review ch_02 annuity

7/31/2019 Review Ch_02 Annuity

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ENGINEERING ECONOMY (GE401)

CHAPTER 2

Factors: How Time andInterest Affect Money

Foundations: Overview

1. F/P and P/F Factors2. P/A and A/P Factors3. F/A and A/F Factors4. Interpolate Factor Values

5. P/G and A/G Factors6. Geometric Gradient7. Calculate i8. Calculate n

F/P and P/F Factors

Basic Derivations: F/P factorF/P Factor To find F given P

Derivation by Recursion: F/P factorF1 = P(1+i)F2 = F1 (1+i)..but:

F2 = P(1+i)(1+i) = P(1+i)2F3 =F2(1+i) =P(1+i) 2 (1+i)= P(1+i) 3

In general:Fn = P(1+i)nFn = P(F/P,i%,n)Present Worth Factor from F/PSince Fn = P(1+i)nWe solve for P in terms of FNP = F{ 1/ (1+i)n} = F(1+i)-nThus:P = F(P/F,i%,n) where


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(P/F,i%,n) = (1+i)-n

Thus, the two factors are:

1. F = P(1+i)n finds the future worth of P;

2. P = F(1+i)-nfinds the present worth from F

P/F factordiscounting back in timeDiscounting back from the future

Example- F/P Analysis

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

Example P/F AnalysisAssume F = $100,000, 9 years from now.What is the present worth of this amount nowif i=15%?


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P/A and A/P Factors

Uniform Series Present Worth and Capital Recovery FactorsAnnuity Cash Flow

Uniform Series Present Worth and Capital Recovery FactorsDesire an expression for the present worth P of a stream ofequal, end of period cash flows A

Uniform Series Present Worth and Capital Recovery FactorsWrite a Present worth expression


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Uniform Series Present Worth and Capital Recovery FactorsThe second equation

Uniform Series Present Worth and Capital Recovery Factors

Setting up the subtraction


Simplifying Eq. [3] further


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This expression will convert an annuity cash flow to anequivalent present worth amount one period to the left of thefirst annuity cash flow.

Capital Recovery Factor A/P, i%, n

F/A and A/F Factors

F/A and A/F Derivations


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Sinking Fund and Series Compound amount factors (A/F andF/A)

A/F Factor

F/A factor from the A/F Factor


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F/A and A/F Derivations

Example:Formosa Plastics has majorfabrication plants in Riyadh and in Jaddh. It is desired toknow the future worth of $1,000,000 invested at the end ofeach year for 8 years, starting one year from now.The interest rate is assumed to be 14% per year.

Sol. Example:

A = $1,000,000/yr; n = 8 yrs, i = 14%/yrF8 = ??

Solution ofExample


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The cash flow diagram shows the annual payments starting atthe end of year 1 and ending in the year the future worth isdesired.

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 yearsfrom now.

Example:How much money must Carol deposit every year starting, l year fromnow at 5.5% per year in order to accumulate $6000 seven years fromnow?

Solution ofExample

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 yearThe A/F factor Value 0f 0.12096 was computed using the A/F factor formula

Interpolation in Interest Tables

Interpolation of Factors All texts on Engineering economy will provide tabulatedvalues of the various interest factors usually at the end of thetext in an appendix

Refer to the back of your text for those tables.

Interpolation of Factors

Typical Format for Tabulated Interest Tables


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Interpolation (Estimation Process)

At times, a set of interest tables may not have the exact interest factorneeded 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-linearfunctions

Hence from 2-5% error may be presentwith interpolation.

An Example

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

7.3% is most likely not a tabulated value in most interesttables

So, one must work with i = 7% and i = 8% for n fixed at 10

Proceed as follows

Basic Setup for InterpolationWork with the following basic relationships


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i = 7.3% using the A/P factor

For 7% we would observe:

COMPOUND PRESENT SINKING COMPOUND CAPITAL IN AMT. FACTORWORTH FUND AMOUNT RECOVERY F/P P/F A/F F/A A/P10 1.9672 0.5083 0.0724 13.8164 0.14238

i = 7.3% using the A/P factor

For i = 8% we observe:

COMPOUND PRESENT SINKING COMPOUND CAPITALN AMT. FACTOR WORTH FUND AMOUNT RECOVERYF/P P/F A/F F/A A/P10 2.1589 0.4632 0.0690 14.4866 0.14903

Estimating for i = 7.3% Form the following relationships


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Final Estimated Factor Value

Observe for i increasing from 7% to 8% the A/P factors alsoincreases.

