Download Basics of Engineering Economy

Lecture slides to accompany
Basics of Engineering Economy
by
Leland Blank and Anthony Tarquin
Chapter 10
Effects of Inflation
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 1
© 2008, McGraw-Hill
All rights reserved
Chapter 10 – Effects of Inflation
PURPOSE
TOPICS
Consider the effects
of inflation in
PW, AW and FW
equivalence
calculations
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 2
• Definition and
impact
• PW adjusted for
inflation
• FW with inflation;
real interest rate;
inflation adjusted
MARR
• AW with inflation
• Spreadsheet usage
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 – Understanding Inflation
• Inflation definition – A decrease in the value of a
currency
• Inflation impact – An increase in the amount of
money required to purchase the same amount of
goods or services over time
• Directly associated with inflation – an increase in the
money supply, i. e., more (government) money is printed
to counteract inflationary impacts
What to do in engineering economy?
• To perform economic comparisons - conversion to
a common basis of money at different times must be
included
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 3
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 – Understanding Inflation
• Constant-value (CV) – Money represents
same purchasing power regardless of
when in time
– Another term used for CV: Today’s dollars
• Future dollars – amount of money in the
future
– Other terms used for future dollars:
Then-current dollars
Inflated dollars
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 4
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 – Understanding Inflation
f - inflation rate per period (year), e.g., 5% per
year
n – number of periods (years) considered
Example: A $100 bill five years from now at 4%
inflation will purchase only $82.20 worth of goods
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 5
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 – Understanding Inflation
Viewed from a future perspective
Future amount = CV amount (1+f)n
Example: If it costs $82.20 today, in five years it will cost
$100 if inflation is 4% per year
82.20(1+0.04)5 = 82.20(1.2167)= $100
Two (equivalent) approaches to adjust for inflation:
1.
Convert all amounts (estimates) to have same value
(purchasing power) by equivalency relations (this uses
CV relation shown above prior to time-value-of-money
calculations)
2.
Change interest rates to account for inflation, as well
as time value of money (this approach follows)
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 6
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 – Understanding Inflation
Consider the erosive impact of even moderate inflation
Save $25,000/year for 40 years of employment (ages 22 to 62)
• Assume 7% per year return on your investments
• Future total amount of retirement portfolio
 No return considered: 25,000(40) = $1 million
 7% return considered
FW = 25,000(F/A,7%,40) = 25,000(199.6351)
= $4.99 million

A return of 7% per year means a lot over time
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 7
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 – Understanding Inflation
• Now, assume 4% per year inflation
• What is the effective purchasing power of your
retirement portfolio 40 years from now?
FW = $1.835 million
This means that purchasing power is $1.835 million, not $4.99 million

Inflation is a real killer of investment returns
Computations discussed later
Conclusions:
 The real return on savings is significantly lower when
inflation is accounted for
 If inflation rate exceeds return rate ( f > i), money loses
purchasing power over time
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 8
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 – Inflation-related Terminology
 i Real (inflation-free) rate – Rate at which
interest is earned or paid when effects of inflation
are removed
 Usually considered the ‘safe investment’ rate. Approximately
3.5% per year, historically, but varies depending upon health of
economy
 if Inflation-adjusted rate – Also called market
rate. Interest rate with effect of inflation
accounted for
 This rate is the one quoted daily, .e.g., 7% per year or effective
7.08%. Combination of real rate i and inflation rate f
 f Inflation rate – Rate of change in value of
currency
 Formulas for and use of these terms follow
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 9
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 - Deflation
• Deflation is opposite of inflation
• Deflation rate is -f in % per year
• Purchasing power of deflated currency is greater
in future than at present
• Sounds good after inflationary period, however …
• There are fewer jobs, less credit and fewer
loans available; overall ‘tighter’ money
situation
• Can disrupt national economies faster and more
severely than an equivalent inflation rate
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 10
© 2008, McGraw-Hill
All rights reserved
Sec 10.1 - Deflation
Equivalency computations use -f, instead of f
Example: Asset costs $10,000 today. If 2% per year
deflation is assumed, future estimated cost of repurchase in
5 years is
FW = 10,000(1-0.02)5 = 10,000(0.9039) = $9,039
International Dumping
■ Import materials (e.g., cars, cement, steel) into a country
from international sources at very low prices
■ Causes temporary deflation in the targeted sector
■ Domestic producers must reduce costs; some go broke, if
financially weak
■ Once competition is beaten, prices return to normal or
above previous levels
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 11
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Adjusted for Inflation
As described earlier, there are two (equivalent) approaches
to adjust for inflation. These are correct for all PW
computations
1. Convert everything to current value amounts by PW
equivalency relations before applying the real interest
rate i in PW computations
2. Determine the inflation-adjusted interest rate if and use it
in all PW calculations
Example: Assume a first cost of $5,000 now (CV amount).
