Lecture no09 unconventional equivalence calculations

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Contemporary Engineering Economics, 6 th edition

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?

0

1 2 3 4

$25,000

$3,000 $5,000

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Check to see if $28,622 is indeed

sufficient.

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

 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withdrawn over 4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

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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 10 th 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 10 th

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

 Excel Solution:

$192,373,928 $9,791,667( / , ,30)

( / , ,30) 22.3965

P A i



15

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Example 3.29: Unconventional Regularity in

Cash Flow Pattern

 Given : Payment series given, i = 10%, and N =

12 years

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Solution

• Equivalence Calculations for a

Skipping Cash Flow Pattern

 Strategy : Since the cash

flows occur every other year,

find out the equivalent

compound interest rate that

covers the two-year period.

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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