1819 EngEcon toledo&VTK

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Engineering Economy
H00K1A
Prof dr ir Liliane Pintelon
2018-2019
Course material
Course
Classes: 1st semester, every Thursday afternoon
Work shops: 1st semester, 4 sessions (on Thursday)
Course
Powerpoints and journal articles on
Handbook (highly recommended)
Blank, L. and Tarquin, A., Engineering Economy,
8th edition, McGraw Hill, Boston, 2018
Evaluation - Exam
Closed book exam
Written part (+/- 25%)
Small questions (insight?!) – multiple choice (guess
correction)/open questions
Oral part with (written) preparation time (+/- 75%)
Exercise(s)
Knowing which method to use and drawing the right
conclusions from the results
Formularium available
Case study – Journal article
Proving “maturity” in the subject
Being able to evaluate a chosen approach
Note: the course is taught and examined in English
https://eng.kuleuven.be/studenten/lijstRM.html
Engineering Economy
Setting the scene
1
Overview
Typical EE problems ‐ EE ? ‐ Since when ? ‐ Why
important to engineers ?
Role of EE in decision making ‐ The decision making process ‐ Ethics and decision making
What costs or (dis)benefits ?
Course content
Note on course material
2
Typical EE problems
Professional examples
A chemical installation needs a new pump. You can buy either a cheap one with a life time of 2 years or a more expensive one with a life time of 10 years. What is the best choice if the company intends to operate the installation for another 8 years? The CEO is preparing performance reports. He wants to relate the costs of spare parts consumption to the value of the installations in questions. How can he obtain this value without going into detailed and lenghty calculations?
3
Typical EE problems
Professional examples (c’d)
The production installation is generating rather high maintenance costs. Should this installation been replaced by a new one now or is it
better to keep the old one (as is or maybe with a small investment in renovation) ?
There is a new technology which could be used for a given production
process. If the company does not go for it, but its competitors do, the company may loose a lot of market share. On the other hand, the new technology is maybe not very reliable yet. Would it be a good idea to
spend some money to gain further insight in the technology and its
possibilities and limitations?
4
Typical EE problems
(Personal examples)
You are buying a house. You have offers from different banks
for your loan. Some banks recommend to pay back a fixed
amount per month, others recommend to pay more interest in the beginning and less in the end. How does this work ? Do you – in the end ‐ pay the same amount of interest? You requested to borrow some money from different banks, but no bank was willing to lend you the amount needed. Now you saw some offers by small companies and private persons in the newspaper. The interest rates stated are given
by week, by month or by year. How to compare these interest rates?? 5
Engineering economy (EE) ?
Some definitions
Engineering economics, previously known as engineering economy, is a subset of economics for application to engineering projects. Engineers seek solutions to problems, and the economic viability of each potential solution is normally considered along with the technical aspects. (Wikipedia, 2009)
Engineering economy involves the systmatic evaluation of the economic merits of proposed solutions to engineering problems. (Sullivan et al., 2009) 6
Engineering economy (EE) ?
Some definitions (c’d)
Fundamentally, engineering economy involves formulating, estimating and evaluating the economic outcome when
alternatives to accomplish a defined purpose are available. Another way to define engineering economy is as a collection of mathematical techniques that simplify economic comparison. (Blank & Tarquin, 2008)
“… the art and science that involves formulating, estimating and evaluating economic outcomes … always concerned wiht the selection and possible execution of alternatives given the economic parameters associated with the project …” (Blank & Tarquin, 2008)
7
Since when ?
Arthur M. Wellington
Justification of railway projects
1870s
E. Paul DeGarma
Holger G. Thuesen
textbooks on Engineering Economy
1942, 1950
Herny R. Towne
The Engineer as Economist
1886
Industrial revolution
°Management science
Arthur Jr Lesser
founding editor or “The Engineering Economist”
1955
8
Why important to Engineers ?
Engineers “design” and create, often they must select between multiple alternatives. Designing involves economic decisions. Engineers must be able to incorporate economic analysis into their creative efforts, but are not responsible for detailed investment financing decisions. Multidisciplinary input needed: e.g. AOC (annual operating costs) is typically a technical input, revenues are a marketing input …
A proper economic analysis for selection and execution is a fundamental aspect of engineering. Engineering economy models help here, but keep in mind: “People make decisions, tools don’t.”
9
Role of EE in decision making
Decision making involves the estimation of future events/outcomes
Engineering economy aids in quantifying past outcomes and forecasting future outcomes
Engineering Economy provides a framework for modeling problems involving:
Money – “cash flows”
Time – times of occurrence of cash flows
Interest rates – time value of money
Measures of economic worth – for selecting an alternative
10
Role of EE in decision making
Time value of money
All firms make use of investment of funds; investments are supposed to earn a return.
Money possesses a “time value”
Interest = manifestation of the time value of money
The “time value” of money is the most important concept in engineering economy
11
Illustration
a)
b)
Calculate the amount deposited 1 year ago to have $1000 now at an
interest rate of 5% per year.
Calculate the amount of interest earned during this time period.
Solution
a) The total amount accrued ($1000) is the sum of the original deposit and the
earned interests. If X is sthe original deposit:
total accrued = original + original*interest rate
$1000 = X+X*0.05  X = $952.38 = original amount
b)
The interest earned is $1000 ‐ $ 952.38 = $47.62
12
The Decision Making Process
1.
2.
3.
4.
5.
6.
7.
Understand the problem – define objectives
Collect relevant data
Define the set of feasible alternatives
Identify the criteria for decision making
Evaluate the alternatives and apply sensitivity analysis
Select the “best” alternative
Implement the alternative and monitor results
All monetary criteria ?
Brainstorming Nominal group technique
“do nothing”
= alternative
15
16
Ethics and decision making
Universal morals or ethics
Fundamental beliefs: stealing, lying, harming or murdering another are wrong
Personal morals or ethics
Beliefs that an individual has and maintains over time; how a universal moral is interpreted and used by each person
Professional or engineering ethics
Formal standard or code that guides a person in work activities and decision making
1‐18
All professional disciplines have a formal code of ethics. National Society of
Professional Engineers (NSPE) maintains a code specifically for engineers;
many engineering professional societies have their own code
1‐19
Money‐related situations with ethical dimensions In the design stage
While the system is operating
Bhopal, 1984
Challenger, 1986
20
21
23
24
Extra: What costs or (dis)benefits ?
This topic is a topic for a course on Management Accounting. Let’s go over what we need here …
CAPEX vs OPEX
Tangibles (€, …) vs intangibles
Standard, historical, predicted costs
Labor costs, material costs, etc.
Direct/indirect costs
Case‐(non)related costs
Current/non‐recurrent costs
Sunk costs
Internal/external costs
Hidden costs
Overhead costs
Out‐of‐pocket/opportunity cost
Fixed/variable costs
Life cycle costs
(etc)
PS: the term “cost” is used to describe costs and (dis)benefits
27
About: Tangibles and intangibles
tangible
• increase in sales or profit
• decrease in information processing cost
• decrease in operating cost
• decrease in required investment
• increased operational ability and efficiency
Illustration
intangible
• new or improved information availability
• improved abilities in computation and analysis
• improved customer service
• improved employee morale
• improved managerial decisions
• improved competitive position
• improved image
Remark: some tangibles are #, but no $ (e.g; accidents)
28
How to compute/estimate costs and benefits ? About: Standard, historical, predicted costs
Point of view: Cost calculation method
About: Labor costs, material costs, …
Point of view: Cost nature About: Case (non)related costs
Be careful to include “all” costs, but beware NOT to take into account non‐relevant costs, also
DON’T take the same costs into account twice …
29
About: Direct/indirect costs
ABC (activity based costing)
Direct: easy to relate “consumption” to “product”
e.g. raw material, labor
Indirect: difficult (or too expensive) to relate “consumption to “product”
e.g. energy, warehouse
About: Overhead costs
Be careful, can be tricky !
Example:
Supervision hours + global
percentage for management
About: Recurrent/non‐recurrent costs
Point of view: Occurence frequency
About: Internal/external costs
Point of view: “who pays ?”
Example:
Environmental fee: charge/km
Insurance & risk
May same straighforward at first, but …
30
About: Fixed/variable costs
output level
output level
cost
cost
output level
cost
cost
cost
Beware:
• Variable costs are not necessarily linear in function of the output level
• Fixed costs are not necessarily constant for any output level
31
output level
output level
Exercise
You are working for a professional organization for warehouse managers. The
organization regularly organizes short courses for their members. For the upcoming short
course there are two options: regular or plus.
Data
Fee for participants: € 200 – Costs for development course material: € 200 Speaker cost: € 300 – Handouts for participant: € 35 – Mailing to announce course
€ 2100
Option 1: “plus”: Renting facility (hotel ): € 2500 – Lunch participant: € 60
Option 2: “regular”: Use in-house room: € 500 – Lunch participant: € 15
Questions:
How many participants do you need to at least cover costs (both options) ?
If 40 participants are expected for option 1 and 25 participants for option 2. How
much profit can you expect (both options) ?
32
•
•
•
•
break even:
p*q=f+v*q
q = number participants
p = prce (income)
f = fixed cost
v = variable cost
short course financial overview
14000
12000
•
•
•
•
p = € 200
f1 = € (200 + 300 + 2500 + 2100)
= € 5100
f2 = € (200 + 300 + 500 + 2100)
= € 3100
v1 = € (35 + 60) = € 95
v2 = € (35 + 15) = € 50
•
 q1 = 49 en q2 = 21 participants
•
euro
10000
8000
income
6000
cost plus
4000
cost regular
2000
0
0
10
20
30
40
50
60
70
number of participants (q)
Sensitvity analysis
•
•
profit1 = € [200 * 40 – 5100 – 95 * 40] = - € 900
profit2 = € [200 * 25 – 3100 – 50 * 25] = € 1650
33
About: Out‐of‐pocket/opportunity costs
An explosion in the chemical plant caused severe damages to the
equipment. The installations had to close down for more than a month
in order to repair all damages. The costs of materials and labor
for these repairs were considerable.
This downtime caused production problems. Penalties were to be paid
to some major customers because of late deliveries. Sales also felt that
they missed some sales because some potential customers preferred to go to their competitors because of the explosion and its bad impact on
the company’s image.
Illustration
34
About: Hidden costs
A hospital is planning to change from a manual pharmacy picking system
to a fully automated picking system, including a state‐of‐the‐art distribution
robot. A consultant helped to gather all necessary data for an economic evalaution:
investment cost in both the hardware and the software for the automated system
(data from supplier), savings in labor costs, maintenance costs and energy costs to be expected and some material costs (packaging). Some costs
for training were also included.
Illustration
Example “hidden cost”: technical and infrastructural investments 35
About: Sunk costs
A company plans to build a new production unit. A detailed study has
been carried out by an engineering firm and some basic construction
works have been carried out (flattening the site). The engineering department is challenging this decision and claims that
renovating the old production unit would – economically – be a better
option.
The manufacturing manager on the other hand considers outsourcing the
production in question. The CEO decides to go ahead with the new production unit seen the fact
that so much money already has been spent.
Illustration
“No use crying over spilled milk”
36
About: CAPEX vs OPEX
CAPEX (capital expenditure)
Costs made for obtaining or upgrading non‐consumable, physical assets, such as installations and buildings
E.g. purchasing cost, commissioning cost
OPEX (operational expenditure)
Costs made to keep asset “usable”
E.g. energy, spare parts, personnel
Seems pretty obvious, but …
Example: Quid major renovation of a distillation plant in CPI ?
Investment or cost ? Will make a difference in financial results !!
Note: EBITDA: earnings before interest, taxes, depreciation and amortization
37
Typical trade‐off
(Fertilizer Focus, 2004)
38
LCC, TCO: related, but not synonyms
About: Life cycle costs
LCC, TCO = {CAPEX, OPEX}
disposal
replacement
specifications
decommissioning
inception
design
maintenance
O&M
modifications
operations
build/buy
commissioning
construction
acquisition
installation/construction
39
40
http://www.assda.asn.au/lifecycle1.html
(source: Pintelon & Van Puyvelde, 2013)
41
(source: Barringer & Associate, 2008)
42
Careful consideration needed! CAPEX vs OPEX
Tangibles (€, …) vs intangibles
Standard, historical, predicted costs
Labor costs, material costs, etc.
Direct/indirect costs
Case‐(non)related costs
Current/non‐recurrent costs
Sunk costs
Internal/external costs
Hidden costs
Overhead costs
Out‐of‐pocket/opportunity cost
Fixed/variable costs
Life cycle costs
(etc)
PS: the term “cost” is used to describe costs and (dis)benefits
44
Course content
②
Classes
①: this class
⑦
③
④
⑫
⑤
⑥: Nov 2, holiday, no class
⑧: Nov 17, Athens, no class
⑨
⑪
⑩
45
Note on course material
Toledo
Powerpoints: for each class on Toledo (beforehand)
Some topics will be elaborated on, background material
will be provided
Handbook
Well explained, many exercises and cases, examples, …
Computational help: Excel (not in classes)
49
Wrap‐up
Ch. 1 ‐ Extra
Setting the scene
What is EE ?
What type of decisions ?
Why important for engineers
Cost classification
Course content
50
Engineering Economy
Basics
1
Overview
Time value of money explored
Factors: How time and interest affect money
Combining factors
Nominal and effective interest rates
2
Time Value of Money: Why ?
All firms make use of investment of funds, these investments are expected to earn a return.
In engineering economy computations it is important to taken into account the
“time value” of money.
Interest is the manifestation of the time value of money.
Time value of money
3
Interest rates and returns
Point of view on interests
borrower’s point of view
interest owed = amount owed now – original amount
lender’s point of view
interest earned = total amount now – original amount
Return on
investment
interest accrued per time unit
* 100%
original amount
interest accrued per time unit
* 100%
rate of return 
original amount
interest rate 
Time value of money
4
Example 1.3 (p 12)
You borrow $10,000 for one full year and must pay back $10,700 at the end of one year
Interest Amount (I) = $10,700 ‐ $10,000=$700 for the year
The $700 represents the return to the lender for this use of his/her funds for one year
Interest rate (i) = 700/$10,000 = 7%/Yr
The interest rate (i) is 7% per year
7% is the interest rate charged to the borrower; 7% is the return rate earned by the lender
Time value of money
5
Equivalence
Different sums of money at different times may be equal in economic value
$106 one
year from now
0 1
Interest rate = 6% per year
$100 now
$100 now is said to be equivalent to $106 one year from now, if the $100 is invested at the interest rate of 6% per year.
Time value of money
6
Simple and compound interest
Simple interest (one period)
= (principal)*(interest rate)
Simple interest (more periods)
= (principal)*(number of periods)*(interest rate)
Compound interest (one period)
= (principal+all accrued interest)*(interest rate)
, 1.8 and 1.9
Time value of money
7
Exercise
I make a deposit of €10000 in a savings account for 5 years. The interest rate is 5 %. How much interest do I earn ?
deposit
interest rate
Every year I take
up the amount of
interest earned.
Time value of money
10000euro
5%
single interest
money interest
in bank earned
Every year I also
deposit the interest
in my account.
compound interest
money interest
in bank earned
after 1 year
10000
500
10000
500
after 2 years
10000
500
10500
525
after 3 years
10000
500
11025
551
after 4 years
10000
500
11576
579
after 3 years
10000
total
500
2500
12155
total
608
2763
8
“Trick”
The time required for an initial amount to double in size with compound interest (rule of 72):
72
estimated n 
i
The time required for an initial amount to double in size with simple interest (rule of 100):
100
estimated n 
i
2X=X+X*n*i/100
2=1+n*i/100
1=n*i/100
n=100/i
Time value of money
9
Terminology and symbols
t represents time
P = a present sum of money at a time designated as t = 0 (now)
F = a future amount of money at some point in time later than t = 0
A = a series of equal, end‐of‐period (i.e. “annual”) cash flows
n = the number of interest periods (study period, horizon)
i = the interest rate or rate of return per time period, in percent
Time value of money
10
Computational help
Tables
See end of book
Time value of money
11
Excel financial functions
Present Value P: =PV(i%,n,A,F)
Future Value F: =FV(i%,n,A,P)
Equal, periodic value: =PMT(i%,n,P,F)
No. of periods:
=NPER((i%,A,P,F)
Compound interest rate: =RATE(n,A,P,F)
Compound interest rate: =IRR(first_cell:last_cell)
Present value of a series: =NPV(i%,second_cell:last_cell) + first_cell
Example
=FV(5%,5,,10000) → (€12767)
Time value of money
12
MARR (minimun attractive rate of return)
Investors expect to earn a return on their investment (commitment of funds) over time
→ in order to be interes ng (profitable) an investment should return funds in excess of the investment amount
→ MARR, also called “hurdle” rate for an investment
The MARR (%) is established by the financial managers of the firm;
it is based on the cost of all types of capital and the allowance for risk, thus always higher than the return for a safe investment !
(see also Ch 10)
Time value of money
13
ROR≥MARR>cost of capital
Time value of money
14
15
Types of financing
Topics in italics will be
elaborated on later in the course.
Equity Financing
the firm uses funds either from retained earnings, new stock issues, or owner’s infusion of money
Debt Financing
the firm borrows funds from outside sources
the cost of debt financing = the interest rate charged on the debt (loan) amounts
Time value of money
16
Combination of debt‐equity financing is possible, it results in a WACC, a weighted average cost of capital
Example
A music system will be financed for 40 % with a credit card at 18% per year and for 60% with a savings account funds earning of 5% per year, then
WACC=0.4*18+0.6*5=12.2% per year
17
Inflation
Topics in italics will be
elaborated on later in the course.
A social‐economic occurrence in which there is more currency competing for constrained goods and services
Where a country’s currency becomes worth less over time thus requiring more of the currency to purchase the same amount of goods or services in a time period
Inflation impacts:
Purchasing Power (reduces)
Operating Costs (increases)
Rate of Returns on Investments (reduces)
Time value of money
18
Taxes
Topics in italics will be
elaborated on later in the course.
Taxes represent a significant negative cash flow to the for‐
profit firm. A realistic economic analysis must assess the impact of taxes, called and AFTER‐TAX cash flow analysis. Not considering taxes is called a BEFORE‐TAX cash flow analysis
A Before‐Tax cash flow analysis (while not as accurate) is often performed as a preliminary analysis
A final, more complete analysis should be performed using an After‐Tax analysis
Both are valuable analysis approaches
Time value of money
19
Cash flows
Cash inflows vs Cash outflows
Cash inflows Money flowing INTO the firm from outside
Revenues, savings, salvage values, etc
Cash outflows Disbursements, money flows OUT
First costs of assets, labor, salaries, taxes paid, utilities, rents, interest, etc
Time value of money
20
Cash flow computation
For many practical engineering economy problems the cash flows must be either:
Assumed known with certainty
Estimated
A range of possible realistic values provided
Generated from an assumed distribution and simulated
Assumption: All cash flows are assumed to occur at the end of an interest period
Tip:
Visualizing cash flows in a cash flow diagram helps your analysis.
Time value of money
21
Beware of conventions
Time value of money
Cash flow diagram
Interest Factors : ANSI Standard Notation
Consists of two cash flow symbols, the interest rate, and the number of time periods
General form: (X/Y,i%,n)
X represents what is unknown ‐ Y represents what is known
i and n represent input parameters; can be known or unknown depending upon the problem
Example: (F/P,6%,20) is read as: the factor needed to find F, given P when the interest rate is 6% and the number of time periods equals 20.
In problem formulation, the standard notation is often used instead of the closed‐form equivalent relations (factor)
Factors
24
29
30
Future value
Present value
Factors F/P and P/F
Single‐payment factors
Objective Derive the present of future worth of a cash flow
Fn
Cash flow diagram
i% / period
0 1 2 3 n‐1 n
P0
Formulas
F  P (1  i )
n
 1 
and P  F 
n 

