time value of money and economic...

Contemporary Engineering Economics Third Canadian Edition, © 2012

Time Value of Money and Economic Equivalence

Lecture No.4 Chapter 3 Contemporary Engineering Economics Third Canadian Edition Copyright © 2012


Chapter Opening Story —Take a Lump Sum or Annual Installments

q  Millionaire Life is a lottery that offers a top prize of $1 million a year for 25 years and 20 prizes of $100,000.

q  Ms. Faye Lepage won the lottery and opted for the lump sum of $17 instead of the annuity of $1 million a year for 25 years.

q  What basis do we compare these two options?


Year Option A (Lump Sum)

Option B (Installment Plan)

0 1 2 3

25

$17M $1M $1M $1M $1M

$1M


What Do We Need to Know?

q  To make such comparisons (the lottery decision problem), we must be able to compare the value of money received at different points in time.

q  To do this, we use interest formulas to place different cash flows received at different times in the same time frame to compare them.


Chapter 3 Objectives

n  What is meant by the time value of money? n  What is the difference between simple interest

and compound interest? n  What is the meaning of economic equivalence

and why do we need it in economic analysis? n  How do we compare different money series by

means of the concept of economic equivalence? n  What are the common types of interest formulas

used to facilitate the calculation of economic equivalence?


Lecture 4 Objectives

n  What is meant by the time value of money? n  What is the difference between simple interest

and compound interest?

Interest: The Cost of Money

n  Market Interest Rate: The interest rate quoted by financial institutions that refers to the cost of money to borrowers or the earnings from money to lenders.


Time Value of Money

n  Money has a time value because a dollar today is worth more than a dollar in the future because the dollar received today can earn interest.

n  Money has both earning power (it can earn more money over time) and purchasing power (loss of value because of inflation) over time.


Elements of Transactions Involving Interest n  Principal - the initial amount of money involving debt

or investments n  Interest Rate - The cost or price of money and is

expressed as a percentage rate per period of time n  Interest Period - A length of time (often a year, but

can be a month, week, day, hour, etc.) that determines how frequently interest is calculated


Elements of Transactions Involving Interest n  Number of Interest Periods – Specified length of

time of the transaction n  Plan for Receipts or Payments – yields a particular

cash flow pattern over a specified length of time n  Future Amount of Money – results from the

cumulative effects of the interest rate over a number of interest periods


Elements of Transactions Involving Interest - Abbreviations n  An – A discrete payment or receipt occurring at the

end of some interest period. n  i - The interest rate per period n  N - The total number of cash flows n  P – A sum of money at time = 0, sometimes referred to

as the present value or present worth n  F – A future sum of money at the end of the analysis

period, sometimes referred to as future value


Elements of Transactions Involving Interest - Abbreviations n  A – An end-of-period payment or receipt in a uniform

series that continues for N periods. This is a special situation where A1= A2 = … = AN

n  Vn – An equivalent sum of money at the end of a specified period n that considers the time value of money. Note that V0 = P and VN = F


Interest Transaction Example

Plan 1 n  Principal amount =

$20,000 n  Loan origination fee =

$200 n  Interest = 9% n  Interest period = 1 year n  Number of interest

periods = 5

Plan 2 n  Principal amount =

$20,000 n  Loan origination fee =

$200 n  Interest = 9% n  Interest period = 1 year n  Number of interest

periods = 5 n  Grace period followed by

a single future payment in year 5



Interest Transaction Example

End of Year Receipts Payments

Plan 1 Plan 2 Year 0 $20,000.00 $200.00 $200.00 Year 1 5,141.85 0 Year 2 5,141.85 0 Year 3 5,141.85 0 Year 4 5,141.85 0 Year 5 5,141.85 30,772.48

The amount of loan = $20,000, origination fee = $200, annual interest rate = 9%

Cash Flow Diagram

n  A cash flow diagram is a graphical summary of the timing and magnitude of a set of cash flows. Upward arrows represent positive flows (receipts) and downward arrows represent negative flows (disbursements)


Cash flow diagram for Plan 1


End-of-Period Convention

§ One of the simplifying assumptions we make in engineering economic analysis is the end-of-period convention, which is the practice of placing all cash flow transactions at the end of an interest period.


Methods of Calculating Interest

n  Simple interest: the practice of charging an interest rate only to an initial sum (principal amount).

n  Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.


