13
Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Embed Size (px)

Citation preview

Page 1: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Lecture No. 23Chapter 7

Contemporary Engineering EconomicsCopyright © 2010

Contemporary Engineering Economics, 5th edition, © 2010

Page 2: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Chapter Opening Story Mr. Clean Takes Car-Wash Gig:

Proctor & Gamble wants to enter car-wash franchise business - $35 billion industry.

Initial investment required by a franchise operator - $500,000.

Average monthly sales - $6,800 or $3,800 profit per month.

At Issue: What is the return on investment on car-wash franchise operation? Contemporary Engineering Economics, 5th edition © 2010

Page 3: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Rate of Return What it is: Interest earned on your invested capital, or commonly known as internal rate of return (IRR) A Simple Example: The interest earned on your savings account is the rate of return on your deposits.

Investopedia® says: IRRs can also be compared against prevailing rates of return in the securities market. If a firm can't find any projects with IRRs greater than the returns that can be generated in the financial markets, it may simply choose to invest its retained earnings into the market.

Contemporary Engineering Economics, 5th edition © 2010

Page 4: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Investing in Wal-Mart StockIn October 1,1970, when Wal-Mart Stores, Inc. went public, an investment of 100 shares cost $1,650. That investment would have been worth $10,053,632 on September 30, 2009. What is the rate of return on this investment?

Contemporary Engineering Economics, 5th edition © 2010

Page 5: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Wal-Mart Investment ProblemGiven:

P = $1,650F = $10,053,632N = 29 years

Find: iFormula to Use:

F = P(1 + i)N

$10,053,632 = $1,650(1 + i)29

i = 25.04%

Cash Flow Diagram

Contemporary Engineering Economics, 5th edition © 2010

$10,053,632

$1,650

19702009

Page 6: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

How Good the Wal-Mart Investment Was and What to Compare with?If you took out $1,650 from your savings account and invested in Wal-Mart stock, you could have

If you did not invest $1,650 in Wal-Mart stock, what could use your money for?

$10,053,632Or equivalent to earning

25.04% interest on your savings account.

If the best you could do was to leave the money in the savings account to earn 6% interest over 29 years, you will have $16,010.

What is the meaning of 6% interest? This will be your opportunity cost rate or minimum return required for any investment.

Contemporary Engineering Economics, 5th edition © 2010

Page 7: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Contemporary Engineering Economics, 5th edition © 2010

In 1970, as long as you could earn more than a 6% interest in another investment opportunity, you would take that investment. Therefore, that 6% is viewed as a minimum attractive rate of return (or required rate of return). This is the interest rate used in NPW analysis. So, to see if the proposed investment is a good one, you adopt the following decision rule.

ROR (25.04%) > MARR(6%)

Page 8: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Contemporary Engineering Economics, 5th edition © 2010

This project will bring in a 15% rate of return on investment.

This project will result in a net surplus of $10,000 in NPW.

Which statement is easier to understand?

Page 9: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Contemporary Engineering Economics, 5th edition © 2010

• Rate of return (ROR) is defined as the interest rate earned on the unpaid (outstanding) balance of an installment loan.

• Example: A bank lends $10,000 and receives annual repayment of $4,021 over 3 years. The bank is said to earn a return of 10% on its loan of $10,000.

Page 10: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Contemporary Engineering Economics, 5th edition © 2010

A = $10,000 (A/P, 10%, 3)= $4,021

Unpaid Return on Unpaid balance unpaid balanceat beg. balance Payment at the end

Year of year (10%) received of year

0123

-$10,000-$10,000

-$6,979-$3,656

-$1,000 -$698 -$366

+$4,021+$4,021+$4,021

-$10,000-$6,979-$3,656

0

A return of 10% on the amount still outstanding at the beginning of each year

Page 11: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Contemporary Engineering Economics, 5th edition © 2010

• Rate of return (ROR) is the break-even interest rate, i*, which equates the present worth of a project’s cash outflows to the present worth of its cash inflows.

• Mathematical Relation:

• Example:

* * *cash inflows cash outflowsPW ( ) PW ( ) PW ( )

0i i i

(10%) 10,000 $4,021( / ,10%,3) 0PW P A

Page 12: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Contemporary Engineering Economics, 5th edition © 2010

• The internal rate of return (IRR) is the interest rate earned on the unrecovered project balance of the investment such that, when the project terminates, the unrecovered project balance will be zero.

•Example: A company invests $10,000 in a computer system which results in equivalent annual labor savings of $4,021 over 3 years. The company is said to earn a return of 10% on its investment of $10,000.

Page 13: Lecture No. 23 Chapter 7 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5 th edition, © 2010

Return on Invested CapitalThe firm earns a 10% rate of return on funds that remain

internally invested in the project. Since the return is internal to the project, we call it internal rate of return.

Contemporary Engineering Economics, 5th edition © 2010

0 1 2 3

Beginning Project Balance -$10,000 -$6,979 -$3,656

Return on Invested Capital -$1,000 -$698 -$366

Payment Received -$10,000 $4,021 $4,021 $4,021

Ending Project Balance -$10,000 -$6,979 -$3,656 $0

Project Balance