Lecture Engineering Economic Analysis New

7/29/2019 Lecture Engineering Economic Analysis New

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Engineering Economics

Fundamentals: EIT Review

Hugh MillerColorado School of MinesMining Engineering DepartmentFall 2008


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Basics

Notation Never use scientific notation

Significant Digits Maximum of 4 significant figures unless the first digit is a 1, in which case a

maximum of 5 sig figs can be used

In general, omit cents (fractions of a dollar)

Year-End Convention Unless otherwise indicated, it is assumed that all receipts and disbursements

take place at the end of the year in which they occur.

Numerous Methodologies for Solving Problems Use the method most easy for you (visualize problem setup)


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Concept of Interest

If you won the lotto, would you rather get $1 Million nowor $50,000 for 25 years?

What about automobile and home financing? What type

of financing makes more economic sense?

Interest: Money paid for the use of borrowed money.

Put simply, interest is the rental charge forusing an

asset over some period of time and then, returning theasset in the same conditions as we received it.

In project financing, the asset is usually money


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Why Interest exist?

Taking the lenders view of point:

Risk: Possibility that the borrower will be unable to pay

Inflation: Money repaid in the future will value less

Transaction Cost: Expenses incurred in preparing theloan agreement

Opportunity Cost: Committing limited funds, a lender willbe unable to take advantage of other opportunities.

Postponement of Use: Lending money, postpones the

ability of the lender to use or purchase goods.

From the borrowers perspective .

Interest represents a cost !


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Simple Interest

Simple Interest is also known as the Nominal Rate of Interest

Annualized percentage of the amount borrowed (principal) whichis paid for the use of the money for some period of time.

Suppose you invested $1,000 for one year at 6% simple rate; at the endof one year the investment would yield:

$1,000 + $1,000(0.06) = $1,060

This means that each year interest gives $60

How much will you earn (including principal) after 3 years?

$1,000 + $1,000(0.06) + $1,000(0.06) + $1,000(0.06) = $1,180

Note that for each year, the interest earned is only calculated over $1,000.Does this mean that you could draw the $60 earned at the end of each year?


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Terms

In most situations, the percentage is not paid at the end of the period, where the interestearned is instead added to the original amount (principal). In this case, interest earned fromprevious periods is part of the basis for calculating the new interest payment.This adding up defines the concept ofCompounded Interest

Now assume you invested $1,000 for two years at 6% compounded annually;

At the end of one year the investment would yield:

$1,000 + $1,000 ( 0.06 ) = $1,060 or $1,000 ( 1 + 0.06 )

Since interest is compounded annually, at the end of the second year the investmentwould be worth:

[ $1,000 ( 1 + 0.06 ) ] + [ $1,000 ( 1 + 0.06 ) ( 0.06 ) ] = $1,124Principal and Interest for First Year Interest for Second Year

Factorizing:

$1,000 ( 1 + 0.06 ) ( 1 + 0.06 ) = $1,000 ( 1 + 0.06 )2 = $1,124

How much this investment would yield at the end of year 3?


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Solving Interest Problems

Step #1: Abstracting the Problem

Interest problems based upon 5 variables:

P, F, A, i, and n

Determine which are given (normally three) and whatneeds to be solved


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Step #2: Draw a Cash Flow Diagram

CashFlow

-

+

P

A1 A2 A3 A4 A5 A6

F

Time

Receipts

Disbursements


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Interest Formulas

The compound interest relationship may generally be expressed as:

F = P (1+r)n (1)

Where F = Future sum of money

P = Present sum of money

r = Nominal rate of interestn = Number of interest periods

Other variables to be introduced later:

A = Series of n equal payments made at the end of each period

i = Effective interest rate per period

Notation: (F/P,i,n) means Find F, given P, at a rate i for n periods

This notation is often shortened to F/P


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Interest Formulas

r = Nominal rate of interest

i = Effective interest rate per period

Nominal Interest is the periodic interest rate times the number ofperiods per year:

Nominal annual interest rate of 12% based upon monthlycompounding means 1% interest rate per month compounded

When the compounding frequency is annually: r = i

When compounding is performed more than once per year, theeffective rate (true annual rate) always exceeds the nominalannual rate: i > r


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Future Value

Example: Find the amount which will accrue at the end of Year 6if$1,500 is invested nowat 6% compounded annually.

