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Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010
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Page 1: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Lecture No. 35Chapter 11

Contemporary Engineering EconomicsCopyright © 2010

Contemporary Engineering Economics, 5th edition, © 2010

Page 2: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Chapter Opening Story

How Much Will It Cost to Send Your Child to College in Year 2015? A year in college cost $17,800 in 2005.Due to inflation, the college expense has been increasing at a rate of 6.5% annually.Then, in 2015 a year in college would cost about $33,413.

Contemporary Engineering Economics, 5th edition, © 2010

College Cost Calculator

Page 3: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Inflation and Economic Analysis What is inflation?

How do we measure inflation?

How do we incorporate the effect of inflation in equivalence calculation?

Contemporary Engineering Economics, 5th edition, © 2010

Page 4: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

What is Inflation?Definition: Inflation is the rate at which the general level of prices and goods and services is rising, and subsequently, purchasing power is falling.

Time Value of Money

Contemporary Engineering Economics, 5th edition, © 2010

Earning PowerPurchasing Power

Earning Power

Investment Opportunity

Purchasing Power

Decrease in purchasing power (inflation)Increase in purchasing power (deflation)

Page 5: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

1990

$100

1990 2010

$100

You could buy 50 Big Macs in year 1990 with $100

You can only buy 28.5 Big Macs in year 2010.

$2.00 / unit $3.50 / unit75% Price change

due to inflation

The $100 in year 2010 has only $57 worth purchasing power of 1990

Page 6: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

2004 2005 2006 2010

$100

2004 2005 2006 2010

$100

You could purchase 63.69 gallons of purified drink water a year ago.

You can now purchase 80 gallons of purifieddrink water.

$1.57 / gallon $1.25 / gallonPrice change due to

deflation

20.38%

Page 7: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Inflation Terminology - IProducer Price Index: a statistical measure of industrial price change,

compiled monthly by the Bureau of Labor Statistics, U.S. Department of Labor

Consumer Price Index: a statistical measure of change, over time, of the prices of goods and services in major expenditure groups—such as food, housing, apparel, transportation, and medical care—typically purchased by urban consumers

Average Inflation Rate (f): a single average rate that accounts for the effect of varying yearly inflation rates over a period of several years.

General Inflation Rate (f ): the average inflation rate calculated based on the CPI for all items in the market basket.

Contemporary Engineering Economics, 5th edition, © 2010

Page 8: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Consumer Price IndexConsumer Price Index (CPI): the CPI compares the cost of a sample “market basket” of goods and services in a specific period relative to the cost of the same “market basket” in an earlier reference period. This reference period is designated as the base period.

CPI (Old measure) – Base Period = 19671967 1002010 649.10 (January)

CPI (New measure) – Base Period (1982-84)1982-84 1002010 216.68 (January)

Contemporary Engineering Economics, 5th edition, © 2010

Page 9: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Contemporary Engineering Economics, 5th edition, © 2010

Page 10: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Average Inflation Rate (f )

Fact: Base Price = $100 (year 0)Inflation rate (year 1) = 4%Inflation rate (year 2) = 8%

Find: Average inflation rate over 2 years?

Step 1: Find the actual inflated price at the end of year 2.

$100 ( 1 + 0.04) ( 1 + 0.08) = $112.32

Step 2: Find the average inflation rate by solving the following equivalence equation.

$100 ( 1+ f)2 = $112.32 f = 5.98%

Contemporary Engineering Economics, 5th edition, © 2010

$100

$112.32

0 1

2

Page 11: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Example 11.1 Average Inflation RateSample Calculation for Average Inflation rate for Gasoline:

Given: P = 127.3, F = 175.3, N = 2009-2000 = 9.

Find: f

Average Inflation Rate

Contemporary Engineering Economics, 5th edition, © 2010

Page 12: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

General Inflation Rate (f)Formula: Calculation:

Given: CPI for 2009 = 213.2, CPI for 2000 = 172.2

Find: f

Contemporary Engineering Economics, 5th edition, © 2010

where

= The general inflation rateCPIn = The consumer price index at the end of period n,CPI0 = The consumer price index for the base period.

f

Page 13: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Example 11.2 Yearly and Average Inflation Rates Year cost data:

Find: Yearly and Average inflation rates

Solution:

Contemporary Engineering Economics, 5th edition, © 2010

Year Cost

0 $504,000

1 538,000

2 577,000

3 629,500

Page 14: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Actual Dollars (An ): Estimates of future cash flows for year n that take into account any anticipated future changes in amount caused by inflationary or deflationary effects.

Constant Dollars (An’ ): Estimates of future cash flows for year n in constant purchasing power, independent of the passage of time (or base period).

Contemporary Engineering Economics, 5th edition, © 2010

Inflation Terminology – II

Page 15: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Finding Actual Dollars Conversion from Constant to Actual

Dollars General inflation rate = 5%

Contemporary Engineering Economics, 5th edition, © 2010

Period Net Cash Flow in Constant $

Conversion Factor

Cash Flow in Actual $

0 -$250,000 (1+0.05)0 -$250,000

1 100,000 (1+0.05)1 105,000

2 110,000 (1+0.05)2 121,275

3 120,000 (1+0.05)3 138,915

4 130,000 (1+0.05)4 158,016

5 120,000 (1+0.05)5 153,154

Page 16: Lecture No. 35 Chapter 11 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010.

Finding Constant Dollars Conversion from Actual to Constant

dollars

Contemporary Engineering Economics, 5th edition, © 2010

Example 11.4 - General inflation rate of 5%

End of period

Cash Flow in Actual $

Conversion at f = 5%

Cash Flow in Constant $

Loss in Purchasing

Power0 -$20,000 (1+0.05)0 -$20,000 0%

1 20,000 (1+0.05)-1 -19,048 4.76

2 20,000 (1+0.05)-2 -18,141 9.30

3 20,000 (1+0.05)-3 -17,277 13.62

4 20,000 (1+0.05)-4 -16,454 17.73


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