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ENGINEERING
ECONOMIC
ANALYSIS, 12/e
Newnan, Lavelle, and Eschenbach
Copyright © 2014 by Oxford University Press
Chapter 3
Interest and Equivalence
Copyright Oxford University Press 2014
Chapter Outline
•
•
•
•
Time Value of Money
Interest Calculations
Cash Flow Equivalence
Single Payment Compound Interest
Formulas
• Nominal and Effective Interest Rates
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Learning Objectives
• Understand the concept of “time value of
money”
• Distinguish between simple and
compound interest
• Understand the concept of “equivalence”
of cash flows
• Solve problems using Single Payment
Compound Interest Formulas
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Computing Cash Flows
• Question: Would you rather
• Receive $1000 today; or
• Receive $1000 10 years from today?
• Answer: Of course today!
• Why?
• I could invest $1000 today to make more money
• I could buy a lot of stuff today with $1000
• Who knows what will happen in 10 years
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Computing Cash Flows
• Because money is more valuable today than in
the future, we need to describe cash receipts
and disbursements at time they occur.
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Example 3-1
Cash flows of 2 payment options
To purchase a new $30,000 machine,
• Pay the full price now minus a 3% discount; or
• Pay $5000 now; $8000 at the end of year 1;
and $6000 at the end of each of the next 4
years
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Example 3-1
Cash flows of 2 payment options
Pay in 5 years
Pay in full
End of
Year
0 (now)
1
2
3
4
5
0
1
End of
Year
0 (now)
1
2
3
4
5
Cash Flow
-$29,100
0
0
0
0
0
2
3
4
5
0
1
Cash Flow
-$5,000
-8,000
-6,000
-6,000
-6,000
-6,000
2
$5,000
$29,100
$8,000
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3
4
5
Example 3-2
Cash flow for repayment of a loan
To repay a loan of $1,000 at 8% interest in 2 years
• Repay half of $1000 plus interest at the end of each year
$1000
Yr Interest Balance Repayment
0
1000
Cash
Flow
1000
1
2
-580
-540
80
40
500
0
500
500
0
1
$580
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2
$540
Time Value of Money
•
•
•
•
Money has purchasing power
Money has earning power
Money is a valuable asset
People are will to pay some charges (interests)
to have money available for their use
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Simple Interest
Interest is computed only on the original sum, and
not on accrued interest
𝑇𝑜𝑡𝑎𝑙 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑒𝑎𝑟𝑛𝑒𝑑 = 𝑃 × 𝑖 × 𝑛
(3-1)
Where P = Principal
i = Simple annual interest rate
n = Number of years
Amount of money at the end of 𝑛 years,
𝐹 =𝑃+𝑃×𝑖×𝑛
Where F = Amount due at the end of n years
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( 3-2)
Example 3-3
Simple Interest Calculation
Loan of $5000 for 5 yrs at simple interest rate of 8%
Total interest earned = $5000(8%)(5) = $2000
Amount due at end of loan = $5000 + 2000 = $7000
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Compound Interest
• Interest is computed on the unpaid balance,
which includes the principal and any unpaid
interest from the preceding period
• Common practice for interest calculation, unless
specifically stated otherwise
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Example 3-4
Compound Interest Calculation
• Loan of $5000 for 5 yrs at interest rate of 8%
Year
Balance at the
Beginning of the year
Interest
Balance at the end
of the year
1
$5,000.00
$400.00
$5,400.00
2
$5,400.00
$432.00
$5,832.00
3
$5,832.00
$466.56
$6,298.56
4
$6,298.56
$503.88
$6,802.44
5
$6,802.44
$544.20
$7,346.64
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Repaying a Debt
Plan #1: Constant Principal
• Repay of a loan of $5000 in 5 yrs at interest rate of 8%
• Plan #1: Constant principal payment plus interest due
Yr
Balance at
the
Beginning
of year
1
Interest
Balance at
the end of
year
Interest
Payment
Principal
Payment
Total
Payment
$5,000.00
$400.00
$5,400.00
$400.00
$1,000.00
$1,400.00
2
$4,000.00
$320.00
$4,320.00
$320.00
$1,000.00
$1,320.00
3
$3,000.00
$240.00
$3,240.00
$240.00
$1,000.00
$1,240.00
4
$2,000.00
$160.00
$2,160.00
$160.00
$1,000.00
$1,160.00
5
$1,000.00
$80.00
$1,080.00
$80.00
$1,000.00
$1,080.00
$1,200.00
$5,000.00
$6,200.00
Subtotal
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Repaying a Debt
Plan #2: Interest Only
• Repay of a loan of $5000 in 5 yrs at interest rate of 8%
• Plan #2: Annual interest payment and principal payment at end of 5 yrs
