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Engineering Economics

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Title: Engineering Economics


1
Engineering Economics
  • Fundamentals EIT Review

Hugh Miller Colorado School of Mines Mining
Engineering Department Fall 2007
2
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)

3
Concept of Interest
  • If you won the lotto, would you rather get 1
    Million now or 50,000 for 20 years?
  • What about automobile and home financing? What
    type of financing makes more sense?
  • Interest Money paid for the use of borrowed
    money.
  • If we put money in the perspective of an asset,
    interest is the rental charge for using an asset
    over some period of time and then, returning the
    asset in the same conditions as we received it.
  • ? In project financing, the asset is usually
    money

4
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
    the loan agreement
  • Opportunity Cost Committing limited funds, a
    lender will be 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 !

5
Simple Interest
  • Simple Interest is also known as the Nominal Rate
    of Interest
  • Annualized percentage of the amount borrowed
    (principal) which is 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 end of 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 get after 3 years?
  • 1,000 1,000(0.06) 1,000(0.06)
    1,000(0.06) 1,180
  • Note that each year the interest are calculated
    only over 1,000.
  • Does it means that you could draw the 60 at the
    end of each year?

6
Terms
  • If the percentage is not paid at the end of the
    period, then, this amount is added
  • to the original amount (principal) to calculate
    the interest for the second term.
  • This adding up defines the concept of
    Compounded 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 investment would be
    worth
  • 1,000 ( 1 0.06 ) 1,000 ( 1 0.06 ) (
    0.06 ) 1,124
  • Principal 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?

7
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
    what needs to be solved

8
Solving Interest Problems
  • Step 2 Draw a Cash Flow Diagram

Receipts
Disbursements
9
Interest Formulas
  • The compound interest relationship may generally
    be expressed as
  • F P (1r)n (1)
  • Where F Future sum of money
  • P Present sum of money
  • r Nominal rate of interest
  • n 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

10
Interest Formulas
  • r Nominal rate of interest
  • i Effective interest rate per period
  • When the compounding frequency is annually r
    i
  • When compounding is performed more than once per
    year, the effective rate (true annual rate)
    always exceeds the nominal annual rate i gt r

11
Future Value
  • Example Find the amount which will accrue at the
    end of Year 6 if 1,500 is invested now at 6
    compounded annually.
  • Method 1 Direct Calculation
  • (F/P,i,n)
  • F P (1r)n
  • Given Find F
  • n
  • P
  • i

12
Future Value
  • Example Find the amount which will accrue at the
    end of Year 6 if 1,500 is invested now at 6
    compounded annually.
  • Method 1 Direct Calculation
  • (F/P,i,n)
  • F P (1r)n
  • Given Find F
  • n 6 years F (1,500)(10.06)6
  • P 1,500 F 2,128
  • r 6.0

13
Future Value
  • Example Find the amount which will accrue at the
    end of Year 6 if 1,500 is invested now at 6
    compounded annually.
  • Method 2 Tables
  • The value of (1i)n (F/P,i,n) has been
    tabulated for various i and n.
  • From the handout, you will find tables with
    interest rates from ½ to 18. The first step is
    to layout the problem as follows
  • F P (F/P,i,n)
  • F 1500 (F/P, 6, 6)
  • F 1500 ( )

14
Future Value
  • Example Find the amount which will accrue at the
    end of Year 6 if 1,500 is invested now at 6
    compounded annually.
  • Method 2 Tables
  • The value of (1i)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

15
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 / (1r)n (2)
  • Example If you will need 25,000 to buy a new
    truck in 3 years, how much should you invest
    now at an interest rate of 10 compounded
    annually?
  • (P/F,i,n)
  • Given Find P
  • F
  • n
  • i

16
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 / (1r)n (2)
  • Example If you will need 25,000 to buy a new
    truck in 3 years, how much should you invest
    now at an interest rate of 10 compounded
    annually?
  • Given Find P
  • F 25,000 P F / (1r)n
  • n 3 years P (25,000) /(1 0.10)3
  • i 10.0 18,783