Simple Interest

( )

( )

1where

= Principal amount = = simple interest rate = number of interest periods = total amount accumulated at the end of period

F P I P iN

PI iP NiNF N

= + = +

§  Simple interest is interest earned on only the principal amount during each interest period. With simple interest, the interest earned during each interest period does not earn additional interest in the remaining periods, even though you do not withdraw it.

Compound Interest

n  With compound-interest, the interest earned in each period is calculated on the basis of the total amount at the end of the previous period. This total amount includes the original principal plus the accumulated interest that has been left in the account.

n  Then, P dollars now is equivalent to: q  P(1+i) dollars at the end of 1 period q  P(1+i)2 dollars at the end of 2 periods q  P(1+i)3 dollars at the end of 3 periods

n  At the end of N periods, the total accumulated value will be F = P(1+i)N



Example 3.1: Compound Interest

n  P = Principal amount n  i = Interest rate n  N = Number of

interest periods n  Example:

q  P = $1,000 q  i = 10% q  N = 3 years

End of Year

Beginning Balance

Interest earned

Ending Balance

0 $1,000

1 $1,000 $100 $1,100

2 $1,100 $110 $1,210

3 $1,210 $121 $1,331


Example 3.1: Compounding Process

$1,000

$1,100

$1,100

$1,210

$1,210

$1,331 0

1

2 3


0

$1,000

$1,331

1 2

3

3$1,000(1 0.10)$1,331

F = +

=

Example 3.1: Cash Flow Diagram


Simple Interest versus Compound Interest


Example 3.2: Comparing Simple with Compound Interest

§  In 1626 the Indians sold Manhattan Island to Peter Minuit of the Dutch West Company for $24.

§  If Minuit had invested the $24 in a bank account that paid 8% interest, how much would it be worth in 2009?

Example 3.2: Solution

383$24(1 0.08) $151,883,149,141,875F = + =

n  Given P = $24, i = 8% per year, and N = 383 years

n  Simple Interest

n  Compound Interest

( )( )$24 1 0.08 383 $759.36F ⎡ ⎤= + =⎣ ⎦


Summary

Compound interest is the practice of charging an interest rate to the initial sum and to any previously accumulated interest that has not been withdrawn from the initial sum. Compound interest is by far the most commonly used system in the real world.



Economic Equivalence




n  What is the meaning of economic equivalence and why do we need it in economic analysis?

n  How do we compare different money series by means of the concept of economic equivalence?


Economic Equivalence

n  The central question in deciding among alternative cash flows involves comparing their economic worth.

n  To determine the economic impact of a series of cash flows completely, we need to know: n  magnitude of the payment n  the direction of the payment (receipt or disbursement) n  the timing of the payment n  the interest rate during the period under consideration


Definition and Simple Calculations

n  Economic equivalence exists between individual cash flows and/or patterns of cash flows that have the same economic effect and could therefore be traded for one another.

n  Even though the amounts and timing of the cash flows may differ, the appropriate interest rate may make them equivalent.


Example 3.3: Equivalence

0 1 2 3 3 4 5

$3000

5

P

0

At an 8% interest, what is the equivalent worth now of $3000 in 5 years?

You are offered the alternative of receiving $3000 at the end of 5 years or P dollars today and you have access to an account that pays an 8% interest. What value of P would make you indifferent between P today and $3000 in 5 years?

=


Example 3.2 - Equivalence

Various dollar amounts that will be economically equivalent to $3,000 in 5 years, given an interest rate of 8%.

0 1 2 3 4 5

P F

$2,042 $3,000 $2,205 $2,382 $2,778 $2,572

5$3,000 $2,042(1 0.08)

P = =+

Equivalence Calculations: General Principles n  Principle 1: Equivalence calculations made to

compare alternatives require a common time basis q  To establish an economic equivalence between

two cash flow amounts, a common base period must be selected.



Example 3.4: Equivalent Cash Flows Are Equivalent at Any Common Point in Time

Equivalence Calculations: General Principles n  Principle 2: Equivalence depends on interest

rates q  The equivalence between two cash flows is a

function of the magnitude and timing of individual cash flows and the interest rate or rates that operate on these flows.


Equivalence Calculations: General Principles n  Principle 3: Equivalence calculations may

require the conversion of multiple payment cash flows to a single cash flow q  Part of the task of comparing alternative cash flow

series involves moving each individual cash flow in the series to the same single point in time and summing these values to yield a single equivalent cash flow.