(1
)
i


!!! “THE” FORMULA !!!
Factors
32
Computation: straightforward
F1  P  iP  P (1  i )
F2  P (1  i )  iP (1  i )  P (1  i  i  i 2 )  P (1  i )2
F3  F2  iF2  (1  i )F2  P (1  i )3
etc.
after n periods : Fn  F  P (1  i )n
Factors
33
Example
Known:
P=$12000; n=24 years; i=8% per year
Unknown:
F
Finding F:
(a) F=P*(1+i)n=$12000*(1+0.08)24=$76094
(b) FV(i%,n,A,P)=FV(8%,24,,12000)=($76094)
Factors
34
Example
How long will it take for $1000 to double if the interest rate is 5% per year ? The n value can be obtained using the P/F or F/P factor
P=F(P/F,i,n) → 1000=2000*(P/F,5%,n) → (P/F,5%,n)=0.5
From the interest tables we see that n is between 14 and 15 years, by interpolation we obtain: n=14.2 years.
Alternatively, we can use the spreadsheet function NPER(5%,0,‐1000,2000) to obtain n=14.2 years.
It is not so much a matter of studying all kinds of different formulas,
but rather of knowing the basics and of working carefully. Excel can be
a powerful help, as are the tables; use them correctly.
Factors
35
36
Present worth factor (uniform series)
Captal recovery factor (uniform series)
Factors P/A and A/P
Uniform series factors
Cash flow diagram
i% per interest period
. . . .
0 1 2 3 n‐2 n‐1 $A per interest period
n
Find P
Required: To find P given A Cash flows are equal, uninterrupted and flow at the end of each interest period
Factors
37
Computing … Factors
Typical computational
procedure for EE formulas
38
Example
How much money should you be willing to pay now for a guaranteed $600 per year for 9 years starting next year, at a rate of return of 16 % per year ?
Cash flow diagram
Computation with
Factors
or …
 (1  i ) n  1 
P  A
  600
n
 i (1  i ) 
 (1  0 .1 6 ) 9  1 
 $2764

9 
 0 .1 6(1  0 .1 6 ) 
41
Typically more solution approaches possible
present
future year amount
worth
(P)
worth
n
A 1/(1+0.16)^n amount (1+0.16)^(9‐n)
0
1
600
0.862 517.241
3.278
2
600
0.743 445.898
2.826
3
600
0.641 384.395
2.436
4
600
0.552 331.375
2.100
5
600
0.476 285.668
1.811
6
600
0.410 246.265
1.561
7
600
0.354 212.298
1.346
8
600
0.305 183.015
1.160
9
600
0.263 157.772
1.000
sum 2763.926
(F)
amount
1967.049
1695.732
1461.838
1260.205
1086.384
936.538
807.360
696.000
600.000
10511.105
future


* 1/(1+0.16)^9
2763.926
present
42
Sinking fund factor
Uniform‐series compound amount factor
Factors A/F and F/A
Cash flow diagram
F = given
i% per interest period
. . . .
0 1 2 3 n‐2 n‐1 A=? per interest period
Find A, given F
Formula
n
P
 1   i (1  i ) n 
AF

n 
n



(1
i
)
(1
i
)
1





i
A F

n
(1

)

1
i


n

(1

i
)
 1
and if A is given and F is to be found
F  A

i


Factors
43
Exercises
a) Formasa Plastics has major fabrication plants in Texas and Hong Kong. The president wants to know the equivalent future worth of a $1 million capital investment each year
for 8 years, starting 1 year from now. Formosa capital
earns at a rate of 14 % per year.
Answer: $ 13 233
b) How much money must Carol deposit every year from now at 5.5 % per year in order to accumulate $ 6000 seven years from now ?
Answer: $ 726 per year
Factors
45
(a)
F = 1000(F/A,14%,8)
= 1000*13.2328
= $ 13233
46
(b)
A = 6000 (A/F, 5.5 %, 7) = 6000 * 0.12098
= $ 725.76 per year
(0.12282+0.11914)/2
=0.12098
49
Gradient factors ‐ 1
Arithmetic gradient series
Definition: Cash flow series that either increases or decreases by a constant amount; the amount of the increase or the decrease is the (arithmetic) gradient G; G can be positive or negative
Cash flow profile
CFn=base amount+(n‐1)G
Formula
Factors
G  (1  i ) n  1
n 

P= 
i  i (1  i ) n
(1  i ) n 
52
53
54
55
Note that …
A1+(n-1)G
A1+(n-2)G
Find P, given gradient cash flow G
A1+2G
A1+G
“Conventional gradient”
Base amount = A1
0
1
2
3
n-1
n
CFn = A1 ± (n‐1)G
Factors
56
Also
Arithmetic‐gradient‐uniform‐series factor (A/G,i,n)
1

n
AG 

n


i
(1
i
)
1


Arithmetic‐gradient future worth factor (F/G,i,n)
 1   (1  i ) n -1  
F  G   
 - n
i
 
 i  
59
some algebra
60
Example (ex. 2.10)
(a) P ?
(b) AT ?
Cash flow diagram
61
62
63
Gradient factors ‐ 2
Geometric gradient series
Definition: Cash flow series that either increases or decreases by a constant percentage from period to period; the uniform rate of change is called the geometric gradient
Cash flow diagram
Formula
Factors
  1  g n 
1  
 
1

i

  gi
Pg  A1 
 ig





64
Start with:
A1
A1 (1  g ) A1 (1  g ) 2
A1 (1  g ) n 1
Pg 


 ... 
(1  i )1
(1  i ) 2
(1  i )3
(1  i ) n
(1)
Factor out A1 out and re‐write
 1
(1  g )1 (1  g ) 2
(1  g ) n 1 


 ... 
Pg  A1 

2
3
(1
i
)
(1
i
)
(1
i
)
(1  i ) n 




(2)
Multiply by (1+g)/(1+i) to obtain Eq. (3 )
Pg
(1 + g )
(1 + g )  1
(1  g ) 1
(1  g ) 2
(1  g ) n  1 
.
..
 A1






(1 + i)
( 1 + i )  (1  i )
(1  i ) 2
(1  i ) 3
(1  i ) n 
(3)
Subtract Eq. (2 ) from Eq. (3 ) to yield
 (1  g ) n
1 
 1+g

Pg 
 1   A1 


n 1
1 i
 1+i

 (1  i )
Solve for Pg and simplify to yield….
In case (g=i), go back to (1), replace g by i and simplify the equation Factors
  1  g n 
1  
 
1

i

  gi
Pg  A1 
 ig





nA1
Pg 
(1  i)
65
Example 2.11
66
  1  g n 
1  
 
1 i  


Pg  A1
gi
 ig





67
Combining factors
Most real life cash flow series do not fit exactly the series for which the basic formulas above were developed. Therefore it is almost always necessary to make combinations.
Combining factors
69
Example
Combining factors
71
Taking closer look …
Cash flow diagram
Total present worth at time 0
Present worth of
uniform annual
series at time other than 0
Present worth of
uniform annual
series at time 0
Combining factors
72
PT  P0  PA
with P0  5000
In year 0
and PA  P * (P / F ,8%,2)
'
A
and P  500 * (P / A,8%,6)
'
A
(1  i )n  1
also (P / A, i %, n ) 
i (1  i )n
thus
Of the shifted series
PA’=present worth of the A series
(numbers below the line)
but also a future worth for the
current time (numbers on above
the line)
PT  5000  500 * (P / A,8%,6) * (P / F ,8%,2) with P / F  1/ (1  i )n
PT  5000  500 * (4.6229) * (0.8573)
PT  $ 6981
Combining factors
73
Example
Combining factors
Example (Ex. 3.8)
74
Type Base amount
($)
Gradient
($)
Period
(years)
Start PI
investment
2000
500
5
Year 1
P2
withdrawal
5000
‐1000
5
Year 6
P3
withdrawal
2000
‐‐‐
2
Year 11
75
Nominal and effective interest rates
Nominal interest rate r
An interest rate that does not include any consideration of compounding(*)
Thus r = interest rate per period * number of periods
Examples
“8 % per year”
“1.5 % per month”
Nominal and effective interest rates
80
(*) interest earned on interest
Effective interest rate ia
The effective interest rate is the actual rate that applies for a stated period of time; it accounts for the compounding of interest during the time period of the corresponding nominal rate.
It is is commonly expressed on an annual basis denoted as “ia”
This is what really matters to properly account for the time value of money
Examples
“4 % per year compounded monthly”
Nominal and effective interest rates
81
In previous examples all interest rates had t and CP = 1 year, so m=1; the interest rates were effective
rates
Interest rates – three time based units
Time Period t – The period over which the interest is expressed (always stated).
Ex: “1% per month”
Compounding Period (CP) – The shortest time unit over which interest is charged or earned.
Ex: “8% per year, compounded monthly”
Compounding Frequency – The number of times m that compounding occurs within time period t.
Ex: “10% per year, compounded monthly” has m = 12
Ex: “1 % per month, compounded monthly” has m=1
and
r % per time period t
r
effective rate per CP 

m compounding periods per t m
Nominal and effective interest rates
82
Nominal and effective interest rates
83
Effective annual interest rates
Nominal and effective interest rates
84
Let
r = nominal interest rate per year
m = number of compounding periods per year
i = effective interest rate per compounding period (CP) = r/m
ia = effective interest rate per year
Then
F=P+iaP=P(1+ia) and also F=P(1+i)m
Which leads to
i a  (1  i )  1
m
85
Example
r
i=r/m
F=P*(1+ia)
Nominal and effective interest rates
ia=(1+i)m‐1
87
Nominal and effective interest rates
88
Effective interest rates for any time period
Compounding period (CP) and payment period (PP) do not necessarily coincide.
Example
A company deposits money each month into an account that pays a nominal interest of 14 % per year, compounded semiannualy,
then
compounding period CP=6 months, payment period PP=1 month
Formula m
r 

effective i   1    1
 m
where
r  nominal interest rate per payment period (PP )
m  number of compounding periods per payment period (CP per PP )
Nominal and effective interest rates
89
90
Example
(p 10)
91
Example
Example (Ex. 4.4)
Nominal and effective interest rates
92
Equivalence relations: PP vs CP
Equivalence concept decision
support EE
PP<CP or PP>CP
Straightforward application of formulas
Different ways of computing
Special case: continuous
compounding
98
Interest rates that vary over time
In practice, interest rates do not stay the same over time unless by contractual obligation. There can exist “variation” of interest rates over time. If required, it is best to build a spreadsheet model. It can be a cumbersome task to perform
Nominal and effective interest rates
115
Example
Example (Ex. 4.13)
Nominal and effective interest rates
116
Note: About the handbook
Every chapter is nicely structured and situated in the course material (course book).
Every chapter has many worked‐out examples. (additional to ones discussed in class).
Every chapter has
Chapter summary
Problems Additional problems and FE exam review questions
Case study
121
Exercises: rather straightforward computations
& somewhat harder ‐ Insights, links … important
see workshops
122
Multiple choice format
(FE=fundamentals of engineering) 125
More/less computational
work, more insight and
critical managemt thinking
required, “case study”
See workshops & classes
SEE WORKSHOP 1
128
130
131
Ch. 1‐4
Wrap‐up
Time value of money: equivalence, simple and compounded interest, MARR, cash flow diagram, WACC
Factors: F/P, P/F, P/A, A/P, A/F, F/A, gradient factors –
Combining factors
Interest rates: nominal, effective, compounding period, payment period, continuous compounding, varying over time
Symbols P, A, F, G, … and Notation (X/Y,i%,n)
Excel built‐in functions – Tables p 581 132
Engineering Economy
Present worth and
annual worth analysis
1
Overview
Formulating alternatives
Analysis – “Criteria”
Present worth (PW) analysis
Annual worth (AW) analysis
2
Formulating alternatives
Process (fig. 5.1)
Formulating alternatives
3
First “filter”
Example
A company is expecting a change in regulations within a few years. This will mean that their current machines will have to be replaced by new ones (solvents, Pb). They can replace the machines now or wait for the new legislation (Do nothing).
Example
Your company needs to boost
production in order to cope with
an increased market demand.
Alternatives
Expand capacity of current production line Build a similar second line
Build a new line using new technology Buy a new line and customize
Example
A hospital wants to invest in new pediatric beds. After collecting information in and outside the hospital and some market research 12 different suppliers were identified. After a first screening 2 suppliers were discarded because of their location and 1 because of non‐compliance with patient safety guidelines.
5
Categories
Mutual exclusive
Every alternative is a viable project. Only one viable project can be selected.
E.g. Pediatric bed example: only one supplier out of 9 potential suppliers will be selected by the economic analysis.
Independent
More than one viable project may be selected. E.g. The safety engineer has come up with a list of several projects to enhance occupational safety beyond what is legally compulsory. More than one project can be executed if selected by the economic analysis. Often however there is a budget limitation. 6
Type of alternatives
Revenue
Each alternative generates costs (disbursement) and revenue (or receipt) cash flows and/or savings.
E.g. A new production line generates costs: initial investment, O&M costs, … and savings: personnel cost (automated line).
Service
Each alternative has only cost cash flows. Revenues or savings are not dependent upon the alternative selected. Sometimes these are also not estimable. E.g. Public projects (tunnel for pedestrians, highway).
7
Present worth analysis
Objective
One alternative: Selection (interesting yes/no ?)
More than one alternative:
Mutual exclusive: classification (most interesting ?)
Independent: selection (all intereresting ones)
What does “interesting” mean ?
An alternative is “interesting” if the PW at MARR is ≥ 0.
financially interesting
Careful
When comparing alternatives, the period considered should be “the same” for all alternatives.
equivalence
8
Equal‐life alternatives
Present worth analysis
9
=3.7908 (table 15)
Present worth analysis
=0.6209 (table 15)
11
Different life‐alternatives
The PW of the alternaties must be compared over the same
number of years and end at the same time, i.e. “equal service requirement”.
Possible approaches
Compare alternatives over a period of time equal to the
least common multiple (LCM) of their lives.
Compare the alternatives using a study period of length n years
(planning horizon), which does not necessariy take into consideration
the useful life of the alternatives.
Present worth analysis
12
Example
13
Present worth analysis
Consider 18 years
Present worth analysis
14
15
16
Future worth analysis
Example
Present worth analysis
17
today
Present worth analysis
18
Present worth analysis
19
Capitalized cost calculation and analysis
Definition
Capitalized cost (CC) is the present worth of an alternative
that will last “forever”.
Used
Public projects such as bridges, dams, irrigation systems, railroads, highways, tunnels, …
Permanent and charitable organization endowments
Formula
A
CC 
i
Present worth analysis
20
Formula derivation
Start from
On the right hand side, divide both numerator and denominator by (1+i)n
If “n” approaches 
the above reduces to:
Present worth analysis
 (1  i)n  1
P  A
n 

i
(1
i
)