Will purchase item during next 4 years. Let
Inflation = 4% per year
f = 4%
Real rate of return = 10% per year
i = 10%
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 12
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Adjusted for Inflation
Approach 1--Use CV dollars and real rate i
CV = PW = $5,000 now
Future costs at f = 4% are
in column (3)
FW = 5,000(1+ f)n
= 5,000(1.04)n
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 13
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Adjusted for Inflation
Approach 1 – Now find equivalent PW of $5,000
CV over 4 years at i = 10%
PW at
real i = 10%
5,000(P/F,10%,t)
Year,
t
Future
cost
in CV $
0
5,000
5,000
1
5,000
4,545
2
5,000
4,132
3
5,000
3,757
4
5,000
3,415
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 14
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Adjusted for Inflation
Effects of 4% inflation and 10% real interest rate over 4 years
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 15
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Adjusted for Inflation
This approach is more commonly used
Approach 2 – Use inflation-adjusted interest
rate if in all PW calculations
 To derive a formula for if, calculate PW from FW amounts
using real i and determine CV amounts using f
 Let if = i + f + if be the inflation-adjusted (market)
interest rate
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 16
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Adjusted for Inflation
 Resulting PW formula
 Interest rate formula
 Requires no conversion to CV amounts to obtain PW
values
Example:
Real i = 8%
Inflation = 6%
Inflation-adjusted (market) rate = 0.08 + 0.06 + (0.08)(0.06)
if = 0.1448 or 14.48%
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 17
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Adjusted for Inflation
From previous example
P = $5,000 is cost now; cost increases to $5,849 in 4 years
 Inflation rate = 4%/year  Real interest rate = 10%/year
Use if to determine equivalent PW now of the cost 4 years in future:
if = 0.10 + 0.04 + (0.10)(0.04) = 14.4%
P = 5,849(P/F,14.4%,4)
= 5,849(0.5838)
= $3,415
Conclusion: At market rate (inflation considered) of 14.4%, paying
$5,849 in four years for something that costs $5,000 now is
equivalent to paying $3,415 now in CV terms
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 18
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Inflation-Adjusted Example #1
Lottery winnings can be taken in 1 of 3 ways.
Which is better financially if f = 4% and i = 6%?
1: $100,000 immediately
2: 8 payments of $15,000 each starting next year
3: 3 payments of $45,000 each in years 0, 4 and 8
Use approach 2 with if: if = 0.06 + 0.04 + (0.06)(0.04) = 10.24%
Plan 1 PW
$100,000
Winner
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
Plan 2 PW
15,000(P/A,10.24%,8)
= $79,329
10 - 19
Plan 3 PW
45,000[(1 +
(P/F,10.24%,4) +
(P/F,10.24%,8)]
= $96,099
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Inflation-Adjusted Example #2
Find PW with and without 11% inflation considered
cont →
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 20
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Inflation-Adjusted Example #2
Use (P/A,g,i,n) = (P/A,12%,interest rate,9) factor for
geometric series to determine PW, where ‘interest rate’
is
1. i = 15% (without inflation)
2. if = 0.15+0.11+(0.15)(0.11) = 27.65% (with inflation)
•
Without inflation considered
PW = -35,000 - 7,000(P/A,15%,4)
- 7,000(P/A,12%,15%,9)(P/F,15%,4)
= $-83,232
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 21
cont →
© 2008, McGraw-Hill
All rights reserved
Sec 10.2 – PW Inflation-Adjusted Example #2
2. Inflation accounted for using if = 27.65%
PW = -35,000 - 7,000(P/A,27.65%,4)
- 7,000(P/A,12%,27.65%,9)(P/F,27.65%,4)
= $-62,436
Conclusions:
 As f and if increase, the P/A and P/F factors decrease, thus making
the PW values smaller
 Moving debt to the future and paying with future (inflated) money
seems financially ‘smart’, BUT …
 If cash is not available in the future, debt-laden people, companies
and countries will suffer and may go bankrupt
 ‘Buy now, pay later’ philosophy can be a dangerous strategy
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 22
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 – FW Adjusted for Inflation
Like PW computations, possible to consider or neglect
