engineering economics ise460 session 3 chapter 3 …engineering economics ise460 session 3 chapter 3...
TRANSCRIPT
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ENGINEERING ECONOMICS ISE460SESSION 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
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ENGINEERING ECONOMICS ISE460SESSION 3
CHAPTER 3 May 28, 2020
Geza P. Bottlik Page 2
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ENGINEERING ECONOMICS ISE460SESSION 3
CHAPTER 3 May 28, 2020
Geza P. Bottlik Page 3
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ENGINEERING ECONOMICS ISE460SESSION 3
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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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ENGINEERING ECONOMICS ISE460SESSION 3
CHAPTER 3 May 28, 2020
Geza P. Bottlik Page 5
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ENGINEERING ECONOMICS ISE460SESSION 3
CHAPTER 3 May 28, 2020
Geza P. Bottlik Page 6
Examples
• See examples of chapter 3 from our text in the accompanying
spreadsheet
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ENGINEERING ECONOMICS ISE460SESSION 3
CHAPTER 3 May 28, 2020
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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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ENGINEERING ECONOMICS ISE460SESSION 3
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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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ENGINEERING ECONOMICS ISE460SESSION 3
CHAPTER 3 May 28, 2020
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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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ENGINEERING ECONOMICS ISE460SESSION 3
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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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ENGINEERING ECONOMICS ISE460SESSION 3
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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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ENGINEERING ECONOMICS ISE460SESSION 3
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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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ENGINEERING ECONOMICS ISE460SESSION 3
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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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ENGINEERING ECONOMICS ISE460SESSION 3
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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)