1

1

 (1  i)n
P  A
i


A
P
i





A
CC 
i
21
Approach (p 139)
Draw a cash flow diagram showing all non‐recurring cash flows and at least two cycles of all recurring cash flows
Find the present worth (P) of all nonrecurring amounts: CC component
Find the equivalent uniform annual worth (A) through one life cycle of all recurring amounts
Add this to all other uniform amounts occurring in years 1 to infinity: uniform annual worth (AW)
Divide this AW by i to obtain the corresponding CC component
Add the CC of the non‐recurring amounts to the CC of the recurring amounts to obtain the overall CC.
Present worth analysis
22
Example
Present worth analysis
25
Suspension bridge:
Initial cost
Right‐of‐way
Inspection and maintenance cost
Concrete resurfacing
Truss bridge:
Initial cost
Right‐of‐way
Inspection and maintenance cost
Sandblasting
Painting
Note: (1) Cost categories are similar but not identical
26
(2) Only costs – nothing on time needed, effectiveness, impact on citizens/businesses, ….
Suspension Bridge Alternative:
0 1 2 3 4 . . . . . 10 11 ……..
$50 Million
(initial)
Annual inspection costs = $35,000/yr=A1
$2 Million
(right‐of‐way)
Resurfacing=$100,000
A2=‐100000(A/F,6%,10)=‐7587
=‐50‐2=‐52 (106)
=
=‐709783
.
=‐52 106‐709783=‐52.71 106
Truss Design:
0 1 2 3 4 5 6 7 8 9 10 11 …..
$25 Million
(initial)
Annual Maintenance costs = $20,000/yr=A1
Sandblast: $‐190,000
A3=‐190000(A/F,6%,10)=‐14415
Paint: $‐40,000A2=‐40000(A/F,6%,3)=‐12564
$15 Million
(right‐of‐way)
=‐25‐15=‐40 (106)
=
.
=‐782983
=‐40 106‐782983=‐40.78 106
Payback period analysis
The payback period estimates the time np to recover the initial investment in a project;
there are two forms
1.
2.
With no interest ‐‐ i = 0% (no‐return payback)
With an assumed interest rate ‐‐ i > 0% (discounted payback analysis)
Formulas for np
t n p
0   P   NCFt
t 1
P
For a uniform NCF: n p 
NCF
t n p
0   P   NCFt ( P / F , i, t )
t 1
(Discounted PBP)
(Undiscounted/no return PBP)
37
Common managerial philosophy is that a shorter payback is preferred to a longer payback period; this is not a good approach from engineering economic point of view: PBP ignores all cash flows after the payback time period; may not use all of the cash flows in the cash flow sequence – myopic approach !
Never use payback analysis as the primary means of making an accept/reject decision on an alternative!
Not a preferred method for final decision making – better as a screening tool
Present worth analysis
38
Example
40
Excel or tables 41
Use predefined formula NPER
=NPER(i,A,P,F)=NPER(0.15,3,‐18,3)
Try to determine n through
trial‐and‐error
42
Example
43
44
Life cycle cost
Life cycle costing (LCC)
Considers the whole life cycle of the equipment, project, etc.
Cost elements: many different ones, not always easy to estimate (esp. if
far in future)
LCC <> TCO (total cost of ownership)
Present worth analysis
45
47
Present worth analysis
It is not unusual to have 75% to 85% of the entire life span LCC committed (locked in) during the preliminary and detail design stages. Watch out with ad hoc “cost‐savings”
(eg use of inferior components)
Actual costs
(maybe higher
than B if design flaws) Improved design
50
Example (ex. 5.9)
Present worth analysis
52
Challenge is not to forget any cost element and to make accurate estimations !
Present worth analysis
53
PW= (0.5+1187+3067)+(6512+20144+13142)=$44822 million
FW=PW(F/P, 18%,10)=$234.6 million
54
“Real life” example
55
Strictly speaking: TCO !
56
Annual worth analysis
Popular analysis technique, because rather “intuitive” and easily understood
Results reported in $ per time period, e.g. $/yr
Eliminates the least common multiple problem associated with the present worth method: Only one life cycle to evaluate Applicable to a variety of engineering economy studies such as:
Asset replacement; Breakeven analysis; Make‐or‐Buy decisions; Studies dealing with manufacturing costs; Economic value added (EVA) analysis
Formula
AW = PW(A/P,i%,n) or AW = FW(A/F,i%,n)
Annual worth analysis
57
Repeatability assumption
Given alternatives with unequal lives, the assumptions are
1. The services provided are needed forever
2. The first cycle of cash flows is repeated for all successive cycles in the
same manner
3. All cash flows will have exactly the same estimated values in every life
cycle.
Similar cycles
Not very reasonable in
an industrial context
Annual worth analysis
At least for the least common
multiple (LCM) of the lives of
the alternatives
Use study period approach
58
Example
(5.2 – 6.1)
Recall: Comparison of two alternatives. T(A)=6 years, T(B)=9 years LCM of 6 and 9 = 18 years; outcome: PWA=$‐45036 and PWB=$‐41384
Annual worth analysis
59
5.2 – 6.1)
Recall:
Comparison of two alternatives. T(A)=6 years, T(B)=9 years LCM of 6 and 9 = 18 years; outcome: PWA=$‐45036** and PWB=$‐41384
Question: Would an AW analysis yield the same result ?
Is already annual worth
Test:
Here in first life cycle: AW=‐15000(A/P,15%,6)+1000(A/F,15%,6)‐3500=$‐7349
Retake **: AW=‐45036(A/P,15%,18)=$‐7349
Answer: Yes
Annual worth analysis
60
Capital recovery and AW values
An economic alternative should have the following cash flow estimates made
Initial investment ‐‐ P
Estimated future salvage value ‐‐ S
Beware: S>0 if there is a market or trade‐in value and S<0 if it will cost money for disposal, S=0 is also possible
Estimated annual operating costs – AOC
Costs for service alternatives. Costs and receipts for revenue alternatives
and
Estimated life ‐‐ n
Interest rate ‐‐ i% (this is usually the MARR)
Annual worth analysis
62
Annual worth (AW)
Is composed of two factors: capital recovery (CR) for the initial investment and the equivalent annual amount (AOC), annual operating or maintenance & operating cost)
AW=‐CR‐AOC
Capital recovery (CR)
Is the equivalent annual amount that the asset (or process) must earn every year to just recover the initial investment plus a stated rate of return over its expected life; any expected salvage value is to be taken into account
CR=‐[P(A/P,i,n)‐S(A/F,i,n)]
Annual worth analysis
63
Example (ex. 6.2)
65
Present worth (PW) of the two separate investment amounts (P0 and P1): 8+5(P/F,12%,1)
Capital recovery CR=‐[P(A/P,i,n)‐S(A/F,i,n)]=capital recovery
= ‐ {[8+5(P/F,12%,1)](A/P,12%,8) ‐ 0.5(A/F,12%,8)}
=$‐2.47
Interpretation: Lockheed Martin should have an equivalent total revenue from the tracker of at least $‐2.47 106 just to recover the initial present worth investment plus the required return of 12% per year. This does not include the annual operating costs AOC of $0.9 106 each year.
Total annual worth AW=‐CR‐AOC=‐2.47‐0.9 or AW=$‐3.37 106 per year. This is the AW for all future life cycles (provided …)
66
Example
(ex. 6.3)
Annual worth analysis
67
68
Mutual exclusive alternatives and AW
Calculate AW at the MARR for each alternative;
If AW≥0, MARR is met or exceeded, the alternative in question is economically justifiable. Select one with lowest AW of costs (service) or highest AW of net incomes (revenue), i.e. select the numerically largest AW alternative.
If AW < 0 at MARR, the (revenue) alternative is not economically justifiable, since initial investment P is not recovered over n years at the required rate of MARR = i% per year
Annual worth analysis
70
Example
72
Initial cost
Salvage value
73
Additional exercise
Problem 6.8
74
FE review problem
76
Ch. 5‐6
Wrap‐up
Formulating alternatives
Process steps. Types.
Present/future worth analysis
PW of (non)equal life alternatives. FW analysis. Capitalized cost (CC). Payback period (PBP). Life cycle
cost (LCC).
Annual worth analysis
Importance. AW calculation. Capital recovery (CR). Alternative selection by AW.
78
! Cash flow diagrams ! Right formulas ! Management interpretation !
Engineering Economy
Rate of return analysis:
Single alternatives – Multiple alternatives
1
Overview
ROR analysis for single alternatives
Basis
Multiple ROR
ROR analysis for multiple alternatives (mutual exclusive)
Two alternatives
More than two alternatives
2
Single alternative
4
Interpretation of a rate of return (ROR)
ROR (rate of return), also called IRR (Internal Rate of Return) method or marginal efficiency of capital method is one of the popular measures of investment worth.
Definition:
ROR is either the interest rate paid on the unpaid balance of a loan, or the interest rate earned on the unrecovered investment balance of an investment such that the final payment or receipt brings the terminal value to exactly equal “0”.
The ROR is found using a PW (present worth) or AW (annual worth) relation. The rate determined is called i*
Single alternative
5
ROR (i*) is the interest rate earned/charged on the unrecovered balance of a loan or investment project. ROR is not the interest rate earned on the original loan amount or investment amount (P)
The i* value is compared to the MARR (minimal attractive rate of return) If i* > MARR, investment is justified
If i* = MARR, investment is justified (indifferent decision)
If i* < MARR, investment is not justified
Mathematically: ‐100%≤i*≤+∞
An i* = ‐100% signals total and complete loss of capital
One can have a negative i* value (feasible) but not less than –100%
All values above i* = 0 indicate a positive return on the investment
Single alternative
6
Example
Single alternative
7
Single alternative
8
Single alternative
9
ROR calculation using a PW or AW equation
Solve for i* either PWcosts/disbursements=PWincomes/receipts
or AWcosts/disbursements=AWincomes/receipts
using either trial and error or Excel There is no closed
formula for the RoR
Trial and error procedure
1.
2.
3.
4.
5.
Draw a cash flow diagram
Set up the appropriate PW equivalence equation and set equal to 0
Select values of i and solve the PW equation
Repeat for values of i until “0” is bracketed, i.e., the equation is balanced
May have to interpolate to find the approximate i* value
Excel functions
Rate(n,A,P,F) or IRR(first_cell:last_cell, guess).
Single alternative
When an A series is given
When the cash flows vary from period
to period
10
Example
Single alternative
11
Single alternative
12
Single alternative
14
RoR method vs PW, AW, FW
When applied correctly, ROR method will always result in a good decision and should be consistent with PW, AW, or FW methods.
However, for some types of cash flows the ROR method can be computationally difficult and/or lead to erroneous decisions
Multiple RoRs
Single alternative
15
Multiple ROR values
A class of ROR problems exist that will possess multiple i* values. This is the case when there is an nonconventional or nonsimple cash flow series, i.e. the net cash flows switch from negative to positive from one year to another, so
that there is more than one sign change.
Single alternative
17
Tests for multiple i* values
Descartes’ rule of signs: the total number of real values i* is always less than or equal to the number of sign changes in the original cash flow series
Cumulative cash flow sign test Norstrom’s criterion – sufficient, not necessary condition:
form the series of cumulative cash flows (CCF): S0, S1, S2, …; if S0 <0 and the sign changes only once in the CCF series, there is a single, real number positive i*
Single alternative
18
Example
Single alternative
19
Single alternative
Descartes
the total number of real values i* is always less than or equal to the number of sign changes in the original cash flow series
Norstrom
form the series of cumulative cash flows S0, S1, S2, …; if S0 <0 and the sign changes only once in the CCF
series, there is a single, real number positive i*
20
21
Removing Multiple i* Values
The result of follow‐up analysis to obtain a single ROR value when multiple, non‐useful i* values are present does not determine the internal rate of return (IROR) for nonconventional net cash flow series. The resulting rate is a function of the additional information provided to make the selected technique work and the accuracy is further dependent upon the reliability of this information. The resulting value will be referred to as external rate of return (EROR) as a reminder that it is different from the IROR discussed before. External = “external to the project “
7‐26
Take the following view: You are the project manager and the project generates cash flows each year. Some years produce positive net cash flow (NCF) and you want to invest the excess money at a good rate of return, in some source external to the project. We will call this the investment rate or reinvestment rate ii. Other years, the NCF will be negative and you must borrow funds from some external source to continue the project. The interest rate you pay should be as small as possible, we will call this the borrowing rate or finance rate ib. 7‐27
Two approaches to determine External ROR (EROR)
(1) Modified ROR (MIRR)
(2) Return on Invested Capital (ROIC) Note that the results of the two methods for rectifying the multiple ROR situation will not be the same, because slightly different additional information is necessary and the cash flows are treated in slightly different fashions from the time value of money view point
(1) Modified ROR Approach (MIRR)
Four step procedure:
Think of WACC
(weighted cost of capital)
Determine PW in year 0 of all negative CF at ib
Determine FW in year n of all positive CF at ii
Calculate EROR = i’ by FW = PW(F/P,i’,n)
If i’ ≥ MARR, project is economically justified Think of MARR
(minimum attractive rate of return)
© 2012 by McGraw‐Hill All Rights Reserved
7‐30
31
MIRR example
For the NCF shown below, find the EROR by the MIRR method if MARR = 9%, ib = 8.5%, and ii = 12%
Year
0 1 2 3
NCF +2000 ‐500 ‐8100 +6800
Solution:
PW0 = ‐500(P/F,8.5%,1) ‐ 8100(P/F,8.5%,2)
= $‐7342
FW3 = 2000(F/P,12%,3) + 6800 = $9610
PW0 (F/P,i’,3) + FW3 = 0
‐7342(1 + i’)3 + 9610 = 0
i’ = 0.939 (9.39%)
Since i’ > MARR of 9%, project is justified© 2012 by McGraw‐Hill All Rights Reserved
7‐32
(2) Return on Invested Capital Approach (ROIC) Measure of how effectively project uses funds that remain internal to project
ROIC rate, i’’, is determined using net‐investment procedure
Extra funds available
Three step Procedure
(1) Develop series of FW relations for each year t using:
Project uses all available funds
Ft = Ft‐1(1 + k) + NCFt
where: k = ii if Ft‐1 > 0 and k = i’’ if Ft‐1 < 0
(2) Set future worth relation for last year n equal to 0 (i.e., Fn= 0); solve for i’’
© 2012 by McGraw‐Hill All Rights (3) If i’’ ≥ MARR, project is justified; otherwise, reject
Reserved
7‐33
ROIC Example
For the NCF shown below, find the EROR by the ROIC method if MARR = 9% and ii = 12%
Year
0 1 2 3
NCF +2000 ‐500 ‐8100 +6800
Solution:
Year 0: F0 = $+2000 F0 > 0; invest in year 1 at ii = 12%
Year 1: F1 = 2000(1.12) ‐ 500 = $+1740 F1 > 0; invest in year 2 at ii = 12%
Year 2: F2 = 1740(1.12) ‐ 8100 = $‐6151 F2 < 0; use i’’ for year 3 Year 3: F3 = ‐6151(1 + i’’) + 6800 Set F3 = 0 and solve for i’’
‐6151(1 + i’’) + 6800 = 0
i’’= 10.55%
Since i’’ > MARR of 9%, project is justified
© 2012 by McGraw‐Hill All Rights Reserved
7‐34
Important Points for EROR
About the computation of an EROR value
EROR values are dependent upon the selected investment and/or borrowing rates
Commonly, multiple i* rates, i’ from MIRR and i’’ from ROIC have different values
About the method used to decide
For a definitive economic decision, set the
MARR value and use the PW or AW method
to determine economic viability of the project
© 2012 by McGraw‐Hill All Rights 7‐35
Reserved
Multiple alternatives
42
Two alternatives
43
Why incremental analysis is needed
Illustration
Assume
MARR=16% ‐ Available budget for investment: $ 90000
Alternative A: investment: $ 50000 – IROR i*=35 %
Alternative B: investment: $ 85000 – IROR i*=29%
At first sight
Alternative A is to be preferred
But
What to do with the excess investment capital ?
Well,
General assumption: excess funds are invested at MARR
Overall RORA=[50000(0.35)+40000(0.16)]/90000=26.6 %
Overall RORB=[85000(0.29)+5000(0.16)]/90000=28.3%
Thus
Alternative B is the most interesting one.
Multiple alternatives
44
Under circumstances, project ROR values do not provide the same ranking of alternatives as do PW, AW and FW analyses. This situation does not occur if we conduct an incremental cash flow ROR.
Equal service life
assumption
Multiple alternatives
45
Interpretation of ROR on extra investment
Guidelines (mutual exclusive projects)
For multiple revenue alternatives, calculate the IROR i*
for each alternative, and eliminate all alternatives that have an i* <MARR. Compare the remaining alternatives incrementally.
If the ROR available through the incremental cash flow equals or exceeds the MARR, the alternative with the extra investment should be selected.
Note
For independent projects, no comparison on the extra investment. The ROR is used to accept all projects with i* >MARR.
Multiple alternatives
46
ROR evaluation using PW:
incremental and breakeven
Procedure for incremental ROR analysis
Multiple alternatives
48
Example
Multiple alternatives
Is the more expensive B justified, after all
it has lower annual costs and a higher salvage
value ... ?
50
Multiple alternatives
51
year
A
B
incremental
cumulative
incremental
0
‐8000
‐13000
‐5000
‐5000
1
‐3500
‐1600
+1900
‐3100
2
‐3500
‐1600
+1900
‐1200
3
‐3500
‐1600
+1900
700
4
‐3500
‐1600
+1900
2600
5
‐3500 ‐1600+2000‐13000
+1900‐11000
‐6500
6
‐3500
‐1600
+1900
‐4600
7
‐3500
‐1600
+1900
‐2700
8
‐3500
‐1600
+1900
‐800
9
‐3500
‐1600
+1900
1100
10
‐3500
‐1600+2000
+1900+2000
5000
52
Multiple alternatives
53
Procedure with Breakeven
Breakeven Interest Rate (Fisherian Intersection Rate) i*(B‐A)
is the interest rate at which the two alternatives are economically equivalent.
More on breakeven in Chapter 13.
Multiple alternatives
54
Example
55
56
ROR using AW
Two methods
Solve AW relation on incremental cash flow series over LCM
Solve equation of respective AW relations – one cycle
60
Example Multiple alternatives
61
(see higher)
AWA=AWB
(1)
(2)
Multiple alternatives
63
Many alternatives
64
Incremental ROR analysis of multiple, mutually exclusive alternatives
Multiple alternatives
66
Example
Multiple alternatives
67
C=challenger
DN=defender
Multiple alternatives
68
Some exercises …
74
81
intermezzo
Single value
IROR: (internal) rate of return; i*
Multiple values
EROR: external rate of return
borrowing rate or finance rate; ib
investment rate or reinvestment rate; ii
MIRR: modified rate of return; i’
ROIC: return on invested capital, i’’
Two alternatives
incremental ROR; ∆i*
48
Will be covered in the work shops
95
Wrap‐up
Chapters 7‐8
ROR analysis: Single alternative
Interpretation
ROR using PW or AW equation
Multiple ROR values
ROR analysis: Multiple alternatives
Two alternatives
Why incremental analysis is necessary – Interpretaionof ROR on the extra investment
ROR using PW: incremental and breakeven
ROR using AW Many alternatives
ROR analysis of multiple, mutual exclusive alternatives
97
Engineering Economy
Making choices: Selecting the basic analysis tool 1
Overview
What is the best method to use ?
If no method is preselected …
If method is predetermined …
2
What is the best method to use ?
Comparing Mutually Exclusive Alternatives by Different Evaluation Methods
Different problem types lend themselves to different engineering economy methods
Different information is available from different evaluation methods
Primary criteria for what method to apply
Speed
Ease of performing the analysis
Note: “Best” method ≠ Only method (e.g. PW and AW)
3
If no method is preselected
Aspects to consider
Evaluation period (→ es mated life)
Type of alternatives (private (income, cost) vs service‐based (cost))
Recommended method
(≠ only possible method)
Series to evaluate (CFs for 1 alternative, incremental CFs for two alternatives)
4
5
If method is predetermined
Aspects to consider
Equivalence relationship (basic: PW, AW – CC, FW, … see chapter 6)
Lives of alternatives – Time period for analysis
PW: LCM of all alternatives
Incremental ROR and B/C: LCM of the two alternatives being compared
AW: analysis over respective alternative lives
Series to evaluate
Estimated cash flows or incremental series
Rate of return (interest rate)
MARR
Decision guideline
Final choice
6
7
Examples
(see handbook)
8
Wrap‐up
Epilogue
What is the best method to use ?
If no method is preselected
If method is predetermined
INSIGHT IN METHODS!
THINK!
17
Engineering Economy
Benefit/cost analysis and public sector economic
Benefit/cost analysis in a company context
Overview Public (and private) sector
Type of projects – B/C ratios ‐ Single and multiple projects
Ethical considerations
Private (and public) sector
Type of projects – C/E analysis – C/B issues
Public/private sector
C/E analysis (CEA) ‐ multiple projects
C/B: MADM ‐ MCDM
Public sector projects
Examples Public sector
Characteristic Public sector
Private sector
Size of investment
Larger Some large, more medium to small
Life estimates
Larger (30‐50+ years)
Shorter (2‐25 years)
Annual cash flow estimates
No profit; costs, benefits and Revenues contribute to disbenefits are estimated
profits; costs are estimated
Funding
Taxes, fees, bonds, private funds
Stocks, bonds, loans, individual owners
Interest rate
Lower Higher, based on market cost of capital
Alternative selection criteria
Multiple criteria
Primarily based on rate of return
Environment of the evaluation
Politically inclined
Primarily economic
Public sector
Note‐1
Costs: estimated expenditures to the government entity for construction, operation and maintenance of the project
Benefits: advantages to be experienced by the owners, the public
Disbenefits: expected undesirable or negative consequences to the owners if the alternative is implemented, e.g. indirect economic disadvantage
It is difficult to estimate and agree upon the economic impact of (dis)benefits for a public sector alternative.
Note‐2
The viewpoint of the public sector analysis must be determined before cost, benefit an disbenefit estimates are made and before the evaluation is formulated and performed. There are several viewpoints for any situation and the different perspectives may alter how a cash flow estimate is classified.
Public sector
Note‐3
Contractor does not share project risk
Fixed price ‐ lump‐sum payment
Cost reimbursable ‐ Cost plus, as negotiated
Contractor shares in project risk
Public‐private partnerships (PPP), such as:
Design‐build projects ‐ Contractor responsible from design stage to operations stage
Design‐build‐operate‐maintain‐finance (DBOMF) projects ‐ Turnkey project with contractor managing financing (manage cash flow); government obtains funding for project
Example
costs
A flood control project will have a first cost of $1.4 million with an annual maintenance
cost of $40,000 and a 10 year life. Reduced flood damage is expected to amount to
$175,000 per year. Lost income to farmers is estimated to be $25,000 per year. At an
interest rate of 6% per year, should the project be undertaken?
benefits
disbenefits
Note‐3: private contractors in public projects
Contractors does not share project risk
Fixed price ‐ lump‐sum payment
Cost reimbursable ‐ cost plus, as negotiated
Contractor shares in project risk
Public‐private partnerships (PPP), such as:
Design‐build projects ‐ Contractor responsible from design stage to operations stage
Design‐build‐operate‐maintain‐finance (DBOMF) projects ‐ Turnkey project with contractor managing financing (manage cash flow); government obtains funding for project
(out‐of‐scope)
Example
(ex. 9.1)
Public sector
Public sector
Citizen of the city
City budget Economic development
C B
D
C B
D
C B
D
cost
benefit
disbenefit
cost
benefit
disbenefit
cost
benefit
disbenefit
1 Annual cost bonds
X
X
X
2 Annual maintenance
X
X
X
3 Annual parks development
X
X
X
4 Annual loss commercial
X
5 Loss of sales tax rebates
X
X
X
X
X
6 Annual income park
X
X
X
7 Savings flood control projects
X
X
X
8 Property damage avoided
X
X
X
Citizen: “maximize quality and wellness of citizens, family and neighborhoods”
City budget: “ensure budget is balanced and sufficient to fund rapidly growing city services”
Economic development: “promote industrial and commercial development (jobs)”
Benefit/cost analysis of a single project
Basically: B/C ratio<1: not economically acceptable, B/C ratio ≥ 1: project acceptable
PW of benefits AW of benefits FW of benefits
B/C 


