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Unformatted text preview: ISyE 3025
Engineering
Economy Course Organization Overview Course
Organization 1 ♦ Remote learning method
♦ How to follow this course
♦ Information sources Welcome ! ♦ Questions and feedback
♦ Text, calculators Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved. Jack R. Lohmann and Gunter P. Sharp
School of Industrial and Systems Engineering
Georgia Institute of Technology Course Organization Remote learning Course Organization 2 ♦ Video lectures on demand
On campus: T1 line
Off campus: 56K modem
Use RealPlayer 8.0 or G2
Typical lecture: 15 – 25 min.
Temporary downloads, can’t save
♦ Course documents on web site
Schedule, Homework
Sample quiz questions
Grades Web site:
www.isye.gatech.edu/~engecon/
Latest announcements
Course schedule items
Administrative items
Text material
Bulletin Board
Grades ISyE 3025 Fall 2002, Learning Cycle #1 How to follow course
Read text 3 Download
lecture material
View lectures
Work homework Take quiz Course Organization Information sources ♦ Quizzes Attend help
sessions Course Organization 4 Contacting us 5 ♦ Web site Bulletin Board
For questions of a
general nature, that may
be of interest to others
♦ Email
engecon@isye.gatech.edu
For questions that
concern only you 1 Course Organization Text, calculators Course Organization 6 ♦ Text: PDF files on WebCT site that
you may download and print for your
own use in ISyE 3025 Quizzes 7 ♦ Brush up on spreadsheets ♦ Quizzes:
Closed book, no note sheets !
Calculators allowed
Every 3 weeks, approximately
(see schedule)
♦ Sample quiz questions from previous
semester on web site
♦ Attend help sessions to
prepare for quizzes Course Organization Course Organization ♦ PowerPoint slide handouts of video
streams available at GT Bookstore
♦ Calculator
Recommend programmable
Financial calculator OK
but avoid cheap ones Grading policy 8 ♦ Take quiz, every 3 weeks, (4 total)
♦ General makeup quiz (Quiz 5),
during final exam week
♦ Highest 3 grades count 30% each
4th highest grade counts 10%
(Student may take 4 or 5 quizzes)
♦ Some opportunity for extra credit HW
♦ Individual makeup quizzes for
illness or schoolsponsored activity Course Organization Web site, email
www.isye.gatech.edu/~engecon/ 10 Course overview 9 ♦ Basic concepts
♦ Time value of money
♦ Economic decision criteria
♦ Multiple alternatives
♦ Income taxes
♦ Inflation
♦ Uncertainty ISyE 3025
Engineering
Economy engecon@isye.gatech.edu Basic
Concepts
Jack R. Lohmann Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved. ISyE 3025 Fall 2002, Learning Cycle #1 School of Industrial and Systems Engineering
Georgia Institute of Technology 2 Basic Concepts Basic Concepts Overview 1 Course focus Course Focus 2 This course is focused
on the principles and
procedures for making
sound economic decisions 3 illustrative problems
Course objectives
4 basic principles Basic Concepts Basic Concepts The Jackpot! 3 “1 million jackpot turns
sour: Lotto winner kept
from selling part of prize and sues” (Newspaper headline) 4 An automobile insurance
company offers customers
the opportunity to either pay
for 6 months insurance in
full now or pay half now and
half in 60 days. The second
option requires a $5 service
fee due now. If the premium
was $150, would you take
them up on their offer? Award: $50,000/yr for 20
yrs ($36,000/yr after taxes)
Deal: Company pays
$241,000 now (after taxes)
for next 10 yrs payments
($500,000 before taxes) Basic Concepts Basic Concepts “Have we got a deal for you!” 5 The Cost of Paying Cash
Investment
$10,000.00
Money Mkt Rate
6.5%
Investment Period 48 mo.
Interest Earned
$2,960.20 “You save
$670.76 to
finance your
car than pay
cash!” The Cost of Financing
Amt Financed
$10,000.00
A.P.R.
10.5%
Term of Loan
48 mo.