Example 3.6: Equivalence Calculations With Multiple Payments n  Two options to pay off 3-year, $1000 loan

which has a 10% interest rate



Option 1 n  Using the equation F = P(1 + i)N and apply it to

each $100 disbursement

Option 2 n  A single disbursement of

$331


0 1 2 3

$100 $100 $100

0 1 2 3

$331 ( ) ( )

( )

2 13

0

$100 1 0.1 $100 1 0.1

$100 1 0.1 $331

F = + + +

+ + =

Equivalence Calculations: General Principles n  Principle 4: Equivalence is maintained

regardless of point of view q  As long as we use the same interest rate in

equivalence calculations, equivalence can be maintained regardless of point of view.



Summary

Economic equivalence exists between individual cash flows and/or patterns of cash flows that have the same worth. Even though the amounts and timing of the cash flows may differ, the appropriate interest rate may make them equivalent.


Development of Interest Formulas




n  What are the common types of interest formulas used to facilitate the calculation of economic equivalence?

The Five Types of Cash Flows n  We identify patterns in cash flow transactions. As a

result, we can develop concise expressions for computing either the present or future worth of the series. The 5 categories of cash flows are:

1.  Single Cash Flow: A single present or future cash flow 2.  Equal (Uniform) Series: A series of cash flows of equal

amounts at regular intervals 3.  Linear Gradient Series: A series of cash flows increasing or

decreasing by a fixed amount at regular intervals. 4.  Geometric Gradient Series: A series of cash flows increasing

or decreasing by a fixed percentage at regular intervals. 5.  Irregular Series: A series of cash flows exhibiting no overall

pattern. However, patterns might be detected for portions of the series.



Five Types of Cash Flows

Single Cash Flow

Equal (Uniform) Series

Linear Gradient Series

Geometric Gradient Series

Irregular (Mixed) Series

Single-Cash-Flow Formulas

n  The simplest case of cash flows.. n  Two commonly used factors relate a single

cash flow in one period to another single cash flow in a different period. They are:

1.  The Compound-Amount Factor: (F/P,i, N)

2.  The Present-Worth Factor: (P/F,i, N)


Compound Amount Factor: (F/P, i, N)

n  The compound amount factor computes the equivalent future value, F, given a present value P, the interest rate is i, and the number of periods N:

F = P(1 + i)N = P (F/P,i,N)


Interest Tables

n  Tables of compound-interest factors allow us to find the appropriate factor for a given interest rate and the number of interest periods. They are included in this text in Appendix A

n  (Using Interest Tables): What is the future worth of $20,000 t deposited today into an account bearing 12% and the number of periods is 15?

(F/P,12%,15) = 5.4736 from table

F = P × (F/P, 12%, 15) = $20,000(5.4736) = $109,472



Example 3.7: Single Amounts – Find F, Given i, N, and P

n  If you had $2,000 now and invested it at 10%, how much would it be worth in 8 years?

$2,000

F = ?

8 0

i = 10%


n  Given: P = $2000, i = 10% per year, and N = 8 years n  Find: F


( )( )( )

8$2,000 1 0.10

$2,000 / ,10%,8

$2,000 2.1436$4,287.18

EXCEL command:=FV(10%,8,0,-2000)=$4,287.20

F

F P

= +

=

=

=

Present Worth Factor: (P/F, i, N)

n  The present worth factor calculates the equivalent present value, P, given a future value, F, interest rate i, and the number of periods N:

P = F/(1 + i)N = F (P/F,i ,N) n  The interest rate i and the P/F factor are also

referred to as the discount rate and discounting factor, respectively.



Example 3.8: Single Amounts – Find P, Given F, i, and N

n  Suppose that $1000 is to be received in five years. At an annual interest rate of 12%, what is the present worth of this amount? $1000

P = ? 5

0 i = 12%


n  Given: F= $1000, i = 12% per year, and N = 5 years n  Find: P


5$1,000(1 0.12)$1,000( / ,12%,5)$567.40

EXCEL command:=PV(12%,5,0,-1000)=$567.40

PP F

−= +

=

=


Example 3.9: Solving for i

n  Suppose you buy a share for $10 and sell it for $20. Then your profit is $10. If that happens within a year, your rate of return is an impressive 100% If it takes five years, what would be the average annual rate of return on your investment?