PW of cos ts
AW of cos ts
FW of cos ts
(1) The conventional B/C approach
benefits  disbenefits B  D

conventional B / C 
cos ts
C
(2) The modified B/C approach
mod ified B / C 
Public sector
benefits  disbenefits  M & O cos ts
initial investment
Note: salvage value = negative cost (denominator)
Benefit (savings)
Cost (initial investment)
Example (ex. 9.2)
Cost (“M&O”)
Public sector
Disbenefit (negative consequence)
Public sector
Alternative selection using incremental B/C analysis for two alternatives
Public sector
Example (ex. 9.4)
Public sector
Public sector
All costs in $
Alternative
A
Alternative
B
∆
(B‐A)
Costs
685000
1030750
345000
Benefits
200000
450000
250000
Disbenefits
500000
400000
‐100000
Incremental B/C=250000/345250=0.72  A
Incremental (B‐D)/C={250000‐(‐100000)}/345250=1.01  B
Incremental B/C analysis of multiple, mutually exclusive alternatives
Given three or more mutually exclusive alternatives; one must be selected
Conduct the pair‐wise ∆B/C analysis (cfr ROR chapter)
The selection rule is:
Choose the largest‐cost alternative that is justified with an incremental B/C ≥ 1 when this alternative has been compared with another justified alternative
Benefits
Direct benefits
Usage cost estimates
Public sector
Example (ex. 9.5)
Public sector
Public sector
Public sector
Public sector
Ethical considerations
Target group
Time impact
Community Money only ? Scope (disbenefits)
Private sector “C/B” projects
Examples
Environment
Occupational safety
Different stakeholders
Automation
DSS‐BI
Chemical company has to choose between alternatives for
cleaning polluted site before selling it.
Manufacturing company wants to invest in oxygen masks for the paint spraying department.
Food company wants to buy new dust masks for their
production workers.
Hospital wants to replace all the current cots by new ones.
Midsized company wants to implement ERP software.
A company is considering to invest in a large scale BI (business intelligence) project (hardware, software, …)
Private sector
Special case: C/E analysis
Cost‐effectiveness analysis: basics
Cost are (of course) in monetary units, Benefits are expressed in non‐monetary, but quantitative measures, e.g. % of risk reduction; % of throughput increase, response time, amount of pollution removed, number of occupational accidents avoided
“tangibles”
Of course aim is to have measures that are: relevant, quantitative/objective, practical
(limited treatment in new book)
Private sector
Maximize “effect”
Effectiveness
budget
threshold
E
K
Minimize “cost”
Private sector
Costs
Example C/E
Three projects were filed for enhancing occupational safety and working conditions beyond what is legally required. Which project represents the best investment ?
Relevant
Quantitative
Objective
Measurable
Project
PW
(euro)
Number of occupational
accidents avoided
Days away from work (accident)
avoided
Project A
100000
10
15
Project B
150000
12
18
Project C
75000
10
12
Private sector
How to choose ?
Example C/E
A hospital service in Brooklyn (NY) is considering some alternatives to reorganize its services: keep the number of ambulances (currently 7) but relocate them to a satellite garage, increase the number of ambulances to 10 and relocate them to a satellite garage, have a number of cruising ambulances (7, 8, or 10) instead of ambulances stationed at a garage. How to choose between these alternatives ?
Private sector
Costs identified
Investment in ambulances + equipment, Investment in garage, Operations and maintenance costs, Personnel (ambulance, garage – supervisor cruising)
Benefits (effects) identified
Quality of healthcare (!), Intervention time {Average time Tavg
‐ Number of quick interventions (within 20 min)}
Calculation of costs
Rather straightforward
Calculation of benefits
With the help of simulation
Waiting time
Private sector
Dispatch delay
Travel time to
Pick up delay
Travel Drop‐off time from delay
(Back to base)
Example C/E
Thresholds: no general rule exists
e.g. Cohen, 2008
CE < $50000 (>$1000000) per life year gained: attractive (unattractive)
Effectiveness
can be measured disease/intervention specific: e.g. LOS, infection rate, mortality
or more general: e.g. QALY
LOS=length‐of‐stay; QALY=quality adjusted life years
C/E – C/B analysis
Example
An automotive plant has investigated four options for a production line for accessories. Besides money (PW) they also wanted to include aspects like speed (capacity), reliability, safety, flexibility (other products) and standardization (equipment components) into their analysis. The results of their analysis was summarized in the table below.
Private sector
Criterion Criterion
weight
Installation option 1
Installation option 2
Installation option 3
Installation option 4
Abs.
score
Rel.
score
Abs.
score
Rel.
score
Abs. score
Rel. score
Abs. score
Rel.
score
Speed
0.32
85
27.2
80
25.6
75
24.0
60
19.2
Reliability
0.24
85
20.4
60
14.4
80
19.2
95
22.8
Safety
0.24
85
20.4
50
12.0
70
16.8
90
21.6
Flexibility
0.12
50
6.0
80
9.6
80
9.6
99
11.9
Standardization
0.08
85
6.8
90
7.2
70
5.6
50
4.0
Total effectiveness
score
80.8
68.8
75.2
79.5
Present worth
(103 euro)
450
250
300
350
Private sector
C/B analysis: issues
Criteria
Multiple criteria
Conflicting criteria No dominance
Important issues
Definition
Measurement Exhaustiveness
Exclusiveness
Private sector
Example
Setting manufacturer of industrial vehicles, painting department
Problem at hand
buying new masks (occupational safety/diseases ‐ solvents)
Analysis
technical risks + legisla on + market study → three alterna ves remaining
Criteria
cost (life time !)
Purchasing cost
O&M cost non‐cost criteria
Easy to move around with
Functionality (protection)
What about these criteria ?
Exhalation (resistance)
Heat build‐up in mask
How to move on ?
Easy to put on
General level of comfort
Air supply (extra filter needed ?)
Readability manual
Robustness
Spare parts replacement
Other (e.g. hygiene, mounting mechanism)
Private sector
Weights – Scoring
Rating/Budget Ranking
Pairwise comparison
Delphi like method
e.g.
Rate according to the importance on a scale from 0‐10 (normalize afterwards)
or
Distribute a budget of 100 point over the criteria according to the importance
Rank (sort) in order of decreasing importance
Iterative panel effort
Likert‐type scales
Very bad
Bad
Private sector
Average
Compare the criteria two by two, each time indicate which is the most important one and also to which degree this holds
Good
Very good
(Numerical values !!)
With N  number of criteria and Ri  rank of criterium
(ranking 1  best )
# criterua
uniformweights
1
wi 
N
7
weights
uniform
criterium
rank
1
2
3
4
5
6
7
rank sumweights
wi 
N  Ri  1
 N  R
N
k 1
k
 1
rank sum rank reciprocal
28
2.593
0.1429
0.250
0.386
0.1429
0.214
0.193
0.1429
0.179
0.129
0.1429
0.143
0.096
0.1429
0.107
0.077
0.1429
0.071
0.064
0.1429
0.036
0.055
1.0000
1.000
1.000
0
0
0
0.4500
0.4000
0.3500
0.3000
weight
rank reciprocal weights
1
Ri
wi 
N 
 k 1 1Rk 
0.2500
uniform
0.2000
rank sum
0.1500
rank reciprocal
0.1000
0.0500
0.0000
1
Private sector
2
3
4
rank
5
6
7
Pairwise comparison
Typically used in AHP
Normalize (add up columns and divide
elements by column sum)
Weights Private sector
Critical reflection
Rating/budget
Ranking Pairwise
comparison
Delphi like
C/E analysis (CEA) ‐ multiple projects
Similar to procedures seen before
CER=cost‐effectiveness ratio (note: in chapter on cost estimation CER=cost estimation relationship)
Example: independent projects
Example: mutual exclusive projects
C/B analysis: MCDM; MADM
Steps
Decomposition step
Alternatives (mutual exclusive !!!)
Criteria
Weights
Scores
Synthesis step
Aggregation of information
Optimization
Conclusion Private sector
Decision support !!
Often one of ‐‐
Multi criteria/atribute decision making
Also e.g. “goal programming”
Decision support models (MCDM)
Mathematical optimization model
Choose one criterion (=objective function) and set min/max requirements for the other criteria (= constraints)
Absolute approach
All alternatives are scored and alternative with the best score is “rank and rate”
determined
E.g. LAM (linear additive method)
E.g. Consensus, TOPSIS (distance functions)
Relative approach
Alternatives are pairwise compared for all criteria
E.g. PROMETHEE E
E.g. AHP (analytic hierarchy process)
Other option: C/E analysis
C
MCDM example: Consensus method
Type: outranking method (absolute)
Start
Basic data
Criterion 1
Criterion 2
Criterion 3
Criterion 4
…
Private sector
i
Alt A
Alt B
...
Utility function
1.00
Utility
0.50
0.00
-0.50
0
1
2
3
4
5
6
7
8
9
10
-1.00
Performance scores
Ideal situation
U2j
+1
A
Alt.1
-1
Alt.2
Alt.4
+1
Alt.3
Worst situation
B
Private sector
-1
U1j
Weight for criterion i
Distance functions
elite

U p u1 j , u 2 j ,..., u mj ; 1 ,...,  m


1 

p
 i 1  u ij 

 
m
i 1

1
p
Utility score of alternative j for criterion i
anti-elite

~
U p u1 j , u 2 j ,..., u mj ; 1 ,...,  m  


p
 i 1  u ij 

  
m
i 1

1
p
1
Distance parameter
Private sector
Example Private sector
Gaussian utility function
(can be different for each criterion)
Linear additive method
Distance parameter
(the larger p, the larger the difference between elite and anti‐elite)
Private sector
Consensus method
MADM: AHP
AHP
Analytic hierarchy process (Saaty) – extension: ANP (analytic network process)
Relative method (note: consensus = absolute method)
Extension: ANP (analytic network process)
Tutorial: http://people.revoledu.com/kardi/tutorial/AHP/index.html
Private sector
Indicator for work order evaluation
Criteria network
(strategic  operational)
Weighting factors
(operational  strategic)
Class scores (1: excellent to 3: improvement needed) Extra‐1: Paper
Paper: The use of alternative solvent purification techniques Cournoyer & Dare – Chemical Health & Safety – July/August 2003, pp 15‐18
ook: http://www.physsci.edu/news/entries/202022.html
Background
University research lab
Lab activities: organic solvent (e.g. benzene) purification to remove moisture and oxygen Reason for C/B analysis
Explosion of solvent distillation still
Student injured (second degree burn wounds)
Material damage of $3.500.000 (2 labs)
Research question
“C/B” of alternatives: column method – purchase ultra‐dry solvents
Technical analysis
Still method: traditional, batch process
Column method: continuous process
Purchasing pure solvents: possible, but expensive
Risk analysis
Risks entailed in organic solvent purification
Legislation
Probability – Severity “Damage”: injuries (death) – material losses
Severity
Comparable for still and column method
Probability Typical problem for working with flammable organic solvents
Different for still and column method
Still: active metal and flammable solvents – ignition sources: heating mantles, vacuum pumps, high temperature water lines –
(safety shutdowns)
Column: no initiators – but stored energy hazards – peroxide accumulation – larger quantities of solvent
…
User validation
Column method: accepted method, O2 concentration mostly OK, problem is Cu cannot be used (Ar, N)
Technical risk (reduction)
Still method
when accident occurred: amount of solvent>legally allowed – fire door blocked open – no sprinklers; But, in normal circumstances: “occasional” for likelihood (1996, 1997)
Severity: “catastrophic”
Residual risk (with controls): “high”
 Management decision needed (case by case)
Column method
Likelihood: “remote” Severity: “high”
Risk: “low”
 Management: acceptable
Costs/Benefits
C/B = business view: “acceptable level may be achieved when the costs of decreasing a given risk further are greater than the costs realized from the occupational exposure to hazardous chemicals”
Problem
No useful numbers published for (lowering) chemical risks Solution
Inspiration from nuclear industry: “$1000‐2000‐10000 per person‐rem”
Costs
Purchasing:
20‐50% more
expensive than regular solvents
Conclusion article
Still method: acceptable, but…
Column method: cost friendly alternative
Purchasing: costly, but recommendable for small amounts
C/B approach in paper
Interesting analysis, dictated by an accident
Elements taken up in the analysis
User requirements
Technical aspects – Risks
Cost elements
Conclusions of the analysis
Some data issues
Recommendations
No closed formula
Decision support, not decision making
Quid situation?
Instructions for use “user knows what he/she is doing”
Blocking safety/fire door
No sprinklers
Volumes chemicals more than needed (allowed) …
Extra‐2: Case discussion
Setting
Medium‐sized company, LiBa
Production line for special effect light bulbs
Recent inspection (governmental regulations): potential problems concerning occupational safety → Threats: paying fines, losing license‐to‐operate)
“Engineering has a number of alternative solutions”
Assignment
Use a MCDM approach to choose the “best” alternative
Alternatives
Sensibilization
Posters and signs
Protection screens
Limited intervention
Control system
Rather extensive intervention
Renovation
Safety as part of
Relocation (international)
Closing down and moving
Outsourcing activities
Social impact
New line
Large investments
Stakeholders
Project engineer
Production
Maintenance
Safety
Management
Trade unions
How to tackle this problem …
Assemble project team, make sure all stakeholders are represented
Have a first look at the options (alternatives), do you want to take up all of them in your analysis?
Determine the criteria
definition (consensus?), measure
exhaustive (not too many!), exclusive (see definitions)? Determine the criteria weights
choose a method to determine the weights
look for a way to “combine” the different stakeholder views Determine the scores
idem
Bring everything together in one MCDM method
which one do you choose? why?
Note on MCDM
Increasing interest Industry & Government & Healthcare
Analysis
Complex – (time consuming)
Data collection
hypotheses – scenarios
Benefits in decision process
Sensitivity analysis
easy – danger for manipulation
Completeness
Tangibles & intangibles
Transparancy
Scientific tool
Acceptance
Involvement stakeholders
C/B report
Sometimes “extra”
Sometimes “funneling”
Role in decision process
Incorporation decision attitude
Decision support: rather
supporting than final conclusion
Risk analysis can be combined
perfectly with MCDM
Scoring & weights (utility funcion)
Tools
Models and software available
Wrap‐up
Chapter 9 Extras
Public sector
Type of projects – B/C ratios ‐ Single and multiple projects
Ethical considerations
Private sector
Type of projects – C/E analysis – C/B issues – MCDM – Consensus method
Engineering Economy
Project Financing
and
Noneconomic Attributes
1
Overview
Cost of capital – MARR
D-E mix
1-2
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
LEARNING OUTCOMES
1. Explain cost of capital and MARR
2. Calculate weighted average cost of capital
3. Estimate cost of debt capital
4. Estimate cost of equity capital
5. Understand high D-E mix and risk
6. Develop weights for multiple attributes
7. Apply weighted attribute method to
alternative evaluations
10-3
© 2012 by McGraw-Hill
All Rights Reserved
Cost of Capital and MARR
Cost of capital is the weighted average interest rate paid based
on debt and equity sources
 Debt capital represents borrowing outside company
 Equity capital is from owners’ funds and retained earnings
MARR is set relative to cost of capital
10-4
© 2012 by McGraw-Hill
All Rights Reserved
Factors Affecting MARR
Project risk: higher risk leads to higher MARR
Investment opportunity: in order to capture perceived opportunity, MARR
may be temporarily lowered
Government intervention: gov’t actions such as tariffs, subsidies, etc.
can cause companies to raise or
lower MARR
Tax structure: rising corporate tax rates lead to higher MARR
Limited capital: as capital becomes limited, MARR increases
Rates at other corporations: competition can cause companies to
raise or lower MARR
10-5
© 2012 by McGraw-Hill
All Rights Reserved
D-E Mix and Weighted Average COC
Debt-to equity (D-E) mix identifies percentages of debt and
equity financing for a corporation
WACC = (% equity)(cost of equity)
+ (% debt)(cost of debt)
This figure
Illustrates WACC
If the percentage of equity capital
from each source is known,
each component of WACC
is separately calculated
10-6
© 2012 by McGraw-Hill
All Rights Reserved
Example: WACC Calculation
A company that specializes in producing cold-weather clothing and
accessories is expanding its ski-jacket and boot-sock manufacturing
facilities. The company plans to borrow $2.5 million at 7% interest, issue
stock worth $4 million worth at 5.9%, and use $1.5 million of retained
earnings at 5.1% to finance the project. Determine the company’s WACC.
Solution:
Equity sources are stock and retained earnings
Total project cost = 4 + 1.5 + 2.5 = $8 million
WACC = 4/8(5.9% + 1.5/8(5.1%) + 2.5/8(7%)
= 6.09%
10-7
© 2012 by McGraw-Hill
All Rights Reserved
Parameters
Cost of debt capital
 Debt capital -- Funds received by borrowing; loans
or issuance of bonds
 Interest to pay to “bank” is actually lower than
stated due tax savings
Cost of equity capital
 Equity capital obtained from 4 possible sources:
 (1) Sale of preferred stock , (2) Use of retained earnings, (3) Owner’s
private capital, (4) Sale of common stock
 No tax advantage or tax savings for equity capital
 Determined using dividends, stock price, …
1-8
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
Debt-Equity Mix and Risk
As proportion of debt capital increases, overall cost of capital decreases
Because interest on debt capital offers a tax advantage
Corporations that become highly leveraged (large D-E mixes) have
increased risk and more difficulty in obtaining project funding
Investors take more risk and lenders are leery to provide funds
Best: balance between debt and equity funding
10-14
© 2012 by McGraw-Hill
All Rights Reserved
Multiple Attribute Analysis
 Attributes other than the economic one are considered in most
alternative selections, e.g., public and service sector projects
 Steps necessary to identify and use multiple attributes are:
Identification of attributes
Determination of importance weight of each attribute
Assignment of a value rating to each attribute
Alternative evaluation using a technique that
accommodates several attributes
(covered in previous chapter in course)
10-15
© 2012 by McGraw-Hill
All Rights Reserved
Summary of Important Points
Cost of capital is weighted average of debt and equity funding
There is a tax savings with debt capital because interest is
deductible; nothing is tax deductible for equity capital
High D-E mixes mean higher risk for lenders and investors;
project funding becomes more problematic
Multiple attribute analysis brings other factors (besides cost)
into the decision-making process
Four different techniques for assigning weights to attributes
Likert scale is good for assigning value ratings to alternatives
Multiple attribute evaluation measure is R j 
10-25
n
WV
j 1
© 2012 by McGraw-Hill
i
ij
All Rights Reserved
Wrap-up
Cost of capital – MARR
D-E mix
1-27
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved
Engineering Economy
Replacement and Retention Decisions
Overview Basics of replacement study
Economic service life (ESL)
Performing a replacement study
No study period specified
Study period specified
Literature !?
Basics of Replacement Study
Why replacing a machine, an installation, a plant, … ?
Reduced Performance:
Wear and tear
Decreasing reliability and productivity
Increasing operating and maintenance costs
Altered Requirements:
New production needs, accuracy, speed, etc.
Obsolescence:
Current assets may be less productive
Not state of the art – need to meet competition
Basics
11‐3
Terminology
Defender Asset:
Currently installed asset;
Challenger Asset:
The potential replacement or “challenging” asset;
Under consideration to replace the defender asset.
Together, Defender (D) and Challenger (C)
Constitute mutually exclusive alternatives;
Select one and reject the other.
Basics
11‐4
AW values are typically calculated
EUAC –Equivalent Uniform Annual Cost
Includes mostly costs estimates
Economic Service Life (ESL)
Not depreciated book value
Not “exotic” trade‐in value
For a given alternative is:
Number of years at which the lowest AW‐of‐cost occurs
Defender First Cost
Initial investment for the defender – P
Current Market Value (MV) now – correct estimate for P to assign to keeping a defender
Basics
Fair market value obtained from professional
appraisers, resellers, ...
11‐5
Actual cost:
Opportunity cost (MV)
Out‐of‐pocket cost (P‐TIV)
Challenger, first cost
Amount of capital that must be recovered when replacing a defender with a challenger
P (normal situation) or P‐(TIV‐MV) (unrealistic trade‐in value)
Sunk cost
Cost that has occurred in the past and cannot be altered
Sunk costs are generally of no use in a before‐tax replacement analysis
Consultant’s viewpoint The analyst assumes that he/she is an outsider (a consultant)
Thus, the analyst owns neither asset and must assign reasonable value to the defender and challenger, and is thus unbiased
It assumes the services provided by the defender can be purchased now by making an “initial investment” equal to the market value of the defender
While it may seem strange to charge an investment cost for keeping one’s own asset (the defender) this is what must occur. Keeping the defender is not free! Why? Because the firm is giving up the opportunity to receive a possible cash flow from selling the current defender.
11‐6
Example
3 years ago
Defender (old system)
Challenger (new system)
P=MV=$70000
P=$100000
AOC=$30000
AOC=$20000
S=$10000
S=$20000
n=3 years
n=10 years
now
Ground leveling system
Purchase price: $120000 ‐
Expected service life: 10 years ‐
Expected salvage value: $ 25000 ‐
Annual operating cost: $ 30000 ‐
‐
Not relevant
Current book value: $ 80000
Expected service life left: 3 years
Expected salvage value: $ 10000
Annual operating cost: $ 30000
Market value appraisal: $ 70000
New system (laser guided)
Purchase price: $ 100000 ($110000 next week)
Trade‐in value current system: $ 70000
Expected service life: 10 years
Expected salvage value: $ 20000
Annual operating cost: $ 20000
Investment Concerns
One must assign an investment cost for KEEPING the defender asset
The appropriate investment cost to assign to the defender asset is:
The current fair market value of the defender at the time the replacement decision is being examined.
Challenger Investment
This is the total investment (Pchallenger) required for a new (challenger) asset that will possibly replace the current defender.
Basics
11‐11
Economic Service Life (ESL)
Definition of ESL is:
Number of years n at which the equivalent annual worth (AW) of costs is minimized considering the most current cost estimates over all possible years that the asset may provide a needed service
ESL is the n where total AW is a minimum for tabulation at i% over a selected range of years.
To find the ESL
Plot n versus total AW, where Capital recovery CR=‐P(A/P,i,n)+S(A/F,i;n)
≈ annual worth of investment
Usually decreasing with time
Total AW = – CR – AW of annual operating costs
Seek the time period that minimizes the Total AW for the range of lives selected
Usually increasing with time
Economic service life (ESL)
11‐12
Complete equation
k
total Ak   P( A / P, i, k )  Sk ( A / F , i, k )   AOC j ( P / F , i,
 j 1
where