Interest Paid
$2,289.44 ??? ISyE 3025 Fall 2002, Learning Cycle #1 Pay Now or Pay Later Course Objectives 6 For you to be able to make wise
economic decisions . . . To be
Proficient in comparing alternatives
Able to see and assess tradeoffs
Able to evaluate justifications
prepared by others
Comfortable using financial
language
Glad you took this course! 3 Basic Concepts Four basic principles Basic Concepts 7 Alternatives: a choice among
two or more things Some aspects certain,
others uncertain Some choices obvious,
some not Monetary & Nonmonetary 1 possible choice: “do
nothing” Monetary: amounts & timing Basic Concepts Basic Concepts 9 3  An economic decision
should be based on the
differences among
alternatives . . . Four basic principles 10 4  An economic decision
should be based on the
objective of making the
“best” use of limited
resources All that is common is
irrelevant to the decision Both monetary and nonmonetary (which also means that the past
is irrelevant, except as a guide to
predict future events, i.e., it is a
“sunk cost”) Basic Concepts Summary 8 2  An economic decision is
no better than the forecasts describing each
alternative
Forecasting . . . not easy! 1  An economic decision is
no better than the alternatives considered . . . Four basic principles . . . Four basic principles 11 ISyE 3025
Engineering
Economy Course focus The Concept of Equivalence
(“Time Value
of Money”) 3 illustrative problems
Course objectives
4 basic principles Jack R. Lohmann
Copyright 1999. Georgia Tech Research Corporation. All rights reserved. ISyE 3025 Fall 2002, Learning Cycle #1 Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved. School of Industrial and Systems Engineering
Georgia Institute of Technology 4 Concept of Equivalence Overview Concept of Equivalence 1 Two viewpoints of “interest” 2 1 • Borrower’s viewpoint:
Money paid for use of
borrowed funds
2 • Investor’s viewpoint:
Return, or capital growth,
from the productive investment of capital Compound interest
& capital growth
“Time Value of Money” Interest rate:
i = Amount accrued/unit time Concept of Equivalence Compound interest: An example Concept of Equivalence 3 What is the capital growth of
$100.00 invested at i = 3% per
year for 3 years?
t
1
2
3 1
1
2
2
3
3
N
N Concept of Equivalence What is the capital growth of
$100.00 invested at i = 3% per
year for 3 years?
$109.27 = 100.00(1.03)3 ISyE 3025 Fall 2002, Learning Cycle #1 Capital
at t
0 t Capital
Interest from
at t
t to t+1
0
$100.00 0.03(100.00) = $3.00
$103.00 0.03(103.00) = $3.09
$106.09 0.03(106.09) = $3.18
$109.27 Our previous example . . . Compound interest: A formula 4 Interest from
t to t+1
P P iP P+iP
= P(1+i)
P(1+i)+iP(1+i)
= P(1+i)2
P(1+i)2+iP(1+i)2
= P(1+i)3
= P(1+i)N iP(1+i)
iP(1+i)2 Concept of Equivalence 5 “Time Value of Money:” Example
+$100
0 1 i = 0.05/yr 2 3 4 6 F=?
5 yrs F = P(1+i)N
= $100(1.05)5
= $127.63 5 Concept of Equivalence Are these sums equivalent? Concept of Equivalence 7 Terminology: “Compounding” +$120 +$100 Calculating an
equivalent
amount money
in the future
amount given
some amount of
money in the
present 0 1 2 3 4 5 yrs
Equivalent at i = 0.03/year?
F = 100(1.03)5 = $115.93
$100 at t = 0 is not equivalent
to $120 at t = 5. Prefer $120
five years from today. Concept of Equivalence Another example . . .
P=? Concept of Equivalence 9 Are these sums equivalent? +$500 +$300
i = 0.06/yr i = 0.06/yr
0 1 2 0 . . . 10 yrs F = P(1+i)N
P = F(1+i)N
P = 500(1.06)10
= $279.20 Calculating an
equivalent
amount money
in the present
amount given
some amount of
money in the
future ISyE 3025 Fall 2002, Learning Cycle #1 1 2 10 +$500 . . . 11 yrs P = 500(1.06)11 = $263.39
$300 at t = 0 is not equivalent
to $500 at t = 11 when i =
0.06/yr. Prefer $300 today. Concept of Equivalence Terminology: “Discounting” 8 Concept of Equivalence 11 One more example . . . Find F
i = 0.1/yr
50
0 50 50 1 2 3 12 F=?
