Single-Payment Factors (P/F, F/P)

Slides:



Advertisements
Similar presentations
Chapter 4 More Interest Formulas
Advertisements

Engineering Economics I
Econ. Lecture 3 Economic Equivalence and Interest Formula’s Read 45-70
Engineering Economics ENGR 3300
Engineering Economics
Engineering Economics Outline Overview
Chapter 2 Solutions 1 TM 661Chapter 2 Solutions 1 # 9) Suppose you wanted to become a millionaire at retirement. If an annual compound interest rate of.
Module 1 – Lecture 4 MONEY TIME RELATIONSHIP Prof. Dr. M.F. El-Refaie.
Borrowing, Lending, and Investing
© 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.
State University of New York WARNING All rights reserved. No part of the course materials used in the instruction of this course may be reproduced in any.
(c) 2002 Contemporary Engineering Economics
(c) 2002 Contemporary Engineering Economics
Interest Formulas – Equal Payment Series
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.
ISU CCEE CE 203 More Interest Formulas (EEA Chap 4)
Exam I Review Exam is closed text book, no loose sheet paper
ISU CCEE CE 203 More Interest Formulas (EEA Chap 4)
Economic System Analysis January 15, 2002 Prof. Yannis A. Korilis.
Equivalence and Compound interest
Interest Formulas (Gradient Series)
Engineering Economics
Single-Payment Factors (P/F, F/P) Fundamental question: What is the future value, F, if a single present worth, P, is invested for n periods at an ROR.
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.
Non-Uniform Cash Flows
ENGINEERING ECONOMICS ISE460 SESSION 3 CHAPTER 3, May 29, 2015 Geza P. Bottlik Page 1 OUTLINE Questions? News? Recommendations – no obligation Chapter.
1 Cash Flow Patterns The “LEGO” blocks of Engineering Economics.
Arithmetic Gradient Factors (P/G, A/G) Cash flows that increase or decrease by a constant amount are considered arithmetic gradient cash flows. The amount.
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.
EGR Single-Payment Factors (P/F, F/P) Example: Invest $1000 for 3 years at 5% interest. F 1 = (1000)(.05) = 1000(1+.05) F 2 = F 1 + F.
TM 661 Engineering Economics for Managers Unit 1 Time Value of Money.
MER Design of Thermal Fluid Systems Econ Lecture 2 Professor Bruno Winter Term 2002.
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.
Construction Accounting & Financial Management, 3/e Steven Peterson © 2013 by Pearson Higher Education, Inc Upper Saddle River, New Jersey All Rights.
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
Equivalence Factors and Formulas Moving Money Around.
Basic Of Engineering Economy
Interest Formulas (Gradient Series)
Practical uses of time value of money factors
NE 364 Engineering Economy
Engineering Economics I
Chapter 2 Factors: How Time and Interest Affect Money
LECTURE 6 NONUNIFORM SERIES
Chapter 2 Factors: How Time and Interest Affect Money
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
Arithmetic Gradient Factors (P/G, A/G)
Equivalence And Compound Interest Course Outline 2
Combining Factors – Shifted Uniform Series
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.
IET 333: Week 3 Jung-woo Sohn
Combining Factors – Shifted Uniform Series
Engineering Economics - Introduction
Mc Factors: How Time and Interest Affect Money Graw Hill CHAPTER II
Let’s work some problems…
Let’s work some problems…
Combining Factors – Shifted Uniform Series
Chapter 3 Combining Factors and Spreadsheet Functions
Presentation transcript:

Single-Payment Factors (P/F, F/P) Example: Invest $1000 for 3 years at 5% interest. F = ? i = .05 $1000 F1 = 1000 + (1000)(.05) = 1000(1+.05) F2 = F1 + F1i = F1 (1+i) = 1000(1+.05)(1+.05) = 1000(1+.05)2 F3 = 1000(1+.05)3 F1 = 1050 F2 = 1102.50 F3 = 1157.63   EGR 312 - 03

Single-Payment Factors (P/F, F/P) Fundamental question: What is the future value, F, if a single present worth, P, is invested for n periods at an ROR of i% assuming compound interest?   $P $? General Solution: F1 = P + Pi = P(1+i) F2 = F1 + F1i = F1 (1+i) = P(1+i)(1+i) = P(1+i)2 … Fn = P(1+i)n   EGR 312 - 03