One then adds the estimated increment to the 7% known value toyield:

The Exact Value for 7.3%

Using a previously programmed spreadsheet model theexact value for 7.3% is:

P/G and A/G Factors

Arithmetic Gradient Factors

In applications, the annuity cash flow pattern is not the onlytype 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


An arithmetic (linear) Gradient is a cash flow series that eitherincreases or decreases by a constant amount over n time periods.

A linear gradient is always comprised of TWO components:



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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

Linear Gradient Example

Example: Linear GradientTypical Negative, Increasing Gradient:G=$50

Example: Linear Gradient

Desire to find the Present Worth of this cash flow


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The G amount is the constant arithmetic change from onetime period to the next.

The G amount may be positive or negative!

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

Two periods to the left of the first gradient cash flow!

Derivation: Gradient Component OnlyFocus Only on the gradient Component

Present Worth Point


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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!Present Worth Point

Gradient Component

Present Worth Point

PW of the Base Annuity is at t = 0PWBASE Annuity=$100(P/A,i%,7)

Present Worth: Linear Gradient


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The present worth of a linear gradient is the present worth ofthe two components: 1. The Present Worth of the Gradient Component and, 2. The Present Worth of the Base Annuity flow Requires 2 separate calculations!

Present Worth: Gradient ComponentThe PW of the Base Annuity is simply the Base Annuity A{P/A, i%, n} factorWhat is needed is a present worth expression for the gradientcomponent cash flow.We need to derive a closed form expression for the gradientcomponent.

Present Worth: Gradient ComponentGeneral CF Diagram Gradient Part Only

To Begin- Derivation of P/G,i%,n

Next Step:Factor out G and re-write as ..

Factoring G out. P/G factor

What is inside of the { }s?

Replace (P/Fs) with closed-form


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Multiply both sides by (1+i)

Mult. Both Sides By (n+1)..

We have 2 equations [1] and [2].

Next, subtract [1] from [2] and work with the resultant equation.

Subtracting [1] from [2]..

The P/G factor for i and N


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Extension The A/G factorSome authors also include thederivation of the A/G factor.A/G converts a linear gradient to an equivalent annuity cashflow.Remember, at this point one is only working with gradient componentThere still remains the annuity component that you must also handleseparately!

The A/G Factor

Convert G to an equivalent A

How to do it

A/G factor using A/P with P/G


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The results follow..Resultant A/G factor

Gradient Example

Consider the following cash flow

Gradient Example- Base Annuity

First, The Base Annuity of $100/period

PW(10%) of the base annuity = $100(P/A,10%,5)

PWBase = $100(3.7908)= $379.08

Not Finished: We need the PW of the gradient componentand then add that value to the $379.08 amount

Focus on the Gradient Component


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We desire the PW of the GradientComponent at t = 0

The Set Up

PW of the Gradient Component

Calculating or looking up theP/G,10%,5 factor yields the following:

Pt=0 = $100(6.8618) = $686.18 for the gradient PW


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Gradient Example: Final Result

PW(10%)Base Annuity = $379.08

PW(10%)Gradient Component = $686.18

Total PW(10%) = $379.08 + $686.18

Equals $1065.26

Note: The two sums occur at t =0 and can be added together concept of equivalence

Example SummarizedThis Cash Flow

Shifted Gradient Example: i =10%

Consider the following Cash Flow

Shifted Gradient Example


1. This is a shifted negative, decreasing gradient.2. The PW point in time is at t = 3 (not t = o)

Shifted Gradient Example


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The base annuity is a $600 cash flow for 3 time periods

Shifted Gradient Example: Base Annuity

PW of the Base Annuity: 2 Steps

Shifted Gradient Example: Gradient

PW of Gradient Component: G = -$50

Geometric Gradient

Geometric Gradients

An arithmetic (linear) gradient changes by a fixed dollar

amount each time period.