Method #1: Direct Calculation

(F/P,i,n)F = P (1+r)n

Given Find F

n =

P =

i =


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Future Value

Example: Find the amount which will accrue at the end of Year 6if$1,500 is invested nowat 6% compounded annually.

Method #1: Direct Calculation

(F/P,i,n)F = P (1+r)n

Given Find F

n = 6 years F = (1,500)(1+0.06)6

P = $ 1,500 F = $ 2,128

r = 6.0 %


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Future Value

Example: Find the amount which will accrue at the end of Year 6if $1,500 is invested now at 6% compounded annually.

Method #2: Tables

The value of (1+i)n = (F/P,i,n) has been tabulated for various i and n.

From the handout, use the table with interest rate of 6% to find theappropriate factor. The first step is to layout the problem as follows:

F = P (F/P,i,n)F = 1500 (F/P, 6%, 6)

F = 1500 ( ) =


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Future Value

Example: Find the amount which will accrue at the end of Year 6if $1,500 is invested now at 6% compounded annually.

Method #2: Tables

The value of (1+i)n = (F/P,i,n) has been tabulated for various i and n.

Obtain the F/P Factor, then calculate F:

F = P (F/P,i,n)

F = 1500 (F/P, 6%, 6)F = 1500 (1.4185) = $2,128


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Present Value

If you want to find the amount needed at present in order to accrue acertain amount in the future, we just solve Equation 1 for P and get:

P = F / (1+r)n (2)

Example: If you will need $25,000 to buy a new truck in 3 years, how muchshould you invest nowat an interest rate of 10% compoundedannually?

(P/F,i,n)

Given Find PF =

n =

i =


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Present Value

If you want to find the amount needed at present in order to accrue acertain amount in the future, we just solve Equation 1 for P and get:

P = F / (1+r)n (2)

Example: If you will need $25,000 to buy a new truck in 3 years, how muchshould you invest nowat an interest rate of 10% compoundedannually?

Given Find PF = $25,000 P = F / (1+r)nn = 3 years P = (25,000) /(1 + 0.10)3

i = 10.0% = $18,783


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Present Value

If you want to find the amount needed at present in order to accrue a

certain amount in the future, we just solve Equation 1 for P and get:

P = F / (1+r)n (2)

Example: If you will need $25,000 to buy a new truck in 3 years, how much shouldyou invest nowat an interest rate of 10% compounded annually?

What is the factor to be used? ____________

Solve:


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Present Value

If you want to find the amount needed at present in order to accrue a

certain amount in the future, we just solve Equation 1 for P and get:

P = F / (1+r)n (2)

Example: If you will need $25,000 to buy a new truck in 3 years, how much shouldyou invest nowat an interest rate of 10% compounded annually?

What is the factor to be used? 0.7513

P = F(P/F,i,n) = (25,000)(0.7513) = $ 18,782


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Present Value

Example: If you will need $25,000 to buy a new truck in 3 years, how much shouldyou invest nowat an interest rate of9.5%compounded annually?

Method #1: Direct Calculation: Straight forward -Plug and Crank

Method #2: Tables: Interpolation

Which table in the appendix will be used? Tables i = 9% & 10%

What is the factor to be used? A13 (9%): 0.7722

A14 (10%): 0.7513

Assume 9.5%: 0.7618

P = F(P/F,i,n) = (25,000)(0.7618) = $ 19,044


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Engineering EconomicsEIT ReviewUniform Series & Effective Interest


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Annuities

Uniform series are known as the equal annualpayments made to an interest bearing accountfor a specified number of periods to obtain a

future amount.

CashFlow

-

+

P

A1 A2 A3 A4 A5 A6

F

Time


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Annuities Formula The future value (F) of a series of payments (A) made during (n)

periods to an account that yields (i) interest:

F = A [ (1+i)n 1 ] (5)i

Where F = Future sum of moneyn = number of interest periods

A = Series of n equal payments made at the end of each periodi = Effective interest rate per period

Derivation of this formula can be found in most engineeringeconomics texts & study guides

Notation: (F/A,i,n) or if using tables F = A (F/A,i,n)


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Example:

What is the future value of a series payments of $10,000 each, for 5years, if deposited into a savings account yielding 6% nominal interestcompounded yearly?