Single-Payment Factors (P/F, F/P) Fundamental questions: What is the future value, F, if a single present worth, P, is invested for n periods at an ROR of i% assuming compound interest? In general, F = P(1+i)n What is the present value, P, if a future value, F, is desired, assuming P is invested for n periods at i% compound interest?   P = F/(1+i)n EGR 312 - 03

Single-Payment Factors (P/F, F/P) Standard Notation: If wanting to know F given some P is invested for n periods at i% interest use – (F/P,i,n) Example: if i = 5%, n = 6 months, ______________ If wanting to know P given some F if P is to be invested for n periods at i% interest use – (P/F,i,n) Example: if i = 7.5%, n = 4 years, ______________   EGR 312 - 03

Single-Payment Factors (P/F, F/P) Standard Notation Equation: To find the value of F given some P is invested for n periods at i% interest use the equation – F = P(F/P,i,n) To find the value of P given some F if P is to be invested for n periods at i% interest use – P = F(P/F,i,n) The compound interest factor tables on pages 727-755 provide factors for various combinations of i and n.   EGR 312 - 03

Single-Payment Factors (P/F, F/P) Example: If you were to invest $2000 today in a CD paying 8% per year, how much would the CD be worth at the end of year four? F = $2000(F/P,8%,4) F = $2000(________) from pg. 739 F = $2721 or, F = $2000(1.08)4 F = $2000(1.3605) draw cfd 1.3605   EGR 312 - 03

Single-Payment Factors (P/F, F/P) Example: How much would you need to invest today in a CD paying 5% if you needed $2000 four years from today? P = $2000(P/F,5%,4) P = $2000(_________) from pg. 736 P = $1645.40 or, P = $2000/(1.05)4 P = $2000/(1.2155) 0.8227   EGR 312 - 03

Uniform Series Present Worth (P/A, A/P) To answer the question: what is P given equal payments (installments) of value A are made for n periods at i% compounded interest? Note: the first payment occurs at the end of period 1. Examples? Reverse mortgages Present worth of your remaining car payments P = ? i   A   EGR 312 - 03

Uniform Series Present Worth (P/A, A/P) To answer the question: what is P given equal payments (installments) of value A are made for n periods at i% compounded interest? Standard Notation: (P/A,i,n)   EGR 312 - 03

Uniform Series Present Worth (P/A, A/P) To answer the related question: what is A given P if equal installments of A are made for n periods at i% compounded interest? Standard Notation: (A/P,i,n) Examples? Estimating your mortgage payment   EGR 312 - 03

Uniform Series Present Worth (P/A, A/P) Example: What is your mortgage payment on a $90K loan if you are quoted 6.25% interest for a 30 year loan. (Remember to first convert to months.) P = $90,000 i = ___________________ n = ______________ A = A = __________ draw cfd i = 6.25/12 = .5208% n = 360 A = 90000[(.005208(1+.005208)360)/((1+.005208)360-1)] A = 554.12 total payments = 199,483.20   EGR 312 - 03

Uniform Series Future Worth (F/A, A/F) To answer the question: What is the future value at the end of year n if equal installments of $A are paid out beginning at the end of year 1 through the end of year n at i% compounded interest? F = ?   i A Example – from quiz EGR 312 - 03

Uniform Series Future Worth (F/A, A/F) Knowing: P = F/(1+i)n Then: and,   EGR 312 - 03

Uniform Series Future Worth (F/A, A/F) Example: If you invest in a college savings plan by making equal and consecutive payments of $2000 on your child’s birthdays, starting with the first, how much will the account be worth when your child turns 18, assuming an interest rate of 6%? A = $2000, i = 6%, n = 18, find F. F = 2000(F/A,6%,18) F = $2000(30.9057) F = $61,811.40 or, draw cfd F = 2000[((1+.06)18-1)/.06] = $61,811.31   EGR 312 - 03

Non-Uniform Cash Flows For example: You and several classmates have developed a keychain note-taking device that you believe will be a huge hit with college students and decide to go into business producing and selling it. Sales are expected to start small, then increase steadily for several years. Cost to produce expected to be large in first year (due to learning curve, small lot sizes, etc.) then decrease rapidly over the next several years.   EGR 312 - 03

Arithmetic Gradient Factors (P/G, A/G) Cash flows that increase or decrease by a constant amount are considered arithmetic gradient cash flows. The amount of increase (or decrease) is called the gradient. $2000 $175 $1500   $150 $1000 $125 $500 $100   0 1 2 3 4 0 1 2 3 4 G = $25 Base = $100 G = -$500 Base = $2000 EGR 312 - 03

