Engineering Economic Analysis
9th Edition
Chapter 3
INTEREST AND EQUIVALENCE
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
1
Economic Decision Components
• Where economic decisions are immediate we need to
consider:
• amount of expenditure
• taxes
• Where economic decisions occur over a considerable
period of time we also need to consider:
• interest
• inflation
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
2
Computing Cash Flows
• Cash flows have:
• Costs (disbursements) > a negative number
• Benefits (receipts) > a positive number
Example 3-1
End of
Year
0
1
2
Cash flow
$ (1,000.00)
$ 580.00
$ 580.00
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
3
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 period
you will receive back your original $100 plus $9
Original amount to be returned = $100
Interest to be returned = $100 x .09 = $9
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
4
Simple Interest
• Interest that is computed only on the original
sum 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
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
5
Future Value of a Loan with
Simple 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?
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
6
Compound Interest
• Interest that is computed on the original
unpaid debt and the unpaid interest
• Total interest earned = In = 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
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
7
Future Value of a Loan with
Compound 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 with
compound interest terms?
• Would a bank?
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
8
Comparison of Simple and Compound
Interest Over Time
Simple and compound interest
Single payment
•If you loaned a friend money
for short period of time the
difference between simple
and compound interest is
negligible.
•If you loaned a friend money
for a long period of time the
difference between simple
and compound interest may
amount to a considerable
difference.
Principal =
Interest =
Check the
table to
see the
difference
over time.
Short or long? When is the $ difference significant?
You pick the time period.
100.00
9.00%
Simple
Compound
Period amount factor amount factor
Find Fs
Given P
Find F Given P
n
Fs/P
F/P
0
100.000
100.000
1
109.000
109.000
2
118.000
118.810
3
127.000
129.503
4
136.000
141.158
5
145.000
153.862
6
154.000
167.710
7
163.000
182.804
8
172.000
199.256
9
181.000
217.189
10
190.000
236.736
11
199.000
258.043
12
208.000
281.266
13
217.000
306.580
14
226.000
334.173
15
235.000
364.248
16
244.000
397.031
17
253.000
432.763
18
262.000
471.712
19
271.000
514.166
20
280.000
560.441
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
9
Four Ways to Repay a Debt
Plan
Repay
Principal
1
3
Equal annual Interest on
installments unpaid balance
End of loan
Interest on
unpaid balance
Equal annual installments
4
End of loan
2
Repay Interest
Compound and
pay at end of
loan
Interest Earned
Declines
Constant
Declines at
increasing rate
Compounds at
increasing rate
until end of loan
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
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Loan Repayment – Four Options
Loan Repayment Option Calculator
This calculator is partially
complete. If you complete
the calculator you can
earn 10 bonus points for
your team.
$5,000 Principal
10.00% Interest rate (enter as .1 for 10%)
10 Years
Plan
1 Enter 1 through 4
Principal payment Equal annual installments
Interest payment EOY on unpaid principal
Amount
owed at the
beginning
of the year
Years
Interest
owed for
that year
Total owed
at the end
of year
Principal
payment
Total end
of year
payment
1
2
3
4
5
6
7
8
9
10
5,000
4,500
4,000
3,500
3,000
2,500
2,000
1,500
1,000
500
500
450
400
350
300
250
200
150
100
50
2,750
5,500
4,950
4,400
3,850
3,300
2,750
2,200
1,650
1,100
550
500
500
500
500
500
500
500
500
500
500
5,000
1,000
950
900
850
800
750
700
650
600
550
7,750
1
2
3
4
5
5,000
5,000
5,000
5,000
5,000
500
500
500
500
500
2,500
5,500
5,500
5,500
5,500
5,500
0
0
0
0
5,000
5,000
500
500
500
500
5,500
7,500
1
2
3
4
5
5,000
4,500
4,000
3,500
3,000
500
450
400
350
300
2,000
5,500
4,950
4,400
3,850
3,300
314
345
380
418
459
1,915
814
814
814
814
814
4,069
1
2
3
4
5
5,000
5,500
6,050
6,655
7,321
500
550
605
666
732
3,053
5,500
6,050
6,655
7,321
8,053
-500
-550
-605
-666
7,321
5,000
0
0
0
0
8,053
8,053
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
1
1
Equivalence
• When an organization is indifferent as to whether it
has a present sum of money now or the assurance of
some other sum of money (or series of sums of
money) in the future, we say that the present sum of
money is equivalent to the future sum or series of
sums.
Each of the plans on the previous slide is
equivalent because each repays $5000 at
the same 10% interest rate.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
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2
Given the choice of these two plans which
would you choose?
