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ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29 , 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter 3

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Page 1: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

ENGINEERING ECONOMICS ISE460SESSION 3

CHAPTER 3, May 29 , 2015

Geza P. Bottlik Page 1

OUTLINE

• Questions?

• News?

• Recommendations – no obligation

• Chapter 3

Page 2: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

ENGINEERING ECONOMICS ISE460SESSION 3

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Geza P. Bottlik Page 2

Overview

• EXCEL DEMO OF CHAPTER 3 MATERIAL

• CHAPTER 3 ESSENTIALS

– FORMULAS

– EXAMPLES 3.25, 3.26, 3.27

– TABLES

Page 3: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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The time value of money

• If you have money, you can either

– Lend it and get interest

– Invest it and hope you get interest and that you get it back

– Invest it and hope you can resell it for more

– Use it to buy something or a service

• In the first two instances, when you ask for it back in the future, you’ll get your money plus what it earned. In other words, it will have grown

• Unfortunately, during the same time period, money in general will have lost some of its value due to inflation

• But if you invest wisely, the net should be higher

• In this chapter we assume that the two effects are combined in the quoted rates

Page 4: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Graphical Representation

Time

Income

Expense

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Graphical Representation - Example

Time

Borrowed Money

Repayment

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Simple Interest

• At 5 % per year (per annum)

Deposit $100

One Year

$5 $5 $5 $5

Page 7: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Compound Interest

• At 5 % per year (per annum), compounded annually

Deposit $100

One Year

$5.00 105*.05=$5.25

110.25*.05=$5.5125

115.76*0.05 = $5.788

Page 8: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Equivalence

• We can compare different cash flows by comparing their

value at the same time

• The choice of time is arbitrary, but is usually the present or

some date in the future

• Equivalence is valid for only one interest rate

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Equivalence - example

• @ 5% compound interest

• $100 held as cash for 3 years will be worth $86.38

• $100 loaned today can be paid back in 3 equal annual

installments of $36.72

• $100 promised to be paid 2 years from now is worth $90.70

today

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Uniform Series

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Linear Gradient

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Geometric Gradient

Page 13: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Irregular Series

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Notation

• An = payment or receipt occurring at the end of period n

• A = a constant payment or receipt at end of each period

• N = number of periods

• i = interest rate per interest period

• P = a sum of money at time zero = Present value or worth

• F = a sum of money at some future date (usually the end of

the time span that you are analyzing

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CHAPTER 3 FORMULAS

Name Formula Excel Function

Single PaymentCompound Amount Factor (compounding) NiPF 1 FV(i, N, 0, P)

Present Worth Factor (discounting) NiFP 1 PV(i, N, 0, F)

Equal Payments

Compound Amount Factor

i

iAF

N 11FV(i, N, A)

Sinking Fund Factor 1

11

i

iFA

N

PMT(i, N, 0, F)

Capital Recovery Factor (Annuity Factor)

1

1

11

N

N

ii

iPA PMT(i, N, P)

Present Worth Factor

N

N

ii

iAP

1

11PV(i, N, A)

Page 16: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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CHAPTER 3 FORMULAS (CONTINUED)

Linear Gradient Series

Present Worth Factor

N

N

ii

iNiiGP

1

112

none

Gradient to equal payment conversion Factor

]11[

11N

N

ii

iNiGA none

Geometric Gradient Series

Present Worth Factor

giiNAP

gigi

igAP

NN

,)1/(

,)1()1(1

1

1 none

Future Worth Factor

giiNAF

gigi

giAF

N

NN

,)1/(

,)1()1(

11

1none

Page 17: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Derivations

• DERIVATION OF SOME FORMULAS

– SINGLE CASH FLOW

– UNIFORM SERIES OF PAYMENTS

– LINEAR GRADIENT SERIES

– GEOMETRIC GRADIENT SERIES

– IRREGULAR SERIES

• GO OVER REMAINING CHAPTER 3 EXAMPLES

Page 18: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Formulas

F – future value

P – present value

N – number of periods

i – interest rate per period

A – payment at the end of each period

G – increase in payment per period starting in the second

A1 – Initial payment

n – Index for periods

g –percentage change from payment to payment

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Formulas – Future value, Present value

iF

P

N

iN

iFP

iF

P

ip

F

ipF

N

N

N

N

1log

log )6

1logF

Plog )5

1 )4

1 )3

)1( )2

)1( )1

Page 20: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Formulas – Payments –the last term in 7) should be raised to N-1

11 )11

11 )10

11 )9

11 )7)8

1....11i1 )8

1....11 )72

2

N

N

N

N

N

i

iFA

i

iAF

iAiF

iAAFiF

iAiAiAF

iAiAiAAF

Page 21: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Formulas – Payments (continued)

11

1 )14

1

11 )13

111

4) into )12

N

N

N

N

NN

i

iiPA

ii

iAP

ii

iAP

Page 22: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Formulas – Gradient Series

Gradient Series

Ni

i

i

GF

ii

iNiGA

ii

iNNiGP

iGjP

iGNiGiG

iAiAiAP

N

N

N

NG

N

j

NG

N

NTOT

11 19)

10) into )18

11

11 18)

)14 into )17

1

11 )17

11 )16

11......1210

1....11 )15

2

1

32

11

21

11

Page 23: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Formulas -- Geometric Gradient Series

• Geometric Gradient Series

N

n

n

N

n

nn

nnn

nnn

nn

i

g

g

AP

igAP

igAP

iAP

gAA

1

1

1

11

11

11

1

1

1 )24

11 23)

:payments allover Summing

11 22)

21) into )20

1 )21

1 )20

Page 24: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Formulas -- Geometric Gradient (continued)

giiNA

gigi

igA

P

x

xxaP

xxaxPP

xxxaxP

xxxaP

i

gx

g

Aa

NN

N

N

N

N

1/

,111

27)

1 cannot x Note 1

)(

)(

)25)26

...( 26)

:by x gmultiplyin

...( )25

1

1

1

1

1

1

1

132

2

1

Page 25: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Formulas -- Geometric Gradient (continued)

giiNA

gigi

igA

F

NN

1

,11

28)

27) into )1

1-N1

1

Page 26: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter

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Tables

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Equivalence at any point in time

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Example 3.6 - solution

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3.25

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3.26

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3.26

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3.26

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Example 3.27

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3.27 – cash flows

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3.27 – part a)

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3.27 part b)

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3.27 part b)

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3.27 part b)

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3.27 part b)