Draw the cash flow diagram.

F = A [ (1+i)n 1 ] =i

Check with Factor Values:

F = A (F/A,i,n)


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Example:

What is the future value of a series payments of $10,000 each, for 5years, if deposited in a savings account yielding 6% nominal interestcompounded yearly?

Draw the cash flow diagram.

F = A [ (1+i)n 1 ] = 10,000 [ (1+0.06)5 1 ] = 10,000 [ 1.3382 1 ]

i 0.06 0.06

= $ 56,370

Checking with Factor Values:

F = A (F/A,i,n)

= 10,000 (F/A, 6%, 5)

= 10,000 ( 5.6371 )

= $ 56,370


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Sinking Fund

We can also get the corresponding value of an annuity (A) during (n)periods to an account that yields (i) interest to be able to get the futurevalue (F) :

Solving for A: A = i F / [ (1+i)n 1 ] (6)

Notation:A = F (A/F,i,n)

Example:

How much money would you have to save annually in order to buy a carin 4 years which has a projected value of $18,000? The savings accountoffers 4.0% yearly interest.


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Sinking Fund

Example:

How much money do we have to save annually to buy a car 4 years fromnow that has an estimated cost of $18,000? The savings account offers4.0 % yearly interest.

A = i F / [ (1+i)n 1 ]

A = (0.04 x 18,000) / [ (1.04)4 -1 ] = 720 / 0.170 = $4,239

A = F (A/F,i,n)

A = (18,000)(A/F,4.0,4) = (18,000)(0.2355) = $4,239


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Present Worth of an Uniform Series

Sometimes it is required to estimate the present value (P) of a series ofequal payments (A) during (n) periods considering an interest rate (i)

From Eq. 1 and 5

P = A [ (1+i)n 1 ] (7)

i (1+i)n

Notation: P = A (P/A,i,n)

Example:What is the present value of a series of royalty payments of $50,000

each for 8 years if nominal interest is 8%?

P =


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Present Worth of an Uniform Series

Example:

What is the present value of a series of royalty payments of $50,000each for 8 years if nominal interest is 8%?

P = A [ (1+i)n 1 ] = 50,000 [ (1+0.08)8 1 ] = 50,000 [ 1.8509 1 ]

i (1+i)n 0.08 (1.08)8 0.1481

= $ 287,300

P = A (P/A,i,n) = (50,000)(P/A,8,8) = (50,000)(5.7466) = $ 287,300


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Uniform Series Capital Recovery

This is the corresponding scenario where it is required to estimate the value of aseries of equal payments (A) that will be received in the future during (n) periodsconsidering an interest rate (i) and are equivalent to the present value of aninvestment (P)

Solving Eq. 7 for A

A = i P (1+i)n

(8)(1+i)n -1

Notation: A = P (A/P,i,n)

Example:If an investment opportunity is offered today for $5 Million, how much must it yield

at the end of every year for 10 years to justify the investment if we want to get a12% interest?

A =


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Uniform Series Capital Recovery

Example:

If an investment opportunity is offered today for $5 Million, how muchmust it yield at the end of every year for 10 years to justify theinvestment if we want to get a 12% interest?

A = i P (1+i)n

= 0.12 x 5 (1+0.12)10

= 0.6 [ 3.1058 ](1+i)n -1 (1.12)10 - 1 2.1058

= 0.8849 Million $ 884,900 per year


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Engineering EconomicsEIT Review

Varying Compounding PeriodsEffective Interest


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Example:

An investment opportunity is available which will yield $1000 peryear for the next three years and $600 per year for the following twoyears. If the interest is 12% and the investment has no terminal

salvage value, what is the present value of the investment?

What is Step #1?


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Example:

An investment opportunity is available which will yield $1000 peryear for the next three years and $600 per year for the following twoyears. If the interest is 12% and the investment has no terminal

salvage value, what is the present value of the investment?