Year
Plan 1
Plan 2
1
2
3
$1400
1320
1240
$400
400
400
4
5
Total
1160
1080
$6200
400
5400
$7000
To make a choice the cash flows must be
altered so a comparison may be made.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
1
3
Technique of Equivalence
• Determine a single equivalent value at a point
in 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 the
two alternatives from the comparable
equivalent values.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
1
4
Repayment Plans
Establish the Interest Rate
Equivalence Calculator
1. Principal outstanding
over time
2. Amount repaid over
time
$5,000
8.00%
5
Plan
1
Principal payment Equal annual installments
Interest payment EOY on unpaid principal
Year
s
Interest
owed for
that year
Amount owed at the beginning of the year
1
2
3
4
5
Totals
5,000
4,000
3,000
2,000
1,000
Interest paid over time
Total owed over time
As an example:
400
320
240
160
80
1,200
1,200
15,000
5,400
4,320
3,240
2,160
1,080
=
8.00%
$4,876.63
9.00%
5
Plan
1
Principal payment Equal annual installments
Interest payment EOY on unpaid principal
i)n
If F = P (1 +
Then i=(F/P)1/n-1
Total owed at
the end of year
Year
s
1
2
3
4
5
Totals
Interest
owed for
that year
Amount owed at the beginning of the year
4,877
3,901
2,926
1,951
975
Total owed at
the end of year
439
351
263
176
88
1,317
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
5,316
4,252
3,189
2,126
1,063
1
5
Application of Equivalence Calculations
Pick an
alternative.
Which would
you choose?
Change the
interest rate.
What
happens at
8%,15%,3%?
Comparing alternatives
Interest rate
Year
0
1
2
3
4
5
6
7
8
9
10
A
$600
$115
$115
$115
$115
$115
$115
$115
$115
$115
$115
P
$1,306.63
A
$212.65
F
$3,389.05
10.00%
Alternative
B
C
-$600
-$850
-$115
-$80
-$115
-$80
-$115
-$80
-$115
-$80
-$115
-$80
-$115
-$80
-$115
-$80
-$115
-$80
-$115
-$80
-$115
-$80
($1,306.63) ($1,341.57)
($212.65)
D
$850
$80
$80
$80
$80
$80
$80
$80
$80
$80
$80
$1,341.57
($218.33)
$218.33
($3,389.05) ($3,479.68)
$3,479.68
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
Present
worth
Annual
worth
Future
worth
1
6
Interest Formulas
• To understand equivalence, the underlying
interest formulas must be analyzed.
• Notation:
I = Interest rate per interest period
n = Number of interest periods
P = Present sum of money (Present worth)
F = Future sum of money (Future worth)
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
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7
Single Payment Compound Interest
Year
Beginning
balance
Interest for
period
Ending
balance
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
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
1
8
Notation for
Calculating 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.
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
1
9
Notation for
Calculating a Present Value
• P=F(1/1+i)n=F(1+i)-n is the
single payment present worth factor.
• Functional notation:
P=F(P/F,i,n)
P=5000(P/F,6%,10)
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
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0
Compound Interest Factors
Examples
F=P(F,i,n)
P=F(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%
n
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
25
30
40
50
60
72
100
Single Amount Factor
Compound Amount Factor
Present Worth Factor
F/P
$1.00
$5,000.00
1.100
$5,500.00
1.210
$6,050.00
1.331
$6,655.00
1.464
$7,320.50
1.611
$8,052.55
1.772
$8,857.81
1.949
$9,743.59
2.144
$10,717.94
2.358
$11,789.74
2.594
$12,968.71
2.853
$14,265.58
3.138
$15,692.14
3.452
$17,261.36
3.797
$18,987.49
4.177
$20,886.24
4.595
$22,974.86
5.054
$25,272.35
5.560
$27,799.59
6.116
$30,579.55
6.727
$33,637.50
10.835
$54,173.53
17.449
$87,247.01
45.259
$226,296.28
117.391
$586,954.26
304.482
$1,522,408.20
955.594
$4,777,969.09
13,780.612 $68,903,061.70
P/F
$1.00
$8,052.55
0.90909
$7,320.50
0.82645
$6,655.00
0.75131
$6,050.00
0.68301
$5,500.00
0.62092
$5,000.00
0.56447
$4,545.45
0.51316
$4,132.23
0.46651
$3,756.57
0.42410
$3,415.07
0.38554
$3,104.61
0.35049
$2,822.37
0.31863
$2,565.79
0.28966
$2,332.54
0.26333
$2,120.49
0.23939
$1,927.72
0.21763
$1,752.47
0.19784
$1,593.15
0.17986
$1,448.32
0.16351
$1,316.66
0.14864
$1,196.96
0.09230
$743.22
0.05731
$461.48
0.02209
$177.92
0.00852
$68.60
0.00328
$26.45
0.00105
$8.43
0.00007
$0.58
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
2
1
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
Effective vs Nominal Interest Comparator
Nominal Interest rate
Effective Interest rate
Number of years
9.00%
9.42%
1.00
@
365
per year
Periods/year
Single Amount Factor
Compound Amount Factor
i
9.00%
0.02%
n
1.00
365
F/P
$1.00
$500.00
1.090
$545.00
1.094
$547.08
Present Worth Factor
P/F
$1.00
$547.08
0.99975
$501.91
0.91394
$500.00
Engineering Economic Analysis - Ninth Edition Newnan/Eschenbach/Lavelle Copyright 2004 by Oxford Unversity Press, Inc.
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2