0

5

$20

$10

i = ?


n  Given: P= $10, F = $20, and N = 5 years n  Find: i


( )

( )

( )( )

5

5

5

$20 $10 1

2 114.87%

Interest Tables:

$20=$10 1

2 / , ,515%

i

ii

i

F P ii

= +

= +

=

+

=

≈

EXCEL command:=RATE(5,0,-10,20)=14.87%

Example 3.10: Single Amounts – Find N, Given P, F, and i n  You have just purchased 100 shares of General Electric

stock at $60 per share. You will sell the stock when its market price has doubled. If you expect the stock price to increase 20% per year, how long do you anticipate waiting before selling the stock?



n  Given: P= $6000, F = $12,000, and i = 20% per year n  Find: N


( ) ( )

( ) ( )

( ) ( )

1 / , ,

$12,000 $6000 1 0.20 $6000 / ,20%,

2 1.20 / ,20%,log2 log1.20

log2log1.203.80 4 years

EXCEL command:=NPER(20%,0,-6000,12000) 3.801784

N

N

N

F P i P F P i N

F P N

F P NN

N

= + =

= + =

= =

=

=

= ≈

=


Example 3.11: Present Value of an Uneven Series by Decomposition Into Single Payments

n  How much do you need to deposit today (P) to withdraw $25,000 at n =1, $3,000 at n = 2, and $5,000 at n =4, if your account earns 10% annual interest?

0 1 2 3 4

$25,000

$3,000 $5,000

P = ?


0

1 2 3 4

$25,000

$3,000 $5,000

P

0

1 2 3 4

$25,000

P1

0

1 2 3 4

$3,000

P2

0

1 2 3 4

$5,000

P4

+ +

1 $25,000( / ,10%,1)$22,727

P P F=

=

2 $3,000( / ,10%,2)$2,479

P P F=

=4 $5,000( / ,10%,4)$3,415

P P F=

=

1 2 4 $28,622P P P P= + + =


Equal Payment Series

n  We often encounter transactions in which a uniform series of payments exists. Rental payments, bond interest payments, and commercial installment plans are based on uniform payment series. Uniform series are: 1.  Compound-Amount Factor: (F/A,i,N) 2.  Sinking-Fund Factor: (A/F, i, N) 3.  Capital Recovery Factor: (A/P, i, N) 4.  Present-Worth Factor: (P/A, i, N)


Uniform Series Compound Amount Factor: (F/A, i, N)

n  Finds F, given A, i and N n  Is used to compute the total amount F that

can be withdrawn at the end of the N periods if an amount A is invested at the end of each period.

( ) ( )

1 1/ , ,

NiF A A F A i N

i

⎡ ⎤+ −⎢ ⎥= =⎢ ⎥⎣ ⎦



Example 3.13: Uniform Series – Find F, Given i, A, and N

n  Suppose you make an annual contribution of $3000 to your savings account at the end of each year for 10 years. If the account earns 7% interest annually, how much can be withdrawn at the end of 10 years?

0 1 2 3 10

F = ?

A


n  Given: A= $3000, i = 7% per year, and N = 10 years n  Find: F


F = $3,000(F / A,7%,10)= $3000(13.8164)= $41,449.20

EXCEL command:=FV(7%,10,-3000,0,0) = $41,449.20

The Sinking Fund Factor: (A/F, i, N)

n  Finds A, given F, i and N n  A sinking fund is an interest-bearing account into which a

fixed sum is deposited each interest period; it is commonly established for the purpose of replacing fixed assets or retiring corporate bonds.


( )( / , , )

1 1N

iA F F A F i Ni

⎡ ⎤⎢ ⎥= =⎢ ⎥+ −⎣ ⎦

Example 3.15 Combination of a Uniform Series and a Single Present and Future Amount

n  You want $5000 five years from now, your father offers to give you $500 now. A part-time job will allow you to make five additional deposits, one at the end of each year. The bank that pays 7% interest, how large must your annual deposits be?




n  Given: F= $5000, P = $500, i = 7% per year, and N = 5 years

n  Find: A n  Amount FC: the worth of your father’s contribution at the

end of year 5 at a 7% interest rate

( )( )( )( )( )( )( )

$5000 ( / ,7%,5)

$5000 500 / ,7%,5 / ,7%,5

$5000 500 1.4026 0.1739

$747.55

CA F A F

F P A F

= −

= −

= −

=

Example 3.16: Comparison of Three Different Investment Plans n  Consider three investment plans at an annual interest

rate of 9.38%: 1.  Investor A. Invest $2000 per year for the first 10 years of your

career. At the end of 10 years, make no further investments, but reinvest the amount accumulated at the end of 10 years for the next 31 years.

