Minggu 2_Interest and Equivalence

  • Upload
    kasmita

  • View
    222

  • Download
    0

Embed Size (px)

Citation preview

  • 8/13/2019 Minggu 2_Interest and Equivalence

    1/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 1

    Engineer ing Econo m ic Ana lys i s9th Edition

    Minggu 2 INTEREST AND EQUIVALENCE

    (Chapter 3)

  • 8/13/2019 Minggu 2_Interest and Equivalence

    2/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 2

    Economic Decision Components

    Where economic decisions are immediate we need toconsider: amount of expenditure

    taxes Where economic decisions occur over a considerable

    period of time we also need to consider: interest inflation

  • 8/13/2019 Minggu 2_Interest and Equivalence

    3/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 3

    Computing Cash Flows

    Cash flows have: Costs (disbursements) > a negative number Benefits (receipts) > a positive number

    Example 3-1

    End ofYear Cash flow

    0 (1,000.00)$1 580.00$2 580.00$

  • 8/13/2019 Minggu 2_Interest and Equivalence

    4/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 4

    Time Value of Money

    Money has value Money can be leased or rented

    The payment is called interest If you put $100 in a bank at 9% interest for one time periodyou will receive back your original $100 plus $9

    Original amount to be returned = $100Interest to be returned = $100 x .09 = $9

  • 8/13/2019 Minggu 2_Interest and Equivalence

    5/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 5

    Simple Interest

    Interest that is computed only on the originalsum or principal

    Total interest earned = I = P x i x n Where

    P present sum of money i interest rate n number of periods (years)

    I = $100 x .09/period x 2 periods = $18

  • 8/13/2019 Minggu 2_Interest and Equivalence

    6/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 6

    Future Value of a Loan withSimple Interest

    Amount of money due at the end of a loan F = P + P i n or F = P (1 + i n )

    Where F = future value

    F = $100 (1 + .09 x 2) = $118

    Would you accept payment with simple interest terms? Would a bank?

  • 8/13/2019 Minggu 2_Interest and Equivalence

    7/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 7

    Compound Interest

    Interest that is computed on the originalunpaid debt and the unpaid interest

    Total interest earned = I n = P (1+i) n - P Where

    P present sum of money i interest rate n number of periods (years)

    I2 = $100 x (1+.09) 2 - $100 = $18.81

  • 8/13/2019 Minggu 2_Interest and Equivalence

    8/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 8

    Future Value of a Loan withCompound Interest

    Amount of money due at the end of a loan F = P(1+i) 1(1+i)2..(1+i) n or F = P (1 + i) n

    Where F = future value

    F = $100 (1 + .09) 2 = $118.81

    Would you be more likely to accept payment withcompound interest terms? Would a bank?

  • 8/13/2019 Minggu 2_Interest and Equivalence

    9/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 9

    Comparison of Simple and CompoundInterest Over Time

    If you loaned a friend moneyfor short period of time thedifference between simpleand compound interest isnegligible.If you loaned a friend moneyfor a long period of time thedifference between simpleand compound interest mayamount to a considerabledifference.

    Short or long? When is the $ difference significant?You pick the t ime per iod.

    Simple and compound interestSingle payment

    Principal = 100.00Interest = 9.00%

    PeriodSimple

    amount factor Compound

    amount factor

    n

    Find FsGiven P

    Fs/P

    Find F Given P

    F/P0 100.000 100.0001 109.000 109.0002 118.000 118.8103 127.000 129.5034 136.000 141.1585 145.000 153.8626 154.000 167.7107 163.000 182.8048 172.000 199.2569 181.000 217.189

    10 190.000 236.73611 199.000 258.04312 208.000 281.26613 217.000 306.58014 226.000 334.17315 235.000 364.24816 244.000 397.03117 253.000 432.76318 262.000 471.71219 271.000 514.16620 280.000 560.441

    Check thetable tosee thedifferenceover time.

  • 8/13/2019 Minggu 2_Interest and Equivalence

    10/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 10

    Four Ways to Repay a Debt

    Plan RepayPrincipal

    Repay Interest Interest Earned

    1 Equal annualinstallments

    Interest onunpaid balance

    Declines

    2 End of loan Interest onunpaid balance

    Constant

    3 Equal annual installments Declines atincreasing rate

    4 End of loan Compound andpay at end ofloan

    Compounds atincreasing rateuntil end of loan

  • 8/13/2019 Minggu 2_Interest and Equivalence

    11/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 11

    Loan Repayment Four Options

    This calculator is partiallycomplete. If you completethe calculator you can

    earn 10 bonus points foryour team.