j )( A / P, i , k )

P  initial investment or current market value ( MV )
Sk  salvage value or market value after k years
AOC j  annual operating cos t for year j
with j  1 to k
Procedure: Year‐by‐year analysis for “k” years – where “k” is given or assumed.
Typical Annual Worth Plot
Seek the n value resulting in minimum
Total AW
Represents the n for
Economic Service Life
Economic service life (ESL)
11‐15
Example
Economic service life (ESL)
11‐16
Solution
Defender Asset;
3 years old now;
Market value now: $13,000;
5‐year study period assumed;
Estimated Future Market Values and AOC’s:
increasing
decreasing
17
Period – by – period analysis
For “k” = 1 year:
S1 = $9000
0 1
P=$13,000
AOC1 = ‐2500
AW(10%)1 = (‐$13,000)(A/P,10%,1) + $9000(A/F,10%,1) ‐2500
= ‐$7800 ( for one year!)
18
For “k” = 2 years:
S2 = $8000
0 1 2
AOC1 = ‐2500
P=$13,000
AOC2 = ‐$2700
AW(10%)2 = (‐13,000)(A/P,10%,2) + 8000(A/F,10%,2)
‐[2500(P/F,10%,1) + 2700(P/F,10%,2)](A/P,10%,2) = ‐$6276/yr for 2 years.
19
For “k” = 3 years:
0 1 S3 = $6000
2 3
AOC1 = ‐2500
P=$13,000
AOC2 = ‐$2700
AOC3 = ‐$3000
AW(10%)3 = (‐13,000)(A/P,10%,3) +6000(A/F,10%,3)
‐[2500(P/F,10%,1) + 2700(P/F,10%,2) + 3000(P/F,10%,3](A/P,10%,3) = ‐$6132/yr for 3 years.
20
A similar analysis for k = 4 and 5 is conducted;
The AW(10)k, K = {1,2,3,4,5} are tabulated as:
Total AWk
k=1: $‐7800
k=2: $‐6276
k=3: $‐6132
Min. Cost Year = 3 years
= Economic Service Life
k=4: $‐6556
k=5: $‐6579
21
Example
Important Points
Marginal costs (MC) are year‐by‐year estimates of the costs to own and operate an asset for one year
Three Components of Marginal Costs:
Cost of ownership (loss in Market Value/yr);
Foregone interest of Market Value at beginning of the year;
AOC for each year
AW of marginal costs = total AW of costs
The analysis is best performed via a spreadsheet model
See example 11.3
Economic service life (ESL)
11‐30
DATA
Performing a Replacement Study
Two approaches
D= defender
C=challenger
ESL=economic service life
AW=annual worth
Performing study
11‐34
I. No study period specified
Example
Solution Defender data
Current Market Value: $15,000;
Future Market Values will decrease by 20%/yr;
Keep for no more than 3 years;
AOC’s: {$4,000,$8,000,$12,000}
Retrofit next year = $16,000;
AOC’sD := {$20,000, $8,000,$12,000} (costs).
Old m/c Challenger data First Cost: $50,000
Future Market Values decreasing by 20%/year;
Retain for no more than 5 years;
AOC’sC := {$5,000,$7,000,$9,000,$11,000,$13,000}
New m/c Assume the interest rate is set at 10%/year.
39
updated
Remark What minimum market value of the defender will make the current challenger economically attractive?
If a high enough market value (trade‐in) is possible for the defender asset, one should take it and go with the challenger immediately!
Break‐even or replacement value (RV)
If the actual market value (trade-in) exceed the breakeven
replacement value, the challenger is the better alternative.
If this is the case, replace now with the challenger!
Performing study
11‐52
Additional Considerations
Several additional aspects of a replacement study
1.
Future‐year replacement decisions;
2.
Opportunity cost vs. Cash flow approaches;
3.
Anticipation of improved future challengers.
The analyst must understand trends, advances, changes in technology, and competitive pressures that often impact the remaining life of a defender and the challenger
Additional considerations
11‐53
II. Specified Study Period
Example
“P” (MV)
P
S
AOC
Option 1
v
Option 2
Truck has a life time of 12 years,
and is 3 years old at the moment
of the study
Remark
If the defender’s remaining life is shorter than the study period …
Focus on upgrading the defender and obtain cost
estimates in order to extend the defender out to the
study period.
These costs become part of retaining the defender
out to the prescribed study period and are used in
the replacement study
Specified study period
11‐66
Remember …
Compare the best challenger to the current defender;
“Best” is defined as the option with the lowest AW of costs over the specified period of time;
Apply the ESL method to find the economic life of the defender and challenger;
For ESL, you need estimates of future market values and annual operating costs for both C and D.
AW is the best method, especially if alternative lives are unequal.
For a fixed study period, apply the traditional AW method for both D and C.
Always focus on the estimates of future costs and market values using sound principles of estimation.
Avoid the trap of incorporating sunk costs into a replacement analysis
11‐67
“sound estimates”
“reason for replacement: reduced performance”
“Replacement” models in literature
I have an asset for which the maintenance costs keep increasing every year and/or the probability of failure keeps increasing with time. What is the best time to replace this asset with a new one ?
If I have a set of identical
devices, should I replace them as a group or as individuals ?
Do I best use a replacement policy based on asset age or based on a fixed period ?
Example
A small welding robot costs € 7000. The maximum useful life is 5 years. The maintenance costs ck are €1000k, with k being the quipment age. From historical data and supplier information, probabilities of unrepairable failures are known to be: p1=0.1, p2=0.1, p3=0.2, p4=0.3 and p5=0.3. Yearly it can be decided to replace the equipment by a new one.
What is the optimum replacement timing ?
Solution
Typically this type of problem (cfr “repair limit”) is solved by renewal theory (a.k.a. replacement theory), provided that the hypotheses for applying this theory hold (AGAN, N=)
The optimum replacement timing is found by minimizing C(n), with
E[cos t ]
C (n) 
E [life]
0
time
72
73
Formulas
Thus
74
Spot the differences
Find the similarities
Exercise
Wrap‐up
Basics of replacement study
Economic service life (ESL)
Performing a replacement study
No study period specified
Study period specified
Literature?!
Critical reflection ! Insight (
Chapter 11
Engineering economy
Independent projects under budget limitation
Overview
Capital rationing among projects
Projects with equal lifes
Projects with unequal lifes
Linear programming model
This chapter uses basic Linear Programming
Overview of Capital Rationing
Capital is a scarce resource; never enough to fund all projects
Each project is independent of others; select one, two, or more projects; don’t exceed budget limit b
‘Bundles’ are collections of independent projects that are mutually exclusive (ME)
For 3 projects, there are 23 = 8 ME bundles, e.g., A, B, C, AB, AC, BC, ABC, Do nothing (DN)
12-3
© 2012 by McGraw-Hill
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All Rights
Capital Budgeting Problem
Each project selected entirely or not selected at all
Budget limit restricts total investment allowed
Projects usually are quite different from each other and have different lives
Reinvestment assumption: Positive annual cash flows reinvested at MARR until end of life of longest‐lived project
12-4
© 2012 by McGraw-Hill
Reserved
All Rights
Capital Budgeting for Equal‐Life Projects
Procedure
Develop ≤ 2m ME bundles that do not exceed budget b
Determine NCF for projects in each viable bundle
Calculate PW of each bundle j at MARR (i)
Note: Discard any bundle with PW < 0; it does not return at least MARR
Select bundle with maximum PW (numerically largest)
12-5
© 2012 by McGraw-Hill
Reserved
All Rights
Example: Capital Budgeting for Equal Lives
Select projects to maximize PW at i = 15% and b = $70,000
Project
Initial investment, $
Annual
NCF, $
Life, years
Salvage value, $
A
‐25,000
+6,000
4
+4,000
B
‐20,000
+9,000
4
0
C
‐50,000
+15,000
4
+20,000
Solution: Five bundles meet budget restriction. Calculate NCF and PW values
Conclusion: Select projects B and C with max PW value
Bundle, j
Projects
1
A
‐25,000
+6,000
2
B
‐20,000
+9,000
0
+5,695
3
C
‐50,000
+15,000
+20,000
+4,261
4
A, B
‐45,000
+15,000
+4,000
+112
5
B, C
‐70,000
+24,000
+20,000
+9,956
6
DN
0
0
0
0
12-6
NCFj0 , $ NCFjt , $
SV, $
PWj , $
+4,000
© 2012 by McGraw-Hill
Reserved
‐5,583
All Rights
Capital Budgeting for Unequal‐Life Projects
LCM is not necessary in capital budgeting; use PW over respective lives to select independent projects
Same procedure as that for equal lives
Example: If MARR is 15% and b = $20,000 select projects
12-7
© 2012 by McGraw-Hill
Reserved
All Rights
Example: Capital Budgeting for Unequal Lives
Solution: Of 24 = 16 bundles, 8 are feasible. By spreadsheet:
Reject with PW < 0
Conclusion: Select projects A and C
12-8
© 2012 by McGraw-Hill
Reserved
All Rights
Capital Budgeting Using LP Formulation
Why use linear programming (LP) approach? ‐‐
Manual approach not good for large number of projects as 2m ME bundles grows too rapidly
Apply 0‐1 integer LP (ILP) model to:
Objective: Maximize Sum of PW of NCF at MARR for projects
Constraints: Sum of investments ≤ investment capital limit
Each project selected (xk = 1) or not selected (xk = 0) LP formulation strives to maximize Z
12-9
© 2012 by McGraw-Hill
Reserved
All Rights
Example: LP Solution of Capital Budgeting Problem
MARR is 15%; limit is $20,000; select projects using LP
PW @
15%, $
6646
-1019
984
-748
LP formulation for projects A, B, C, D labeled k = 1, 2, 3, 4 and b = $20,000 is:
Maximize: 6646x1 ‐ 1019x2 + 984x3 ‐ 748x4
Constraints:
8000x1 + 15,000x2 +8000x3 + 8000x4 ≤ 20,000
x1, x2, x3, and x4 = 0 or 1
12-10
© 2012 by McGraw-Hill
Reserved
All Rights
Different Project Ranking Measures
Possible measures to rank and select projects:
PW (present worth) – previously used to solve capital
budgeting problem; maximizes PW value)
IROR (internal ROR) – maximizes overall ROR; reinvestment assumed at IROR value PWI (present worth index) – same as PI (profitability index); provides
most money for the investment amount over life of the project, i.e., maximizes ‘bang for the buck’. PWI measure is:
Projects selected by each measure can be different
since each measure maximizes a different parameter
12-12
© 2012 by McGraw-Hill
Reserved
All Rights
Wrap‐up
Capital rationing among projects
Projects with equal lifes
Projects with unequal lifes
Linear programming model
Chapter 12
Engineering Economy
Breakeven and payback analysis
1
Overview
For a single project
With two or three alternatives
(Spreadsheet tool: Excel’s Solver)
2
Breakeven analysis
Performed to determine the value of a variable or parameter of a project or alternative that make to elements “equal”
Illustration
That specific replacement value of the defender in a replacement study that makes the challenger an equally good choice
That sales figure that will allow to cover all costs made; if we sell less we have a loss; if we sell more we make a profit
…
Tools
Direct solution (one factor). Trial and error (multiple factors). Spreadsheets (using PV, FV, RATE, IRR, NPV, PMT, and NPER and Goal Seek or Solver in Excel).
Estimates are considered to be certain
→Part of sensitivity analysis
→Can be complemented with e.g. simula ons
Introduction
3
Breakeven Analysis for a Single Project
Given P, F, A, i, n
If all of the parameters (variables) shown above are known except one, then the unknown parameter can be calculated or approximated
A breakeven value can be determined by setting PW, FW, or AW = 0 and solve for or approximate the unknown parameter
Single project
1‐4
Cost – Revenue Model Approach
A popular application of Breakeven (BE) is where cost‐
revenue‐volume relationships are studied
Define cost and revenue functions and assume some linear or non‐linear cost or revenue relationships
One objective: Find a parameter value ‐‐ termed QBE ‐‐ that will minimize costs or maximize profits Concepts: fixed costs, variable costs, total costs
Single project
1‐5
• see
See also first class
6
Total Costs
Total Cost (TC) = Fixed Costs (FC) + Variable Costs (VC)
TC = FC + VC
Profit (P) = Revenue (R) – Total Cost (TC)
P = R – TC = R – (FC + VC)
Single project
1‐9
Cost – Revenue Relationships
Linear Models and Non‐linear models are used as approximations to reality
Single project
1‐10
11
Breakeven point
The breakeven (BE) point QBE is the point where the revenue and total cost relationships intersect
For non‐linear relations, it is possible to have more than one QBE point
Revenue and total cost relationships tend to be static in nature
May not truly reflect reality of the dynamic firm
However, the breakeven point(s) can be useful for planning purposes
Single project
1‐12
Example: One Project Breakeven Point
A plant produces 15,000 units/month. Find breakeven level if FC = $75,000 /month, revenue is $8/unit and variable cost is $2.50/unit. Determine expected monthly profit or loss.
Solution: Find QBE and compare to 15,000; calculate Profit
QBE = 75,000 / (8.00‐2.50) = 13,636 units/month
Production level is above breakeven Profit
Profit = R – (FC + VC) = rQ – (FC + vQ) = (r‐v)Q – FC
= (8.00 – 2.50)(15,000) – 75,000
= $ 7500/month
13-6
Breakeven analysis with more than one alternative
Given two alternatives (assume mutually exclusive)
Need to determine a common variable or economic parameter common to both alternatives
Parameter could be Interest rate, First cost (investment), Annual operating cost, etc.
More than 1 alternative
1‐17
Example: Two Alternative Breakeven Analysis
Perform a make/buy analysis where the common variable is X, the number of units produced each year. AW relations are:
AWmake = ‐18,000(A/P,15%,6) +2,000(A/F,15%,6) – 0.4X
AW, 1000 $/year
Breakeven value of X
AWbuy
8
7
AWbuy = ‐1.5X
6
Solution: Equate AW relations, solve for X
5
AWmake
4
‐1.5X = ‐4528 ‐ 0.4X
X = 4116 per year
3
2
1
If anticipated production > 4116,
select make alternative (lower variable cost)
0
1 2 3 4 5
X, 1000 units per year
13-8
Three Alternative Analysis
If three alternatives are present, compare the alternatives pair‐wise, or
Use a spreadsheet model to plot the present worth or annual worth over a specified range of values.
A typical three‐alternative BE plot might look like ….
More than 1 alternative
1‐22
Using Excel’s SOLVER for Breakeven Analysis
SOLVER is one of many built‐in Excel analysis tools;
SOLVER has been designed to aid in more complex forms of “goal seeking” and performing “what‐if” evaluations of properly constructed models.
For a properly constructed model SOLVER will require that the analyst:
Specify a target cell (the objective);
Identify one or more changing cell(s) that will have to change to achieve the desired target cell value
Self‐study
SOLVER
1‐24
Remember
BE analysis can be a form of sensitivity analysis
Breakeven point for a variable X is normally expressed as: Units per time period; Hours per month; etc.
At exactly breakeven (QBE) one is indifferent regarding a project
Typical breakeven models are:
Linear
Non‐linear
Two or more alternatives can be compared using breakeven analysis
Complex models can be evaluated using Excel’s SOLVER tool
1‐28
Exercise
29
Real‐life example
31
Wrap‐up
Chapter 13
For a single project
With two or three alternatives
[Spreasheet tool: Excel’s Solver]
33
Engineering Economy
Effects of Inflation
1
Overview
Impact of inflation
Rates: i, if, f
PW (FW, AW) with inflation considered
2
Impact of Inflation
Inflation is an increase in the amount of money necessary to obtain the same amount of product or service before the inflated price was present.
Impact because the value of the currency changes downward in value
Opposite: deflation
Firms should set their MARR rate to:
Cover the cost of capital;
Cover or buffer the inflationary aspects perceived to exist;
Account for risk.
1‐3
Equating “Value”
Money in time period t1 can be related to money in time period t2 by the following:
Dollars t1 =
Dollars t 2
inflation rate between t1 and t 2
1‐5
The Inflation rate ‐‐ f
The inflation rate, f, is a percent per time period; stated in a manner similar to interest rates.
Example: If f = 5.0% per year inflation, $100 today requires $105 to buy the same amount next year.
Inflation is on top of the interest rate
1‐6
The Basic Inflation Relationship
Let n represent the amount of time between t1 and t2 then . . . .
Future Dollars = Today’s dollars(1+f)n
Dollars in period t1 are termed:
Constant‐value dollars or today’s dollars
Dollars in time period t2 are termed:
Future dollars or then‐current dollars.
1‐7
Example
Assume a firm desires to purchase an asset that costs $209,000 in today’s dollars.
Assume a reasonable inflation rate of, say, 4% per year
In 10 years, that same piece of equipment would cost:
$209,000(1.04)10 = $309,371