50
4 yrs F = 50(1.1)3 + 50(1.1)2 +
50(1.1)1 + 50 = $232.05 6 Concept of Equivalence Concept of Equivalence One more example . . . Find P 13 P= 1 2 50(1.1)1 3 14 232.05
158.49 i = 0.1/yr
50 50 50 50 P = ? i = 0.1/yr
50 50 50 50
0 F and P equivalent? 0 4 yrs 1 2 3 4 yrs Is $158.49 at t = 0 equivalent
to $232.05 at t = 4? 50(1.1)2 +
+
50(1.1)3 + 50(1.1)4 = $158.49 Yes! 158.49(1.1)4 = $232.05 Concept of Equivalence Concept of Equivalence Summary 15 Summary 16 “Time Value of Money”
Compound interest
& capital growth
Two views of “interest”
F = P(1+i)N ISyE 3025
Engineering
Economy Equivalence depends on:
1  amount of money;
2  their timing; and
3  an interest rate Equivalence Formulas Equivalence
Formulas
(sometimes called
Equivalence Factors)
Jack R. Lohmann Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved. 3 examples:
compounding,
discounting, both School of Industrial and Systems Engineering
Georgia Institute of Technology ISyE 3025 Fall 2002, Learning Cycle #1 Overview 1 Purpose
4 classic equivalence models
Single cash flow
Uniform cash flow series
Arithmetic gradient cash
flow series
Geometric gradient cash
flow series 7 Equivalence Formulas Purpose 2 Equivalence Formulas A cash flow convention 3 Several cash flow conventions exist: this course will
use endofperiod models To facilitate
performing
equivalence
calculations $10,000 01 2 N1 N Yr, Qtr Yr, Mo, Wk, Day, etc
“EndofYear Convention” Equivalence Formulas Equivalence Formulas Four classic equivalence models
$ Time 4 Single
cash flow $ $ T0 T1 T2 T3 . . . TN1 TN
T0 = past, present, or future
TN = N periods after T0
F = P(1+i)TNT0
= P(1+i)N
= P(F/P,i,N)
Appendix Equivalence Formulas Equivalence Formulas Single cash flow: An example
100
3 4 F=?
56 5 F=? P Uniform
series
Arithmetic
gradient
series
Geometric
gradient
series $ Single cash flow: The model 78 What future amount F at
time t = 8 is equivalent to
100 at t = 3 if i = 0.05/year?
F = 100(F/P,0.05,5)
= 100(1.276)
= 127.60 ISyE 3025 Fall 2002, Learning Cycle #1 6 Single cash flow: The model 7 F P=? T0 T1 T2 T3 . . . TN1 TN
P = F(1+i)T0TN
= F(1+i)N
= F(P/F,i,N)
Appendix 8 Equivalence Formulas Equivalence Formulas Single cash flow: Another example
P=? 8 Uniform cash flow series: The model
AAA 500 3 2 1 0 1 . . . 6 7 9 A A F=? T0 T1 T2 T3 . . . TN1 TN What amount P at t =  3 is
equivalent to 500 at t = 7 if
i = 0.06/year? F = A(1+i)N1+A(1+i)N2+ . . . +A(1+i)+A
Algebra
[(1+i)1]
F = A[(1+i)N1+(1+i)N2+ . . . +(1+i)+1]
[(1+i)1]
F[(1+i)1]= A[(1+i)N+(1+i)N1 +(1+i)N2 + . . . +(1+i)
(1+i)N1 (1+i)N2  . . . (1+i)1] P = 500(P/F,0.06,10)
= 500(0.5584)
= 279.20 i Equivalence Formulas Uniform cash flow series: The model
AAA Equivalence Formulas 10 50 A = F[i/{(1+i)N1}]
= F(A/F,i,N) Equivalence Formulas AAA 50 Arithmetic gradient series: The model A = P[i(1+i)N/{(1+i)N1}]
= P(A/P,i,N) ISyE 3025 Fall 2002, Learning Cycle #1 13 (N1)G
(N2)G P P = A(1+i)1+A(1+i)2+ . . . +A(1+i)N
P = A[{(1+i)N1}/i(1+i)N] Algebra
= A(P/A,i,N) 50 Equivalence Formulas 12 AA T0 T1 T2 T3 . . . TN1 TN 50 01
2
3 4 yrs
What future amount F at t = 4
is equivalent to $50 each year
for the next four years if i =
0.1/year?
F = 50(F/A,0.1,4)
= 50(4.641)
= 232.05 F = A[{(1+i)N1}/i ]
= A(F/A,i,N) Uniform cash flow series: The model 11 F=? AAF T0 T1 T2 T3 . . . TN1 TN P Uniform cash flow series: Example G 2G A T0 T1 T2 T3 . . . TN1 TN
P = G[{(1+i)NiN1]/[i2(1+i)N]
= G(P/G,i,N)
A = G[{(1+i)NiN1]/[i(1+i)N1]
= G(A/G,i,N) 9 Equivalence Formulas Equivalence Formulas An example
Maintenance costs
01
2
34 14 Service contract
01
2
34  3000
A=?
 4000
 5000
 6000
P=? An example: Another way
Maintenance costs
01
2
34 Service contract
01
2
34  3000
A=?