    $5,000 Principal10.00% Interest rate (enter as .1 for 10%)

    10 YearsPlan 1 Ent er 1 through 4

    Principal payment Equal annual installmentsInterest payment EOY on unpaid principal

    Years

    Amountowed at thebeginningof the year

    Interestowed forthat year

    Total owedat the end

    of year Principal payment

    Total endof year

    payment

    1 5,000 500 5,500 500 1,0002 4,500 450 4,950 500 9503 4,000 400 4,400 500 9004 3,500 350 3,850 500 8505 3,000 300 3,300 500 800

    6 2,500 250 2,750 500 7507 2,000 200 2,200 500 7008 1,500 150 1,650 500 6509 1,000 100 1,100 500 600

    10 500 50 550 500 5502,750 5,000 7,750

    1 5,000 500 5,500 0 5002 5,000 500 5,500 0 5003 5,000 500 5,500 0 5004 5,000 500 5,500 0 5005 5,000 500 5,500 5,000 5,500

    2,500 5,000 7,500

    1 5,000 500 5,500 314 8142 4,500 450 4,950 345 8143 4,000 400 4,400 380 8144 3,500 350 3,850 418 8145 3,000 300 3,300 459 814

    2,000 1,915 4,069

    1 5,000 500 5,500 -500 02 5,500 550 6,050 -550 03 6,050 605 6,655 -605 04 6,655 666 7,321 -666 05 7,321 732 8,053 7,321 8,053

    3,053 5,000 8,053

    Loan Repayment Option Calculator

  • 8/13/2019 Minggu 2_Interest and Equivalence

    12/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 12

    Equivalence (1)

    When an organization is indifferent as to whether ithas a present sum of money now or the assurance ofsome other sum of money (or series of sums of

    money) in the future, we say that the present sum ofmoney is equivalent to the future sum or series ofsums.

    Each of the plans on the previous slide isequivalent because each repays $5000 atthe same 10% interest rate.

  • 8/13/2019 Minggu 2_Interest and Equivalence

    13/32

    13

    Equivalence (2)

    Example : Four plans for repayment of $5000in five years at 8% interest :- Plan 1 : At end of each year pay $1000

    principle plus interest due.

    (1) (2) (3) (4) (5) (6)

    YearAmount owed

    at beginning ofyear

    Interest owedfor that year

    Total owed atend of year

    Principlepayment

    Total end ofyear payment

    [8% x (2)] [(2) + (3)}

    1 5000 400 5400 1000 1400

    2 4000 320 4320 1000 1320

    3 3000 240 3240 1000 1240

    4 2000 160 2160 1000 1160

    5 1000 80 1080 1000 1080

    Total 1200 5000 6200

  • 8/13/2019 Minggu 2_Interest and Equivalence

    14/32

    14

    Equivalence (3)- Plan 2 : Pay Interest due at end of each year

    and principal at end of five years(1) (2) (3) (4) (5) (6)

    Year

    Amountowed at

    beginning ofyear

    Interestowed forthat year

    Total owedat end of

    year

    Principle

    payment

    Total end ofyear

    payment

    [8% x (2)] [(2) + (3)}

    1 5000 400 5400 0 400

    2 5000 400 5400 0 400

    3 5000 400 5400 0 400

    4 5000 400 5400 0 400

    5 5000 400 5400 5000 5400

    Total 2000 5000 7000

  • 8/13/2019 Minggu 2_Interest and Equivalence

    15/32

    15

    Equivalence (4)- Plan 3 : Pay in five equal end-of-year

    payments.(1) (2) (3) (4) (5) (6)

    YearAmount owedat beginning

    of year

    Interestowed for

    that year

    Total owedat end of

    year

    Principle

    payment

    Total end ofyear

    payment[8% x (2)] [(2) + (3)}

    1 5000 400 5400 852 1252

    2 4148 331 4479 921 1252

    3 3227 258 3485 994 12524 2233 178 2411 1074 1252

    5 1159 93 1252 1159 1252

    Total 1260 5000 6260

  • 8/13/2019 Minggu 2_Interest and Equivalence

    16/32

    16

    Equivalence (5)- Plan 4 : Pay interest and principal at the end

    of periods.(1) (2) (3) (4) (5) (6)

    YearAmount owedat beginning

    of year

    Interestowed for

    that year

    Total owedat end of

    year

    Principle

    payment

    Total end ofyear

    payment[8% x (2)] [(2) + (3)}

    1 5000 400 5400 0 0

    2 5400 432 5832 0 0

    3 5832 467 6299 0 04 6299 504 6803 0 0

    5 6803 544 7347 5000 7347

    Total 2347 5000 7347

  • 8/13/2019 Minggu 2_Interest and Equivalence

    17/32

    17

    Equivalence (6) ratio = total interest paid /total amount owed

    at the beginning of year.