2.  Investor B. Do nothing for the first 10 years. Then start investing $2000 per year for the next 31 years.

3.  Investor C. Invest $2000 per year for the entire 41 years.

n  All investments are made at the beginning of each year; the first deposit will be made at the beginning of age 25 (n=0) and you want to calculate the balance at the age of 65 (n = 41).



Example 3.16: Comparison of Three Different Investment Plans



n  Investor A: n  Investor B:

n  Investor C:

F65 = $2,000(F / A,9.38%,10)(1.0938)

Balance at the end of 10 years! "###### $######

$33,845% &###### '###### (F / P,9.38%,31)

= $545,216

F65 = $2,000(F / A,9.38%,31)$322,159

! "#### $#### (1.0938)

= $352,377

F65 = $2,000(F / A,9.38%,41)$820,620

! "#### $#### (1.0938)

= $897,594

Capital Recovery Factor: (A/P, i, N) n  Finds A, given P, i and N n  Commonly used to determine the revenue

requirements needed to address the upfront capital costs for projects.

n  The A/P factor is referred to as the annuity factor and indicates a series of payments of a fixed, or constant, amount for a specified number of periods

( )(1 ) / , ,(1 ) 1

N

N

i iA P P A P i Ni

⎡ ⎤+= =⎢ ⎥+ −⎣ ⎦


Example 3.17: Uniform Series – Find A, Given P, i, and N n  BioGen Company has borrowed $250,000 to purchase

laboratory equipment for gene splicing. The loan carries an interest rate of 8% per year and is to be repaid in equal installments over the next six years. Compute the amount of the annual installment.




n  Given: P = $250,000, i = 8% , and N = 6 years n  Find: A

$250,000( / ,8%,6)$250,000(0.2163) $54,075

EXCEL command:=PMT(8%,6,-25000) $54,075

A A P=

=

=

=


Example 3.18 – Deferred Loan Repayment

§  In Example 3.17, suppose that BioGen wants to negotiate with the bank to defer the first loan repayment until the end of year 2 (but still desires to make six equal installments at 8% interest). If the bank wishes to earn the same profit, what should be the annual installment, also known as deferred annuity?



n  Given: P = $250,000, i = 8% , N = 6 years, but the first payment occurs at the end of year 2.

n  Find: A Step 1 Step 2 $270,000( / ,8%,6)

$58,401A A Pʹ′ =

=

' $250,000( / ,8%,1)$270,000

P F P=

=

Present-Worth Factor: (P/A, i, N) n  Finds P, given A, i and N n  Answers the question “What would you have to

invest now in order to withdraw A dollars at the end of each of the next N periods?”

( )

( )( )

1 1/ , ,

1

N

N

iP A A P A i N

i i

⎡ ⎤+ −⎢ ⎥= =⎢ ⎥+⎣ ⎦


Example 3.19: Uniform Series– Find P, Given A, i, and N n  Let us revisit the Millionaire Life example. Suppose that

you had selected the annual payment option. Let’s see how your decision stands up against the $17 million cash prize option. If you could invest your money at 8% interest, what is the present worth of the 25 future payments at $1,000,000 per year for 25 years?




n  Given: A = $1,000,000, i = 8%, and N = 25 years n  Find: P

$1,000,000( / ,8%,25)$1,000,000(10.6748) $10,674,800

EXCEL command:=PV(8%,25,-1000000) $10,674,776.19

P P A=

=

=

=


n  One common pattern of variation in cash flows occurs when each cash flow in a series increases (or decreases) by a fixed amount. The cash flow diagram produces an ascending (or descending) straight line. Linear gradient series are:

1.  Present Worth Factor: Linear Gradient (P/G,i,N) 2.  Gradient-to-Equal-Payment Series Conversion

Factor (A/G,i,N),



Important Characteristics: 1.  The cash flow in period 1 is zero. 2.  The cash flows in periods 2 through N increase at a

constant amount.


Present Worth Factor: Linear Gradient (P/G,i,N)

n  Finds P, given G, i and N n  Is used when it is necessary to convert a

gradient series into a present value cash flow.