17
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 / (1r)n (2)
  • Example If you will need 25,000 to buy a new
    truck in 3 years, how much should you invest
    now at an interest rate of 10 compounded
    annually?
  • What is the factor to be used? ____________
  • Solve

18
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 / (1r)n (2)
  • Example If you will need 25,000 to buy a new
    truck in 3 years, how much should you invest
    now at 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

19
Present Value
  • Example If you will need 25,000 to buy a new
    truck in 3 years, how much should you invest
    now at an interest rate of 9.5 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

20
Engineering EconomicsEIT ReviewUniform Series
Effective Interest
21
Annuities
  • Uniform series are known as the equal annual
    payments made to an interest bearing account for
    a specified number of periods to obtain a future
    amount.

22
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 (1i)n 1 (5)
  • i
  • Where F Future sum of money
  • n number of interest periods
  • A Series of n equal payments made at the end
    of each period
  • i Effective interest rate per period
  • Derivation of this formula can be found in most
    engineering economics texts study guides
  • Notation (F/A,i,n) or if using tables F A
    (F/A,i,n)

23
Example
  • What is the future value of a series of savings
    of 10,000 each for 5 years if deposited in a
    savings account yielding 6 nominal interest
    compounded yearly?
  • Draw the cash flow diagram.
  • F A (1i)n 1
  • i
  • Check with Factor Values
  • F A (F/A,i,n)

24
Example
  • What is the future value of a series of savings
    of 10,000 each for 5 years if deposited in a
    savings account yielding 6 nominal interest
    compounded yearly?
  • Draw the cash flow diagram.
  • F A (1i)n 1 10,000 (10.06)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

25
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 future
    value (F)
  • Solving for A A i F / (1i)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 car in 4 years which has a
    projected value of 18,000? The savings account
    offers 4.0 yearly interest.

26
Sinking Fund
  • Example
  • How much money do we have to save annually to
    buy a car 4 years from now that has an estimated
    cost of 18,000? The savings account offers 4.0
    yearly interest.
  • A i F / (1i)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

27
Present Worth of an Uniform Series
  • Sometimes it is required to estimate the present
    value (P) of a series of equal payments (A)
    during (n) periods considering an interest rate
    (i)
  • From Eq. 1 and 5
  • P A (1i)n 1 (7)
  • i (1i)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

28
Present Worth of an Uniform Series
  • 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 A (1i)n 1 50,000 (10.08)8 1
    50,000 1.8509 1
  • i (1i)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

29
Uniform Series Capital Recovery
  • This is the corresponding scenario where it is
    required to estimate the value of a series of
    equal payments (A) that will be received in the
    future during (n) periods considering an interest
    rate (i) and are equivalent to the present value
    of an investment (P)
  • Solving Eq. 7 for A

  • A i P (1i)n (8)

  • (1i)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 a 12 interest?
  • A

30
Uniform Series Capital Recovery
  • 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 a 12 interest?
  • A i P (1i)n 0.12 x 5 (10.12)10
    0.6 3.1058
  • (1i)n -1 (1.12)10 - 1
    2.1058
  • 0.8849 Million ? 884,900 per year

31
Engineering EconomicsEIT ReviewVarying
Compounding PeriodsEffective Interest
32
Solving Interest Problems
  • Example
  • An investment opportunity is available which
    will yield 1000 per year for the next three
    years and 600 per year for the following two
    years. 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?