    Plan Total Interestpaid

    Total amountowed at the

    beginning of yearratio

    1 $ 1200 $ 15000 0,08

    2 2000 25000 0,08

    3 1260 15767 0,08

    4 2347 29334 0,08

    From our calculations, we more easily see why the repaymentplans require the payment of different total sums of money,yet are actually equivalent to each other.

  • 8/13/2019 Minggu 2_Interest and Equivalence

    18/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 18

    Given the choice of these two plans which

    would you choose?Year Plan 1 Plan 2

    1 $1400 $4002 1320 4003 1240 4004 1160 4005 1080 5400

    Total $6200 $7000

    To make a choice the cash flows must bealtered so a comparison may be made.

  • 8/13/2019 Minggu 2_Interest and Equivalence

    19/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 19

    Technique of Equivalence

    Determine a single equivalent value at a pointin time for plan 1.

    Determine a single equivalent value at a point

    in time for plan 2.

    Both at the same interest rate.

    Judge the relative attractiveness of thetwo alternatives from the comparableequivalent values.

  • 8/13/2019 Minggu 2_Interest and Equivalence

    20/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 20

    Repayment PlansEstablish the Interest Rate

    1. Principal outstandingover time

    2. Amount repaid overtime

    As an example:

    If F = P (1 + i)n

    Then i=(F/P) 1/n-1

    $5,0008.00%5

    Plan 1Principal payment Equal annual

    Interest payment EOY on unpai

    Year s Amount owed at the beginning o f the year

    Interestowed forthat year

    Total owed atthe end of year

    1 5,000 400 5,4002 4,000 320 4,3203 3,000 240 3,2404 2,000 160 2,1605 1,000 80 1,080

    Totals 1,200

    Interest paid over time 1,200Total owed over time 15,000

    $4,876.639.00%

    5Plan 1

    Principal payment Equal annualInterest payment EOY on unpai

    Year s Amount owed at the beginning o f the year

    Interestowed forthat year

    Total owed atthe end of year

    1 4,877 439 5,3162 3,901 351 4,2523 2,926 263 3,1894 1,951 176 2,1265 975 88 1,063

    Totals 1,317

    Equivalence Calculator

    = 8.00%

  • 8/13/2019 Minggu 2_Interest and Equivalence

    21/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 21

    Application of Equivalence Calculations

    Interest rate 10.00%

    Year A B C D0 $600 -$600 -$850 $8501 $115 -$115 -$80 $802 $115 -$115 -$80 $803 $115 -$115 -$80 $804 $115 -$115 -$80 $805 $115 -$115 -$80 $806 $115 -$115 -$80 $807 $115 -$115 -$80 $808 $115 -$115 -$80 $809 $115 -$115 -$80 $8010 $115 -$115 -$80 $80

    P $1,306.63 ($1,306.63) ($1,341.57) $1,341.57 Present

    worth

    A $212.65 ($212.65) ($218.33) $218.33 Annualworth

    F $3,389.05 ($3,389.05) ($3,479.68) $3,479.68 Futureworth

    Alternative

    Comparing alternatives

    Pick analternative.Which wouldyou choose?

    Change theinterest rate.Whathappens at8%,15%,3%?

  • 8/13/2019 Minggu 2_Interest and Equivalence

    22/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 22

    Interest Formulas

    To understand equivalence, the underlyinginterest formulas must be analyzed.

    Notation:I = Interest rate per interest periodn = Number of interest periodsP = Present sum of money (Present worth)F = Future sum of money (Future worth)

  • 8/13/2019 Minggu 2_Interest and Equivalence

    23/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 23

    Single Payment Compound Interest

    Year BeginningbalanceInterest for

    periodEndingbalance

    1 P iP P(1+i)

    2 P(1+i) iP(1+i) P(1+i) 2

    3 P(1+i) 2 iP(1+i)2 P(1+i) 3

    n P(1+i)n-1

    iP(1+i)n-1

    P(1+i)n

    P at time 0 increases to P(1+i) n at the end of time n.Or a Future sum = present sum (1+i) n

  • 8/13/2019 Minggu 2_Interest and Equivalence

    24/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 24

    Notation forCalculating a Future Value

    Formula:F=P(1+i) n is the

    single payment compound amount factor. Functional notation:

    F=P(F/P,i,n) F=5000(F/P,6%,10)

    F =P(F/P) which is dimensionally correct.

  • 8/13/2019 Minggu 2_Interest and Equivalence

    25/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 25

    Notation forCalculating a Present Value

    P=F(1/1+i) n=F(1+i) -n is thesingle payment present worth factor .

    Functional notation:P=F(P/F,i,n) P=5000(P/F,6%,10)

  • 8/13/2019 Minggu 2_Interest and Equivalence

    26/32

  • 8/13/2019 Minggu 2_Interest and Equivalence

    27/32

  • 8/13/2019 Minggu 2_Interest and Equivalence

    28/32

    28

    Single Payment Formulas (4)example :

    An investor (owner) has an option to purchase a tract ofland that will be worth $ 10.000 in six years. If the valueof the land increases at 8% each year, how much souldthe investor be willing to pay now for this property ?

