CE 203
EEA Chap 3
Interest and Equivalence
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Time Value of Money
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The use of money has value
– Somebody will pay you to use your money
– You will pay others to use their money
Interest is “Rent ” for the use of money
When decisions involve “cash flows” over a
considerable length of time, economic
analysis should include the effects of:
– Interest
– Inflation
Pre-Course EEA Assessment

$1000 in savings account at 10% for 1 year?
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$1000 per year for 30 years at 10%?
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For typical “investment plan”, $1000 per
year for 30 years at 10% yields $164,494
(see formula for “sinking fund”)
Your time value of money?
Would you rather have $100 now or
–
–
–
–
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$100 a year from now?
$110 a year from now?
$120 a year from now?
$150 two years from now?
Simple Interest
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Interest computed only on the original sum
of money or “principal”
Total interest earned = I = P x i x n where
P = principal or present sum of money
i = interest rate per period
n = number of periods
Example: $1000 borrowed at 8% for two
years, simple interest
I = $1000 x .08 x 2 = $160
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Simple Interest
Future Value, F, of a Loan, P

F = P + P x in = P (1 + in) = P [1 +i(period)]

Example: $1000 borrowed at 8% for
five years, simple interest
F = $1000 (1 + .08(5)) = $1400
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Compound Interest
Interest on original sum + on interest
Example $1000 @ 10% per year, F
F1 (first year) = $1000 + $1000(0.1)
or $1000 (1 + 0.1) = $1100
F2 (second year) = $1100 + $1100(0.1)
or $1000(1+0.1)(1+0.1) = $1210
or $1000(1+0.1)2
F3 = $1000(1+0.1)3;
Fn = $1000(1+0.1)n
General: Fn = P(1+i)n
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Compound Interest
Interest computed on the original sum and
any unpaid interest
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Period
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
n
P(1 + i)2
P(1 + i)n-1
iP(1 + i)2
iP(1 + i)n-1
P(1 + i)3
P(1 + i)n
Compound Interest

Future Value, F, = P (1 + i)n
– P = principal or present sum of money
– i = interest rate per period
– n = number of periods (years, months, …)

Example: $1000 borrowed at 8% for five
years, compound interest, all of principal and
interest repaid in 5 years:
F = $1000 (1 +.08)5 = $1469.33
F = P (F/P, i, n) = $1000 (F/P, .08, 5) =
$1469.00 (see EEA p. 576)
Note: Simple interest
yield would be $1400
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Compound Interest

Total interest earned = In = P (1 + i)n - P
P = principal or present sum of money
i = interest rate
n = number of periods