Step #1

What are we trying to solve? P

What are the known variables? A, i, n


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Step #2: Draw a Cash Flow Diagram

Receipts

Time

PV (?)

A1 A2 A3

A4 A5

A1 + A2 + A3 = $1000

A4 + A5 = $600


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Step #2: Draw a Cash Flow Diagram Method #1

Time

A1

A2 A3A1

A1 = A2 = A3 = A4 = A5 = $600 A1 = A2 = A3 = $400

A2 A3 A4 A5

Time+

P = A1 (P/A1, i, n1) + A2 (P/A2, i, n2)= ($600)(P/A1, 12, 5) + ($400)(P/A2, 12, 3)= $3,124


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Step #2: Draw a Cash Flow Diagram Method #2

Time

A1

A4 A5

A1 = A2 = A3 = $1000 A4 = A5 = $600

A2 A3

Time+

P = A1 (P/A1, i, n1) + A2 (P/A2, i, n2)(P/F, i, n3)= ($1000)(P/A1, 12, 3) + ($600)(P/A2, 12, 2)(P/F, 12, 3)= $3,124

P


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Varying Payment and Compounding Intervals

Thus far, problems involving time value of money haveassumed annual payments and interest compoundingperiods

In most financial transactions and investments, interest

compounding and/or revenue/costs occur at frequenciesother than once a year (annually)

An infinite spectrum of possibilities

Sometimes called discrete, periodic compounding

In reality, the economics of project feasibility are simplycomplex annuity problems with multiple receipts &disbursements


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Compounding Frequency

Compounding can be performed at any interval(common: quarterly, monthly, daily)

When this occurs, there is a difference between nominaland effective annual interest rates

This is determined by:

i = (1 + r/x)x 1

where: i = effective annual interest rate

r = nominal annual interest ratex = number of compounding periods per year


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Example: If a student borrows $1,000 from a financecompany which charges interest at a compoundrate of 2% per month:

What is the nominal interest rate:

r = (2%/month) x (12 months) = 24% annually

What is the effective annual interest rate:

i = (1 + r/x)x

1

i = (1 + .24/12)12 1 = 0.268 (26.8%)


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Nominal and EffectiveAnnual Rates of Interest

The effective interest rate is the rate compounded once a year which isequivalent to the nominal interest rate compoundedxtimes a year

The effective interest rate is always greater than or equal to the nominalinterest rate

The greater the frequency of compounding the greater the differencebetween effective and nominal rates. But it has a limit ContinuousCompounding.

Frequency Periods/year Nominal Rate Effective Rate

Annual 1 12% 12.00%Semiannual 2 12% 12.36%Quarterly 4 12% 12.55%Monthly 12 12% 12.68%Weekly 52 12% 12.73%Daily 365 12% 12.75%

Continuously 12% 12.75%


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It is also important to be able to calculate the effectiveinterest rate (i) for the actual interest periods to be used.

The effective interest rate can be obtained by dividing thenominal interest rate by the number of interest paymentsper year (m)

i = (r/m)

where: i = effective interest rate for the periodr = nominal annual interest rate


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When Interest Periods Coincide with Payment Periods

When this occurs, it is possible to directly use theequations and tables from previous discussions (annualcompounding)

Provided that:

(1) the interest rate (i) is the effective rate for the period

(2) the number of years (n) must be replaced by the

total number of interest periods (mn), where mequals the number of interest periods per year


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Example: An engineer plans to borrow $3,000 from his companycredit union, to be repaid in 24 equal monthlyinstallments. The credit union charges interest at therate of 1% per month on the unpaid balance. Howmuch money must the engineer repay each month?

A = P (A/P, i, mn) = i P (1+i)n

(1+i)n -1

A = ($3000) (A/P, 1%, 24) = $141.20


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Example: An engineer wishes to purchase an $80,000 lakeside lot(real estate) by making a down payment of $20,000 andborrowing the remaining $60,000, which he will repay on amonthly basis over the next 30 years. If the bank chargesinterest at the rate of 9% per year, compounded monthly,how much money must the engineer repay each month?

i = (r/m) = (0.095/12) = 0.00792 (0.79%)

A = P (A/P, i, mn) = i P (1+i)n(1+i)n -1

A = ($60000) (A/P, 0.79%, 360) = $504.50

Total amount repaid to the bank?