    P = 10.000 (P/F,8%,6)= 10.000 (0,6302)= $6.302

    P = ?

    F = 10.000

    n = 6i = 0,08

    1 2 3 4 5 6

  • 8/13/2019 Minggu 2_Interest and Equivalence

    29/32

    MENCARI F DIKETAHUI P

    Rumus F = P (1+i) n Dengan Excel F =FV(i%,n,,-P) Dengan tabel F = P(F/P,i,n )

    Contoh:Bila sejumlah $500 didepositokan di bank, berapakah

    jumlahnya 3 tahun yang akan datang kalau bunga bank 6%per tahun?

    F = 500 (F/P,6%,3)= 500 (1,191)= $ 595,5

    F = FV(6%,3,,-500) = ????P = 500

    F = ?

    n = 3i = 0,06

    1 2 3

    29

  • 8/13/2019 Minggu 2_Interest and Equivalence

    30/32

    MENCARI P DIKETAHUI F

    Rumus Dengan Excel P =PV(i%,n,,-F) Dengan tabel P = F(P/F,i,n ) Contoh:Seorang investor dapat membeli tanah yang nilainya akan menjadi $

    10.000 dalam 6 tahun mendatang. Bila nilai tanah naik sebesar 8%setiap tahun, berapakah harga yang harus dibayar investor ini saatsekarang?

    P = 10.000 (P/F,8%,6) Dengan Excel:= 10.000 (0,6302) P=PV(8%,6,,-10000) = ????

    = $6.302

    1

    n P F i

    P = ?

    F = 10.000

    n = 6i = 0,08

    1 2 3 4 5 6

    30

  • 8/13/2019 Minggu 2_Interest and Equivalence

    31/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 31

    Examples

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

    F=$5000 i=0.10 n=5 P=?

    F=P(1+i) n=$5000(1+0.10) 5=$5000(1.611)=$8055

    F=P(F/P,10,5)=$5000(1.611) =$8055

    P=F(P/F,10,5)=$8055(.62092)=$5000

    10.00%

    n0 $1.00 $5,000.00 $1.00 $8,052.551 1.100 $5,500.00 0.90909 $7,320.502 1.210 $6,050.00 0.82645 $6,655.003 1.331 $6,655.00 0.75131 $6,050.004 1.464 $7,320.50 0.68301 $5,500.005 1.611 $8,052.55 0.62092 $5,000.006 1.772 $8,857.81 0.56447 $4,545.457 1.949 $9,743.59 0.51316 $4,132.238 2.144 $10,717.94 0.46651 $3,756.579 2.358 $11,789.74 0.42410 $3,415.07

    10 2.594 $12,968.71 0.38554 $3,104.6111 2.853 $14,265.58 0.35049 $2,822.3712 3.138 $15,692.14 0.31863 $2,565.7913 3.452 $17,261.36 0.28966 $2,332.5414 3.797 $18,987.49 0.26333 $2,120.4915 4.177 $20,886.24 0.23939 $1,927.7216 4.595 $22,974.86 0.21763 $1,752.4717 5.054 $25,272.35 0.19784 $1,593.1518 5.560 $27,799.59 0.17986 $1,448.3219 6.116 $30,579.55 0.16351 $1,316.6620 6.727 $33,637.50 0.14864 $1,196.9625 10.835 $54,173.53 0.09230 $743.2230 17.449 $87,247.01 0.05731 $461.4840 45.259 $226,296.28 0.02209 $177.9250 117.391 $586,954.26 0.00852 $68.6060 304.482 $1,522,408.20 0.00328 $26.4572 955.594 $4,777,969.09 0.00105 $8.43100 13,780.612 $68,903,061.70 0.00007 $0.58

    Compound Amount Factor

    F/P

    Single Amount Factor

    Compound Interest Factors

    Present Worth Factor

    P/F

  • 8/13/2019 Minggu 2_Interest and Equivalence

    32/32

    Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc. 32

    18% Compounded Monthly

    18% interest: Assume a yearly rate if not stated Compounded monthly: Indicates 12 periods/year [18%/year] / [12months/year] = 1.5% / month

    Nominal Interest rate 9.00% @ 365 Periods/year Effective Interest rate 9.42% per year

    Number of years 1.00

    i n $1.00 $500.00 $1.00 $547.089.00% 1.00 1.090 $545.00 0.99975 $501.910.02% 365 1.094 $547.08 0.91394 $500.00

    Effective vs Nominal Interest Comparator

    Compound Amount Factor

    F/P

    Single Amount Factor Present Worth Factor

    P/F