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Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

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Page 1: Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

Lecture No. 9Chapter 3

Contemporary Engineering EconomicsCopyright © 2010

Contemporary Engineering Economics, 5th edition, © 2010

Page 2: Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

Composite Cash Flows

Situation 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annual deposits of $100 in your savings account which earns a 10% annual interest, what equal annual amount (A) can be withdrawn over 4 subsequent years? Situation 2: What value of A would make the two cash flow transactions equivalent if i = 10%?

Contemporary Engineering Economics, 5th edition, © 2010

Page 3: Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

Establishing Economic Equivalence

Method 1: at n = 0 Method 2: At n = 4

Contemporary Engineering Economics, 5th edition, © 2010

Page 4: Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

Example 3.26 Cash Flows with Subpatterns

Given: Two cash flow transactions, and i = 12%

Find: C

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.

Contemporary Engineering Economics, 5th edition, © 2010

Page 5: Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

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

Contemporary Engineering Economics, 5th edition, © 2010

Page 6: Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

Cash Flows with Missing Payments

Given: Cash flow series with a missing payment, i = 10% Find: P

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.

Contemporary Engineering Economics, 5th edition, © 2010

Page 7: Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

Example 3.28 Calculating an Unknown Interest Rate

Given: Two payment options

Find: i at which the two options are equivalent

Excel Solution:

Contemporary Engineering Economics, 5th edition, © 2010

Option 1: Take a lump sum payment in the amount of $167M. Option 2: Take the 26-installment option.

Page 8: Lecture No. 9 Chapter 3 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010

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

Find: P

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

Solution:

Equivalence Calculations for a skipping cash flow pattern

Contemporary Engineering Economics, 5th edition, © 2010

Actually, $10,000 payment occurs every other year for 12 years at 10%.We can view this same cash flow seriesas having $10,000 payment occurs every periodat an interest rate of 21% over 6 years.