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When Interest Periods are Smaller than Payment Periods

When this occurs, the interest may be compoundedseveral times between payments.

One widely used approach to this type of problem is todetermine the effective interest rate for the given interestperiod, and then treat each payment separately.


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Example: Approach #1

An engineer deposits $1,000 in a savings account at the end of each year. Ifthe bank pays interest at the rate of 6% per year, compounded quarterly, howmuch money will have accumulated in the account after 5 years?

Effective Interest Rate: i = (6%/4) = 1.5% per quarter

F = P (F/P,i,mn)

F = $1000(F/P,1.5%,16) + $1000(F/P,1.5%,12) + $1000(F/P,1.5%,8) +$1000(F/P,1.5%,4) +$1000(F/P,1.5%,0)

Using formulas or tables: F = $5,652


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Another approach, often more convenient, is to calculatean effective interest rate for the given payment period,and then proceed as though the interest periods and the

payment periods coincide.

i = (1 + r/x)x 1


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Example: Approach #2

An engineer deposits $1,000 in a savings account at the end of each year. Ifthe bank pays interest at the rate of 6% per year, compounded quarterly, howmuch money will have accumulated in the account after 5 years?

i = (1 + r/x)x 1 = (1 + 0.06/4)4 1 = 0.06136 (6.136%)

F = $1,000 (F/A,6.136%,5)

Using formula: F = $5,652


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When Interest Periods are Larger than Payment Periods

When this occurs, some payments may not have been deposited for anentire interest period. Such payments do not earn any interest during thatperiod.

Interest is only earned by those payments that have been deposited orinvested for the entire interest period.

Situations of this type can be treated in the following manner:

Considerall deposits that were made during the interest period to have beenmade at the endof the interest period (i.e., no interest earned during the period)

Consider all withdrawals that were made during the interest period to have been

made at the beginning of the interest period (i.e., earning no interest) Then proceed as though the interest periods and the payment periods coincide.


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Example: A person has $4,000 in a savings account at the beginning of acalendar year; the bank pays interest at 6% per year, compoundedquarterly. Given the transactions presented in the following table (nextslide), find the account balance at the end of the calendar year.

Effective Interest Rate (i) = 6%/4 = 1.5% per quarter


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Example:Date Deposit Withdrawal Effective Date

Jan. 10 $175 Jan. 1st (beginning 1st Q)

Feb. 20 $1,200 Mar. 31th (end of 1st Q)

Apr. 12 $1,800 April 1st (beginning 2nd Q)

May 5 $65 June 30 (end of 2

nd

Q)May 13 $115 June 30 (end of 2nd Q)

May 24 $50 April 1st (beginning 2nd Q)

June 21 $250 April 1st (beginning 2nd Q)

Aug. 10 $1,600 Sept. 30 (end of 3rd Q)

Sept. 12 $800 July 1st (beginning 3rd Q)

Nov. 27 $350 Oct. 1 (beginning 4th Q)

Dec. 17 $2,300 Dec. 31 (end of 4th Q)

Dec. 29 $750 Oct. 1 (beginning 4th Q)


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Example: A person has $4,000 in a savings account at the beginning of acalendar year; the bank pays interest at 6% per year, compoundedquarterly. Given the transactions presented in the following table (nextslide), find the account balance at the end of the calendar year.

F = ($4000-$175)(F/P,1.5%,4) + ($1200-$2100)(F/P,1.5%,3) +

($180-$800)(F/P,1.5%,2) + ($1600-$1100)(F/P,1.5%,1) + $2300

Using formula: F = $5,287


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Continuous Compounding

Continuous Compounding can be thought of as a limitingcase example, where the nominal annual interest rate isheld constant at r, the number of interest periods becomes

infinite, and the length of each interest period becomesinfinitesimally small.