Yr
Balance at
the
Beginning
of year
Interest
Balance at
the end of
year
1
$5,000.00
$400.00
$5,400.00
$400.00
$0.00
$400.00
2
$5,000.00
$400.00
$5,400.00
$400.00
$0.00
$400.00
3
$5,000.00
$400.00
$5,400.00
$400.00
$0.00
$400.00
4
$5,000.00
$400.00
$5,400.00
$400.00
$0.00
$400.00
5
$5,000.00
$400.00
$5,400.00
$400.00
$5,000.00
$5,400.00
$2,000.00
$5,000.00
$7,000.00
Subtotal
Interest
Payment
Principal
Payment
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Total
Payment
Repaying a Debt
Plan #3: Constant Payment
• Repay of a loan of $5000 in 5 yrs at interest rate of 8%
• Plan #3: Constant annual payments
Yr
Balance at
the
Beginning
of year
Interest
Balance at
the end of
year
1
$5,000.00
$400.00
$5,400.00
$400.00
$852.28
$1,252.28
2
$4,147.72
$331.82
$4,479.54
$331.82
$920.46
$1,252.28
3
$3,227.25
$258.18
$3,485.43
$258.18
$994.10
$1,252.28
4
$2,233.15
$178.65
$2,411.80
$178.65
$1,073.63
$1,252.28
5
$1,159.52
$92.76
$1,252.28
$92.76
$1,159.52
$1,252.28
$1,261.41
$5,000.00
$6,261.41
Subtotal
Interest
Payment
Principal
Payment
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Total
Payment
Repaying a Debt
Plan #4: All at Maturity
• Repay of a loan of $5000 in 5 yrs at interest rate of 8%
• Plan #4: All payment at end of 5 years
Yr
Balance at
the
Beginning
of year
Interest
Balance at
the end of
year
1
$5,000.00
$400.00
$5,400.00
$0.00
$0.00
$0.00
2
$5,400.00
$432.00
$5,832.00
$0.00
$0.00
$0.00
3
$5,832.00
$466.56
$6,298.56
$0.00
$0.00
$0.00
4
$6,298.56
$503.88
$6,802.44
$0.00
$0.00
$0.00
5
$6,802.44
$544.20
$7,346.64
$2,346.64
$5,000.00
$7,346.64
$2,346.64
$5,000.00
$7,346.64
Subtotal
Interest
Payment
Principal
Payment
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Total
Payment
A Closer Look at the 4 Repayment
• Differences:
• Repayment structure (repayment amounts at various
points in time)
• Total payment amount
• Similarities:
• All interest charges were calculated at 8%
• They all achieved the same purpose of repaying the
loan within 5 years
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Equivalence
• If a firm believes 8% was reasonable, it would
have no preference about whether it received
$5000 now or was paid by any of the 4
repayment plans.
• The 4 repayment plans are equivalent to one
another and to $5000 now at 8% interest
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Equivalence
• ratio = total interest paid /total amount owed
at the beginning of year.
Plan
Total Interest
paid
Total amount
owed at the
beginning of year
ratio
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 repayment
plans require the payment of different total sums of money,
yet are actually equivalent to each other.
21
Use of Equivalence in
Engineering Economic Studies
• Using the concept of equivalence, one can
convert different types of cash flows at different
points of time to an equivalent value at a
common reference point
• Equivalence is dependent on interest rate
Copyright Oxford University Press 2014
Given the choice of these two plans which
would you choose?
Year
1
2
3
4
5
Total
Plan 1
Plan 2
$1400
1320
1240
1160
1080
$6200
$400
400
400
400
5400
$7000
To make a choice the cash flows must be altered so a
comparison may be made.
23
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.
24
Single Payment
Compound Interest Formulas
Notation:
𝑖 = interest rate per compounding period
𝑛 = number of compounding periods
𝑃 = a present sum of money
𝐹 = a future sum of money
Single Payment Compound Amount Formula
𝐹 = 𝑃(1 + 𝑖)𝑛
𝐹 = 𝑃(𝐹 𝑃 , 𝑖, 𝑛)
Find F, given P, at 𝑖, over 𝑛
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(3-3)
(3-4)
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
26
Example 3-5 Single Payment
Compound Interest Formulas
$500 were deposited in a saving account (pays 6%
compounded annually) for 3 years
F=?
F = P(1+i)n = 500(1+0.06)3
= $595.50
0
1
i=6%
P=500
2
3
Or
F = P(F/P, i, n) = 500(F/P, 6%, 3)
= 500(1.191) = $595.50
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Single Payment
Compound Interest Formulas
Notation:
𝑖 = interest rate per compounding period
𝑛 = number of compounding periods
𝑃 = a present sum of money
𝐹 = a future sum of money
Single Payment Present Worth Formula
𝑃 = 𝐹(1 + 𝑖)−𝑛
𝑃 = 𝐹(𝑃 𝐹, 𝑖, 𝑛)
Find P, given F, at 𝑖, over 𝑛
Copyright Oxford University Press 2014
(3-5)
(3-6)
Example 3-6 Single Payment
Compound Interest Formulas
Wish to have $800 at the end of 4 years, how much
should be deposited in an account that pays 5% annually?
F=800
P = F(1+i)-n = 800(1+0.05)-4
= $658.16
0
1
2
i=5%
P=?
