lecture engineering economic analysis new
TRANSCRIPT
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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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Solving Interest Problems
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
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+
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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Solving Interest Problems
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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Solving Interest Problems
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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Solving Interest Problems
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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Solving Interest Problems
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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Solving Interest Problems
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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Compounding Frequency
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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Compounding Frequency
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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When Interest Periods Coincide with Payment Periods
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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When Interest Periods Coincide with Payment Periods
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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When Interest Periods are Smaller than Payment Periods
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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When Interest Periods are Smaller than Payment Periods
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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When Interest Periods are Smaller than Payment Periods
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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When Interest Periods are Larger than Payment Periods
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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When Interest Periods are Larger than Payment Periods
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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When Interest Periods are Larger than Payment Periods
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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Continuous Compounding
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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Continuous Compounding
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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Uniform Gradient Series
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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Uniform Gradient Series
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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Uniform Gradient Series
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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Geometric Gradient Series
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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Geometric Gradient Series
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