Lecture no09 unconventional equivalence calculations

Irregular Payment Series and
Unconventional Equivalence Calculations
Lecture No. 9
Chapter 3
Contemporary Engineering Economics
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Example 3.23: Uneven Payment Series
How much do you
need to deposit today
(P) to withdraw $25,000
at n = 1, $3,000 at n = 2,
and $5,000 at n = 4, if
your account earns 10%
annual interest?
$25,000
$3,000

$5,000

0
1

2

3

4

P
Contemporary Engineering Economics, 6 th edition
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Check to see if $28,622 is indeed
sufficient.
0

1

2

3

4

Beginning
Balance

0

28,622

6,484.20

4,132.62

4,545.88

Interest
Earned
(10%)

0

2,862

648.42

413.26

454.59

Payment

+28,622

−25,000

−3,000

0

−5,000

Ending
Balance

$28,622

6,484.20

4,132.62

4,545.88

0.47

Rounding error.
It should be “0.”
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Example 3.25: Future Value of an Uneven
Series with Varying Interest Rates
 Given: Deposit series as given over 5 years

 Find: Balance at the end of year 5

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Solution

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Composite Cash Flows
 Situation 1: If you make
4 annual deposits of $100
in your savings account,
which earns 10% annual
interest, what equal
annual amount (A) can be
withdrawn over 4
subsequent years?
 Situation 2: What value
of A would make the two
cash flow transactions

equivalent if i = 10%?

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Establishing Economic Equivalence
Method 1: At n = 0

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Method 2: At n = 4

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Example 3.26: Cash Flows with Sub-patterns
 Given: Two cash flow transactions, and i =
12%

 Find: C

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Solution
 Strategy: First select the
base period to use in
calculating the equivalent
value for each cash flow
series (say, n = 0). You can
choose any period as your
base period.

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Example 3.27: Establishing a College Fund
 Given: Annual college expenses = $40,000 a year
for 4 years, i = 7%, and N = 18 years

 Find: Required annual contribution (X)

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Solution
Strategy: It
would be
computationally
efficient if you
chose n = 18 (the
year she goes to
college) as the
base period.

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Cash Flows with Missing Payments
 Given: Cash flow series with a missing payment, i = 10%

 Find: P

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Solution
 Strategy: Pretend that we
have the 10th missing
payment so that we have a
standard uniform series. This
allows us to use (P/A,10%,15)
to find P. Then, we make an
adjustment to this P by
subtracting the equivalent
amount added in the 10th
period.

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Example 3.28: Calculating an Unknown Interest
Rate
 Given: Two payment options
o Option 1: Take a lump sum payment in the amount of $192,373,928.
o Option 2: Take the 30-installment option ($9,791,667 a year).

 Find: i at which the two options are equivalent

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Solution
$192,373,928  $9,791,667(P / A, i ,30)
(P / A, i ,30)  22.3965

 Excel Solution:

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Example 3.29: Unconventional Regularity in
Cash Flow Pattern
 Given: Payment series given, i = 10%, and N =
12 years

 Find: P

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Solution
 Strategy: Since the cash
flows occur every other year,
find out the equivalent
compound interest rate that
covers the two-year period.

• Equivalence Calculations for a
Skipping Cash Flow Pattern

Solution
Actually, the $10,000 payment occurs every other year
for 12 years at 10%.
We can view this same cash flow series as having a
$10,000 payment that occurs every period
at an interest rate of 21% over 6 years.

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