Lectures in Engineering Economy
Prof. Corrado lo Storto
DIEG, Dept. of Economics and Engineering Management
School of Engineering, University of Naples Federico II
email: corrado.lostorto@unina.it
phone: 081-768.2932
Major issues
 What is MARR?
 What is the cost of capital?
 What is the relation between cost of capital and financing funds?
 How can we measure the cost of capital?
 MARR, IRR and the cost of capital
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Minimum requirements of acceptability (MARR)
The interest rate or discount rate that should be used to estimate cash
flows for several competing alternatives is the minimum
attractive/acceptable rate of return (MARR).
MARR is also called the cost of capital.
The determination of MARR is generally controversial and difficult.
An easy method to compute for determining what is alleged to be the
minimum rate of return is to determine the rate of cost of each source of
funds and to weight these by the proportion that each sources
constitutes of the total.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
MARR: example
If 1/3 of the capital of a firm is borrowed at 6% and the remainder of its
capital is equity earning 12%, then the alleged minimum rate of return is
1
2
 6% +
 12% = 10%
3
3
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
MARR and capital budgeting
MARR determination is a task of capital budgeting.
Capital budgeting is a critical function that takes place at the highest
level of management. It can be defined as:
“The series of decisions by individual economic units as to how much
and where resources will be obtained and expended for future use,
particularly in the production of future goods and services”.
However, many decisions at the lower level in the management
hierarchy affect those proposals competing in the overall capital budget.
For example, before a major project is considered in top management
capital budgeting process, usually many sub-alternatives of design and
technical specifications are considered and the related decisions have
been made.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
The scope of Capital Budgeting
The scope of capital budgeting addresses
1. how the money is acquired and from what sources
2. how individual capital project opportunities (and combination of
opportunities) are identified and evaluated
3. how minimum requirements of acceptability are set
4. how final project selections are made, and
5. how post-mortem reviews are conducted
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
MARR calculation
There are different school of thought relatively to how MARR can be calculated.
1.
if particular projects are to be undertaken using borrowed funds, then the
minimum rate of return should be based on the rate of cost of those borrowed
funds alone;
2.
the minimum rate of return should be based on the cost of equity funds alone,
on the grounds that a firm tends to adjust its capitalization structure to the
point at which the real costs of new debt and new equity capital are equal (see
E. Solomon “The management of corporate capital, The Free Press, NY);
3.
Modigliani and Miller developed a theory that asserts that the average cost of
capital to any firm is completely independent of its capital structure and is
strictly the capitalization rate of future equity earnings;
4.
another way to determine the minimum rate of return is to consider it as an
opportunity cost in the “capital rationing” perspective. Capital rationing
describes what is necessary when there is a limitation of funds relative to
prospective proposals to use the funds. This limitation may be either internally
or externally imposed. Its parameter is often expressed as a fixed sum of
capital, but when the prospective returns from the investment proposals
together with the fixed sum of capital available to invest are known, then the
parameter can be expressed as a minimum acceptable rate of return, or cut-off
rate.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
MARR calculation
Ideally, the cost of capital by the opportunity cost principle can be
determined by ranking prospective projects according to a ladder of
profitability and then establishing a cut-off point where the capital is
used on the better projects.
The rate of return earned by the last project before the cut-off point is
the cost of capital or minimum rate of return by the opportunity cost
principle.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
MARR calculation
45
Prospective rate of return
40
35
30
25
20
15
10
5
0
1
2
3
4
5
Amount of money invested
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
MARR standards
It is not uncommon for firms to set two or more MARR levels according to
risk categories.
1. High risk (MARR=40%)
New products, new business, acquisitions, joint venture
2. Moderate risk (MARR=25%)
Capacity increase to meet forecasted sales
3. Low risk (MARR=15%)
Cost improvements, make versus buy, capacity increase to meet
existing order
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
MARR standard determination
To determine MARR standards, the firm could rank prospective projects in
each risk category according to the prospective rates of returns and
investment amounts.
After tentatively deciding how much investment capital should be allocated
to each risk category, the firm could then determine the MARR for each risk
category for a single category.
In theory, it would be desirable that a firm invest additional capital as long
as the return from the capital were greater than the cost of obtaining that
capital. In such a case, the opportunity cost would be equal to the marginal
cost of the last capital used.
