the Actual Worth of a Long-Term Contract

Beginning Contract

of Season

0 2001 21,000,000$ 2,000,000$ 23,000,000$ 1 2002 21,000,000 2,000,000 23,000,0002 2003 21,000,000 2,000,000 23,000,0003 2004 21,000,000 2,000,000 23,000,0004 2005 25,000,000 2,000,000 27,000,0005 2006 25,000,000 25,000,0006 2007 27,000,000 27,000,0007 2008 27,000,000 27,000,0008 2009 27,000,000 27,000,0009 2010 27,000,000 27,000,000

Salary

Prorated

Signing Bonus Annual Payment

Total

P M P F M P F M P F

M

$23 ( / , ) $23 ( / , ) $27 ( / , )

$215.

6%,1 6%,2 6%,9

75

. . .


25

Equal Payment Series

A

0 1 2 3 4 5 N-1 N

F

P


26

Equal Payment Series Compound Amount Factor

F Ai

iA F A i N

N

( )

( / , , )

1 1

Example 4.13:• Given: A = $3,000, N = 10 years, and i = 7%• Find: F• Solution: F = $3,000(F/A,7%,10) = $41,449.20

0 1 2 3N

F

A


27

Sinking Fund Factoris an interest-bearing account into which a fixed sum is deposited each interest period; it is commonly established for the purpose of replacing

fixed assets.

Example 4.15:• Given: F = $5,000, N = 5 years, and i = 7%• Find: A• Solution: A = $5,000(A/F,7%,5) = $869.50

0 1 2 3N

F

A

A Fi

i

F A F i N

N

( )

( / , , )

1 1


28

Sinking fund

Sinking fund: 1) A fund accumulated by periodic deposits and reserved exclusively for a specific purpose, such as retirement of a debt or replacement of a property. 2) A fund created by making periodic deposits (usually equal) at compound interest in order to accumulate a given sum at a given future time for some specific purpose.


29

Capital Recovery Factor

Example 4.16:• Given: P = $250,000, N = 6 years, and i = 8%• Find: A• Solution: A = $250,000(A/P,8%,6) = $54,075

1 2 3N

P

A

0

A Pi i

i

P A P i N

N

N

( )

( )

( / , , )

1

1 1


30

Capital Recovery Factor (Annuity Factor)

• Annuity: 1) An amount of money payable to a recipient at regular intervals for a prescribed period of time out of a fund reserved for that purpose. 2) A series of equal payments occurring at equal periods of time. 3) Amount paid annually, including reimbursement of borrowed capital and payment of interest.

• Annuity factor: The function of interest rate and time that determines the amount of periodic annuity that may be paid out of a given fund.


31

Equal Payment Series Present Worth FactorPresent worth: The equivalent value at the present, based on time value

of money.

Example 4.18:• Given: A = $32,639, N = 9 years, and i = 8%• Find: P• Solution: P = $32,639(P/A,8%,9) = $203,893

1 2 3N

P

A

0

P Ai

i i

A P A i N

N

N

( )

( )

( / , , )

1 1

1


32

Linear Gradient SeriesEngineers frequently meet situations involving periodic payments that increase or decrease by a constant amount (G) from period to period.

refer to book, Equation 4.17

P Gi i iN

i i

G P G i N

N

N

( )

( )

( / , , )

1 1

12

P


33

Gradient Series as a Composite Series


34

Example 4.20

$1,000$1,250 $1,500

$1,750$2,000

1 2 3 4 50

P =?

How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure?


35

Method 1:

$1,000$1,250 $1,500

$1,750$2,000

1 2 3 4 50

P =?

$1,000(P/F, 12%, 1) = $892.86$1,250(P/F, 12%, 2) = $996.49$1,500(P/F, 12%, 3) = $1,067.67$1,750(P/F, 12%, 4) = $1,112.16$2,000(P/F, 12%, 5) = $1,134.85

$5,204.03


36

Method 2:

P P A1 000 12%,5

604 80

$1, ( / , )

$3, .

P P G2 12%,5

599 20

$250( / , )

$1, .

P

$3, . $1, .

$5,204

604 08 599 20


37

Geometric Gradient SeriesMany engineering economic problems, particularly those relating to

construction costs, involve cash flows that increase over time, not by a constant amount, but rather by a constant percentage (geometric), called

compound growth.

PA

g i

i gi g

NA i i g

N N

1

1

1 1 1

1

( ) ( )

/ ( ),

, if

if


38

Geometric Gradient: Find P, Given A1,g,i,N

• Given:g = 7%

i = 12%

N = 5 years

A1 = $54,440

• Find: PUsing equation 4.26 in the

text, we obtain

P=$222,283


39

Summary

• Money has a time value because it can earn more money over time.

• Economic equivalence exists between individual cash flows and/or patterns of cash flows that have the same value. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal.

• The purpose of developing various interest formulas was to facilitate the economic equivalence computation.

(c) 2002 contemporary engineering economics 1 chapter 4 time is money interest: the cost of money...

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