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## CHAPTER 2

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**CHAPTER 2**Factors: How Time and Interest Affect Money**1. Foundations: Overview**• F/P and P/F Factors • P/A and A/P Factors • F/A and A/F Factors • Interpolate Factor Values • P/G and A/G Factors • Geometric Gradient • Calculate i • Calculate “n” • Spreadsheets**CHAPTER 2: Section 1**F/P and P/F Factors**Fn**…………. n P0 2.1 Basic Derivations: F/P factor • F/P FactorTo find F given P To Find F given P Compound forward in time**2.1 Derivation by Recursion: F/P factor**• F1 = P(1+i) • F2 = F1(1+i)…..but: • F2 = P(1+i)(1+i) = P(1+i)2 • F3 =F2(1+i) =P(1+i)2 (1+i) = P(1+i)3 In general: Fn = P(1+i)n Fn = P(F/P,i%,n)**2.1 Present Worth Factor from F/P**• Since Fn = P(1+i)n • We solve for P in terms of FN • P = F{ 1/ (1+i)n} = F(1+i)-n • Thus: P = F(P/F,i%,n) where (P/F,i%,n) = (1+i)-n • Thus, the two factors are: • F = P(1+i)n finds the future worth of P; • P = F(1+i)-n finds the present worth from F**Fn**…………. n P 2.1 P/F factor – discounting back in time • Discounting back from the future P/F factor brings a single future sum back to a specific point in time.**F = ??**0 1 2 3 P=$1,000 i=10%/year Example 2.1 - F/P Analysis • Example: P= $1,000;n=3;i=10% • What is the future value, F? F3 = $1,000[F/P,10%,3] = $1,000[1.10]3 = $1,000[1.3310] = $1,331.00**F9 = $100,000**………… 0 1 2 3 8 9 P= ?? Example 2.2 – P/F Analysis • Assume F = $100,000, 9 years from now. What is the present worth of this amount now if i =15%? i = 15%/yr P0 = $100,000(P/F, 15%,9) = $100,000(1/(1.15)9) = $100,000(0.2843) = $28,430 at time t = 0**CHAPTER 2: Section 2**P/A and A/P Factors**P = ??**………….. 1 2 3 .. .. n-1 n 0 2.2 Uniform Series Present Worth and Capital Recovery Factors • Annuity Cash Flow $A per period**P = ??**0 1 2 3 n-1 n A = given 2.2 Uniform Series Present Worth and Capital Recovery Factors • Desire an expression for the present worth – P of a stream of equal, end of period cash flows - A**2.2 Uniform Series Present Worth and Capital Recovery**Factors • Write a Present worth expression [1] Term inside the brackets is a geometric progression. Mult. This equation by 1/(1+i) to yield a second equation**2.2 Uniform Series Present Worth and Capital Recovery**Factors • The second equation [2] To isolate an expression for P in terms of A, subtract Eq [1] from Eq. [2]. Note that numerous terms will drop out.**2.2 Uniform Series Present Worth and Capital Recovery**Factors • Setting up the subtraction [2] - [1] = [3]**2.2 Uniform Series Present Worth and Capital Recovery**Factors • Simplifying Eq. [3] further n**2.2 Uniform Series Present Worth and Capital Recovery**Factors • This expression will convert an annuity cash flow to an equivalent present worth amount one period to the left of the first annuity cash flow.**2.2 Capital Recovery Factor A/P, i%, n**The present worth point of an annuity cash flow is always one period to the left of the first A amount • Given the P/A factor Solve for A in terms of P Yielding…. A/P,i%,n factor**CHAPTER 2: Section 3**F/A and A/F Factors**$F**………….. N 2.3 F/A and A/F Derivations • Annuity Cash Flow 0 Find $A given the Future amt. - $F $A per period**2.3 Sinking Fund and Series Compound amount factors (A/F and**F/A) • Take advantage of what we already have • Recall: • Also: Substitute “P” and simplify!**2.3 A/F Factor**• By substitution we see: • Simplifying we have: • Which is the (A/F,i%,n) factor**2.3 F/A factor from the A/F Factor**• Given: • Solve for F in terms of A**$F**………….. N 2.3 F/A and A/F Derivations • Annuity Cash Flow 0 Find $F given the $A amounts $A per period**2.3 Example 2.5**• Formosa Plastics has major fabrication plants in Texas and Hong Kong. • It is desired to know the future worth of $1,000,000 invested at the end of each year for 8 years, starting one year from now. • The interest rate is assumed to be 14% per year.**2.3 Example 2.5**• A = $1,000,000/yr; n = 8 yrs, i = 14%/yr • F8 = ??