In practice, the amount of capital actually invested is more limited due to
risk and conservative money policies. Hence, the opportunity cost is higher
than the marginal cost of the capital.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
MARR standard determination
Since managers cannot truly know the opportunity cost for capital in any
given period (such as a budget year), it is useful to proceed as if the MARR
for the upcoming period were the same as in the previous period.
In addition to the normal difficulty of projecting the profitability of future
projects and acceptability to permit the approval of favored, even if
economically undesirable, projects or classes of projects.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Minimum Acceptable Rate of Return
• Lower limit for investment
acceptability
• Criterion set by organization
• Before-tax calculations
• Depends on the cost-ofcapital (expense for acquiring
funds)
• Determined by:
– Organization
circumstances and goals
– Project risk
(high-risk requires higher
return rate)
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Cost of Capital – What is it?
Cost of capital is the risk-adjusted discount rate (k) to be used in
computing a project’s NPV.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Methods of Financing
• Equity Financing – Capital is
coming from either retained
earnings or funds raised
from an issuance of stock
Capital Structure
• Debt Financing – Money
raised through loans or by
an issuance of bonds
• Capital Structure – Well
managed firms establish a
target capital structure and
strive to maintain the debt
ratio
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Debt
Equity
Equity Financing
 Flotation (discount) Costs: the
expenses associated with
issuing stock
 Types of Equity Financing:
•
Retained earnings
•
Common stock
•
Preferred stock
Retained earnings
+
Preferred stock
+
Common stock
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Debt Financing
 Bond Financing:
•
Incur floatation cost
•
Pay only interests at the end of
each period (usually semiannually)
•
Pay the entire principal (face value)
in a lump sum when the bond
matures
 Term Loan:
•
Involve an equal repayment
arrangement.
•
May incur origination fee
•
Allow terms to be negotiated
directly between the borrowing
company and a financial institution
Bond Financing
+
Term Loans
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Cost of Capital
 Cost of Equity (ie) –
Opportunity cost associated
with using shareholders’
capital
 Cost of Capital (k) – Weighted
average of ie and id
Cost of Debt
Cost of capital
 Cost of Debt (id) – Cost
associated with borrowing
capital from creditors
Cost of Equity
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Calculating the Cost of Equity
 Cost of Retained Earnings (kr)
 Cost of issuing New Common
Stock(ke)
 Cost of Preferred Stock (kp)
 Cost of equity: weighted
average of kr ke, and kp
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Cost of Retianed
Earnings
Cost of Issuing
New Stock
Cost of Issuing
Preferred Stock
Method 1: Calculating Cost of Equity
Based on Financing Sources
ie  (cr / c e )k r  (c c / c e )k e
(cp / c e )k p
Where Cr = amount of equity financed
from retained earnings,
Cc = amount of equity financed from
issuing new stock,
Cp = amount of equity financed from
issuing preferred stock, and
Ce = Cr + Cc + Cp
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Example: Determining the Cost of Equity
Source
Amount
Interest
Rate
Fraction of Total Equity
Retained
earnings
$1 M
20.50%
0.167
New common
stock
$4 M
22.27%
0.666
Preferred
stock
$1 M
10.08%
0.167
ie  (0.167)(0.205)  (0.666)(0.2227)  (0.167)(0.1008)
 19.96%
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Method 2: Calculating Cost of Equity based on CAPM
• The cost of equity is the risk-free cost of debt (20 year U.S.
Treasury Bills around 7%) plus a premium for taking a risk as to
whether a return will be received.
• The premium is the average return on the market, S&P 500,
(i.e., 12.5%) less the risk-free cost of debt. This premium is
multiplied by beta, a measure of stock price volatility.
• Beta quantifies risk and is an approximate measure of stock
price volatility. It measures one firm’s stock price compared
(relative) to the market stock prices as a whole.
• A number greater than one means that the stock is more
volatile than the market on average; a number less than one
means that the stock is less volatile than the market on average.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Method 2: Calculating Cost of Equity based on CAPM
The following formula quantifies the cost of equity (ie).
i e  rf   [rM  rf ]
where rf = risk free interest rate (commonly referenced to U.S.
Treasury bond yield)
rM = market rate of return (commonly referenced to average return on
S&P 500 stock index funds)
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Terminology relative to Common Stock
1.Common Stock: Ownership shares in a publicly held corporation
2.Secondary Market: Market in which already issued securities are
traded by investors
3.Dividend: Periodic cash distribution from the firm to the
shareholders
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Valuing Common Stocks
• Key concept: expected return rate
• Expected Return Rate: The percentage yield that an
investor forecasts from a specific investment over a set
period of time
• “Market Capitalization Rate”
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Market Capitalization Rate
•
•
•
DIVt: dividend during year t
Pt: stock price at end of year t
Using a single-period model:
DIV1  P1  P0
r
P0
•
•
•
DIV1/P0: Dividend yield
P1-P0: Capital gain
(P1-P0)/P0: Capital appreciation
r
DIV1 P1  P0