Inflation is not an interest rate or rate of return consideration !!
1‐8
Impact The impact of inflation can be significant.
From the previous example we see that even at a modest 4% rate of inflation, the future impact on cost can and is significant!
The previous example does not consider the time value of money.
A proper engineering economy analysis should consider both inflation and time value of money.
1‐9
Three Important Rates
Real or inflation‐free interest rate
Denoted as “i”.
Inflation‐adjusted interest rate – market rate
Denoted as “if ”
Inflation rate
Denoted as “f”
1‐10
Real or Inflation‐free Interest Rate ‐ i
Rate at which interest is earned;
Effects of any inflation have been removed;
Represents the actual or real gain received/charged on investments or borrowed funds.
1‐11
PW adjusted for inflation
In prior chapters, present worth was calculated assuming that all cash flows were in constant value dollars;
Example: $5,000 (now) inflated at 4% per year with a interest of 10% per year.
verification
1‐12
Derivation of a Combined Interest Rate
Start with:
1
PF
(1  i ) n
Real interest rate
Assume F is a future dollar amount with inflation built in.
Then F
1
P
(1  f ) n (1  i ) n
1
PF
(1  i  f  if ) n
Inflation rate
Or 1
PF
 F ( P / F , i f , n)
n
(1  i f )
if
Market rate
i
1‐14
FW adjusted for inflation
Solving F=f(i,if,n)
• There are four different interpretations for future worth Current value
calculations.
1. Actual future $$;
2. Purchasing power of future $$ stated in constant‐value $$;
3. Future $$ required at t = n to maintain t = 0 purchasing power;
4. $$ at t = n to maintain purchasing power and earn a stated interest rate of i% per time period.
1‐15
17
Example
P = $1,000 now and the market rate of interest (if) is 10% per year. If n = 7, what is the future value of the $1,000 now?
F = $1,000(F/P,10%,7) = $1,948.
Remember, the market rate of interest includes both the inflation rate and the discount rate.
1‐19
Example Find the real interest rate for a market rate of 10% and inflation at 4%.
0.10  0.04
Solution  0.0577  5.77%
i
1  0.04
Purchasing power of $1000 (cv), with a market rate of 10% and inflation at 4%?
F = $1,000(F/P,5.77%,7) = $1481
Inflation of 4%/year has reduced to a real rate that is
less than 6% per year!
1‐24
Example
P = $1,000 now
Inflation of 4% per year
F = $1,000(1.04)7 = 1,000(F/P,4%,7) = $1316
26
Example
Given P = $1000
inflation rate f = 4%
Real interest rate i of 5.77%;
Calculate if as;
if = 0.0577 + 0.04 + 0.0577(0.04) = 0.10
Then:
F = $1000(F/P,10%,7) = $1948
$1948 seven years out is equivalent to $1,000 now with a real return of 5.77% per year and inflation at 4% per year.
1‐28
Importance of Inflation Impacts
Most countries: inflation is from 2% to 8% per year;
Some countries with weak currencies, political instability, poor balance of payments can have hyperinflation (as high as 100% per year).
1‐30
When Hyperinflation is Present
Spend money almost immediately;
Money looses value quickly;
Very difficult to perform engineering economy calculations in a hyper‐inflated economy;
Future values are unreliable and,
Future availability of investment capital is very uncertain.
1‐31
Capital Recovery Calculations Adjusted for Inflation
With inflation present:
Current dollars invested in a productive asset must be recovered over time with future inflated dollars.
With loss of future purchasing power, future dollars will have less buying power than current dollars;
More dollars will be required to recover a present investment is a productive asset.
1‐32
33
AW
40
42
Chapter 14
Wrap‐up
Impact of inflation
Rates: i, if, f
PW (FW, AW) with inflation considered
45
Engineering Economy
Cost estimation
Overview Cost estimation: what ?
Need. Cost types. Approaches (↑↓). Accuracy. Extra: Learning curve.
Cost estimation: some techniques
“Per unit” method. Cost indexes. Cost‐capacity equations (CER). Factor method (CER). Or professional organizations …
Indirect costs: Traditional rates and allocation vs Activity based costing.
Cost estimation: what ?
For the most part Engineering economy analysis tends to be “cost driven”. Future costs are very critical to the analysis of a project.
Engineers generally have the responsibility of cost estimation. Revenue generation generally comes from marketing/sales areas.
Questions to be resolved:
What cost components must be estimated?
What approaches will be utilized?
How accurate should the estimates be?
What estimation techniques will be utilized?
Introduction
Accuracy issue
Estimation software
Two major cost types
initial or first cost P Delivered‐equipment cost
Installed‐equipment cost
Such as equipment cost, delivery charges, installation costs, insurance coverage, initial training
annual operating costs AOC (M&O costs)
Such as direct labor, direct materials, energy, maintenance, rework & rebuild
Annual Operating Costs
Maintenance & Operating costs
Introduction
Approaches
Bottom‐up Approach
Treats the final cost as an independent (output) variable and the associated costs as input or dependent variables.
Design‐to‐Cost Approach
Treats the competitive cost as an input variable and the associated cost estimates as the output variables.
Introduction
Management
accounting
Introduction
Bottom‐up Approach
Treats the final cost as an output variable and the associated costs as input variables.
Cost components are first identified;
Cost elements are estimated;
Summed to = total direct cost; (DC)
Factor in estimated indirect costs. (TIDC)
Factor in required margin or profit.
Price = DC + TIDC + Profit.
Traditional costing approach;
Applied in industries where pricing the product or the service is not the dominant factor in competition .
Introduction
! Value engineering !
Design‐to‐Cost Approach
Treats the competitive cost as an input variable and the associated cost estimates as the output variables.
Applied in the early stages of new or enhanced product design.
Price estimates are conducted to set “target” values.
Detailed design and equipment needs are not yet defined. Becoming more popular especially in high competition markets;
Places greater emphasis on the accuracy of the estimation process;
Target cost estimates must be realistic;
Design to the target price;
Common practice in Eastern cultures (Japan and other Asian countries).
Introduction
Illustration: remote control
Illustration: facility design
Accuracy needed
No estimate is intended to be “exact”
But must be reasonable and accurate enough to provide a robust economic analysis
In preliminary design phase:
Estimates are viewed as “first cut” estimates (go – no go)
Serve as inputs to the project initial budget
The “unit method” is often applied here
“Per unit” approaches should fall into the 5% to 15% range for total costs.
Estimates outside of the 15% range are not very reliable!
Introduction
General guidelines for accuracy
Conceptual/Feasibility stage – order‐of‐magnitude estimates are in range of ±20% of actual costs
Detailed design stage ‐ Detailed estimates are in range of ±5% of actual costs
Characteristic curve of accuracy vs. time to make estimates
Work breakdown structure
Physical work elements
Functional work elements
(Sullivan)
Learning curves
or unit production cost
A learning curve is a line displaying the relationship between unit production time and the cumulative number of units produced. Learning (or experience) curve theory has a wide range of application in the business world.
e.g. In airplane industry (origin), in manufacturing1,2 and distribution, healthcare, etc.
Types
Individual learning: “practice makes perfect”
Organizational learning: “practice makes perfect” plus changes in administration, equipment and product design
Learning curve theory is based on three assumptions:
1.
2.
3.
1
The amount of time required to complete a given task or unit of a product will be less each time the task is undertaken.
The unit time will decrease at a decreasing rate.
The reduction in time will follow a predictable pattern.
for new tasks – trade‐off in JIT systems with short runs
2 also in machine learning
Typical formula
Also: learning rate, improvement rate
Assuming a constant percentage reduction in input resources each time the output quantity is doubled
Example
Illustration
“(Per) Unit method”
Examples
Number of units
Introduction
Unit cost factor
Example (p 390) Cost indexes
A cost index is a ratio of the cost of something today to its cost sometime in the past. As such, a cost index is always a dimensionless number.
Examples
Well‐known daily life example: CPI (consumer price index)
Common industrial examples: Plant cost index (chemical engineering), Marshall & Swift index (equipment)
Measuring the cost of living
Example: Belgium
http://economie.fgov.be/nl/statistieken/cijfers/
economie/consumptieprijzen/consumptieprijsi
ndexen/#dwtable
FOD Economie, K.M.O., Middenstand & Energie
Statistiek en Economische Informatie
Eenheid Consumptieprijzen
Simon Bolivarlaan 30 - 1000 Brussel
Tel. : 02/277.51.11 - Fax : 02/277.96.15
Autom. antwoordapparaat : 02/277.56.40
http://economie.fgov.be
http://statbel.fgov.be
e-mail : [email protected]
Formula
 It 
Ct  C0  
 I0 
where
Ct  estimated cost at present time
C0  estimated cost at time t0
I t  index value at time t
I 0  index value at time t 0
Bit outdated example, but are still updated
Chemical Engineering
Engineering News Record
Marshall & Swift
(Blank & Tarquin, 2018)
Example
Engineering News Record
Example
100.17=56.70/56.60; 100; 56.40/56.60=99.65
Note that
In an index different (sub)components (with different weights) are taken into account.
The base period is a selected time when the index is defined with a basis value of 100 (or 1).
The index each period (year) is determined as the cost divided by the base year cost and multiplied by 100.
Future index values may be forecast using simple extrapolation or more refined techniques such as time‐series analysis.
Refer to the articles on the next pages to obtain more insights in cost indexes.
Example: Read this article to gain more insight in how indices are composed
Chemical Engineering Plant Cost Index (CE PCI)
Structure
Evolution: °1963, only some cosmetic revisions (names) changed till
1982: major revisions like reduction of number of components (110→66) and change in productivity factor and in 2002: major revision, not in structure, but in underlying details
Values
Not a monotonically
increasing curves,
but ups & downs !
Weight revisions
Normalization
Seamless transition between old and new indices Example:
ENR Construction Cost Index (CCI) – ENR Building Cost Index (BCI)
Both for construction cost, but difference in labor component: CCI more suited if labor costs are high proportion of total costs, BCI more applicable for structures
Data
20‐US cities unweighted average Not including e.g. Arizona and Florida – Canadian version for Montreal and Toronto
Base year: 1913
For materials “spot prices” from a single source for all materials tracked
No seasonal adjustment (! Careful if period of analysis is less than 1 year)
Verifiability: published
Evolution
Makeup of index components, e.g. 1996: price structural steel →beams
Revisions, if necessary
CCI and BCI do not capture all factors influencing construction costs
Included are cement, lumber, structural steel and labor
Not all factors influencing project costs: snapshot of cost trends !
Example: Read this article to gain insight in the “local” character of an index Focus on methodology and concept,
not on numerical values
Belgian chemical engineering plant cost index
Idea: need for index because of lack of historical data and of large variety of equipment – estimation, quality not so good as cost engineering calculations or cost estimation software, but faster !
A lot of indices are available
Chemical industry
Average process industries
(cement, chemical, clay, glass, paint, paper, petroleum, rubber)
General industry
Petroleum refineries
Rapid growth in activities in CPI in mid and late 1970s
(fast increase) – 1980s over‐
supply of constructors (price drops)
Large difference
Inflation dropped,
thus lower increase rate
after 1981
Research question: need for “Belgian” index (BCEPC)
Methodology
Start with a simple index (few components) and add components until it is a good fit with the official American CEPC
•
•
•
•
General
2 parameters: labor + material
3 parameters: + productivity
4 parameters: + inflation
5 parameters: + crude oil price
Finetune the new cost index for the Belgian situation on hand (company !) taking into account the fact that construction projects in Belgium also include non‐Belgian inputs
Company‐
specific
country
weight
Belgium 67 %
Netherlands
16 %
Germany 8 %
France
6 %
UK
3 %
USA
0%
Implementation
Compare new, improved index with current practice in CPI
Company Index
1
Langfrist
2
In‐house (German)
3
Nelson‐Farrar
4
2‐parameter – UK model
5
Individual indices only
Analyse what the impact of the new index will be on established management ratios
* Maintenance ratio =(total maintenance cost per year)/capital replacement value
* Stores ratio=(MRO inventory value)/ capital replacement value
Capital replacement value = Equipment replacement value
Belgian chemical engineering plant cost index (c’d)
maintenance ratio (%) based on
old cost index
new cost index
3.8
4.9
3.3
4.1
3.0
3.6
year
X
X+1
X+2
stores ratio (%) based on
old cost index
new cost index
1.0
1.3
0.9
1.1
0.8
0.9
Trend is similar for old and new index, numerical values however are quite different !
3 .5
Used for benchmarking
Beware of value of benchmarks !
stores ra tio (%)
3
2 .5
2
1 .5
world class
1
0 .5
0
U1
U2
U3
U4
U5
U6
U7
U8
units
Very low !
Too high ?
Briefly describe how cost indices work …
What do you see as benefits of cost indices for management ?
What challenges do you see ? (Parametric methods)
Cost‐estimating relationships (CER)
Design variables (speed, weight, thrust, physical size, etc.) for plants, equipment, and construction are determined in the early design stages. Cost estimating relationships (CER) use these design variables to predict costs. A CER is thus generically different from the cost index method, because the index is based on the cost history of a defined quantity and quality of a variable.
Examples
Cost‐capacity equation
Factor method
Statistical and other
mathematical models
are used!
Cost‐capacity equations
One of the most widely used CER models
Cost‐capacity equations are also called power law and sizing models.
Sources like e.g. Chemical Engineers’ Handbook, US Formula
x
Straight line on
log‐log paper
Q 
C2  C1  2 
 Q1 
where
C1  cost at capacity Q1
Environment Protection Agency, handbooks, professional journals, consultants, equipment manufacturers, …
C2  cost at capacity Q2
Average x = 0.6
If 0<x≤1: economies of scale
If x=1: linear relationship
If x>1: diseconomies of scale
x  correlating exponent
Assume C1=1000, Q1=1000 and Q2=2000, then C2?
x
1
x
1/ 2
 Q2 
 Q2 
C2  C1    C1    2C1  2000  linear relationship
 Q1 
 Q1 
Q 
Q 
C2  C1  2   C1  2 
 Q1 
 Q1 
x
 21 / 2 C1  1.73C1  1730  economies of scale
2
Q 
Q 
C2  C1  2   C1  2   22 C1  4C1  4000  di sec onomies of scale
 Q1 
 Q1 
Cost‐capacity equations can be time adjusted using cost indexes.
Formula
  Q  x  I 
C2 ,t  C1,0  2   t 
  Q1   I 0 
Example
Factor method
Originally developed for estimating total plant costs;
Identify the major equipment items then multiply by certain factors;
Premise:
Major equipment costs are readily available;
Using the proper factors, then a rapid estimate of total plant cost can be achieved.
Factors: called “Lang factors” (1947).
Formula
CT  hCE
where
CT  total plant cost
CE  total cost of major equipment
h  overall cost factor or sum of individual cost factors
Example
Recall: Direct & indirect costs
manufacturing
Direct and indirect costs
Recall Direct costs – identifiable with product, function, process
Indirect costs – general administration, computer services, quality, safety, taxes, security, …
Lang showed that direct cost factors and indirect cost factors can be combined into one overall factor for some types of plants: 3.1 for solid process plants; 3.63 for solid‐fluid process plants and 4.74 for fluid process plants.
Subsequent refinements of the factor method have led to the development of separate factors for direct and indirect cost components.
formulas
n
h  1   fi
i 1
where
f i  factor for each cost component
direct & indirect fi
indirect cost factor
applied to total
direct cost
i  1 to n components, including indirect cost
n
 

