Contemporary Engineering Economics, 4 th edition, © 2007 Unconventional Equivalence Calculations...

Contemporary Engineering

Economics, 4th edition, © 2007

Unconventional Equivalence Calculations

Lecture No. 9Chapter 3Contemporary Engineering EconomicsCopyright © 2006



Equivalent Present Worth Calculation – Brute Force Approach using Only P/F Factors



$50

$100 $100 $100

$150 $150 $150 $150$200

Group 1 $50( / ,15%,1)

$43.48

P P F

Group 2 $100( / ,15%,3)( / ,15%,1)

$198.54

P P A P F

Group 3 $150( / ,15%,4)( / ,15%,4)

$244.85

P P A P F

Group 4 $200( / ,15%,9)

$56.85

P P F

$43.48 $198.54 $244.85 $56.85

$543.72

P

01 2 3 4 5 6 7 8 9

Equivalent Present Worth Calculation – Grouping Approach



Unconventional Equivalence Calculations – A Personal Savings ProblemSituation 1: If you make

4 annual deposits of $100 in your savings account which earns a 10% annual interest, what equal annual amount (A) can be withdrawn over 4 subsequent years?



Unconventional Equivalence Calculations – An Economic Equivalence Problem Situation 2:

What value of A would make the two cash flow transactions equivalent if i = 10%?



Method 1: Establish the Economic Equivalence at n = 0



Method 2: Establish the Economic Equivalence at n = 4



Multiple Interest Rates

$300$500

$400

5% 6% 6% 4% 4%

Find the balance at the end of year 5.

0

12

34 5

F = ?



Solution1:

$300( / ,5%,1) $315

2 :

$315( / ,6%,1) $500 $833.90

3:

$833.90( / ,6%,1) $883.93

4 :

$883.93( / , 4%,1) $400 $1,319.29

5 :

$1,319.29( / , 4%,1) $1,372.06

n

F P

n

F P

n

F P

n

F P

n

F P



Cash Flows with Missing Payments

P = ?

$100

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Missing paymenti = 10%



Solution

P = ?

$100

01 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Pretend that we have the 10th

Payment in the amount of $100i = 10%

$100Add $100 tooffset the change



Approach

P = ?

$100

01 2 3 4 5 6 7 8 9 10 11 12 13 14 15

i = 10%

$100

Equivalent Cash Inflow = Equivalent Cash Outflow



$100( / ,10%,10) $100( / ,10%,15)

$38.55 $760.61

$722.05

P P F P A

P

P

Equivalence Relationship



Unconventional Regularity in Cash Flow Pattern

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

C C C C C C C

i = 10%

Payments are made every other year



Approach 1: Modify the Original Cash Flows

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

i = 10%

A A A A A A A A A A A A A A

$10,000( / ,10%,14)

$1,357.46

A A P



Relationship Between A and C

$10,000

01 2 3 4 5 6 7 8 9 10 11 12 13 14

i = 10%

A A A A A A A A A A A A A A

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

C C C C C C C

i = 10%



C

A =$1,357.46

A A

i = 10%

$10,000( / ,10%,14)

$1,357.46

( / ,10%,1)

1.1

2.1

2.1($1,357.46)

$2,850.67

A A P

C A F P A

A A

A

Solution



Approach 2: Modify the Interest Rate Idea: Since cash flows occur every other

year, let's find out the equivalent compound interest rate that covers the two-year period.

How: If interest is compounded 10% annually, the equivalent interest rate for two-year period is 21%.

(1+0.10)(1+0.10) = 1.21



Solution

$10,000

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14

C C C C C C C

i = 21%

$10,000( / , 21%,7)

$2,850.67

C A P

1 2 3 4 5 6 7



Example 3.25 – At What Value of C would Make the Two Cash Flows Equivalent?

1

2

$100( / ,12%,2) $300( / ,12%,3) $932.55

( / ,12%,2) ( / ,12%,2)( / ,12%,1) 3.6290

3.6290 $932.55

$256.97

V F A P A

V C F A C P A P F C

C

C



Example 3.26 Establishing a College Fund



Solution: Establish the Economic Equivalence at n = 18



Example 3.27 Calculating an Unknown Interest Rate with Multiple Factors



Establish an economic Equivalence at n =7



Linear Interpolation to Find an Unknown Interest Rate



Linear Interpolation

( / , ,7)1

( / , ,13)

( / , ,7)

( / , ,13)

6% 0.9482

? 1.0000

7% 1.0355

1 0.94826% (7% 6%)

1.0355 0.9482

6.5934%

F A i

P A i

F A ii

P A i

i



Using the Goal Seek Function to Find the Unknown Interest rate

Contemporary Engineering Economics, 4 th edition, © 2007 Unconventional Equivalence Calculations...

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