P0
P0
 Market capital -   Dividend

  
ization
rate

  yield
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
  Capital 
  

appreciati
on
 

Valuing Common Stocks
Example: Company X stock is selling at $100 a share. Expected
dividends over the next year $5. Expected price in a year $110.
DIV1  P1  P0 5  110  100
r

 .15  15%
P0
100
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Valuing Common Stocks
Example: Company X stock.
Expected dividends over the next
year $5. Expected price in a year
$110. Given that equally risky
investments have a capitalization
rate of r=15%, what is the price
of the stock today?
P0 
Equilibrium condition of capital
markets:
All securities in the same risk
class are priced to offer the same
expected rate of return
DIV1  P1
1 r
$5  $110
P0 
 $100
1  .15
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Dividends vs. Capital Gains

Beginning of 1st yr: investor buys at P0 based on
• dividends for 1st yr, and
• sell price at end of 1st yr
DIV1
P
P0 
 1
1 r 1 r

End of 1st yr: another investor is willing to buy at P2
P1 

Combining the two:
P0 
DIV2
P
 2
1 r 1 r
DIV1 DIV2
P2


1  r 1  r 2 1  r 2
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Dividends vs. Capital Gains
DIV1
P
 1
1 r 1 r

DIV3
DIVt
DIV1 DIV2




...


t
1  r 1  r 2 1  r 3
t 1 1  r 
P0 
Question: Which of the following is the value of a stock is equal
to?
1.The discounted PV of the sum of next period’s dividend plus
next period’s stock price, or
2.The discounted PV of all future dividends
Answer: Both
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Dividend Discount Model
• Finite horizon of H periods:
H
P0  
t 1
DIVt
PH

, H  1,2,...
t
H
1  r  (1  r )
• Infinite horizon:

P0  
t 1
DIVt
1  r t
• Criticism against model:
“shortsighted investors do not
care about long-run stream of
dividends”
• But: an investor who wants to
cash out early must find another
investor who is willing to buy
 Dividend-discount model holds
even if investors have short-term
horizons
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Valuation of Different Types of Stocks
Comment:
 “Growth” typically refers to EPS (earnings per share)
 The examples assume that DIV grow at the same rate as EPS
1. Zero Growth
DIV1  DIV2  DIV3  ...
P0 
DIV1
DIV1

 ...
2
1  r 1  r 
P0 
DIV1
r
[Perpetuity Formula]
2. Constant Growth (rate g)
DIV1 , DIV1 (1  g), DIV1 (1  g)2 , ...
P0 
DIV1 DIV1 (1  g)

 ...
2
1r
1  r 
P0 
DIV1
rg
[Growing Perpetuity Formula]
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Dividend Discount Model
1. Where does g (growth rate) come from?
2. Where does r (return rate) come from?
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Where Does g Come From?
• Growth: earnings next year are
higher than this year
• Payout Ratio: fraction of
earnings paid out as dividends
• Net investment necessary for
growth
• Plowback or Retention Ratio:
fraction of earnings retained
by the firm
• Part of this year’s earnings are
retained (not paid out as
dividends) and reinvested
• Growth rate
 Plowback   Return on retained 
  

g  
earnings
 ratio  

Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Formula for Firm’s Growth Rate
• DIV: dividends per share
• EPS: earnings per share
 Payout 
 Plowback  DIV