The effective annual interest rate in continuouscompounding is expressed by the following equation:

i = limm[(1 + r/m)m 1] = er- 1


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Example: A savings bank is selling long-term savingscertificates that pay interest at the rate of 7 % per year,compounded continuously. What is the actual annual

yield of these certificates?

i = er 1 = e0.075 1 = 0.0779 (7.79%)


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Discrete payments:If interest is compounded continuously but payments are

made annually, the following equations can be used:

F/P = ern A/P = (er 1) / (1 e-rn)

P/F = e-rn P/A = (1 e-rn) / (er 1)

F/A =(ern 1) / (er 1) A/F =(er 1) / (ern 1)

Where: n = the number of years

r = nominal annual interest rate


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Continuous Compounding (Discrete payments)

Example:A savings bank offers long-term savingscertificates at 7 % per year, compounded continuously.If a 10-year certificate costs $1,000, what will be its valueupon maturity?

F = P x (F/P,r,n) = P x ern

F = ($1,000) x e(0.075)(10) = $2,117


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Continuous Compounding (Discrete payments)

If interest is compounded continuously but payments aremade (x) times per year, the previous formulas remainvalid as long as r is replaced by r/x and with n beingreplaced by nx.

Example:A person borrows $5,000 for 3 years, to berepaid in 36 equal monthly installments. The interestrate is 10% per year, compounded continuously. Howmuch must be repaid at the end of each month?

(A/P,r/x,nx) (A/P,10/12,36)

A = (P) [(er 1) / (1 e-rn)]

= ($5,000) [(e0.10/12 1) / (1 e-(0.10/12)(12x3))]

= $161.40


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Gradient Series

Thus far, most of the course discussion has focused on uniform-series problems

A great many investment problems in the real world involve theanalysis of unequal cash flow series and can not be solved with theannuity formulas previously introduced

As such, independent and variable cash flows can only be analyzedthrough the repetitive application of single payment equations

Mathematical solutions have been developed, however, for twospecial types of unequal cash flows:

Uniform Gradient Series

Geometric Gradient Series


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A Uniform Gradient Series (G) exists when cash flows either increase ordecrease by a fixed amount in successive periods.

In such cases, the annual cash flow consists of two components:(1) a constant amount (A1) equal to the cash flow in the first period

(2) a variable amount (A2) equal to (n-1)G

As such: AT = (A1) + (A2)

A2 = G [(1/i) (n/i)(A/F,i,n)]

where: [(1/i) (n/i)(A/F,i,n)] is called the uniform gradient factorand is written as (A/G,i,n)

Therefore: AT = (A1) + G(A/G,i,n))


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Example: An engineer is planning for a 15 year retirement. In order tosupplement his pension and offset the anticipated effects of inflation andincreased taxes, he intends to withdraw $5,000 at the end of the first year,and to increase the withdrawal by $1,000 at the end of each successiveyear. How much money must the engineer have in this account at the start

of his retirement, if the money earns 6% per year, compounded annually?

Want to Find: PGiven: A1, G, i, and n

T = 0

P

1

$5000

$6000

2 14 15

$18000

$19000

$7000

$8000

3 4


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Example:

AT = (A1) + G(A/G,i,n)

A2 = G(A/G,i,n) = $1000 (A/G,6%,15) = $1000 (5.926) = $5926

AT = $5000 + $5926 = $10,926

P = AT (P/A,i,n) = $10,926 (P/A,6%,15) = $10,926 (9.7123) = $106,120


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Since receipts and expenditures rarely increase or decrease every periodby a fixed amount, Uniform Gradient Series (G) problems have limitedapplicability

With Geometric Gradients, the increase or decrease in cash flows betweenperiods is not a constant amount but a constant percentage of the cash flow

in the preceding period.

Like Uniform Gradients, Geometric Gradients limited applicability but aresometimes used to account for inflationary cost increases

AK = A (1 + j)K-1

Where: j equals the percent change in the cash flow between periodsA is the cash flow in the initial periodAK is the cash flow in any subsequent period


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Present Value of the series equals:

P = A (1 + j)-1 [(1 + j)/(1 + i)]-K

For i = j: P = (n x A ) / (1 + i)

For i j: P = A [1(1-j)n (1+i)-n] / (i - j)

Nomenclature: P = A(P/A,i,j,n)

n

K = 1

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