ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3 …ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3...

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ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3 May 28, 2020 Geza P. Bottlik Page 1 OUTLINE Questions? News, Experiences? Chapter 3 How do we determine the value of money at any given time? We need to know the interest rate and the elapsed time the appropriate formula. This chapter has all those formulas We represent all problems with a graph of money vs. time Next homework due 6/1/20 Quiz on 6/2/20

Transcript of ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3 …ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3...

Page 1: ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3 …ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3 May 28, 2020 Geza P. Bottlik Page 4 Basic principles • The value of money changes

ENGINEERING ECONOMICS ISE460SESSION 3

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OUTLINE

• Questions?

• News, Experiences?

• Chapter 3

– How do we determine the value of money at any given time?

– We need to know the interest rate and the elapsed time the appropriate formula. This chapter has all those formulas

– We represent all problems with a graph of money vs. time

• Next homework due 6/1/20

• Quiz on 6/2/20

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Basic principles

• The value of money changes with time

– Depends on the length of time

– Depends on a rate called interest, discount rate, rate of return

• Two different amounts of money at two different times are equivalent if

we know the interest rate and the length of time

• All equations are based on one single simple equation:

• Future value = Present value *(1 + interest rate per period)^Number of

periods

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Examples

• See examples of chapter 3 from our text in the accompanying

spreadsheet

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CHAPTER 3 FORMULASFormula Excel Function

Single Payment

Compound 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

N11

FV(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)

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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

1none

Future Worth Factor

giiNAF

gigi

giAF

N

NN

,)1/(

,)1()1(

11

1none

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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

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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

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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

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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

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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

N

G

N

N

TOT

1119)

10)into)18

11

1118)

)14into)17

1

11)17

11)16

11......1210

1....11)15

2

1

32

1

1

2

1

1

1

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Formulas -- Geometric Gradient Series

• Geometric Gradient Series

N

n

n

N

n

nn

nn

n

n

nn

n

n

i

g

g

AP

igAP

igAP

iAP

gAA

1

1

1

1

1

1

1

1

1

1

1

1)24

1123)

:paymentsallover Summing

1122)

21)into)20

1)21

1)20

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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)

1cannot x Note1

)(

)(

)25)26

...(26)

:by xgmultiplyin

...()25

1

1

1

1

1

1

1

132

2

1

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Formulas -- Geometric Gradient (continued)

giiNA

gigi

igA

F

NN

1

,11

28)

27)into)1

1-N

1

1

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Tables

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Example 3.26

• The two cash flows below are equivalent at an interest rate of

12%, compounded annually. Determine the value of C

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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)