  1  
 
ratio
ratio



 EPS
• ROE: return on equity
 Return on retained

earnings

 Plowback   Return on retained 
  

g  
ratio
earnings

 

 DIV 
g  1 
  ROE
 EPS 

  ROE

Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Where Does r Come From?
Two methods:
1. Risk analysis
2. Dividend discount model
r
DIV1
g
P0
Dividend Discount Model:
• Assume constant growth
DIV1
DIV1
P0 
r 
g
rg
P0
• DIV, P0: publicly available
information
• g: estimate
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Example: Estimating Growth and Return Rates
Example: Krueger Enterprises has 1
million shares outstanding. Firm
just reported earnings of $2 million.
It plans to retain 40% of earnings.
Historical ROE has been 16%, a
figure that is expected to continue
in the future. The stock is selling at
$10. What is the rate of return?
• Growth rate
 Plowback 
  ROE  (0.40)(0.16)  6.4%
g  
ratio


• Earnings a year from now
$2,000,000(1  0.064)  $2,128,000
• Dividends
$2,128,000(1  0.40)  $1,276,800
• Dividends per share
DIV1 
$1,276,800
 $1.28
1,000,000
• Return rate
r
DIV1
$1.28
g
 0.064  19.2%
P0
$10
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Estimating Growth and Return Rates
Assumptions/Approximations:
•
Return on retained earnings
equal to historical ROE
•
Plowback ratio remains the
same
•
Constant growth
Extreme Cases:
•
No-dividend firm
•
g close to r : infinite-growth
Suggestions:
1. Estimate r for an entire
industry – averaging reduces
estimation error
2. A high estimate of the growth
rate (g close to r):
•
might be correct for the next
few years
•
cannot be sustained forever
firm
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Example: Dividends vs. Growth Opportunities
Example: Krueger Enterprises
forecasts to pay a $5.00 dividend
next year, which represents 100%
of its earnings. This will provide
investors with a 12% expected
return. Instead, the firm decides to
plow back 40% of the earnings at
the firm’s current return on equity
of 20%. What is the value of the
stock before and after the plowback
decision?
• No Growth:
P0 
DIV1
$5

 $41.67
r
0.12
• With Growth:
g  (0.4)(0.2)  8%
DIV1  $5(0.6)  $3
P0 
DIV1
$3

 $75
r  g 0.12  0.08
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Practice Problem – Cost of Equity
Alpha Corporation needs to raise $10 million for plant modernization.
Alpha’s target capital structure calls for a debt ratio of 0.4, indicating
that $6 million has to be financed from equity.
Alpha is planning to raise $6 million from the financial market
Alpha’s Beta is known to be 1.8, which is greater than 1, indicating the
firm’s stock is perceived more riskier than market average.
The risk free interest rate is 6%, and the average market return is 13%.
Determine the cost of equity to finance the plant modernization.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Solution
?
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Calculating the Weighted (after-tax) Cost of Capital
k 
i d C d i eC e

V
V
Cd= Total debt capital(such as bonds) in dollars,
Ce=Total equity capital in dollars,
V = Cd+ Ce,
ie= Average equity interest rate per period considering
all equity sources,
id = After-tax average borrowing interest rate per
period considering all debt sources, and
k = Tax-adjusted weighted-average cost of capital.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Example: calculation of the Cost of Capital
• Given: Cd = $4 million, Ce = $6 million, V= $10 millions,
id= 6.92%, ie=19.96%
• Find: k
0.0692(4) 0.1996(6)
k 