Example: $1000 borrowed at 8% for five
years, compound interest
I = $1000 (1 +.08)5 - $1000 = $469.33
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Simple vs. Compound Interest
Future value, F, for P = $1000 at 8%
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Periods
F/P, simple i
F/P, compound i
1
$1080
$1080
2
$1160
$1166
3
$1240
$1260
4
$1320
$1360
5
$1400
$1469
10
$1800
$2159
15
$2200
$3172
20
$2600
$4661
Specification of Interest Rate, i
1) “8%” - assumed to mean per year and
compounded annually
2) “8% compounded quarterly” - 2% per
each 3 months, compounded every 3
months
3) “8% compounded monthly” – 2/3% each
month, compounded every month
4) “8%” = .08 in equations NB1000
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In-class Example
Would you rather have $5000 today or
$35,000 in 25 years if the interest rate
was 8% compounded yearly/ monthly…
1) yearly?
F = P(F/P, .08, 25) = 5000 (6.848) = $34, 240
2) monthly?
F = P (F/P, .006667, 300) = $36,700
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Four Ways to Repay a Debt
Plan
Principal
repaid …
Interest paid
…
Trend on
interest
earned
1
in equal
annual
installments
on unpaid
balance
declines
2
at end of loan
3
in equal annual installments
4
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on unpaid
balance
at end of loan
constant
declines at
increasing rate
at end of loan
increases at
(compounded) increasing rate
Loan Repayment Plan 1 ($5000, 5 years, 8%)
Owed at
Annual
Year beginInterest
ning
Total
Principal
owed at
Paid
end
Total
payment
1
$5000
$400
$5400
$1000
$1400
2
3
$4000
$3000
$320
$240
$4320
$3240
$1000
$1000
$1320
$1240
4
5
$2000
$1000
$160
$80
$1200
$2160
$1080
$1000
$1000
$5000
$1160
$1080
$6200
Bank loans with yearly payments:
Principal repaid in equal installments
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Loan Repayment Plan 2 ($5000, 5 years, 8%)
Owed at
Annual
Year beginInterest
ning
Total
Princiowed at
pal Paid
end
Total
payment
1
$5000
$400
$5400
$0
$400
2
3
$5000
$5000
$400
$400
$5400
$ 5400
$0
$0
$400
$400
4
5
$5000
$5000
$400
$400
$2000
$ 5400
$ 5400
$0
$5000
$5000
$400
$5400
$7000
Interest only loans:
used in bonds and international loans
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Loan Repayment Plan 3 ($5000, 5 years, 8%)
Owed at
Annual
Year beginInterest
ning
Total
Princiowed at
pal Paid
end
Total
payment
1
$5000
$400
$5400
$852
$1252
2
3
$4148
$3227
$331
$258
$4479
$ 3485
$921
$994
$1252
$1252
4
5
$2233
$1159
$178
$93
$2000
$ 2411
$ 1252
$1074
$1159
$5000
$1252
$1252
$6260
Equal annual installments:
Auto/home loans (but usually monthly payments)
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Loan Repayment Plan 4 ($5000, 5 years, 8%)
Owed at
Annual
Year beginInterest
ning
Total
Princiowed at
pal Paid
end
Total
payment
1
$5000
$400
$5400
$0
$0
2
3
$5400
$5832
$432
$467
$5832
$ 6299
$0
$0
$0
$0
4
5
$6299
$6803
$504
$544
$2347
$ 6804
$ 7347
$0
$5000
$5000
$0
$7347
$7347
Interest and principal repaid at end of loan:
Certificates of deposit (CD’s) and IRA’s
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Loan repayment plans 1-4
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Are all equivalent to $5000 now in
terms of time value of money,
May not be equally attractive to
loaner or borrower
Have different cash flow diagrams
“Equivalent in nature but different in
structure”
Equivalence
The present sum of money is
equivalent to the future sum(s)
(from our perspective), if..
..we are indifferent as to whether we
have a quantity of money now or the
assurance of some sum (or series of
sums) of money in the future (with
adequate compensation)
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Equivalence


