Engineering Economy Factors

8/10/2019 Engineering Economy Factors

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Chapter 2

Factors:

How Time and Interest Affect Money


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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 WorthFactor 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 Diagrambasic format

0 1 2 3 n-1 n

P0

Fn

i% / period


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Basic Derivation: F/P factor

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Fn= P0(1+i)n(F/P, i%, n) factor:

P0= Fn1/(1+i)n(P/F, i%, n) factor:


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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

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)

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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

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P/F factordiscounting 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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ExampleP/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 flow

to an equivalent present worth amount one

period to the left of the first annuity cash flow.

(1 ) 1 0

(1 )

n

n

iP A for i

i i

/ %,P A i n factor

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Capital Recovery Factor (CRF) = A/P factor

Given the P/A factor

(1 ) 1 0

(1 )

n

n

iP A for i

i i

(1 )

(1 ) 1

n

n

i iA 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 firstA amount.

CRF calculates the equivalent uniform annual worthA over n years for a given Pinyear 0, when the interest rate is i.


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Example

A maker of micro-electromechanical systems [MEMS], believes it canreduce 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 of15% per year?

(A/P, 15%, 4)Using Tables (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 eachyear.

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Example

V-Tek Systems is a manufacturer of vertical compactors, and it is examining itscash flow requirements for the next 5 years. The company expects to replace officemachines and computer equipment at various times over the 5-year planningperiod. 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?

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

Take advantage of what we already have

Recall:

Also:

1

(1 )nP F

i

(1 )

(1 ) 1

n

n

i iA P

i

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Modern context of a Sinking Fund

In modern finance, a sinking fund is a method enabling an organization to set 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 stockscan be retired. For the organization that is retiring debt, it has the benefit that theprincipal 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 isdue.

In some US states, Michigan for example, school districts may ask the voters toapprove a taxation for the purpose of establishing a Sinking Fund. The StateTreasury 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 Fundwas a device used in Great Britain in the 18thcentury to reduce national debt.

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Example

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

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Arithmetic Gradient Factors

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An arithmetic (linear) Gradient is a cash flow series that eitherincreases 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


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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 the GradientComponentand,

2. The Present Worth of the Base Annuity flow

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 estimatedthat a total of $500,000 will be deposited at the beginning of next

year into an account for the repair of old and safety-questionablebridges throughout the three-county area. Further, they estimate thatthe deposits will increase by $100,000 per year for only 9 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 ifcounty 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 decimal

form 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

LetA1= the first cash flow in the series

0 1 2 3 4 .. n-1 n

A1

A1(1+g)A1(1+g)

2A1(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 periodi

The constant rate of changeg

No. of time periodsn

The starting cash flowA1


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Pg /A Equation

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In summary, the engineering economy relation and factorformulas to calculate Pgin 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., 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 behigh 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.

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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 PTis

continued


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continued *


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Payment Periods (PP), Interest Periods (IP) and

Compounding Periods (CP)

PPhow often cash flows occur IPhow often the interest is incurred CPhow often interest in compounded If PP = CP, no problem concerning effective i rate

PP IP or PP CP or IP CPThen there is a problem


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Payment Periods (PP), Interest Periods (IP) and

Compounding Periods (CP)

If MARR of the cash flow below is 6% per monthcompounded semiannually then:


PP=1 month, IP=1 month, CP=6 months


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Nominal and Effective interest Rates


In an effective interest rate: IP = CP.

All factors are using effective interest rate.

If IP CP then this is a nominal interest rate.


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Nominal and Effective interest Rates


Important rules:

If nominal interest rate is divided or multiplied by anynumber the value of the interest and the IP period are

multiplied or divided. Example:

6% per month compounded semiannually x 6 =

36 per semi annual compounded semi annually.


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Effective Interest Rate Formula

i = effective rate per some stated period, e.g., quarterly,annually

r = nominal rate for same time period

m = frequency of compounding per same time period


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Effective Interest Rate

Example:Find i per year, if m = 4 for quarterlycompounding, and r = 12% per year

Stated period for i is YEAR

i = (1 + 0.12/4)4 - 1 = 12.55%

rEffective i = (1+ ) 1

m

m


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Nominal and Effective RatesNominal

r = rate/period periods

Example:Rate is 1.5% per month.Determine nominal rate perquarter, year, and over 2 years

Qtr: r = 1.5 3 mth = 4.5%

Year: r = 1.5 12 mth = 18%

= 4.5 4 qtr = 18%

2 yrs: r =1.5

24 mth = 36%= 18 2 yrs = 36%

Effective

Example:Credit card rate is 1.5% per monthcompounded monthly. Determine effectiverate per quarter and per year

Period is quarter:

r = 1.5 3 mth = 4.5%

m = 3

i = (1 + 0.045/3)31 = 4.57%per quarter

Period is year: r = 18% m = 12

i = (1 + 0.18/12)12- 1) = 19.6%per year

rEffective i = (1+ ) 1m

m


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Effective Interest Rate

As m , continuous compounding is approached

effective i = (r1)

Example:r = 14% per year compoundedcontinuously

i = ( 0.14- 1) = 15.03% per year


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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 with i = 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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irate is unknown

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A class of problems may deal with all of the parametersknowexcept 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


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Example: i unknown

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Assume one can invest $3000 now in a venture in anticipation ofgaining $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) = 1.66670.20

i = 1.10761 = 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 double in 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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WHAT A DIFFERENCE THE YEARS AND COMPOUNDINTEREST CAN MAKE

Real World Situation - Manhattan Island purchase.

It is reported that Manhattan Island in New York waspurchased 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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Engineering Economy Factors