10
10
 14.74%
Comments: This 14.74% would be the (marginal) cost of capital that a
company with this financial structure would expect to pay to raise $10
million.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
The marginal cost of capital
The marginal cost of capital is defined as the cost of obtaining another
dollar of new capital. The marginal cost rises as more and more capital
is raised during a given period.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
The CAPM approach
The CAPM approach involves the following steps:
1. Estimate the risk free rate rf (i.e., the Treasury Bill rate)
2. Estimate the stock’s beta coefficient, b, which is an index of systematic
(or nondiversifiable) market risk
3. Estimate the rate of return on the market portfolio (such as the
Standard & Poor’s 500 Stock Composite Index or Dow Jones 30
Industrials)
4. Estimate the required rate of return on the firm’s stock using the CAPM
equation
k e  rf  b rm - rf 
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
The CAPM approach: example
Assuming that rf is 7%, b is 1.5, and rm is 13%, then
k e  rf  b rm - rf  = 7% + 1.5 13% - 7% = 16%
This 16% cost of common stock can be viewed as consisting of a 7% risk
free rate plus a 9% risk premium which reflects that the firm stock price is
1.5 times more volatile than the market portfolio due to factors affecting
nondiversifiable or systematic risk.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Calculating Cost of Capital using CAPM
• Cost of Debt
Cost of debt = debt interest rate (1 - tax rate)
• Cost of Equity
Cost of Equity = Risk free return + Risk Premium
=Rf   [RM  Rf ]
where Rf  risk free return (U.S. Treasury Bills)
RM  Average rate of return on market
 = stock price volatility
Cost of Capital = (cost of debt) x (% of capital from debt)
+ (cost of equity) x (% of capital from equity)
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Return Rates
1.
Minimum Acceptable Rate of Return (MARR)
Lowest level of return that makes an investment acceptable
2.
Internal Rate of Return (IRR)
Discount rate that makes PW of an investment zero
3.
External Rate of Return (ERR)
Return Rate that can be obtained for an investment under
current economic conditions
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Internal Rate of Return
Internal (or discounted cash flow)
Rate of Return:
Interest rate at which the PW of
the cash flow generated by the
project is zero
Caution!
1. IRR calculation
2. Interpretation of IRR for
evaluation of mutually
exclusive projects
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Calculation of the IRR (Single Project)
• Find discount rate i, such that
the PW of the cash flow is
equal to zero
• Polynomial equation on i
• Existence of multiple solutions
i*
PW (i )  0 
C0 
CN
C1
C2

...

0
(1  i )1 (1  i ) 2
(1  i ) N
• Cn: receipt (positive),
disbursement (negative), or
net benefit (receipt minus
disbursement) at period n
• C0: initial cost
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Multiple IRR’s
Descartes' “Rule of Signs”
There can be as many different solutions to a polynomial
equation, as the number of changes of sign in the
polynomial
PW (i )  0 
C0 
CN
C1
C2

...

0
1
2
N
(1  i ) (1  i )
(1  i )
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Simple and Non-simple Investments
1.Simple Investment:
only one change of sign
Period
Cash
Flow
0
Period
Cash
Flow
-1000
0
-1000
1
-200
1
200
Yes
2
500
2
-500
Yes
3
500
3
500
Yes
4
500
4
500
Unique IRR
Sign
Change
2.Non-Simple Investment:
multiple changes of sign
Yes
Possibly multiple IRR’s
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Sign
Change
IRR Calculation: Single Simple Investment
$150,000
i
$650
0
1
5
 $80,000
Exercise: Buying land
 $80,000  $650( P / A, i,5)
 $150,000( P / F , i,5)  0
Method 1: Manual trial-and-error
Answer: IRR=14.04%
PW(i)
0.00% $73,250.00
50.00% -$59,118.11
25.00% -$29,099.97
10.00% $15,602.21
20.00% -$17,774.47
15.00% -$3,244.59
12.00%
$7,457.13
14.00%
$136.80
14.50% -$1,575.95
14.10%
-$209.34
14.05%
-$36.50
14.04%
-$1.87
14.03%
$32.77
i low
i up
0.00%
50.00%
25.00%
10.00%
20.00%
15.00%
12.00%
14.00%
14.50%
14.10%
14.05%
14.04%
14.03%
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
IRR Calculation: Single Simple Investment
Exercise: Buying land
$80.00
 $80,000  $650( P / A, i,5)
 $150,000( P / F , i,5)  0
PW(i)
0.00% $73,250.00
50.00% -$59,118.11
25.00% -$29,099.97
10.00% $15,602.21
20.00% -$17,774.47
15.00% -$3,244.59
12.00%
$7,457.13
14.00%
$136.80
14.50% -$1,575.95
14.10%
-$209.34
14.05%
-$36.50
14.04%
-$1.87
14.03%
$32.77
i low
i up
0.00%
50.00%
25.00%
10.00%
20.00%
15.00%
12.00%
14.00%
PW(i) (thousands)
i
$60.00
$40.00
$20.00
$0.00
0%
10%
20%
30%
40%
-$20.00
-$40.00
-$60.00
i
14.50%
14.10%
14.05%
14.04%
14.03%
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
50%
60%
IRR Calculation: Single Simple Investment
$150,000
$650
0
1
 $80,000
Ex. 5.1: Buying land
5
i
PW(i)
 $80,000  $650( P / A, i,5)
 $150,000( P / F , i,5)  0
Method 2: Manual trial-and-error
using factor tables and
interpolation
14%
(14+x)%
15%
$137.00
0
-$3244.10
x
1 x