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A GEOMETRIC gradient changes by a fixed percentage eachtime period.

We define a UNIFORM RATE OF CHANGE (%) for each timeperiod

Define g as the constant rate of change in decimal form by

which amounts increase or decrease from one period to thenext

Geometric Gradients: Increasing

Typical Geometric Gradient Profile

Let A1 = the first cash flow in the series

Geometric Gradients: Decreasing

Typical Geometric Gradient Profile

Let A1 = the first cash flow in the series

Geometric Gradients: Derivation

First Major Point to Remember:

A1 does NOT define a Base Annuity;

There is no BASE ANNUITY for a Geometric Gradient!

The objective is to determine the Present Worth one periodto the left of the A1 cash flow point in timeRemember: The PW point in time is one period to the left ofthe first cash flow A1!Geometric Gradients: Derivation

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

Geometric Gradients: Starting


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Pg = The Ajs time the respective (P/F,i,j) factor

Write a general present worthrelationship to find Pg.

Now, factor out the A1 value and rewrite as..

Geometric Gradients

Subtract (1) from (2) and the result is..

Geometric Gradients

Geometric Gradient P/A factor


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This is the (P/A,g,i,n) factor and is valid if g is not equal to i.

Geometric Gradient P/A factor

Note: If g = i we have a division by 0 undefined.

For g = i we can derive the closed form PW factor for this special case.

We substitute i for g into the Pg relationship to yield:

Geometric Gradient: i = g Case

Geometric Gradients: Summary

Geometric Gradient: Notes

The geometric gradient requiresknowledge of:

A1, i, n, and g

There exist an infinite number of combinations for i, n, and g: Hence

one will not find tabulated tables for the (P/A,g,i,n) factor.


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Geometric Gradient: Notes

You have to calculated either from the closed form for eachproblem or apply a pre-programmed spreadsheet model to find theneeded factor value

No spreadsheet built-in function for this factor!

Geometric Gradient: Example

Assume maintenance costs for a particular activity will be$1700 one year from now.

Assume an annual increase of 11% per year over a 6-yeartime period.

Geometric Gradient: Example

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.

Geometric Gradient Example (+g)

g = +11% per period; A1 = $1700; i = 8%/yr

Solution

P = $1700(P/A,11%,8%,7)

Need to calculate the P/A factor from the closed-form expressionfor a geometric gradient.

From a spreadsheet we see:

Geometric Gradient ( -g ) Consider the following problem with a negative growth rate g.


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We simply apply a g value = -0.10

Geometric Gradient (-g value)

Determination of an Unknown Interest Rate

When the i rate is unknown

A class of problems may deal with all of the parameters knowexcept 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

Example: i unknown

Assume on 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

two cash flows?


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Example: i unknown

The Cash Flow Diagram is

Example: i unknown

For i unknown

In general, solving for i in a time value formulation is notstraight forward.

More often, one will have to resort to some form of trial anderror approach as will be shown in future sections.

A sample spreadsheet model for this problem follows.

Example of the IRR function


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Determination of Unknown Number of Years

Unknown Number of Years

Some problems require knowing the number of time periods requiredgiven the other parameters

Example:

How long will it take for $1,000 to double in value if the discountrate is 5% per year?

Draw the cash flow diagram as.


i = 5%/year; n is unknown!


Solving we have..

Fn=? = 1000(F/P,5%,x):


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2000 = 1000(1.05)x

Solve for x in closed form


Solving we have..(1.05)x = 2000/1000

Xln(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 to a mass $2,000(have a little more that $2,000)

No. of Years NPER function

From Excel one can formulate as:

Chapter Summary

This chapter presents the fundamental time value of moneyrelationships common to most engineering economic analysiscalculations

Derivations have been presented for:

Present and Future Worth- P/F and F/P

Annuity Cash flows P/A, A/P, F/A and A/F

Gradients P/G, A,G and P/A,g,i,n

One must master these basic time value of money relationships inorder to proceed with more meaningful analysis that can impactdecision making.

These relationships are important to you professionally and in yourpersonal lives.


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Master these concepts!!!

review ch_02 annuity

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