Used to make a meaningful
engineering economic analysis
Apply by finding equivalent value at
a common time for all alternatives
– value now or “Present Worth”
– value at some logical future time or
“Future Worth”
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Assume the same time value of money
(interest rate) for all alternatives
You borrow $8000 to help pay for senior year
at ISU – the bank offers you two repayment
plans for paying off the loan in four years:
Year
Plan 1
payment
Plan 2
payment
1
$0
$0
2
3
4
Total
$0
$0
$10,400
$10,400
$3300
$3300
$3300
$9900
Which would you choose?
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Comparing Economic Alternatives
“Technique of Equivalence” requires that we
1)Determine our “time value of money”
2)Determine a single equivalent value at a
selected time for Plan 1
3) Determine a single equivalent value at
the same selected time for Plan 2
4) Compare the two values
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Technique of Equivalence
Cash flows can be compared by calculating their
total Present Worth (“now value”) or their total
Future Worth (at some time in the future)
– Future Worth, F = P (1 + i)n
– Solving the previous equation for P gives
Present Worth, P = F (1 + i)-n
» P = some present amount of money
» F = some future amount of money
» i = interest rate per time period
(appropriate time value of money)
» n = number of time periods
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Technique of Equivalence
To compare payment plans, “move” all
amounts to now (Present Worth)
or to 4 years hence (Future Worth)
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Year
Plan 1
payment
Plan 2
payment
1
$0
$0
2
3
4
Total
$0
$0
$10,400
$10,400
$3300
$3300
$3300
$9900
Technique of Equivalence – Present Worth
Assume our “time value of money” is 7% APR
PW?
Present Worth:
Payment Plan 1
P = F(1 + i)-n
1
2
3
4
0
P4 = 10,400(1 + .07)-4 = 7,934.11
Total PW
= $7,934.11
Or P = $10,400 (P/F, .07, 4) = .7629 (10,400) = $7934.16
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$10,400
Technique of Equivalence – Present Worth
Assume our “time
value of money” is
7% APR
Present Worth:
Payment Plan 1
P = F(1 + i)-n
PW?
1
CCEE
3
4
0
P2 = 3300(1 + .07)-2 = 2882.35
P3 = 3300(1 + .07)-3 = 2693.78
P4 = 3300(1 + .07)-4 = 2517.55
Total PW
= $8093.68
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2
$3300 $3300 $3300
Technique of Equivalence – Present Worth
Which would you choose?
$7934.11
Plan 1
$8093.68
1
2
3
4
0
Plan 2
1
2
3
4
0
$3300 $3300 $3300
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$10,400
Technique of Equivalence – Future Worth
Assume our “time value of money” is 7% APR
Future Worth:
Payment Plan 1
F = P(1 + i)n
1
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3
4
0
F4 = 10,400(1 + .07)0 = 10,400
Total FW
2
= $10,400.00
$10,400
Technique of Equivalence – Future Worth
Assume our “time value of money” is 7% APR
1
Future Worth:
Payment Plan 2
P = F(1 + i)-n
F2 = 3300(1 + .07)2
F3 = 3300(1 + .07)1
F4 = 3300(1 + .07)0
Total FW
CCEE
3
4
0
=
3778.17
=
3531.00
=
3300.00
= $10609.17
Again, larger than Plan 1
ISU
2
$3300 $3300 $3300
FW?
In-class Example
You deposit $3000 in an account that earns
5%, compounded daily.
How much could you withdraw from the
account at the end of two years?
F = 3000 (F/P, .05/365, 730) = $3,315.49
How much could you withdraw if the
compounding were monthly?
F = 3000 (F/P, .05/12, 24) = $3,314.82
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In-class Example
You are considering two designs for a wastewater
treatment facility.
Plan 1: Costs $1,700,000 to construct and will
have to be replaced every 20 years.
Plan 2: Costs $2,100,000 to construct and will
have to be replaced every 30 years.
Which is the better of the two designs?
Assume 7% APR and neglect inflation and
operating, maintenance and disposal costs.
Sketch CFD’s for each plan and compare.
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How to compare two schemes with
different lives
One plant lasts 20y and the other 30y
 Could we make a comparison over 20y?
 Could we make a comparison over 30y?
 What would be a decent period over
which to compare?
The smallest common denominator,
i.e. in this case 60y.
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Technique of Equivalence – Present Worth
Based on dollars only, which is the better plan?
$2.254 M
(if provision is made for final replacement, $ 2.283M)
10
Plan 1
20
30
40
50
0
$1.7 M
$1.7 M
$1.7 M
P = $ 1.7 + $ 1.7 (1.07)-20 + $ 1.7 (1.07)-40
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60
repeats
[$1.7 M]*
[ + $ 1.7 (1.07)-60]*
•This term will only apply when it is a requirement to replace at the end, not normally and it
has been excluded from this analysis. You could get to 60 years without a final replacement.
Technique of Equivalence – Present Worth
Based on dollars only, which is the better plan?
$2.376 M
10
Plan 2
= P = 2.1 + 2.1(1.07)-30
20
30
40
$2.1 M
Other option was $2.254 M
CCEE
60
0
$2.1 M
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50
+ the 60y replacement, which does not apply
Not
required
repeats