137 3244.10
x  0.04
Answer: IRR=14.04%
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
IRR Calculation: Single Simple Investment
$150,000
$650
0
5
1
 $80,000
Exercise: Buying land
 $80,000  $650( P / A, i,5)
 $150,000( P / F , i,5)  0
Capital
O&M
Other
Salvage,
Period Investment Revenue Expenses Expenses Other Inc.
0
-80000
0
0
0
0
1
0
1500
-850
0
0
2
0
1500
-850
0
0
3
0
1500
-850
0
0
4
0
1500
-850
0
0
5
0
1500
-850
0
150000
Inflation Rate: 0.00%
IRR=14.04%
Method 3: Computer program
• Excel: IRR() function
• Numerical solution of equation
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Cash
Flow
-80000
650
650
650
650
150650
IRR Comparisons – SI (Simple Investments)
Exercise: Subassemblies of product
purchased for $71 apiece. Demand
350 units/yr. for 3 yrs. Equipment to
produce: initial cost $21,000, no
salvage. Production costs $18,500
1st yr, $12,250 each of last two yrs.
Buy or make?
 “Buy” annual cost:
350($71)=$24,850
 “Make” annual net savings:
C1=$24,850-$18,500=$6350
C2=$24,850-$12,250=$12,600
C3=$24,850-$12,250=$12,600
Period
Method:
Find IRR
Compare to MARR
Cash
Flow
0
-$21,000
1
2
3
$6350
$12,600
$12,600
IRR: i* that solves
 $21,000 
$6350 $12,600 $12,600


0
(1  i ) (1  i ) 2
(1  i)3
i*=20.91%
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
IRR Comparisons – SI
Ex. 5.3 (cont):
$12.00
$10.00
$8.00
PW(i) (thousands)
IRR=20.91%
Comparison to MARR:
MARR<=20.91%: Accept
MARR>20.91%: Reject
$6.00
$4.00
$2.00
$0.00
-$2.00
-$4.00
0%
10%
20%
30%
40%
MARR < IRR (Accept) MARR > IRR (Reject)
-$6.00
i
PW (i )  $21,000 
$6350 $12,600 $12,600


