CTC 475 Review Uniform Series Find F given A

Slides:



Advertisements
Similar presentations
Nominal and Effective Interest Rates Lecture slides to accompany
Advertisements

CTC 475 Review Interest/equity breakdown What to do when interest rates change Nominal interest rates Converting nominal interest rates to regular interest.
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 4-1 Lecture slides to accompany Engineering Economy 7 th edition Leland Blank Anthony Tarquin.
Econ. Lecture 3 Economic Equivalence and Interest Formula’s Read 45-70
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 3-1 Lecture slides to accompany Engineering Economy 7 th edition Leland Blank Anthony Tarquin.
CTC 475 Review Uniform Series –Find F given A –Find P given A –Find A given F –Find A given P Rules: 1.P occurs one period before the first A 2.F occurs.
(c) 2002 Contemporary Engineering Economics
(c) 2002 Contemporary Engineering Economics
LECTURE 6 NONUNIFORM SERIES Prof. Dr. M. F. El-Refaie.
EGR 403 Capital Allocation Theory Dr. Phillip R. Rosenkrantz
Chapter 2 Factors: How Time and Interest Affect Money
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford University Press, Inc.1 Engineering Economic Analysis.
Interest Formulas (Gradient Series) Lecture No.6 Professor C. S. Park Fundamentals of Engineering Economics Copyright © 2005.
Single-Payment Factors (P/F, F/P)
CTC 475 Review Cost Estimates Job Quotes (distributing overhead) – Rate per Direct Labor Hour – Percentage of Direct Labor Cost – Percentage of Prime (Labor+Matl)
Economic System Analysis January 15, 2002 Prof. Yannis A. Korilis.
The terms ‘nominal’ and ‘effective’ enter into consideration when the compounding period (i.e. interest period) is less than one year. A nominal interest.
CTC 475 Review Gradient Series –Find P given G –Find A given G Rules: 1.P occurs two periods before the first G 2.n equals the number of cash flows + 1.
Equivalence and Compound interest
Interest Formulas – Equal Payment Series
1 EGGC4214 Systems Engineering & Economy Lecture 4 Interest Formulas.
Slide Sets to accompany Blank & Tarquin, Engineering Economy, 6 th Edition, 2005 © 2005 by McGraw-Hill, New York, N.Y All Rights Reserved 2-1 Developed.
Irregular Payment Series and Unconventional Equivalence Calculations
Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.
Financial Mathematics 1. i = interest rate (per time period) n = # of time periods P = money at present F = money in future –After n time periods –Equivalent.
A shifted uniform series starts at a time other than year 1 When using P/A or A/P, P is always one year ahead of first A When using F/A or A/F, F is in.
CTC 475 Review Simple vs Compound Interest Find F given P –(equation and factor form) Find P given F –(equation and factor form) Find i given P, F and.
MER Design of Thermal Fluid Systems Econ Lecture 2 Professor Bruno Winter Term 2002.
1 Equivalence Between Two Cash Flows Step 1: Determine the base period, say, year 5. Step 2: Identify the interest rate to use. Step 3: Calculate equivalence.
Chapter 4 More Interest Formulas
Basic Of Engineering Economy
Chapter 4: The Time Value of Money
Unconventional Equivalence Calculations
Contemporary Engineering Economics
Practical uses of time value of money factors
Chapter 2 Time Value of Money
Interest Formulas for Single Cash Flows
Irregular Payment Series and Unconventional Equivalence Calculations
Example 2.14:Powerball Lottery
Chapter 2 Factors: How Time and Interest Affect Money
Nominal and Effective Interest Rates Lecture slides to accompany
Chapter 3 Combining Factors and Spreadsheet Functions
Shifted Uniform Series
LECTURE 6 NONUNIFORM SERIES
Chapter 4: The Time Value of Money
Nominal and Effective Interest Rates Lecture slides to accompany
Chapter 2 Factors: How Time and Interest Affect Money
Engineering Economics
Engineering Economic Analysis
Chapter 3 Combining Factors and Spreadsheet Functions
Chapter 2 Factors: How Time and Interest Affect Money
Chapter 2 Factors: How Time and Interest Affect Money
Let’s work some problems…
Engineering Economy [3] Combining Factors Examples
Chapter 4: The Time Value of Money
Combining Factors – Shifted Uniform Series
CTC 475 Review Interest/equity breakdown
Note that P is Two Periods Ahead of the First G
Example 1: Because of general price inflation in the economy, the purchasing power of the Turkish Lira shrinks with the passage of time. If the general.
Geometric Gradients Change by the Same Percentage Each Period
IET 333: Week 3 Jung-woo Sohn
Combining Factors – Shifted Uniform Series
Mc Factors: How Time and Interest Affect Money Graw Hill CHAPTER II
Varying Interest Rates
Chapter 4: The Time Value of Money
CTC 475 Review Simple vs Compound Interest Find F given P
Combining Factors – Shifted Uniform Series
Chapter 3 Combining Factors and Spreadsheet Functions
Contemporary Engineering Economics
CTC 475 Review Gradient Series Find P given G Find A given G Rules:
Presentation transcript:

CTC 475 Review Uniform Series Find F given A Find P given A (and deferred withdrawal) Find A given F Find A given P (and deferred withdrawal) Rules: P occurs one period before the first A F occurs at the same time as the last A n equals the number of A cash flows

Gradient Series and Geometric Series CTC 475 Gradient Series and Geometric Series

Objectives Know how to recognize and solve gradient series problems Know how to recognize and solve geometric series problems

Gradient Series Cash flows start at zero and vary by a constant amount G G=? EOY Cash Flow 1 $0 2 $200 3 $400 4 $600 5 $800

Gradient Series Tools Find P given G Find A given G Converts gradient to uniform There is no “find F given G” Find “P/G” and then multiply by “F/P” or Find “A/G” and then multiply by “F/A”

Gradient Series Rules (differs from uniform/geometric) P occurs 2 periods before the first G n = the number of cash flows +1 (or---the first n cash flow is zero)

Find A given G (n=???) EOY Cash Flow 1 2 G 3 2G 4 3G 5 4G EOY 1 2 G 3 2G 4 3G 5 4G EOY Cash Flow 1 A 2 3 4 5

Find P given G (Pure Gradient) How much must be deposited in an account today at i=10% per year compounded yearly to withdraw $100, $200, $300, and $400 at years 2, 3, 4, and 5, respectively? P=G(P/G10,5)=100(6.862)=$686

Find P Uniform +Gradient How much must be deposited in an account today at i=10% per year compounded yearly to withdraw $1000, $1100, $1200, $1300 and $1400 at years 1, 2, 3, 4, and 5, respectively? This is not a pure gradient (doesn’t start at $0) ; however, we could rewrite this cash flow to be a gradient series with G=$100 added to a uniform series with A=$1000 P=1000(P/A10,5)+100(P/G10,5)

Uniform + Gradient EOY Cash Flow 1 A=$1000 2 3 4 5 EOY Cash Flow 1 1 A=$1000 2 3 4 5 EOY Cash Flow 1 G=$0 2 G=$100 3 G=$200 4 G=$300 5 G=$400

Combinations Uniform + a gradient series Uniform – a gradient series (like previous example) Uniform – a gradient series (see next slide for example)

Find P Uniform–Gradient What deposit must be made into an account paying 8% per yr. if the following withdrawals are made: $800, $700, $600, $500, $400 at years 1, 2, 3, 4, and 5 years respectively. P=800(P/A8,5)-100(P/G8,5)

Uniform-Gradient EOY Cash Flow 1 A=$800 2 3 4 5 EOY Cash Flow 1 G=0 2 1 A=$800 2 3 4 5 EOY Cash Flow 1 G=0 2 A=$100 3 A=$200 4 A=$300 5 A=$400

Example What must be deposited into an account paying 6% per yr in order to withdraw $500 one year after the initial deposit and each subsequent withdrawal being $100 greater than the previous withdrawal? 10 withdrawals are planned. P=$500(P/A6,10)+$100(P/G6,10) P=$3,680+$2,960 P=$6,640

Example An employee deposits $300 into an account paying 6% per year and increases the deposits by $100 per year for 4 more years. How much is in the account immediately after the 5th deposit? Convert gradient to uniform A=100(A/G6,5)=$188 Add above to uniform A=$188+$300=$488 Find F given A F=$488(F/A6,5)=$2,753

Geometric Series Cash flows differ by a constant percentage j. The first cash flow is A1 Notes: j can be positive or negative geometric series are usually easy to identify because there are 2 rates; the growth rate of the account (i) and the growth rate of the cash flows (j)

Tools Find P given A1, i, and j Find F given A1, i, and j

Geometric Series Rules P occurs 1 period before the first A1 n = the number of cash flows

Geometric Series Equations (i=j) P=(n*A1) /(1+i) Note: inside of the front cover of your book shows equation as A1*(n/(1+i)) F=n*A1*(1+i)(n-1) Note: inside of the front cover of your book does not have this equation but F=P(1+i) so the above equation can be derived

Geometric Series Equations (i not equal to j) P=A1*[(1-((1+j)n*(1+i)-n)/(i-j)] F=A1*[((1+i)n-(1+j)n)/(i-j)]

Geometric Series Example How much must be deposited in an account in order to have 30 annual withdrawals, with the size of the withdrawal increasing by 3% and the account paying 5%? The first withdrawal is to be $40,000? P=A1*[(1-((1+j)n*(1+i)-n)/(i-j)] A1=$40,000; i=.05; j=.03; n=30 P=$876,772

Geometric Series Example An individual deposits $2000 into an account paying 6% yearly. The size of the deposit is increased 5% per year each year. How much will be in the fund immediately after the 40th deposit? F=A1*[((1+i)n-(1+j)n)/(i-j)] A1=$2,000; i=.06; j=.05; n=40 F=$649,146

Next lecture Changing interest rates Multiple compounding periods in a year Effective interest rates