3
4
Or
P = F(P/F, i, n) = 800(P/F, 5%, 4)
= 800(0.8227) = $658.16
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Example 3-7 Single Payment
Compound Interest Formulas
Tabulate the future value factor for interest rates of
5%, 10%, and 15% for n’s from 0 to 20 (in 5’s).
18,000
15%
16,000
n
0
5
10
15
20
5%
10%
15%
1.000 1.000 1.000
1.276 1.611 2.011
1.629 2.594 4.046
2.079 4.177 8.137
2.653 6.727 16.367
14,000
12,000
10,000
8,000
10%
6,000
4,000
5%
2,000
0,000
0
5
10
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15
20
25
Example 3-8 Single Payment
Compound Interest Formulas
$100 were deposited in a saving account (pays 6%
compounded quarterly) for 1 years
F=?
iqtr =1.5%, n = 4 quarters
F = P(1+i)n = 100(1+0.015)4
0
1
2
i=1.5%
P=100
3
4
= $106.14
F = P(F/P, i, n) = 100(F/P, 1.5%, 4)
= 100(1.061) = $106.10
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Nominal and Effective
Interest Rates
Notation:
𝑟 = Nominal interest rate per year without considering the
effect of any compounding
𝑖 = Effective interest rate per compounding period
𝑖𝑎 = Effective annual interest rate taking into account the
effect of compounding
𝑚 = Number of compounding periods per year
𝑟
𝑟 𝑚
𝑖 = , 𝑖𝑎 = (1 + ) − 1 = 1 + 𝑖
𝑚
𝑚
For Continuous Compounding
𝑖 = 𝑒𝑟 − 1
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𝑚
−1
(3-7)
(3-8)
(3-9)
Example 3-9 Nominal and
Effective Interest Rates
If a credit card charges 1.5% interest every month, what
are the nominal and effective interest rates per year?
𝑟 = 12 × 1.5% = 18%
𝑟 𝑚
0.18 12
𝑖𝑎 = (1 + ) − 1 = (1 +
) − 1 = 0.1956
𝑚
12
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
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Example 3-10 Application of
Nominal and Effective Interest Rates
“If I give you $100 today, you will write me a check
for $120, which you will redeem or I will cash on your
next payday.”
a) Bi-weekly interest rate = ($120-100)/100 = 20%
Nominal annual rate = 20% * 26 = 520%
b) Effective annual rate
26
𝑟 𝑚
520%
𝑖𝑎 = (1 + ) − 1 = 1 +
− 1 = 113.48
𝑚
26
c) End-of-the-year balance
𝐹 = 𝑃(1 + 𝑖)𝑛 = 100(1 + 0.20)26 = $11,448
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Nominal and Effective Interest
(Table 3-3)
Nominal
Rate
Effective Annual Rate when compounded
Yearly
Semiannually
Quarterly
Monthly
Daily
Continuously
1%
1%
1.0025%
1.0038%
1.0046%
1.0050%
1.0050%
2%
2%
2.0100%
2.0151%
2.0184%
2.0201%
2.0201%
3%
3%
3.0225%
3.0339%
3.0416%
3.0453%
3.0455%
4%
4%
4.0400%
4.0604%
4.0742%
4.0808%
4.0811%
5%
5%
5.0625%
5.0945%
5.1162%
5.1267%
5.1271%
6%
6%
6.0900%
6.1364%
6.1678%
6.1831%
6.1837%
8%
8%
8.1600%
8.2432%
8.3000%
8.3278%
8.3287%
10%
10%
10.2500%
10.3813%
10.4713% 10.5156%
10.5171%
15%
15%
15.5625%
15.8650%
16.0755% 16.1798%
16.1834%
25%
25%
26.5625%
27.4429%
28.0732% 28.3916%
28.4025%
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Continuous Compounding
Interest Formulas
Effective annual interest rate
𝑟
𝑖𝑎 = 𝑒 − 1
(3-9)
Single Payment Compound Amount
𝐹 = 𝑃 𝐹 𝑃 , 𝑟, 𝑛 = 𝑃(𝑒 𝑟𝑛 )
(3-15)
Single Payment Present Worth
𝑃 = 𝐹[𝑃 𝐹, 𝑟, 𝑛] = 𝐹(𝑒 −𝑟𝑛 )
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(3-16)
Example 3-11 Application of
Continuous Compounding
How much would be in the account for $2000
deposit in a bank that pays 5% nominal interest,
compounding continuously?
𝐹 = 𝑃 𝑒 𝑟𝑛 = 2000𝑒 (0.05)(2) = $2210.34
or
𝑖𝑎 = 𝑒 𝑟 − 1 = 𝑒 0.05 − 1 = 5.1271%
𝐹 = 𝑃(1 + 𝑖)2 = 2000(1 + 5.1271%)2 = $2210.34
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Example 3-13 Application of
Continuous Compounding
How long will it take for money to double at 10%
nominal interest, compounding continuously?
𝐹 = 𝑃 𝑒 𝑟𝑛 = 𝑒 (0.10)(𝑛) = 2
0.10 𝑛 = 𝑙𝑛2 = 0.693
𝑛 = 6.93 𝑦𝑒𝑎𝑟𝑠
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