inflation rate f on top of real interest rate i
 Let P = $1,000; n = 7; find F under different conditions
 Market return if = 10%
 Inflation rate f = 4%
Cases 1 and 4: Actual future amount accumulated at
market rate if (both purchasing power and return included)
F = P(F/P,if,n) = 1,000(F/P,10%,7) = $1,948
Interpretation: This case correctly projects amount of capital
needed in future to overcome ongoing inflation
plus make a stated return on the investment
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 23
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 – FW Adjusted for Inflation
Case 2: Purchasing power maintained, but no
inflation considered requires use of real interest
rate i
 Usually the market (inflation-adjusted) rate if and
inflation rate f are estimated
 Determine real interest rate i by dividing market
rate by inflation factor (1+f)
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 24
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 – FW Adjusted for Inflation
Case 2 (cont):
 Let P = $1,000; n = 7; find F
 Market return if = 10%
 Inflation rate f = 4%
 Therefore, real return i = (0.10 – 0.04)/(1.04) = 5.77%
F = 1,000(F/P,5.77%,7) = $1,481
Interpretation: This case correctly estimates future purchasing
power with inflation effects removed.
An inflation rate of 4% reduces the future 10% return amount that
can be purchased from $1,948 to $1,481 -- a 24% reduction
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 25
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 – FW Adjusted for Inflation
Case 3: Future amount with no return uses only
the inflation rate f
 Let P = $1,000; n = 7; find F
 Inflation rate f = 4%
F = P(F/P,f%,n) = 1,000(F/P,4%,7) = $1,316
Interpretation: This case correctly estimates the future amount
needed to keep up with inflation only.
A price of $1,000 grows by 31+% in 7 years at 4% per year
inflation
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 26
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 – Inflation-Adjusted MARR
 Most corporations determine MARR using both
inflation and a return to cover capital increases
and expected return
 This is if used in cases 1 and 4
 Inflation-adjusted MARR is:
MARRf = i + f + if
 The real rate of return i is the corporate return
requirement at or above the following:
 ‘safe’ investment, which is usually ~3.5%, or
 cost of capital
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 27
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 – Inflation-Adjusted MARR
Example:
Cost of capital = 10%
Required return = 3%
Inflation rate projected = 4%
Real return i = 13%
 Perform PW, FW and AW computations at
MARRf = 0.13 + 0.04 + (0.13)(0.04) = 17.52%
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 28
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 – FW Inflation-Adjusted Example
• Alternatives: purchase now for $200,000, or in 3 years at
an estimated $340,000
• MARR is 12% and inflation averages 6.75% per year
INFLATION CONSIDERED
WITHOUT INFLATION
MARR = 12%
MARRf = 12% + 6.75% + 12%(6.75%) =
19.56%
Purchase now
FW = -200,000(F/P,19.56%,3)
= $-341,812
Purchase now
FW = -200,000(F/P,12%,3)
= $-280,986
Purchase later
Purchase later
FW = $-340,000
FW = $-340,000
Buy now is more economic
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
Buy later is slightly more economic
10 - 29
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 - Hyperinflation
 Inflation usually averages 2% to 8% per year
 Political and/or financial instability,
overspending, serious trade imbalance can
increase inflation dramatically
 Inflation rates above 50%, 75% and 100% per
year are considered hyperinflation
 FW estimates skyrocket as inflation increases
become larger
 But, future estimates are so uncertain that a
dependable economic analysis can not be
performed
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 30
© 2008, McGraw-Hill
All rights reserved
Sec 10.3 - Hyperinflation
Previous example: P = $200,000
MARR = 12% per year
If f = 10%/month = 120%/ year (without considering
compounding of inflation)
MARRf = 12% + 120% + 12%(120%) = 146.4%
Future equivalent cost estimate with inflation is huge
FW = -200,000(F/P,146.4%,3) = $-2.99 million
• Yet, estimated cost in 3 years is unpredictable
• What is to be done from the economic perspective????