( )

( )( )

2

1 1/ , ,

1

N

N

i i iNP G G P G i N

i i

⎡ ⎤+ − −⎢ ⎥= =⎢ ⎥+⎣ ⎦



$1,000 $1,250

$1,500 $1,750 $2,000

1 2 3 4 5 0

P =?

The maintenance costs for a truck during the first year will be $1000 and are expected to increase at a rate of $250 per year. The firm wants to set up a maintenance account that earns 12% annual interest. How much does the firm have to deposit in the account now?

Example 3.20: Linear Gradient – Find P, Given A1, G, i, and N



( ) ( )( ) ( )

1 2 1 / ,12%,5 / ,12%,5

$1000 3.6048 $250 6.397 $5204

P P P A P A G P G= + = +

= + =

§  Given: A1 = $1000, G = $250, i = 12%, and N = 5 years §  Find: P

Gradient-to-Equal-Payment Series Conversion Factor, (A/G, i, N)

n  Finds A, given G, i and N n  Is used used when it is necessary to

convert a gradient series into a uniform series of equal cash flows.

( )

( )( )

1 1/ , ,

1 1

N

N

i iNA G G A G i N

i i

⎡ ⎤+ − −⎢ ⎥= =

⎢ ⎥⎡ ⎤+ −⎢ ⎥⎣ ⎦⎣ ⎦



Example 3.21: Linear Gradient – Find A, Given A1, G, i, and N



n  Given: A1 = $1000, G = 300, i = 10%, and N = 6 years n  Find: A

$1,000 $300( / ,10%,6)$1,000 $300(2.22236)$1,667.08

A A G= +

= +

=


Example 3.22: Declining Linear Gradient Series – Find F, Given A1, G, i, and N



n  Given: A1 = $1200, G = 200, i = 10%, and N = 5 years n  Find: F

F = F1 !F2

= A1(F / A,10%,5)! $200(P /G,10%,5)

Equivalent Present Worth at n = 0! "### $###(F / P,10%,5)

= $1,200(6.105)!$200(6.862)(1.611)= $5,115

Geometric Gradient Series

n  A series of cash flows that increase or decrease by a constant percentage each period

n  Price changes caused by inflation are a good example of a geometric gradient series. We use g to designate the percentage change in a payment from one period to the next.

n  Geometric gradient series are: 1.  Present-Worth Factor: (P/A1,g,i,N)


Types of Geometric Gradient Series

Geometric Gradient Series Present Worth Factor: (P/A1, g, i, N)

n  The present worth of a geometric series is:

n  Where A1 is the cash flow value in year 1 and g is the growth rate.

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Example 3.23: Geometric Gradient – Find P, Given A1, g, i, and N Current System n  Because of leaks, the

compressor is expected to run 70% of the time that the plant will be in operation during the upcoming year.

n  This will require 260 kWh of electricity at a rate of $0.05/kWh. (Plant runs 250 days a year, 24 hours per day.)

n  With current air delivery system, the compressor run time will increase by 7% per year for the next five years

New System n  Can replace all of the old

piping now at a cost of $28,570.

n  The compressor will still run the same number of days; however, it will run 23% less because of the reduced air pressure loss.

n  No annual increase.




n  Given: current power consumption, g = 7%, i = 12%, and N = 5 years

n  Find: A1, P

Power cost A1( )= % of day operating! days operating per year

! hours per day ! kWh ! $/kWh

= 70%( ) ! 250 days year ! 24 hours day

! 260 kWh( ) ! $0.05/kWh( )= $54,600 (Lets say $54,440)


Example 3.23: Geometric Gradient – Find P, Given A1, g, i, and N



1

5 5

$54,440( / ,7%,12%,5)

1 (1 0.07) (1 0.12)$54,4400.12 0.07

$222,283

$54,440(1 0.23)( / ,12%,5)$41,918.80(3.6048)$151,109

Old

New

P P A

P P A

−

=

⎡ ⎤− + += ⎢ ⎥−⎣ ⎦=

= −

=

=

§ The net cost for not replacing the old system now is $71,174 ( = $222,283 - $151,109). Since the new system costs only $28,570, the replacement should be made now.

