Ch2 Lecture Notes

8/2/2019 Ch2 Lecture Notes

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Chapter 2

Factors:How Time and Interest Affect Money

ENGINEERING ECONOMY Sixth Edition

Blank and Tarquin


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Learning Objectives

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1. F/P and P/F Factors (Single Payment Factors)2. P/A and A/P Factors (Uniform Series Present Worth

Factor and Capital Recovery Factor)

3. F/A and A/F Factors (Sinking Fund Factor and Uniform-Series Compound Amount Factor)

4. P/G and A/G Factors (Arithmetic Gradient Factors)

5. Geometric Gradient

6. Calculate i (unknown interest rate)

7. Calculate n (number of years)


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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

0 1 2 3 n-1 n

P 0

Fn

i% / period


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Basic Derivation: F/P factor

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Fn = P 0(1+i) n (F/P, i%, n) factor:

Excel: =FV(i%, n, ,P)

P 0 = F n1/(1+i) n (P/F, i%, n) factor:

Excel: =PV(i%, n, , F)


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Derivation: F/P factor Find F given P F1 = P + Pi =P(1+i) F2 = F1 + F1i = F1(1+i)..or F

2= 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)

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Present Worth Factor from F/P

SinceFN = 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

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P/F factor discounting back in time

Discounting back from the future

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P

Fn

N.

P/F factor brings a single futuresum back to a specific point intime.


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Example- F/P Analysis

Example: P= $1,000;n=3;i=10% What is the future value, F?

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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


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Example P/F Analysis

Assume F = $100,000, 9 years from now. What is the presentworth of this amount now if i =15%?

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0 1 2 3 8 9

F9 = $100,000

P= ??

i = 15%/yr

P0 = $100,000(P/F, 15%,9) = $100,000(1/(1.15) 9)

= $100,000( 0.2843 ) = $28,426 at time t = 0


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Uniform Series Present Worth and Capital Recovery Factors


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Uniform Series Present Worth

This expression will convert an annuity cash flowto an equivalent present worth amount oneperiod to the left of the first annuity cash flow.

(1 ) 10

(1 )

n

n

iP A for i

i i

/ %,P A i n factor 4/15/2012 11


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Capital Recovery Factor (CRF) = A/P factor

Given the P/A factor

(1 ) 10

(1 )

n

n

iP A for i

i i

(1 )

(1 ) 1

n

n

i i A P

i

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Solve for A in terms of P

Yielding.

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.

CRF calculates the equivalent uniform annual worth Aover n years for a givenP in year 0, when the interest rate isi .


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Modified HW Problem 2.5

A maker of microelectromechanical systems [MEMS], believes it careduce product recalls if it purchases new software for detectingfaulty parts. The cost of the new software is $225,000.

How much would the company have to save each year for 4 years torecover its investment if it uses a minimum attractive rate of return o15% per year?(A/P, 15%, 4) p. 745 Table 19 (interest rate i=15%) column: A/P,row: n=4 A/P factor = 0.35027Company would have to save$225,000 x0.35027 = $78,810.75each year.

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HW Problem 2.12*

Comparison of Table with equation: V-Tek Systems is a manufacturer of vertical compactors, and it is examining i

cash flow requirements for the next 5 years. The company expects to replace officmachines and computer equipment at various times over the 5-year planninperiod. 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 presenworth of the planned expenditures at an interest rate of 10% per year?

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Sinking Fund and Series Compound Amount Factors(A/F and F/A)

Take advantage of what we already haveRecall:

Also:

1

(1 ) nP F

i

(1 )

(1 ) 1

n

n

i i A P

i

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Modern context of a Sinking Fund

In modern finance, a sinking fund is a method enabling an organization toset asidemoney over time to retire its indebtedness. More specifically, it is a fund into whichmoney can be deposited, so that over time its preferred stock, debentures or stockcan be retired. For the organization that is retiring debt, it has the benefit that thprincipal of the debt or at least part of it, will be available when due. For thcreditors, the fund reduces the risk the organization will default when the principadue.

In some US states, Michigan for example, school districts may ask the voters tapprove a taxation for the purpose of establishing a Sinking Fund. The StatTreasury Department has strict guidelines for expenditure of fund dollars with tpenalty for misuse being an eternal ban on ever seeking the tax levy again.

Historical Context: ASinking Fundwas a device used in Great Britain in the 18thcentury to reduce national debt.

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HW Problem 2.23

Southwestern Moving and Storage wants to have enoughmoney to purchase a new tractor-trailer in 3 years. If theunit will cost $250,000, how much should the companyset 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 = 0.30505A= $250,000 x 0.30505 = $76262.50

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Interpolation (Estimation Process)

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At times, a set of interest tables may not have the exactinterest factor needed for an analysis

One may be forced to interpolate between two tabulatedvalues

Linear Interpolation is not exact because:

The functional relationships of the interest factors arenon-linear functions

From 2-5% error may be present with interpolation.


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Example 2.7

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Assume you need the value of the A/P factor for i = 7.3% andn = 10 years.