33
Solving Interest Problems
  • Example
  • An investment opportunity is available which
    will yield 1000 per year for the next three
    years and 600 per year for the following two
    years. 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? PV
  • What are the known variables? A, i, n

34
Solving Interest Problems
  • Step 2 Draw a Cash Flow Diagram

PV (?)
Receipts
A1
A2
A3
A4
A5
Time
A1 A2 A3 1000
A4 A5 600
35
Solving Interest Problems
  • Step 2 Draw a Cash Flow Diagram Method 1

P A1 (P/A1, i, n1) A2 (P/A2, i, n2)
(600)(P/A1, 12, 5) (400)(P/A2, 12, 3)
3,124
A1
A2
A3
A4
A5
A2
A3
A1

Time
Time
A1 A2 A3 A4 A5 600
A1 A2 A3 400
36
Solving Interest Problems
  • Step 2 Draw a Cash Flow Diagram Method 2

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
A1
A2
A3
A4
A5

Time
Time
A1 A2 A3 1000
A4 A5 600
37
Varying Payment and Compounding Intervals
  • Thus far, problems involving time value of money
    have assumed annual payments and interest
    compounding periods
  • In most financial transactions and investments,
    interest compounding and/or revenue/costs occur
    at frequencies other than once a year (annually)
  • An infinite spectrum of possibilities
  • Sometimes called discrete, periodic compounding
  • In reality, the economics of mine feasibility are
    simply complex annuity problems with multiple
    receipts disbursements

38
Compounding Frequency
  • Compounding can be performed at any interval
    (common quarterly, monthly, daily)
  • When this occurs, there is a difference between
    nominal and effective annual interest rates
  • This is determined by
  • i (1 r/x)x 1
  • where i effective annual interest rate
  • r nominal annual interest rate
  • x number of compounding periods per year

39
Compounding Frequency
  • Example If a student borrows 1,000 from a
    finance company which charges interest at
    a compound rate 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)

40
Nominal and EffectiveAnnual Rates of Interest
  • The effective interest rate is the rate
    compounded once a year which is equivalent to the
    nominal interest rate compounded x times a year
  • The effective interest rate is always greater
    than or equal to the nominal interest rate
  • The greater the frequency of compounding the
    greater the difference between effective and
    nominal rates. But it has a limit ? Continuous
    Compounding.
  • 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 8 12
    12.75

41
Compounding Frequency
  • It is also important to be able to calculate the
    effective interest rate (i) for the actual
    interest periods to be used.
  • The effective interest rate can be obtained by
    dividing the nominal interest rate by the number
    of interest payments per year (m)
  • i (r/m)
  • where i effective interest rate for the
    period
  • r nominal annual interest rate

42
When Interest Periods Coincide with Payment
Periods
  • When this occurs, it is possible to directly use
    the equations and tables from previous
    discussions (annual compounding)
  • 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
    m equals the number of interest periods per year

43
When Interest Periods Coincide with Payment
Periods
  • Example An engineer plans to borrow 3,000 from
    his company credit union, to be repaid in 24
    equal monthly installments. The credit union
    charges interest at the rate of 1 per month on
    the unpaid balance. How much money must the
    engineer repay each month?
  • A P (A/P, i, mn) i P (1i)n

  • (1i)n -1
  • A (3000) (A/P, 1, 24) 141.20

44
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 and
    borrowing the remaining 60,000, which he will
    repay on a monthly basis over the next 30
    years. If the bank charges interest at the
    rate of 9½ 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 (1i)n

  • (1i)n -1
  • A (60000) (A/P, 0.79, 360) 504.50
  • Total amount repaid to the bank?

45
When Interest Periods are Smaller than Payment
Periods
  • When this occurs, the interest may be compounded
    several times between payments.
  • One widely used approach to this type of problem
    is to determine the effective interest rate for
    the given interest period, and then treat each
    payment separately.

46
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. If the bank pays
    interest at the rate of 6 per year, compounded
    quarterly, how much 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

47
When Interest Periods are Smaller than Payment
Periods
  • Another approach, often more convenient, is to
    calculate an 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

48
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. If the bank pays
    interest at the rate of 6 per year, compounded
    quarterly, how much 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

49
When Interest Periods are Larger than Payment
Periods
  • When this occurs, some payments may not have been
    deposited for an entire interest period. Such
    payments do not earn any interest during that
    period.
  • Interest is only earned by those payments that
    have been deposited or invested for the entire
    interest period.
  • Situations of this type can be treated in the
    following manner
  • Consider all deposits that were made during the
    interest period to have been made at the end of
    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.