(1  i ) (1  i ) 2
(1  i)3
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
IRR Comparisons of Mutually Exclusive Investments – SI
 Mutually Exclusive projects
 Projects with equal lives
 Incremental IRR comparisons
 Incremental IRR: equivalent to PW and AW methods
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
IRR Comparisons of Mutually Exclusive Investments – SI
Example: Projects X and Y
PW comparison at 10%:
PWX  $1000
Year
0
1
2
3
4
PW(10% )
IRR
X
Project
Y
X-Y
-$1,000
$100
$350
$600
$850
-$1,000
$1,000
$200
$200
$200
$0
-$900
$150
$400
$650
$411.52
23.27%
$361.25
34.26%
$50.27
12.81%
 [$100  $250( A / G ,10,4)]( P / A,10,4)
 $411.52
PWY  $1000
 [$1000  $200( P / A,10,3)]( P / F ,10,1)
 $361.25
Individual IRR’s:
MARR=10%.
IRR X  23.27%
Which should be chosen?
IRRY  34.26%
Incremental IRR:
IRR X Y  12.81%
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
IRR Comparisons of Mutually Exclusive Investments – SI
Incremental IRR Comparison:
1.IRRX-Y>MARR: choose X
$10
$8
3.IRRX-Y=MARR: X and Y equivalent
$6
In example:
IRRX-Y=12.81%>10%
choosing X over Y provides
incremental IRR of 12.81% -higher than 10% MARR
PW(i) (hundreds)
2.IRRX-Y<MARR: choose Y
$4
$2
$0
0%
-$2
10%
20%
30%
Y
X
Neither
-$4
Rate of Return, i
X
Y
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
40%
X-Y
50%
Exercise: IRR Comparisons of Mutually Exclusive
Investments – SI
Example (cont): What would have
happen if we used Y-X for the
incremental IRR comparison?
Year
X
0
-$1,000
1
$100
MARR=10%
2
$350
3
$600
4
$850
Project
Y
-$1,000
$1,000
$200
$200
$200
1.
2.
3.
4.
5.
IRRY-X<0
IRRY-X<MARR
IRRY-X=-IRRX-Y
IRRY-X=IRRX-Y
None of the above
?
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
IRR Comparisons of Mutually Exclusive Investments – SI
Example (cont):
$300
0
1
2
3
4
PW(10% )
IRR
X
-$1,000
$100
$350
$600
$850
-$1,000
$1,000
$200
$200
$200
$411.52
23.27%
$361.25
34.26%
X-Y
$0
-$900
$150
$400
$650
$200
Y-X
$0
$900
-$150
-$400
-$650
$50.27 -$50.27
12.81% 12.81%
$100
PW(i)
Year
Project
Y
$0
0%
10%
20%
30%
-$100
-$200
-$300
Return Rate, i
Answer: IRRY-X=IRRX-Y
PWX-Y(i) = -PWY-X(i), all i
PWX-Y(i)=0 and PWY-x(i)=0
have the same roots
X-Y
Y-X
Which is better: X or Y?
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
40%
50%
Caution: Lending or Borrowing?
Year
Project
Y
X
$300
X-Y
Y-X
$200
PW(10% )
IRR
-$1,000
$100
$350
$600
$850
-$1,000
$1,000
$200
$200
$200
$411.52
23.27%
$361.25
34.26%
$0
-$900
$150
$400
$650
$0
$900
-$150
-$400
-$650
$50.27 -$50.27
12.81% 12.81%
$100
PW(i)
0
1
2
3
4
$0
0%
10%
20%
30%
-$100
-$200
-$300
Return Rate, i
Rules for IRR
X-Y
Y-X
1. Lending:
IRR > cost-of-capital (i.e., MARR)
 Y-X: borrowing
2. Borrowing:
IRR < cost-of-capital (i.e., MARR)
 X-Y: lending
(pure investment)
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
40%
50%
IRR Comparisons: Lending or Borrowing?
$50.27
12.81%
$0
$900
-$150
-$400
-$650
$8
PW(i) (hundreds)
$0
-$900
$150
$400
$650
Y-X
$10
$6
$4
$2
$0
0%
-$2
10%
ERR<IRRX-Y
20%
30%
ERR>IRRX-Y
Rate of Return, i
Y
50%
$8
$6
$4
$2
$0
-$2
-$4
-$4
X
40%
-$50.27
12.81%
$10
PW(i) (hundreds)
X-Y
X-Y
X-Y: lending
• ERR<12.81%: accept X-Y
Prefer X over Y
• ERR>12.81%: reject X-Y
Prefer Y over X
0%
10%
ERR<IRRY-X
20%
30%
ERR>IRRY-X
Rate of Return, i
X
Y
Y-X
Y-X: borrowing
• ERR<12.81%: reject Y-X
Prefer X over Y
• ERR>12.81%: accept Y-X
Prefer Y over X
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
40%
50%
Incremental IRR Comparisons
Rule:
Compare alternatives in increasing order of
required investment
 Works in most cases of interest
 But not always!
 “Lending or borrowing?”
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Rule for Incremental IRR Comparisons
Example: Two designs for a product.
10-yr study period. MARR=10%
$350
$300
PW at 10%
IRR
Design C
-$300,000
$66,000
C-A
-$130,000
$22,000
$100,361
22.47%
$105,541
17.68%
$5,180
10.92%
$250
PW(i) (thousands)
Initial Cost
Net Benefit/yr
Design A
-$170,000
$44,000
$200
$150
$100
$50
$0
0%
IRRA>10%: Design A acceptable
IRRC-A>10%: Design C preferred
over Design A
