Factors: How Time and Interest Affect Money ENGINEERING ECONOMY Sixth Edition Blank and Tarquin Chapter 2 Factors: How Time and Interest Affect Money

Learning Objectives F/P and P/F Factors (Single Payment Factors) P/A and A/P Factors (Uniform Series Present Worth Factor and Capital Recovery Factor) F/A and A/F Factors (Sinking Fund Factor and Uniform-Series Compound Amount Factor) P/G and A/G Factors (Arithmetic Gradient Factors) Geometric Gradient Calculate i (unknown interest rate) Calculate “n” (number of years) 11/14/2018

Single Payment Factors (F/P and P/F) Objective: Derive factors to determine the present or future worth of a cash flow Cash Flow Diagram – basic format Fn i% / period 0 1 2 3 n-1 n P0

Basic Derivation: F/P factor Fn = P0(1+i)n →(F/P, i%, n) factor: Excel: =FV(i%, n, ,P) P0 = Fn1/(1+i)n →(P/F, i%, n) factor: Excel: =PV(i%, n, , F) 11/14/2018

Derivation: F/P factor Find F given P F1 = P + Pi = P(1+i) F2 = F1 + F1i = F1(1+i)…..or F2 = P(1+i) + P(1+i)i = P(1+i)(1+i) = P(1+i)2 F3 = F2+ F2 i = 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) 11/14/2018

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 P = F(P/F,i%,n) where (P / F, i%, n) = (1+i)-n 11/14/2018

P/F factor – discounting back in time Discounting back from the future P Fn N …………. P/F factor brings a single future sum back to a specific point in time. 11/14/2018

What is the future value, F? Example- F/P Analysis Example: P= $1,000;n=3;i=10% What is the future value, F? 0 1 2 3 P=$1,000 F = ?? i=10%/year F3 = $1,000 [F/P,10%,3] = $1,000[1.10]3 = $1,000[1.3310] = $1,331.00 11/14/2018

Example – P/F Analysis Assume F = $100,000, 9 years from now. What is the present worth of this amount now if i =15%? F9 = $100,000 i = 15%/yr 0 1 2 3 8 9 ………… P= ?? P0 = $100,000(P/F, 15%,9) = $100,000(1/(1.15)9) = $100,000(0.2843) = $28,426 at time t = 0 11/14/2018

Uniform Series Present Worth and Capital Recovery Factors 11/14/2018

Uniform Series Present Worth 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. 11/14/2018

Capital Recovery Factor (CRF) = A/P factor CRF calculates the equivalent uniform annual worth A over n years for a given P in year 0, when the interest rate is i. 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 11/14/2018

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Modified HW Problem 2.5 A maker of microelectromechanical systems [MEMS], believes it can reduce product recalls if it purchases new software for detecting faulty parts. The cost of the new software is $225,000. How much would the company have to save each year for 4 years to recover its investment if it uses a minimum attractive rate of return of 15% per year? (A/P, 15%, 4) p. 745 – Table 19 (interest rate i=15%)  column: A/P, row: n=4 A/P factor = 0.35027 Company would have to save $225,000 x 0.35027 = $78,810.75 each year. 11/14/2018

HW Problem 2.12* Comparison of Table with equation: V-Tek Systems is a manufacturer of vertical compactors, and it is examining its cash flow requirements for the next 5 years. The company expects to replace office machines and computer equipment at various times over the 5-year planning period. Specifically, the company expects to spend $9000 two years from now, $8000 three years from now, and $5000 five years from now. What is the present worth of the planned expenditures at an interest rate of 10% per year? 11/14/2018

Sinking Fund and Series Compound Amount Factors (A/F and F/A) Take advantage of what we already have Recall: Also: 11/14/2018

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Modern context of a Sinking Fund In modern finance, a sinking fund is a method enabling an organization to set aside money over time to retire its indebtedness. More specifically, it is a fund into which money can be deposited, so that over time its preferred stock, debentures or stocks can be retired. For the organization that is retiring debt, it has the benefit that the principal of the debt or at least part of it, will be available when due. For the creditors, the fund reduces the risk the organization will default when the principal is due. In some US states, Michigan for example, school districts may ask the voters to approve a taxation for the purpose of establishing a Sinking Fund. The State Treasury Department has strict guidelines for expenditure of fund dollars with the penalty for misuse being an eternal ban on ever seeking the tax levy again. Historical Context: A Sinking Fund was a device used in Great Britain in the 18th century to reduce national debt. 11/14/2018

HW Problem 2.23 Southwestern Moving and Storage wants to have enough money to purchase a new tractor-trailer in 3 years. If the unit will cost $250,000, how much should the company set aside each year if the account earns 9% per year? (A/F, 9%, 3) or n = 3, F = $250,000, i = 9% Using Table 14 (pg 740), the A/F = 0.30505 A= $250,000 x 0.30505 = $76262.50 11/14/2018

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 From 2-5% error may be present with interpolation. 11/14/2018

7.3% is likely not a tabulated value in most interest tables Example 2.7 Assume you need the value of the A/P factor for i = 7.3% and n = 10 years. 7.3% is likely not a tabulated value in most interest tables So, one must work with i = 7% and i = 8% for n fixed at 10 Proceed as follows: 11/14/2018

Basic Setup for Interpolation Work with the following basic relationships 11/14/2018

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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 always has TWO components: The gradient component The base annuity component The objective is to find a closed form expression for the Present Worth of an arithmetic gradient 11/14/2018

Linear Gradient Example A1+n-1G A1+n-2G Assume the following: A1+2G A1+G 0 1 2 3 n-1 N This represents a positive, increasing arithmetic gradient 11/14/2018