50
When Interest Periods are Larger than Payment
Periods
  • Example A person has 4,000 in a savings
    account at the beginning of a calendar year the
    bank pays interest at 6 per year, compounded
    quarterly. Given the transactions presented in
    the following table (next slide), find the
    account balance at the end of the calendar year.
  • Effective Interest Rate (i) 6/4 1.5 per
    quarter

51
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. 30th (end of 1st
    Q)
  • Apr. 12 1,500
    April 1st (beginning 2nd Q)
  • May 5 65 June 30 (end of
    2nd 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)

52
When Interest Periods are Larger than Payment
Periods
  • Example A person has 4,000 in a savings
    account at the beginning of a calendar year the
    bank pays interest at 6 per year, compounded
    quarterly. Given the transactions presented in
    the following table (next slide), find the
    account balance at the end of the calendar year.
  • F (4000-175)(F/P,1.5,4) (1200-1800)(F/P,1
    .5,3)
  • (180-800)(F/P,1.5,2) (1600-1100)(F/P,1.5,1
    ) 2300
  • Using formula F 5,601

53
Continuous Compounding
  • Continuous Compounding can be thought of as a
    limiting case example, where the nominal annual
    interest rate is held constant at r, the number
    of interest periods becomes infinite, and the
    length of each interest period becomes
    infinitesimally small.
  • The effective annual interest rate in continuous
    compounding is expressed by the following
    equation
  • i limm?8(1 r/m)m 1 er - 1

54
Continuous Compounding
  • Example A savings bank is selling long-term
    savings certificates that pay interest at the
    rate of 7 ½ per year, compounded continuously.
    What is the actual annual yield of these
    certificates?
  • i er 1 e0.075 1 0.0779 (7.79)

55
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

56
Continuous Compounding (Discrete payments)
  • Example A savings bank offers long-term savings
    certificates at 7 ½ per year, compounded
    continuously. If a 10-year certificate costs
    1,000, what will be its value upon maturity?
  • F P x (F/P,r,n) P x ern
  • F (1,000) x e(0.075)(10) 2,117

57
Continuous Compounding (Discrete payments)
  • If interest is compounded continuously but
    payments are made (x) times per year, the
    previous formulas remain valid as long as r is
    replaced by r/x and with n being replaced by nx.
  • Example A person borrows 5,000 for 3 years, to
    be repaid in 36 equal monthly installments. The
    interest rate is 10 per year, compounded
    continuously. How much 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

58
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 the analysis of unequal cash flow
    series and can not be solved with the annuity
    formulas previously introduced
  • As such, independent and variable cash flows can
    only be analyzed through the repetitive
    application of single payment equations
  • Mathematical solutions have been developed,
    however, for two special types of unequal cash
    flows
  • Uniform Gradient Series
  • Geometric Gradient Series

59
Uniform Gradient Series
  • A Uniform Gradient Series (G) exists when cash
    flows either increase or decrease 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 factor
  • and is written as (A/G,i,n)
  • Therefore AT (A1) G(A/G,i,n))

60
Uniform Gradient Series
  • Example An engineer is planning for a 15 year
    retirement. In order to supplement his pension
    and offset the anticipated effects of inflation
    and increased 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 successive year. 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 P Given A1, G, i, and
    n

19000
18000
8000
7000
6000
5000
T 0
1
2
14
15
3
4
P
61
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

62
Geometric Gradient Series
  • Since receipts and expenditures rarely increase
    or decrease every period by a fixed amount,
    Uniform Gradient Series (G) problems have limited
    applicability
  • With Geometric Gradients, the increase or
    decrease in cash flows between periods 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 are sometimes used to
    account for inflationary cost increases
  • AK A (1
    j)K-1
  • Where j equals the percent change in the
    cash flow between periods
  • A is the cash flow in the initial period
  • AK is the cash flow in any subsequent period

63
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 (1i)-n / (i - j)
  • Nomenclature P A (P/A,i,j,n)

n
K 1
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