**2.3 Example 2.5**Solution: The cash flow diagram shows the annual payments starting at the end of year 1 and ending in the year the future worth is desired. Cash flows are indicated in $1000 units. The F value in 8 years is F = l000(F/A,14%,8) =1000( 13.23218) = $13,232.80 = 13.232 million 8 years from now.**2.3 Example 2.6**• How much money must Carol deposit every year starting, l year from now at 5.5%per year in order to accumulate $6000 seven years from now?**2.3 Example 2.6**• Solution • The cash How diagram from Carol's perspective fits the A/F factor. • A= $6000 (A/F,5.5%,7) = 6000(0.12096) = $725.76 per year • The A/F factor Value 0f 0.12096 was computed using the A/F factor formula**CHAPTER 2: Section 4**Interpolation in Interest Tables**2.4 Interpolation of Factors**• All texts on Engineering economy will provide tabulated values of the various interest factors usually at the end of the text in an appendix • Refer to the back of your text for those tables.**2.4 Interpolation of Factors**• Typical Format for Tabulated Interest Tables**2.4 Interpolation (Estimation Process)**• At times, a set of interest tables may not have the exact interest factor needed for an analysis • One may be forced to interpolate between two tabulated values • Linear Interpolation is not exact because: • The functional relationships of the interest factors are non-linear functions • Hence from 2-5% error may be present with interpolation.**2.4 An Example**• Assume you need the value of the A/P factor for i = 7.3% and n = 10 years. • 7.3% is most likely not a tabulated value in most interest tables • So, one must work with i = 7% and i = 8% for n fixed at 10 (Tables 12 & 13) • Proceed as follows:**2.4 Basic Setup for Interpolation**• Work with the following basic relationships**2. 4 i = 7.3% using the A/P factor**• For 7% we would observe: (A/P,7%,10) = 0.14238**2. 4 i = 7.3% using the A/P factor**• For i = 8% we observe: (A/P,8%,10) = 0.14903**2. 4 Estimating for i = 7.3%**• Form the following relationships**2.4 Final Estimated Factor Value**• Observe for i increasing from 7% to 8% the A/P factors also increases. • One then adds the estimated increment to the 7% known value to yield:**2.4. The Exact Value for 7.3%**• Using a previously programmed spreadsheet model the exact value for 7.3% is:**CHAPTER 2 Section 5**P/G and A/G Factors**2.5 Arithmetic Gradient Factors**• In applications, the annuity cash flow pattern is not the only type of pattern encountered • Two other types of end of period patterns are common • The Linear or arithmetic gradient • The geometric (% per period) gradient • This section presents the Arithmetic Gradient**2.5 Arithmetic Gradient Factors**• An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a constant amount over n time periods. • A linear gradient is always comprised of TWO components:**2.5 Arithmetic Gradient Factors**• The Two Components are: • The Gradient component • The base annuity component • The objective is to find a closed form expression for the Present Worth of an arithmetic gradient**A1+(n-1)G**A1+(n-2)G A1+2G A1+G 0 1 2 3 n-1 N 2.5 Linear Gradient Example • Assume the following: This represents a positive, increasing arithmetic gradient**2.5 Example: Linear Gradient**• Typical Negative, Increasing Gradient: G=$50 The Base Annuity = $1500**2.5 Example: Linear Gradient**• Desire to find the Present Worth of this cash flow The Base Annuity = $1500**2.5 Arithmetic Gradient Factors**• The “G” amount is the constant arithmetic change from one time period to the next. • The “G” amount may be positive or negative! • The present worth point is always one time period to the left of the first cash flow in the series or, • Two periods to the left of the first gradient cash flow!**(n-1)G**(n-2)G +2G G 0 1 2 3 n-1 N 2.5 Derivation: Gradient Component Only • Focus Only on the gradient Component “0” G Removed Base annuity**2.5 Present Worth Point…**• The Present worth point of a linear gradient is always: • 2 periods to the left of the “1G” point or, • 1 period to the left of the very first cash flow in the gradient series. DO NOT FORGET THIS!