7.3% is likely not a tabulated value in most interest tables

So, one must work withi = 7% and i = 8%for n fixed at 10 Proceed as follows:


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Basic Setup for Interpolation

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Work with the following basic relationships


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Arithmetic Gradient Factors

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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 Woof an arithmetic gradient


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Linear Gradient Example

Assume the following:

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0 1 2 3 n-1 N

A1+G

A1+2G

A1+n-2G

A1+n-1G

This represents a positive, increasing arithmetic gradient


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Example: Linear Gradient

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Arithmetic Gradient Factors

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The G amount is the constant arithmetic change fromone time period to the next.

The G amount may be positive or negative.

The present worth point is always one time period to theleft of the first cash flow in the series or,

Two periods to the left of the first gradient cash (G) flow.


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Present Worth Point

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0 1 2 3 4 5 6 7

$100$200

$300

$400

$500

$600

$700

X

The Present Worth Point of theGradient


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Present Worth: Linear Gradient

The present worth of a linear gradient is the presentworth of the two components:

1. The Present Worth of theGradientComponentand,

2. The Present Worth of theBase Annuityflow

Requires 2 separate calculations.

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Present Worth: Gradient Component(Example 2.10 *)

Three contiguous counties in Florida have agreed to pool taxresources already designated for county-maintained bridgerefurbishment. At a recent meeting, the county engineers estimatethat a total of $500,000will be deposited at the end of next year into

an account for the repair of old and safety-questionable bridgethroughout the three-county area. Further, they estimate that thedeposits willincrease by $100,000 per year for only9 yearsthereafter, then cease.

Determine the equivalent (a) present worth and (b) annual seriesamounts if county funds earn interest at a rate of 5% per year .

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Example 2.10 (b) *

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Determine the equivalent annual series amounts if county funds earn interest at a rate of 5% per year .


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Equations for P/G and A/G

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Geometric Gradients

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An arithmetic (linear) gradient changes by a fixed dollaramount each time period.

A GEOMETRIC gradient changes by a fixed percentage

each time period.We define a UNIFORM RATE OF CHANGE (%) for eachtime period

Define g as the constant rate of change in decimalform by which amounts increase or decrease from oneperiod to the next


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cash flow diagrams for geometric gradient series

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Geometric Gradients: Increasing

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Typical Geometric Gradient Profile

Let A1 = the first cash flow in the series

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

The series starts in year 1 at an initial amount A1, not considered a base

amount as in the arithmetic gradient.


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Geometric Gradients

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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


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Pg /A Equation

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In summary, the engineering economy relation and factor formulas to calculate Pg in period t = 0 for a geometricgradient series starting in period 1 in the amount A1 and

increasing by a constant rate of g each period are


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Engineers at SeaWorld, a division of Busch Gardens, Inc., havecompleted an innovation on an existing water sports ride tomake it more exciting. The modification costs only $8000 andis expected to last 6 years with a $1300 salvage value for the

solenoid mechanisms. The maintenance cost is expected to behigh at $1700 the first year , increasing by 11% per yearthereafter. Determine the equivalent present worth of themodification and maintenance cost. The interest rate is 8%

per year.

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$1700 $1700(1.11) 1

$1700(1.11) 2$1700(1.11) 3

$1700(1.11) 5

0 1 2 3 4 5 6

PW(8%) = ??

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 worthof the future expenses at time t = 0.

First, draw a cash flow diagram to represent the model.

continued


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Cash flow diagram

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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

continued


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continued *


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i rate is unknown

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A class of problems may deal with all of theparametersknowexcept the interest rate.

For many application-type problems, this can become adifficult task

Termed, rate of return analysis In some cases:

i can easily be determined

In others,trial and error must be used


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Example: i unknown

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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 twocash flows?

0 1 2 3 4 5

$3,000

$5,000

F = P(1+i) n (1+i) 5 = 5,000/3000 = 1.6667

(1+i) = 1.6667 0.20

i = 1.1076 1 = 0.1076 = 10.76%


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Unknown Number of Years

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Some problems require knowing the number of time periodsrequired given the other parameters

Example:

How long will it take for $1,000 to doublein value if the discount rateis 5% per year?

Draw the cash flow diagram as.

0 1 2 . . . . . . . n

P = $1,000

Fn = $2000

i = 5%/year; n is unknown!


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Unknown Number of Years

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Solving we have..

0 1 2 . . . . . . . n

P = $1,000

Fn = $2000

(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


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Formulas and factors derived and applied in this chapter performequivalence calculations for present, future, annual, and gradientcash flows. Capability in using these formulas and their standardnotation manually and with spreadsheets is critical to complete an

engineering economy study. Using these formulas and spreadsheet functions, you can convert

single cash flowsintouniform cash flows, gradients into presentworths, and much more.

You can solve for rate of return i or time n. A thorough understanding of how to manipulate cash flows using th

material in this chapter will help you address financial questions inprofessional practice as well as in everyday living.

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WHAT A DIFFERENCE THE YEARS AND COMPOUINTEREST CAN MAKE

Real World Situation -Manhattan Island purchase.It is reported that Manhattan Island in New York wapurchased for the equivalent of $24 in the year 1626. In the year 2001, the 375th anniversary of thepurchase 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)

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Ch2 Lecture Notes