Arithmetic Gradient Factors (P/G, A/G) Equivalent cash flows: => + Note: the gradient series by convention starts in year 2. $175 $150 $75 $125 $100 $100 $50 $25   0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 G = $25 Base = $100   EGR 312 - 03

Arithmetic Gradient Factors (P/G, A/G) To find P for a gradient cash flow that starts at the end of year 2 and end at year n: or P = G(P/G,i,n) where (P/G,i,n) = $nG $2G $G 0 1 2 3 … n $P   EGR 312 - 03

Arithmetic Gradient Factors (P/G, A/G) To find P for the arithmetic gradient cash flow: P1 = _____________ P2 = _____________ P = Base(P/A, i, n) + G(P/G, i, n) = _________ $175 P = ? P2 = ? $150 P1 = ? $75 $125 $100 $100 $50  + $25   0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 i = 6% P1 = 100 (P/A, 6%, 4) = 100(3.4651) = 346.51 P2 = 25 (P/G, 6%, 4) = 25 (4.9455) = 123.64 P = Base(P/A, i, n) + G(P/G, i, n) = 346.51 + 123.64 = 470.15   EGR 312 - 03

Arithmetic Gradient Factors (P/G, A/G) To find P for the declining arithmetic gradient cash flow: P1 = _____________ P2 = _____________ P = Base(P/A, i, n) - G(P/G, i, n) = _________ P1 = ? $2000 $2000 P2 = ? $1500 $1500 $1000 $1000 -  $500 $500   0 1 2 3 4 0 1 2 3 4 0 1 2 3 4 i = 10% P1 = 2000(P/A, 10%, 4) =2000 (3.1699) =6339.80 P2 = 500( P/G, 10%, 4) = 2189.05 P = 4150.75   EGR 312 - 03

Arithmetic Gradient Factors (P/G, A/G) To find the uniform annual series, A, for an arithmetic gradient cash flow G: A = G(P/G,i,n) (A/P,i,4) = G(A/G,i,n) Where (A/G,i,n) = $nG $2G $G 0 1 2 3 … n 0 1 2 3 … n $A   EGR 312 - 03

Geometric Gradient Factors (Pg/A) A Geometric gradient is when the periodic payment is increasing (decreasing) by a constant percentage: A1 = $100, g = 0.1 A2 = $100(1+g) A3 = $100(1+g)2 An = $100(1+g)n-1 $133 $121 $110 $100   0 1 2 3 4   EGR 312 - 03

Geometric Gradient Factors (Pg/A) To find the Present Worth, Pg, for a geometric gradient cash flow G: $Pg $133 $121   $110 $100   0 1 2 3 4 EGR 312 - 03

Determining Unknown Interest Rate To find an unknown interest rate from a single-payment cash flow or uniform-series cash flow, the following methods can be used: Use of Engineering Econ Formulas Use of factor tables Spreadsheet (Excel) a) =IRR(first cell: last cell) b) =RATE(n,A,P,F)   EGR 312 - 03

Determining Unknown Interest Rate Example: The list price for a vehicle is stated as $25,000. You are quoted a monthly payment of $658.25 per month for 4 years. What is the monthly interest rate? What interest rate would be quoted (yearly interest rate)? Using factor table: $25000 = $658.25(P/A,i,48)  (P/A,i,48) = _________ i = ________ (HINT: start with table 1, pg. 727) 0r _______ annually 37.974 = (P/A,i,48) i = 1% from table 4, pg 730 0r 12 % annually   EGR 312 - 03

Determining Unknown Interest Rate Example (cont’d) Using formula: Use calculator solver or Excel trial and error method to find i.   EGR 312 - 03

Determining Unknown Number of Periods (n) To find an unknown number of periods for a single-payment cash flow or uniform-series cash flow, the following methods can be used: Use of Engineering Econ. Formulas. Use of factor tables Spreadsheet (Excel) a) =NPER(i%,A,P,F)   EGR 312 - 03

Determining Unknown Number of Periods (n) Example: Find the number of periods required such that an invest of $1000 at 5% has a future worth of $5000. P = F(P/F,5%,n) $1000 = $5000(P/F,5%,n) (P/F,5%,n) = ______________ n = ___________________ (P/F,5%,n) = 0.2 n ~ 33 periods EGR 312 - 03