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

ENGINEERING ECONOMICS ISE460SESSION 3

CHAPTER 3 May 28, 2020

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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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ENGINEERING ECONOMICS ISE460SESSION 3

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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)