CT  CE 1   f i (1  f I )
i 1

 
where
f I  indirect cost factor
f i  factors for direct cost components
Examples
Example 1
Example 2
Direct cost factors
Indirect cost factors
“Issue: Indirect costs”
Traditional indirect cost rates and allocation
indirect cos t rate 
estimated indirect cos ts
estimated basis level
Activity‐based costing
Define cost pools. Usually these are support functions.
Identify cost drivers. These help to trace costs to the cost pools.
Management Accounting
Why ABC makes sense …
Setting: 3rd party warehouse
Assume warehousing costs (building maintenance, equipment, packaging materials, personnel, …) are € 300000 per year
There are three customers, each of the
customers has a contract for 1500 pallet position
(including m2 needed and all incoming/outgoing
movements.
It seems logical to split the € 300000 evenly over the three customers (€ per pallet position) or not?
Only logical when the three customers have similar activities. If e.g. for customers A and B receptions and shipments are always in full pallets, but for customer C only the receipts
are in full pallets and the shipping occurs in individual boxes, it seems fair that C pays more for the warehousing services. A more detailed analysis on warehousing activities is needed then.
Exercises
Have a look at it
Wrap‐up
Chapter 15
Extras Cost estimation: what ?
Need. Cost types. Approaches (↑↓), learning curve.
Accuracy. Cost estimation: some techniques
“Per unit” method. Cost indexes. Cost‐capacity equations (CER). Factor method (CER). Or professional organizations …
Indirect costs: Traditional rates and allocation vs Activity based costing.
Engineering Economy
Depreciation methods
1
Overview Terminology
Depreciation methods
Depletion methods 2
Terminology
Depreciation is the reduction in value over time of an asset
(due to wear & tear, use, deterioration, obsolescence, …) . The method used to depreciate an asset is a way to account for the decreasing value of the asset to the owner and to represent the diminishing value (amount) of capital funds invested in it.
The annual depreciation amount Di does not represent an actual cash flow, nor does it necessarily reflect the actual use pattern of the asset during ownership.
Terminology
3
Book depreciation and tax depreciation
Book depreciation: used by a corporation or business for internal financial accounting
Indicates the reduced investment in an asset based upon the usage pattern and expected useful life of an asset Classical, internationally accepted methods: straight line, declining balance, sum‐of‐year digits
Tax depreciation: used in tax calculations per government regulations.
In the USA and many industrialized countries, the annual tax depreciation is tax deductible, i.e. it is subtracted from income when calculating the amount of taxes due each year. However the tax depreciation amount must be calculated using a government‐
approved method.
Terminology
4
First Cost or Unadjusted Basis ‐ B
Initial purchase price + all costs incurred in placing the asset in service
Book Value ‐ BV
Remaining undepreciated capital investment on the accounting books
(B ‐ sum of all depreciation claimed in the past)
Recovery Period – n
Depreciable life of the asset in question – often set by law
Market Value ‐ MV
Amount realized by sale on the open market
Salvage Value ‐ S
Estimated trade‐in value or market value at the end the asset’s useful life
Depreciation Rate ‐ dt
The fraction of the first cost removed by depreciation each year
Personal Property
All property except real estate use in the pursuit of profit or gain
Real Property
Real estate and improvements, buildings and certain structures
Note 1: Half‐year convention: assets are placed in service or disposed of in midyear
Note2: Land is Real Property, but by law is NOT depreciable for tax purposes
Terminology
5
Depreciation methods
Single line (SL)
Declining balance (DB) and double declining balance (DDB)
Sum‐of‐years digits (SYD)
Modified accelerated cost recovery system (MACRS)
Depreciation
7
Straight line (SL) deprecation
Book value decreases linearly with time. Depreciation rate d=1/n is the same each year of recovery period n
Formulas
SLN(B,S,n)
Dt  ( B  S )d 
BS
d
BV t  B  tD t
where
t  year ( t  1,2 ,..., n )
Dt  annual depreciation charge
Book value after
t years of service
B  first cost or unadjusted basis
S  estimated salvage value
n  recovery period
Depreciation
d  d t  depreciation rate 1 / n
8
Example (16.1) Depreciation
9
Declining balance (DB) and double declining balance (DDB) depreciation
Accelerated depreciation method; also known as fixed percentage or uniform percentage method
Annual depreciation is determined by multiplying the book value at the beginning of the year by a fixed percentage d, thus the depreciation amount decreases every year.
DB does not directly use the estimated salvage value; DB has its own implied salvage value.
The pure DB method will never depreciate an asset down to a “0” Depreciation
10
Double declining
Max. depr. rate by law:
balance
d max
2

n
Depreciation for year t:
Dt  ( d ) BVt 1
Actual depr. rate for year t:
d t  d (1  d )t 1
If BVt‐1 not known, apply:
Book Value amounts (two ways):
BVt  B (1  d )t
BVt  BVt 1  Dt
Implied Salvage Value:
impS  BVn  B (1  d ) n
Implied d for S >0:
1/ n
S
implied d  1   
B
Dt  dB (1  d )t 1
DDB(B,S,n,t,d)
Depreciation
16‐11
Example
Depreciation
12
Depreciation
13
Depreciation
14
Sum‐of‐years digits (SYD) depreciation
Classical accelerated depreciation technique that removes much of the basis in the first one‐third of this recovery period; however write‐off in not as rapid as for DDB
n  t 1
Formulas
D 
(BS )
t
SUM
j n
SUM   j 
j 1
n( n  1 )
2
t( n  t / 2  0.5 )
(BS )
SUM
n  t 1
dt 
SUM
BVt  B 
Depreciation
SYD(B,S,n,t)
16
Example
B‐S=25000‐4000=21000
n‐t+1=8‐2+1
SUM=8*9/2=36 Depreciation
17
18
Modified accelerated cost recovery system (MACRS)
By US Federal Tax Law, all assets placed in service and eligible for depreciation MUST use the current MACRS methods of calculation of depreciation amounts. Tax Law permits states to have their own respective depreciation methods for state income tax purposes (complicating factor)
Formula
Dt  d t B
From table ! Depreciation
19
Depreciation
20
Example
Depreciation
22
Depreciation
23
Depreciation
24
MACRS recovery periods
Alternative depreciation system
General depreciation system
Depreciation
25
Compared …
26
Depletion methods
Depreciation is applied to assets that can be replaced.
Depletion applies to (natural) resources that are not easily replaced, like: timber, mineral deposits, oil and gas, etc.
Cost and percentage depletion
Percentage depletion
percentage depletion amount  percentage* gross income from property
Cost depletion or factor depletion
first cos t
pt 
resource capacity
Examples 16.5 – 16.6
Depletion
27
28
Remember
Depreciation may be determined for internal company records (book depreciation) or for income tax purposes (tax depreciation).
Depreciation is a book method by which the capital investment in tangible property is recovered. Depreciation does not result in actual cash flows directly – rather, tax savings are the result of depreciation. The annual depreciation amount is tax deductible, which can result in actual cash flow changes.
In the US, the MACRS method is the only one allowed for tax depreciation.
Depletion methods are used to recover investment in the extraction or harvesting of natural resources.
16‐29
Wrap‐up
Chapter 16
Terminology
Depreciation methods
Depletion methods 31
Engineering Economy
After‐tax economic analysis
1
Overview Book focuses on USA !!
Terminology
Before‐Tax and After‐Tax Cash Flow
Taxes and Depreciation method
After‐Tax analysis
2
Income Tax Terms and Relations (Corporations)
Income taxes are real cash flow payments to governments levied against income and profits. The (noncash) allowance of asset depreciation is used in income tax computations.
Two fundamental relations: NOI and TI
Net operating income = gross revenue – operating expenses NOI = GI – OE
(only actual cash involved)
NOI is also call EBIT (earnings before interest and taxes)
Taxable income = gross revenue – operating expenses – depreciation
TI = GI – OE – D
(involves noncash item)
Note: All terms and relations are calculated for each year t,
but the subscript is often omitted for simplicity
17‐3
© 2012 by McGraw-Hill
Reserved
All Rights
Tax Terms and Relations ‐ Corporations
Gross Income GI or operating revenue R ‐‐ Total income for the tax year realized from all revenue producing sources
Operating expenses OE ‐‐ All annual operating costs (AOC) and maintenance & operating (M&O) costs incurred in transacting business; these are tax deductible; depreciation not included here
Income Taxes and tax rate T ‐‐ Taxes due annually are based on taxable income TI and tax rates, which are commonly graduated (or progressive) by TI level.
Taxes = tax rate × taxable income
= T × (GI – OE – D)
Net operating profit after taxes NOPAT – Money remaining as a result of capital invested during the year; amount left after taxes are paid.
NOPAT = taxable income – taxes = TI – T × (TI)
= TI × (1 – T)
17‐4
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Average and Effective Tax Rates
Marginal tax rates change as TI increases. Calculate an average tax rate using:
total taxes paid
taxes
Average tax rate = = taxable income
TI
To approximate a single‐figure tax rate that combines local (e.g., state) and federal rates calculate the effective tax rate Te
Te = local rates + (1‐ local rates) × federal rate
Then, Taxes = Te × TI
17‐6
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Cash Flow After Taxes (CFAT)
NCF is cash inflows – cash outflows. Now, consider taxes and deductions, such as depreciation
Cash Flow Before Taxes (CFBT)
CFBT = gross income – expenses – initial investment + salvage value
= GI – OE – P + S
A negative TI value is Cash Flow After Taxes (CFAT)
CFAT = CFBT – taxes
= GI – OE – P + S – (GI – OE – D)(Te)
considered a
tax savings
for the project
Once CFAT series is determined, economic evaluation using any method is performed the same as before taxes, now using estimated CFAT values
17‐9
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Effects on Taxes of Depreciation Method and Recovery Period
Goal is to minimize PW of taxes, which is equivalent to maximizing PW of depreciation
Compared …
DEPRECIATION METHOD
All methods have the same amount of total taxes due
RECOVERY PERIOD
All lengths have the same amount of total taxes due
Accelerated depreciation methods result in lower PWtaxes
Shorter recovery periods result in lower PWtaxes
General observation for SL, DDB and MACRS methods:
MACRS PWtaxes < SL PWtaxes < DDB PWtaxes
General goal: use shortest (MACRS) recovery period allowed
(Note: with same single tax rate, recovery period and salvage value)
17‐10
(Note: with same single tax rate, depreciation method and salvage value)
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12
13
Depreciation Recapture (DR) and Capital Gain (CG)
DR, also called ordinary gain, in year t occurs when an asset is sold for more than its BVt
DR = selling price – book value = SP – BVt
CG occurs when an asset is sold for more than its unadjusted basis B (or first cost P)
CG = selling price – basis = SP – B
CL occurs when an asset is sold for less than its current BVt
CL = book value – selling price = BVt – SP 17‐11
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Effects of DR, CG and CL on TI and Taxes
CG = SP1 ‐ B
DR = SP2 ‐ BV CL = BV – SP3
Update of TI relation: TI = GI –
OE – D + DR + net CG – net CL
17‐12
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Example: Depreciation Recapture
A laser‐based system installed for B = $150,000 three years ago can be sold for SP = $180,000 now. Based on 5‐year MACRS recovery, BV3 = $43,200. GI for year is $800,000 and annual operating expenses average $50,000. Determine TI and taxes if Te = 34% and the system is sold now.
Solution: Depreciation recapture (DR) and capital gain (CG) are present
DR = SP‐BVt = 180,000 – 43,200 = $136,800 CG = SP‐B = 180,000 – 150,000 = $30,000
MACRS D3 = 0.192(150,000) = $28,800
TI = GI – OE – D + DR + CG = 800,000 – 50,000 – 28,800 + 136,800 + 30,000
= $888,000
Taxes = TI × Te = 888,000 × 0.34 = $301,920
Note: If not sold now, taxes = (800,000 – 50,000 – 28,800) × (0.34) = $245,208
17‐13
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After‐Tax Evaluation Use CFAT values to calculate PW, AW, FW, ROR, B/C or other
measure of worth using after‐tax MARR
Same guidelines as before‐tax; e.g., using PW at after‐tax MARR:
One project: PW ≥ 0, project is viable
Two or more alternatives: select one ME alternative with best (numerically largest) PW value
For costs‐only CFAT values, use + sign for OE, D, and other savings
and use same guidelines Remember: equal‐service requirement for PW‐based analysis
ROR analysis is same as before taxes, except use CFAT values:
One project: if i* ≥ after‐tax MARR, project is viable
Two alternatives: select alternative with ∆i* ≥ after‐tax MARR for incremental CFAT series
17‐14
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Approximating After‐Tax ROR Value
To adjust a before‐tax ROR without details of after‐tax analysis,
an approximating relation is:
After‐tax ROR ≈ before‐tax ROR × (1 – Te)
Example: P = $‐50,000
GI – OE = $20,000/year
n = 5 years D = $10,000/year Te = 0.40
Estimate after‐tax ROR from before‐tax ROR analysis
Solution: Set up before‐tax PW relation and solve for i*
0 = ‐ 50,000 + 20,000(P/A,i*%,5)
i* = 28.65%
After‐tax ROR ≈ 28.65% × (1 – 0.40) = 17.19% (Note: Actual after‐tax analysis results in i* = 18.03%)
17‐15
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Example: After‐Tax Analysis
Asset: B = $90,000
Per year: R = $65,000
S = 0
n = 5 years
OE = $18,500 D = $18,000
Effective tax rate: Te = 0.184
Find ROR (a) before‐taxes, (b) after‐taxes actual and (c) approximation
Solution: (a) Using IRR function, i* = 43%
(b) Using IRR function, i* = 36%
(c) By approximation: after‐tax ROR = 43% × (1 – 0.1840) = 35%
17‐16
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After‐Tax Replacement Analysis
Consider depreciation recapture (DR) or capital gain (CG), if challenger is selected over defender
Can include capital loss, if trade occurs at very low (‘sacrifice’) trade‐in for defender
An after‐tax analysis can reverse the selection compared to before‐tax analysis, but more likely it will provide information about differences in PW, AW or ROR value when taxes are included
Apply same procedure as before‐tax replacement evaluation once CFAT series is estimated 17‐17
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Economic Value Added (EVA)TM Analysis
Definition: The economic worth added by a product or service
from the perspective of the consumer, owner or investor
In other words, it is the contribution of a capital investment to
the net worth of a corporation after taxes
Example: The average consumer is willing to pay significantly more for
potatoes processed and served at a fast-food restaurant as fries
(chips) than as raw potatoes in the skin from a supermarket.
Value‐added analysis is performed in a different way than CFAT
analysis, however…
Selection of the better economic alternative is the same for EVA and
CFAT analysis, because it is always correct that … AW of EVA estimates = AW and CFAT estimates
TM At this time, the term EVA is a registered trademark of Stern Stewart & Co.
17‐19
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International Tax Structures
Tax related questions for internationally located projects concentrate on items such as:
Depreciation methods approved by host country
Capital investment allowances
Business expense deductibility
Corporate tax rates
Indirect tax rates – Value‐Added Tax and Goods and Service Tax (VAT/GST)
Rules and laws vary considerably from country to country
17‐21
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Summary of International Corporate Tax Rates
Tax Rate Levied on TI, %
≥ 40
For These Countries
United States, Japan
35 to < 40
Pakistan, Sri Lanka
32 to < 35
France, India, South Africa
28 to < 32
Australia, United Kingdom, Canada, New Zealand, Spain, Germany, Mexico
24 to < 28
China, Indonesia, South Korea, Israel
20 to < 24
Russia, Turkey, Saudi Arabia
< 20
Singapore, Hong Kong, Taiwan, Chile, Ireland, Iceland, Hungary
Source: KPMG Corporate and Indirect Tax Rate Survey, 2010 17‐22
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Summary of Important Points
For a corporation’s taxable income (TI), operating expenses and asset depreciation are deductible items
Income tax rates for corporations and individuals are graduated by increasing
TI levels
CFAT indirectly includes (noncash) depreciation through the TI computation
Depreciation recapture (DR) occurs when an asset is sold for more than the book value; DR is taxed as regular income in all after‐tax evaluations
After‐tax analysis uses CFAT values and the same guidelines for alternative selection as before‐tax analysis
EVA estimates extra worth that an alternative adds to net worth after taxes; it mingles actual cash flows and noncash flows
A VAT system collects taxes progressively on unfinished goods and services;
different than a sales tax system where only end users pay
17‐25
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Exercises
30
32
17.36
34
Wrap‐up
Chapter 17
Terminology
Before‐Tax and After‐Tax Cash Flow
Taxes and Depreciation method
After‐Tax analysis
37
Engineering Economy
Sensitivity analysis Decision making under risk and under uncertainty
Staged decisions
1
Overview Structure of chapter
is not following the handbook (extras !)
Basics Risk – (Un)Certainty
Decision making under uncertainty
Decision trees
Basics – VOPI, VOII, VOC
Case study
Simulation
2
Basics
Chapter 18
(8th edition)
Sensitivity analysis
Parameter: a variable or factor for which an estimated or stated value is necessary
Sensitivity analysis: an analysis to determine
how a measure of worth (e.g. PW) changes when one or more parameters vary over a selected range of values
Alternatives
Pessimistic (P), most likely (ML), optimistic (O) estimate
Expected value calculation
3
Expected value – standard deviation
Probability distribution (discrete, continuous)
Random sample
Selection in a random fashion of n values for a variable from a population with assumed or known probability distribtion
4
Simulation analysis
uses random samples from the probability distribution of selected variables for alternative evaluation using a measure of worth
Monte Carlo sampling is a commonly used simulation approach
5
7
Illustration
Data (in €)
Cash flows (present worth)
CF in
probability CF i n
50000
30 %
60000
40 %
70000
30 %
CF out
50000
70000
probability CF out
45 %
55 %
Model
PW=CF in - CF out
Parameters and their cumulative distribution
CF in (€)
cum. prob.
CF out (€)
50000
30%
50000
60000
70%
70000
70000
100%
cum. prob.
45%
100%
Random generator
(use RN – random numbers between 0 en 99) and corresponding parameters
CF in
RN1
CF out
RN2
50000
50000
029
044
60000
70000
3069
4599
70000
7099
8
Results (simulation runs)
Run
RN1
1
46
2
30
3
14
4
35
5
9
6
19
7
72
8
20
9
75
10
16
…
CF in
60000
60000
50000
60000
50000
50000
70000
50000
70000
50000
RN2
81
8
88
21
73
77
1
46
97
43
CF uit
70000
50000
70000
50000
70000
70000
50000
70000
70000
50000
PW
-10000
+10000
-20000
+10000
-20000
-20000
+20000
-20000
0
0
Average
Std dev
Reliable result needed!
(confidence intervals)
9
Risk – (Un)certainty
Deterministic (no variation !) estimates for all parameters
Certainty For some parameters there is a probability distribution available for the estimates.
Expected value or simulation approach
Decision making
“Probabilities” are unknown
Uncertainty
Risk 10
Decision making under uncertainty
Chances are not known for the identified states of nature (values) of some parameters.
It is possible to agree that each state of nature is equally likely to occur (cfr risk)
Other methods exist; e.g. Laplace criterion, Minimax (Wald) approach, Hurwicz criterion, Savage criterion, …
Certainty ?!
Uncertainty
Risk 12
Θj with j=1, …n
Assume
What can happen … we don’t know anything about the chances, we can’t influence anything
The choices we have,
i.e. investment alternatives
ai with i=1, …m
The outcome vij: what willl happen if we decided for 13
a given action i and a specific state of nature j occurs Criterion
Decision rule
Wald – maximin return
Hurwicz – maximax return
optimism‐pessimism index
Savage – minimax regret
Laplace – insufficient
reason
j
i
sk=security level
ok=optimism level
rij=regret
ρk=index
15
16
17
18
19
These criteria may very well all suggest a different cause of action.
Example
vij
θ1
θ2
θ3
θ4
2
2
0
1
1
1
1
1
0
4
0
0
1
3
0
0
20
i=1, ... m
j=1, ... n
vij
si
θ1
θ2
θ3
θ4
2
2
0
1
0
1
1
1
1
1
0
4
0
0
0
1
3
0
0
0
Wald ‐ maximin:
For each action: worst possible outcome = security level si
Choose action with largest security level: a2
22
i=1, ... m
j=1, ... n
vij
θ1
θ2
θ3
2
2
0
1
1
0
1
θ4
si
1
0
oi
1
1
1
1
4
0
0
0
4
3
0
0
0
3
2
Hurwicz ‐ maximax
Optimism level oi = best consequence that can result if ai is chosen
Choose action with best oi : a3
or
User preference (optimistic‐pessimistic) with the use of weighted sum introducing α
α*0+(1‐ α)*2; α*1+(1‐ α)*1; α*0+(1‐ α)*4; α*0+(1‐ α)*3
2*(1‐ α); 1; 4*(1‐ α); 3*(1‐ α)
Als α <3/4: a3
23
i=1, ... m
j=1, ... n
vij
θ1
θ2
θ3
θ4
2
2
0
1
1
1
1
1
0
4
0
0
1
3
0
0
Savage – minimax regret
rij
Compute regret
θ1
θ2
θ3
θ4
2
2‐2=0
2
4‐2=2
0
1‐0=0
1
1‐1=0
1
2‐1=1
1
4‐1=3
1
1‐1=0
1
1‐1=0
0
2‐0=2
4
4‐4=0
0
1‐0=0
0
1‐0=1
1
2‐1=1
3
4‐3=1
0
1‐0=1
0
1‐0=1
Compute Max rij
Find i with Min Max rij
24
i=1, ... m
j=1, ... n
vij
θ1
θ2
θ3
θ4
2
2
0
1
1
1
1
1
0
4
0
0
1
3
0
0
¼*(2+2+0+1)=5/4
¼*(1+1+1+1)=1
¼*(0+4+0+0)=1
¼*(1+3+0+0)=1
Laplace
* Equal probabilities !!
25
27
Certainty Decision trees
?!
Uncertainty
Risk When to use decision trees
More than one stage of alternative selection
Selection of an alternative at one stage that leads to another stage
Probability estimates for each outcome
Estimates of economic value (cost or revenue) for each outcome Measure of worth (expected value of ‐) as the selection criterion, e.g. E(PW)
28
You decide
Basic concept
“Nature” decides
Build tree
Solve tree
29
Steps to take
Build up the tree structure from the left (final decision to make) to the right, including each possible decision and outcome. A square represents a decision node with the possible alternatives indicated on the branches. A circle represents a probability node with the possible outcomes and estimated probabilities on the branches. Since outcomes always follow decisions, a treelike structure results.
Fill out all quantitative information on the tree.
Economic value (payoff) for each branch of the decision tree. Cost, revenue or benefit. Expressed in PW, AW or FW.
Probabilities for each of the mutually exclusive outcomes emanating from each decision node. These probabilities must sum to 1 for each set of outcomes (branches) that are possible from a given decision
Solve the decision tree (foldback), going from the right to the left
Start at the top right node.
Calculate the expected value for each decision alternative
E[decision]=∑ (outcome es mate) p(outcome)
At each decision node, select the best E( decision ) value
Continue moving to the left of the tree back to the root in order to select the best alternative
Trace the best decision path back through the tree
30
Example Probability – not sure what will happen
Sequence of two decisions
(staged decisions)
32
Probability
nodes
Decision nodes
33
Example
34
43
CFBT=cash flow before tax
Exercise
A company needs to take a decision concerning product M9 which has been developed in one of its R&D labs. The decision is about either putting M9 on a test market or to stop all activities for M9. Launching M9 on a test market would cost around € 100K. Based on historical data the company knows that only about 30% of products tested on the market prove to be successful.
If M9 would be a success on the test market, another decision needs to be made, i.e. how large should the production plant for M9 be designed, if any at all. A small plant would cost € 150K to build and produce 2000 units per year. A large plant would cost € 250K and produce 4000 units per year.
Marketing estimates a 40% probability that competitors put a similar product on the market, this would of course influence the selling price. Assuming that everything produced will be sold the following estimates are made
Unit price (€)
With competition
No competition
Large plant Small plant
20
35
50
65
The commercial life time of M9 is estimated to be 7 years. Yearly operating costs for both plants are equal, being € 50K.
Would you recommend to continue or to abandon the M9 project?
45
Value of control
Decision trees
Value of perfect information
Informational analysis:
How much are we willing to pay for …? Value of imperfect information
50
Value of perfect information (VOPI)
VOPI on a variable is calculated as
Expected value with perfect information – Expected value without perfect information
VOPI is an upper bound on the value of additional research to improve the probability assessment on a uncertain variable
In a more complicated problem, the variables can be ranked according to VOPI, providing guidance for additional research.
51
Value of imperfect information (VOII)
The value of imperfect information (VOII) can also be determined with decision analysis. An example of imperfect information is e.g. a forecast report with a known accuracy of 80%.
This is a more complex calculation and requires the use of Bayesian updating.
52
Value of control (VOC)
The value of control determines the upper bound on the value of controlling an uncertainty.
Value of Control is calculated as
Expected Value With Control
‐ Expected Value Without Control
This value can be used to gauge the cost effectiveness of new alternatives.
53
Example: Informational analysis
Orange Grower’s Decision Problem:
Frost could occur overnight
Frost protection costs money
Total crop loss if frost occurs without protection measures in place
54
Helps analyzing and structuring
the decision problem Uncertainty Impact Decision Influence on
Net benefit of frost protection decision
= Crop value – Frost protection cost 55
Data
Frost protection cost = 25
Value of undamaged crop = 100
Value of crop if frost occurs, but with frost protection = 75
Value of crop if frost occurs, no frost protection = 0
Probability(frost)=0.4; Probability(no frost)=0.6
Tree building
56
Tree solving
57
Value of perfect information
Tree structure is different !
Probabilities are still there
(nature !)
Initial problem
58
80% accuracy
Value of imperfect information
P ( Frost | " Frost ") 
P(" Frost " | Frost ) * P ( Frost )
0.32

