Minggu 2_Interest and Equivalence
Transcript of Minggu 2_Interest and Equivalence
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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)
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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
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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$2 580.00$
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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
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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
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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?
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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
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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?
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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.
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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
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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
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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.
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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
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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
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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
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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
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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.
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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.
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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.
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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%
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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%?
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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)
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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
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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.
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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)
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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
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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
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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
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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
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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