 4000
 5000
 6000 i = 0.08/yr P =  3000(P/A,.08,4)
 1000(P/G,.08,4)
=  14,586 15 i = 0.08/yr A =  3000 1000(A/G,.08,4)
=  4404 A = 14,586(A/P,.08,4) =  4404 Equivalence Formulas Equivalence Formulas Geometric gradient series: The model
P A1(1+g)N1 A1(1+g)2
A1(1+g)1
A1 T0 T1 16 A1 P g>0 (1+g)N1 A1(1+g)2
A1(1+g)1
A1 g=0
g<0 T0 T1 T2 T3 TN1 TN P = [A1/(1+i) + A1(1+g)/(1+i) …
+ A1(1+g)N1/(1+i)N] 17 g>0
g=0
g<0 T2 T3 TN1 TN
A1[{1(1+g)N(1+i)N}/{ig}]
for i ≠ g P= A1N(1+i)1 for i = g
= A1(P/A1,i,g,N) Equivalence Formulas An example Geometric gradient series: The model Equivalence Formulas 18 An example 19 F=? Assume you plan to save
10% of your salary each year
and invest it in an account at
8%/year, how much would
you accumulate in the
account in 10 years? 6000(1.06)9
6000(1.06)2
6000(1.06)
6000 Assume your current salary
is $60,000, and raises are
expected at the rate of
6%/year. P = 6000(P/A1,.08,.06,10)
= 6000(8.5246)
= 51,148 ISyE 3025 Fall 2002, Learning Cycle #1 P 0 1 2 3 … 10 F = 51,148(F/P,.08,10)
= 132,662 10 Equivalence Formulas Equivalence Formulas Summary 20 Purpose
4 classic equivalence models
Single cash flow
(F/P,i,n) & (P/F,i,n)
Uniform cash flow series
(F/A,i,n), (P/A,i,n),
(A/F,i,n), (A/P,i,n) Summary 21 4 classic equivalence models
Arithmetic gradient
cash flow series
(P/G,i,n)
Geometric gradient
cash flow series
(P/A1,i,g,n) Interest Rates ISyE 3025
Engineering
Economy Overview Interest Rates 1 Time scale conversions
“Nominal” versus
“Effective” rates Jack R. Lohmann
Copyright 1999. Georgia
Tech Research Corporation.
All rights reserved. School of Industrial and Systems Engineering
Georgia Institute of Technology Interest Rates Interest Rates Time scale conversion: Example
If you were offered i4 = 0.04
per quarter year, what semiannual rate, i2, would make
you in different between
i4 = 0.04 and i2?
F
P
0
0 1 23
1
1 Year 4 qtr yrs
2 half yrs ISyE 3025 Fall 2002, Learning Cycle #1 2 Time scale conversion: Example
P
0
0 3 F 1 23
1
1 Year 4 qtr yrs
2 half yrs F = P(1+i4)4 = P(1+i2)2
i2 = (1+i4)4/21
= (1.04)21
= 0.0816 11 Interest Rates Interest Rates Time scale conversions: Formula
F P
0
0 4 1 2 ...
1 ...
1 Year Nominal vs. Effective Rates 5 Nominal Annual Interest
Rate, r =MiM
Credit cards often charge
1.5%/mo for unpaid
balances and report it
as “18% compounded
monthly,” i.e., 12(0.015) =
0.18. M2
M1 iM1 = (1+iM2)M2/M11 Interest Rates Interest Rates Nominal vs. Effective Rates 6 Another view of “effective” interest 7 A bank offers loans of $2500 for
30 months at an advertised interest rate of 2%/mo. The monthly
payments are computed as
follows: Interest = 0.02(2500)30 =
$1500; Processing Fee = $50;
Total Owed = 2500+1500+50
= $4050; Monthly Payment =
4050/30 = $135. What is the
effective annual interest rate? Effective Annual Interest Rate:
i = (1+iM)M1
= (1.015)12  1
= 0.196 Interest Rates Interest Rates Solution
+2500
30 mos
... 135
2500 = 135(P/A,i12,30)
i12 = 0.036 per month
i = (1.036)12  1 = 0.529/year 8 Summary 9 Time scale conversions
iM1 = (1+iM2)M2/M11
Nominal vs effective
interest rates
r =MiM
i = (1+iM)M1
i should include all costs
Copyright 1999. Georgia Tech Research Corporation. All rights reserved. ISyE 3025 Fall 2002, Learning Cycle #1 12 ...
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This note was uploaded on 09/07/2009 for the course ISYE 3025 taught by Professor Lee during the Spring '09 term at Georgia Institute of Technology.
 Spring '09
 Lee

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