Example 3.24: Geometric Gradient – Find A1, Given F, g, i, and N n  Given: F = 1,000,000, g = 6%, i = 8%, and N = 20 years n  Find: A1




( )( )( )

1 1

1

1

/ ,6%,8%,20 / ,8%,20

72.6911

$1,000,00072.6911

$13,757

F A P A F P

A

A

=

=

=

=


Summary

Cash flow diagrams are visual representations of cash inflows and outflows. They are particularly useful for helping us detect which of the following five patterns of cash flow: single payment, uniform series, linear gradient series, geometric gradient series, and irregular series.



n  How can you use out unconventional solution methods to solve cash flow problems that do not conform to the neat patterns for which there are interest formulas?

Unconventional Equivalence Calculations

n  Two categories of problems that demand unconventional treatment are composite (mixed) cash flows and problems in which we must determine the interest rate implicit in a financial contract.



Composite Cash Flows – Brute Force Approach

$50

$100 $100 $100

$150 $150 $150 $150 $200

0 1 2 3 4 5 6 7 8 9

$543.72

Composite Cash Flows – Grouping Approach

$50

$100 $100 $100

$150 $150 $150 $150 $200

Group 1 $50( / ,15%,1)

$43.48

P P F=

=

Group 2 $100( / ,15%,3)( / ,15%,1)

$198.54

P P A P F=

=

Group 3 $150( / ,15%,4)( / ,15%,4)

$244.85

P P A P F=

=

Group 4 $200( / ,15%,9) $56.85P P F= =

$43.48 $198.54 $244.85 $56.85$543.72

P = + + +

=

0 1 2 3 4 5 6 7 8 9



Example 3.25: Cash Flows with Subpatterns

1

2

$100( / ,12%,2) $300( / ,12%,3) $932.55( / ,12%,2) ( / ,12%,2)( / ,12%,1) 3.6290

3.6290 $932.55$256.97

V F A P AV C F A C P A P F CCC

= + =

= + =

=

=

§ The two cash flows in figure below are equivalent at an interest rate of 12% compounded annually. Determine the unknown value C. (Easiest to equate them at year 2)


Example 3.26: Establishing a College Fund § A couple with a newborn wants to save for their child’s university

expenses when she is 18. The university fund that pays 7% annual interest. The parents estimate that an amount of $40,000 per year will be required for four years. Determine the equal annual amounts the couple must save for 18 years.


Example 3.26: Solution:

18 $40,000 $40,000( / ,7%,3)$144,972

V P A= +

=

Economic equivalence of $40,000 withdrawls on 18th birthday

Economic equivalence of $X deposits on 18th birthday

18 ( / ,7%,18)33.9990

V X F AX

=

=33.9990 $144,972

$4264XX=

=

Solution

Determining an Interest Rate to Establish Economic Equivalence

n  Thus far, we have assumed that a typical interest rate is given.

n  When making some investments in financial assets, such as stocks, you may want to know the rate of growth (or rate of return) at which your asset is appreciating over the years.

n  To find the rate that is implicit in single payments, annuities, and uneven series of payments, a trial-and-error procedure or computer software is be used.


Example 3.27: Calculating an Unknown Interest Rate with Multiple Factors n  As a grand prize winner, you may choose between a $1

million cash prize paid immediately or $100,000 per year for 20 years. you decide to accept the 20 annual installments of $100,000. Suppose that you are considering the following two options: q  Option 1: You save your winnings for the first seven years and

then spend every cent of the winnings in the remaining 13 years. q  Option 2: You do the reverse, spending for seven years and then

saving for 13 years.

q  What interest rate makes the two options economically equivalent?


Example 3.27: Calculating an Unknown Interest Rate with Multiple Factors

Option 1 Option 2


7 $100,000( / , ,7)V F A i=7 $100,000( / , ,13)V P A i=

Ignore the arrow at N=20. We are comparing the two V7’s.


Example 3.27: Linear Interpolation

$100,000( / , ,7) $100,000( / , ,13)( / , ,7) 1( / , ,13)

F A i P A iF A iP A i

=

=

We equate the two values:

Using the interest tables, we need to resort to a trial-and-error method:

We approximate the interest rate by linear interpolation:

( ) 1 0.94826% 7% 6% 6.5934%1.0355 0.9482

i −⎡ ⎤= + − =⎢ ⎥−⎣ ⎦


Example 3.27: Linear Interpolation to Find an Unknown Interest Rate


Example 3.27: Using the Goal Seek Function to Find the Unknown Interest rate


Summary

Two categories of problems that demand unconventional treatment are composite (mixed) cash flows and problems in which we must determine the interest rate implicit in a financial contract.

time value of money and economic...

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