Conclusion: with MARR=10%, use
Design C
5%
10%
15%
20%
-$50
-$100
Return Rate, i
A
C
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
C-A
25%
30%
It works…but not always!
Example: MARR=10%
Wrong:
• PWX(10%)=$511.52
Project
Y
Year
X
0
1
2
3
4
-$900
$100
$350
$600
$850
-$1,000
$1,000
$200
$200
$200
-$100
$900
-$150
-$400
-$650
27.73%
34.26%
19.92%
IRR
Y-X
• PWY(10%)=$361.25
What went wrong?
 Y-X is borrowing
•
Sum of cash flows: -$400
•
Current balance for each
period:
Least investment: X
IRRX>10%: Accept X
IRRY-X>10%: Choose Y over X
Year
X
0
1
2
3
4
-$900
$100
$350
$600
$850
Project
Y
Y-X
-$1,000
$1,000
$200
$200
$200
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
-$100
$900
-$150
-$400
-$650
CB(Y-X)
-$100
$780.08
$785.50
$542.01
$0.00
IRR calculation: Single non-simple investment
• Non-Simple: multiple sign
changes in cash flow
• PW(i)=0 might have multiple
roots i*
• Pure or Mixed Investments
• CB(i*)t: Current balance at
period t, with discount rate i*
1.Pure Investment
– CB(i*)t<=0, all t
– all project receipts
reinvested internally to
cover expenditures
– IRR=i*
2.Mixed Investment
– CB(i*)t>0, some t
– how are positive cash flows
reinvested?
– IRR not known
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
IRR Calculation: Single Non-Simple Investment
Ex. 5.4: Cash flows in table
0
1
2
3
4
5
6
Cash
Flow
$ 3,000
0
-10,000
2,000
2,000
2,000
2,000
$1,000
$800
$600
PW(i)
Year
$400
$200
$0
0%
PW (i)  $3000  $10,000( P / F , i,2)
-$200
 $2000( P / A, i,4)( P / F , i,2)
0
-$400
10%
20%
30%
40%
Return Rate, i
Two roots:
1. i*=9.58%
2. i*=50.84%
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
50%
60%
IRR Calculation: Single Non-Simple Investment
Ex. 5.4: Cash flows in table
Year
Cash
Flow
0
1
2
3
4
5
6
$ 3,000
0
-10,000
2,000
2,000
2,000
2,000
PW (i)  $3000  $10,000( P / F , i,2)
 $2000( P / A, i,4)( P / F , i,2)
0
Two roots:
1. i*=9.58%
Interpretation
IRR=9.58%: positive balances
invested at rate 9.58%
IRR=50.84%: positive cash
flows invested at rate 50.84%
 Realistic to assume IRR=i*?
Real Practice:
 Positive balances invested
externally
 Real IRR depends on external
rate of return (ERR)
 Typically ERR=MARR
2. i*=50.84%
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Non-Simple Investments
 Explicit Investment Rate Method:
•
ERR to limited part of cash flow
•
Minimal “disturbance”
•
Eliminate sign changes
 Historical External Rate of Return (HERR) Method:
•
ERR to all positive cash flows
 Project Balance Method (PBM):
•
ERR to positive balances
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Summary
 Methods of financing:
1. Equity financing uses retained earnings or funds raised from an
issuance of stock to finance a capital.
2. Debt financing uses money raised through loans or by an issuance of
bonds to finance a capital investment.
 Companies do not simply borrow funds to finance projects. Wellmanaged firms usually establish a target capital structure and strive
to maintain the debt ratio when individual projects are financed.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Summary
 The selection of an appropriate MARR depends generally
upon the cost of capital—the rate the firm must pay to
various sources for the use of capital.
 The cost of the capital formula is a composite index
reflecting the cost of funds raised from different sources.
The formula is
k 
i d C d i eC e

,
V
V
where V  C d  C e
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Summary
The marginal cost of capital is defined as the cost of obtaining another
dollar of new capital. The marginal cost rises as more and more capital
is raised during a given period.
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto
Summary
MARR:
• Lower limit for investment
acceptability
• Criterion set by organization
• Before-tax calculations
• Depends on the cost-ofcapital (expense for acquiring
funds)
• Determined by:
– Organization
circumstances and goals
– Project risk
(high-risk requires higher
return rate)
Engineering Economy/cost of capital and MARR/ 2005 /prof. corrado lo storto