Example: Linear Gradient 11/14/2018

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 (G) flow. 11/14/2018

The Present Worth Point of the $700 $600 $500 $400 $300 $200 $100 X 0 1 2 3 4 5 6 7 The Present Worth Point of the Gradient 11/14/2018

Present Worth: Linear Gradient The present worth of a linear gradient is the present worth of the two components: 1. The Present Worth of the Gradient Component and, 2. The Present Worth of the Base Annuity flow Requires 2 separate calculations. 11/14/2018

Present Worth: Gradient Component (Example 2.10*) Three contiguous counties in Florida have agreed to pool tax resources already designated for county-maintained bridge refurbishment. At a recent meeting, the county engineers estimated that a total of $500,000 will be deposited at the end of next year into an account for the repair of old and safety-questionable bridges throughout the three-county area. Further, they estimate that the deposits will increase by $100,000 per year for only 9 years thereafter, then cease. Determine the equivalent (a) present worth and (b) annual series amounts if county funds earn interest at a rate of 5% per year. 11/14/2018

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Example 2.10 (b)* Determine the equivalent annual series amounts if county funds earn interest at a rate of 5% per year. 11/14/2018

Equations for P/G and A/G 11/14/2018

Geometric Gradients An arithmetic (linear) gradient changes by a fixed dollar amount each time period. A GEOMETRIC gradient changes by a fixed percentage each time period. We define a UNIFORM RATE OF CHANGE (%) for each time period Define “g” as the constant rate of change in decimal form by which amounts increase or decrease from one period to the next 11/14/2018

cash flow diagrams for geometric gradient series 11/14/2018

Geometric Gradients: Increasing Typical Geometric Gradient Profile Let A1 = the first cash flow in the series The series starts in year 1 at an initial amount A1, not considered a base amount as in the arithmetic gradient. 0 1 2 3 4 …….. n-1 n A1 A1(1+g) A1(1+g)2 A1(1+g)3 A1(1+g)n-1 11/14/2018

Geometric Gradients For a Geometric Gradient the following parameters are required: The interest rate per period – i The constant rate of change – g No. of time periods – n The starting cash flow – A1 11/14/2018

Pg /A Equation In summary, the engineering economy relation and factor formulas to calculate Pg in period t = 0 for a geometric gradient series starting in period 1 in the amount A1 and increasing by a constant rate of g each period are 11/14/2018

Engineers at SeaWorld, a division of Busch Gardens, Inc Engineers at SeaWorld, a division of Busch Gardens, Inc., have completed an innovation on an existing water sports ride to make it more exciting. The modification costs only $8000 and is expected to last 6 years with a $1300 salvage value for the solenoid mechanisms. The maintenance cost is expected to be high at $1700 the first year, increasing by 11% per year thereafter. Determine the equivalent present worth of the modification and maintenance cost. The interest rate is 8% per year. 11/14/2018

continued Assume maintenance costs will be $1700 one year from now. Assume an annual increase of 11% per year over a 6-year time period. If the interest rate is 8% per year, determine the present worth of the future expenses at time t = 0. First, draw a cash flow diagram to represent the model. 0 1 2 3 4 5 6 PW(8%) = ?? $1700 $1700(1.11)1 $1700(1.11)2 $1700(1.11)3 $1700(1.11)5 11/14/2018

continued Cash flow diagram Solution: The cash flow diagram shows the salvage value as a positive cash flow and all costs as negative. Use Equation [2.24] for g ≠ i to calculate Pg. The total PT is 11/14/2018

* continued 11/14/2018

i – rate is unknown A class of problems may deal with all of the parameters know except the interest rate. For many application-type problems, this can become a difficult task Termed, “rate of return analysis” In some cases: i can easily be determined In others, trial and error must be used 11/14/2018

Example: i unknown Assume one can invest $3000 now in a venture in anticipation of gaining $5,000 in five (5) years. If these amounts are accurate, what interest rate equates these two cash flows? $5,000 0 1 2 3 4 5 F = P(1+i)n (1+i)5 = 5,000/3000 = 1.6667 (1+i) = 1.66670.20 i = 1.1076 – 1 = 0.1076 = 10.76% $3,000 11/14/2018

Unknown Number of Years Some problems require knowing the number of time periods required given the other parameters Example: How long will it take for $1,000 to double in value if the discount rate is 5% per year? Draw the cash flow diagram as…. Fn = $2000 i = 5%/year; n is unknown! 0 1 2 . . . . . . ……. n P = $1,000 11/14/2018

Unknown Number of Years P = $1,000 Fn = $2000 Solving we have….. (1.05)x = 2000/1000 X ln(1.05) =ln(2.000) X = ln(1.05)/ln(2.000) X = 0.6931/0.0488 = 14.2057 yrs With discrete compounding it will take 15 years 11/14/2018

You can solve for rate of return i or time n. Formulas and factors derived and applied in this chapter perform equivalence calculations for present, future, annual, and gradient cash flows. Capability in using these formulas and their standard notation manually and with spreadsheets is critical to complete an engineering economy study. Using these formulas and spreadsheet functions, you can convert single cash flows into uniform cash flows, gradients into present worths, and much more. You can solve for rate of return i or time n. A thorough understanding of how to manipulate cash flows using the material in this chapter will help you address financial questions in professional practice as well as in everyday living. 11/14/2018

WHAT A DIFFERENCE THE YEARS AND COMPOUND INTEREST CAN MAKE Real World Situation - Manhattan Island purchase. It is reported that Manhattan Island in New York was purchased for the equivalent of $24 in the year 1626. In the year 2001, the 375th anniversary of the purchase of Manhattan was recognized. F = P (1+i)n = 24 (1+ 0.06)382 = 111,443,000,000 (2008) F = P + Pin = 24 + 24(0.06)382 = $550.08 (simple interest) 11/14/2018