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 31
© 2008, McGraw-Hill
All rights reserved
Sec 10.4 - AW Adjusted for Inflation
 Include inflation in AW computations because:
 Capital must be recovered with future, inflated dollars
 Less buying power in future means more money
needed to recover present investments plus a return
To find future inflated amount needed per year –
Use market rate if to determine A, given P
If future amount is known, fewer annual dollars are needed,
since their current buying power is greater –
Again, use if to determine A, given F
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 32
© 2008, McGraw-Hill
All rights reserved
Sec 10.4 - AW Adjusted for Inflation
Example:
•
•
•
•
Spend $1,000,000 now for 3-year IT service contract
Expect 4% return over cost of capital of 8.5%
Inflation averages 6% per year
What annual amount is needed to recover cost?
Expected real return = 4% + 8.5% = 12.5%
MARRf = 12.5% + 6% + 12.5%(6%) = 19.25%
Required annual recovery amount:
AW = 1 million(A/P,19.25%,3) = $469,159 / year
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 33
© 2008, McGraw-Hill
All rights reserved
Sec 10.4 - AW Adjusted for Inflation
For comparison purposes only, assume contract
provider agreed to one payment of $1 million
after 3 years. What is AW now?
 Again, MARRf = 19.25%
 Use A/F factor to find AW
AW = 1 million(A/F,19.25%,3) = $276,659
There is a large reduction of ~$192,000 per year
to recover the (fixed) $1 million expenditure after 3 years
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 34
© 2008, McGraw-Hill
All rights reserved
Sec 10.4 - AW Inflation-Adjusted Example
• For retirement, desire is to invest equal amounts
annually for 5 years to maintain purchasing power of
$10,000 today
• Expected (market) return is 10%
• Assume inflation is relatively high at 8%
Step 1: Required total after 5 years when f = 8% (case 3):
F = 10,000(F/P,8%,5) = $14,693.28
Step 2: Annual amount for 5 years at if = 10% (cases 1, 4):
A = 14,693.28(A/F,10%,5) = $2406.76
Note: real return is only i = (0.10-0.08)/(1.08) = 1.85%/year
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 35
© 2008, McGraw-Hill
All rights reserved
Sec 10.4 – Inflationary Period Reactions
THINGS THAT OCCUR WHEN INFLATION INCREASES
• More needed annually to recover capital investments
+ required return
• Lenders tend to increase interest rates to individuals
(credit cards, mortgages) and corporations (loans)
• People make lower payments on cards and loans
(money is used to buy other, necessary items to live)
• Lenders need more money to cover their higher costs
of making loans
• Bankruptcies increase
• Spiraling effects of inflation can lead to national
recession
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 36
© 2008, McGraw-Hill
All rights reserved
Sec 10.5 – Spreadsheet Usage
Lottery winnings can be taken in 1 of 3 ways
1: $100,000 immediately
2: 8 payments of $15,000 each starting next year
3: 3 payments of $45,000 each in years 0, 4 and 8
Assume f = 4% and i = 6% per year
Questions about the 3 plans:
cont →
A. Best plan based on PW values?
B. FW in 8 years with inflation considered?
C. FW in 8 years in terms of today’s purchasing power?
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 37
© 2008, McGraw-Hill
All rights reserved
Sec 10.5 – Spreadsheet Usage
A. Calculate if = 10.24%, enter cash flows and
use NPV function (row 13); select Plan 1
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 38
cont →
© 2008, McGraw-Hill
All rights reserved
Sec 10.5 – Spreadsheet Usage
B. Determine FW with inflation considered using FV
function at 10.24% (row 16 – note minus sign)
C. Determine FW without inflation considered using FV
function at 6% (row 18- note minus sign)
As expected, Plan 1 has best PW and FW values
Note: The impact of inflation is very clear. For example,
•
•
Plan 3 generates equivalent value in 8 years of
~$209,600.
However, this will purchase only about ~ $153,100
of goods in terms of today’s dollars
Slide to accompany Blank and Tarquin
Basics of Engineering Economy, 2008
10 - 39
© 2008, McGraw-Hill
All rights reserved