 0.73
P(" Frost " | Frost ) * P( Frost )  P(" Frost " | No frost ) * P ( No frost ) 0.32  0.12
59
80% accuracy
Bayes
P ( Frost | " Frost ") 
P(" Frost " | Frost ) * P ( Frost )
0.32

 0.73
P(" Frost " | Frost ) * P( Frost )  P(" Frost " | No frost ) * P ( No frost ) 0.32  0.12
60
80% accuracy
Bayes
P ( Frost | " Frost ") 
P(" Frost " | Frost ) * P ( Frost )
0.32

 0.73
P(" Frost " | Frost ) * P( Frost )  P(" Frost " | No frost ) * P ( No frost ) 0.32  0.12
61
New tree
62
Value of control
63
Case study
Interesting case study – full text available on Toledo, please read 66
Case study: Underground pipeline
Case source
AIChE
Problem at hand
Potential leakage of flammable and toxic material in old pipelines
Action alternatives
Do nothing
Replace old pipeline
Install automated shut‐off valves
Truck feed stock to the plant
67
Brainstorm
First feasibility check of alternatives  elimination of unrealistic alternatives
do nothing
replace current (aging) pipeline
insert automated shutoff valves
truck feedstock to plant
List of uncertainties (to be taken into account in decision)
risk of pipeline failure
whether a failure would result in a slow leak or a rupture
the amount of chemical released due to failure
the risk of ignition if there is a rupture
the cost to clean up a chemical release from the pipeline
68
Preliminary list with concerns and their priorities
cost of modifying current pipeline or installing new pipeline
cost of cleaning up a spill
worker safety
public safety
public concern
Purely economic decision
Model
value measure for decision
= NPV (net present value), i.e. PW (present worth)
Additional assumptions and simplifications




horizon = 10 years
E[NPV]=E[PW]
Salvage value old pipeline after 10 years = 0
p(ignition) and cost (fire)  f(amount chemical released)
69
Constructie van influence diagram
Legend
Decision to take
Value measure
(criterium)
Dependency, influence
Uncertainties (independent of alternative)
70
Constructie van influence diagram
Legend
Decision to take
Value measure
(criterium)
Dependency, influence
Uncertainties (independent of alternative)
71
Deterministic phase
Calculate the value function
total cost = CA +CC +C F -V, where CC =CG *Q
with
C A  cost of pipeline action
CC  cost of cleanup
CF  cost of fire
V  value of remaining life of new pipeline after 10 years
CG  cleanup cost per gallon
Q  quantity released due to failure
Collect numerical values for all variables (certain and uncertain)
72
Calculate the nominal value (expected value) for all alternatives  best nominal alternative
NPV for each alternative under nominal scenario
Do nothing
Replace pipeline
Insert valves
$ 1.96 M
$ 2.73 M
$ 2.04 M
Carry out a sensitivity analysis and visualise in a tornado‐diagram 
variables for which further study is recommended
at nominal setting
in further analysis
73
Probabilistic phase
Determine all probability functions for the appropriate variables Build and solve decision tree(*)  best alternative
$ 1.95 M
$ 2.42 M
$ 1.52 M
75
Compute the value of information en the value of control
upperbound on cost for
monitoring system to control
size of leak (limited to 250 gallons)
= $ 1.2 M
upperbound on cost study
about remaining life current pipeline
= $ 111000
77
Conclusions
From this rough study
If immediate action is required and there is no possibility for further study  install automated shut‐off valves
If no immediate action is necessary
 additional research to obtain more information on the failure probability of the current pipeline (years to failure) ‐ max. cost = $ 111000 A monitoring system that would limit a slow leak to 250 gallon needs to be investigated ‐ max. cost = $ 1.2 M real life
78
Conclusions (c’d)
real life
No immediate action required: ultrasonic study of current pipeline
(cost $ 20000)  info to model
Investigate “monitoring system”  add to list of feasible alternatives
Refine estimation of probabilities on "type of failure"  info to model
Uncertainty on system effectiveness and amount of release (prevented by early warning) resp.  add to model
Update analysis and formulate final recommendation on “best alternative”
79
Wrap‐up
Chapters 18 & 19
Extras Risk – (Un)Certainty
Decision making under uncertainty
Decision trees
Basics – VOPI, VOII, VOC
Case study
Simulation
88
Course material
Blank & Tarquin, 2018
Work shops
[email protected]
Course content
7
Will be provided
Evaluation – Exam
Closed book
Written part (+/‐ 25%)
Small questions (insight?!) – multiple choice (guess
correction)/open questions
Oral part with (written) preparation time (+/‐ 75%)
Exercise(s)
Knowing which method to use and drawing the right conclusions from the results
Formularium available
Case study – Journal article
Proving “maturity” in the subject
Being able to evaluate a chosen approach
As discussed in first class (see “Introduction” slides on Toledo)
https://eng.kuleuven.be/studenten/lijstRM.html
FEATURE
The use of alternative solvent
purification techniques
A recent accident at the University of California, Irvine involved the purification of an organic
solvent using a solvent still resulting in a fire in the lab. A graduate student was seriously burned
and $3.5 million in property damage was incurred. In this report, lessons learned from this
accident are used to improve the safety of solvent purification operations. Column methods of
aprotic solvent purification processes, as well, as the purchasing of ultra-dry organic synthesis
solvents are evaluated. Residual risks associated with the solvent still operation are compared
with the alternative purification methods. While severity remains unchanged using the column
method, employing the column method reduces the likelihood of an accident. For most
applications, the column method and the purchase of ultra-dry solvents remove moisture and
oxygen at least to the same level as the solvent still method. This is not without additional costs.
Column systems have higher upfront capital costs and ultra-dry solvents cost 20–50% more
than certified solvents. Cost–benefit analysis argues that the additional costs make alternative
purification methods acceptable. In conclusion, column methods of aprotic solvent purification
processes and the purchasing of ultra-dry organic synthesis solvents are, in many applications, a
cost-effective alternative to solvent stills for producing a moisture- and oxygen-free product.
By Michael E. Cournoyer
and Jeffrey H. Dare
INTRODUCTION
In order to obtain satisfactory results in
many syntheses involving air moisture
sensitive reactions, it may be necessary
to purify solvents to remove reactive
impurities such as water, other protic/
acidic materials, or atmospheric contaminants such as oxygen. A commonly
employed purification method is the
solvent still, which involves the
reflux/distillation of an organic solvent
in the presence of a dehydrating/deoxygenating reagent. A recent accident at
the University of California, Irvine
(UCI) involved the purification of an
organic solvent using a solvent still.1
The consequence of the accident was
a fire in the lab that seriously burned a
graduate student, and caused $3.5
million in property damage. An accident of this magnitude warrants an
Michael E. Cournoyer, Ph.D., and
Jeffrey H. Dare are affiliated with
Nuclear Materials Technology
Division, Los Alamos National
Laboratory, Los Alamos, NM 87545
(Tel.: 505-665-7616; fax: 505-6657170; e-mail: [email protected]).
1074-9098/03/$30.00
doi:10.1016/S1074-9098(03)00055-8
evaluation of this traditional method
of purification to assess the adequacy
of the control measures currently in
place and to consider alternative methods of producing a quality solvent. In
1996, the column method of solvent
purification was introduced which
eliminates hazards associated with distillation.2 This report consists of a critical evaluation of solvent still and
column methods of aprotic solvent purification processes, as well, justification
for the purchasing of ultra-dry organic
synthesis solvents.
Since the sodium benzophenone
ketyl solution is among the most common solvent still methods to prepare
pure, anhydrous, oxygen-free solvents,
it will be used as an example. Typically,
an organic solvent is refluxed in the
presence of sodium and benzophenone in an inert atmosphere, as shown
in Figure 1. The reactive metal removes
moisture from the solvent, and the
ketyl intermediate that forms upon
reaction of the ketone and the metal
acts as an oxygen scavenger. The blue
color of benzophenone ketyl is used as
an indicator that the solvent is ready
for use. Usually the reflux process is
continued for several hours to ensure
solvent purity.
Alternatives to the traditional solvent still include the column method
in which dry nitrogen or argon is used
to force commercially purified solvents
through a stainless steel column. The
column is packed with activated alumina and copper catalysts to remove
trace amounts of water (Figure 2).
Alternatively, the solvent must be run
through a bubble degassing process
when not suited for copper catalyst
oxygen removal. Last, if only small
quantities of pure anhydrous solvent
are needed, it is sometimes considered
more cost effective to buy these solvents directly from a supplier.
TECHNICAL VALIDITY
The technical validity of the solvent
purification process is the most important criteria of the review process. The
column method and the purchase of
ultra-dry solvents must remove moisture and oxygen at least to the same
The technical
validity of the solvent
purification process
is the most important
criteria of the
review process.
ß Division of Chemical Health and Safety of the American Chemical Society
Elsevier Science Inc. All rights reserved.
15
level as the solvent still method. Qualitatively, both the column method and
the purchase of ultra-dry solvents meet
the tests for assessing oxygen and protic contaminants by using the sodium
benzophenone-ketyl test or titanocene
dichloride/zinc dust test. Quantitative
values are shown in Table 1.3
RISK
The risk associated with the purification of an organic solvent is a function
of the likelihood and potential severity
of injury, harm, incurred liability,
damage or loss. As the accident at
UCI clearly illustrated, the primary consequences (severity) associated with
purifying flammable solvents are injuries and property losses due to fire and
explosions. It should be noted that the
severity remains unchanged using the
column method.
The main difference in risk between
the solvent still and column method is
in the likelihood of an accident. Procedures reacting active metals and flammable liquids, such as sodium and most
organic solvents increase the likelihood
that a fire or an explosion will occur.
This is not limited to the purification
procedure, but extends to the quenching of spent sodium for disposal, as well.
The likelihood of an explosion or a fire
increases further with the nearness of
the following ignition sources: heating
mantles, vacuum pumps, and water
lines from condensers and the high temperatures needed for distillation. Stills
with automatic controls that shut down
the system conditions, such as loss of
cooling or overheating of the still pot,
do enhance the safety of the distillation
operation greatly.
While the column method has none
of these fire or explosion initiators, it
introduces some hazards of its own.
Since the columns are pressurized
(5–50 psi), a stored energy hazard is
present. Peroxides may accumulate on
the columns, as well, introducing an
explosive hazard. In addition, the column method does involve larger quantities of solvent.
Figure 1. A typical solvent still.
Figure 2. A commercially available column system.
Table 1. Quality of Purified Solvent
Impurities
Method
Solvent still
Column
Purchasing
16
Water (ppm)
Oxygen (ppb)
Peroxides (ppm)
10
<14
5
100
10000
<100
<1
<2
<1
DISCUSSION
With several prominent universities
and companies using the column
Chemical Health & Safety, July/August 2003
method to purify organic solvents, the
ppm level of oxygen is not an issue with
most air and moisture sensitive reactions. On the other hand, the copper
catalyst is incompatible with some solvents including tetrahydrofuran and
methylene chloride. Oxygen must be
removed from these solvents by purging
them with dry nitrogen or argon. Thus,
column method is a valid substitute for
the solvent still method within limits.
The main goal of
implementing a
process improvement
is to decrease the
risk to an acceptable
level associated
with abnormal
occurrences from
the purification of an
organic solvent.
The main goal of implementing a
process improvement is to decrease
the risk to an acceptable level associated with abnormal occurrences from
the purification of an organic solvent.
The first issue that this raises is whether
solvent stills with their current set of
controls are processes that can be run
with an acceptable level of risk. An
objective method of determining risk
with hazardous material operation
may be obtained using the risk matrix
from the paper titled ‘‘A Risk Determining Model for Hazardous Material
Operations,’’ as shown in Table 2.5
Clearly, the severity of the accident that
occurred at the UCI is rated ‘‘Catastrophic’’ due to the magnitude of the
property damage. It should be noted
that several factors contributed to the
immediate consequences of the accident, including the following:
The quantities of flammable and
combustible liquids present in the
building at the time of the fire were
in excess of the quantities allowed
for B-2 occupancy by the 1979
Building Code.
Table 2. Risk Determination Matrix
Likelihood
Severity
Catastrophic
Critical
Moderate
Negligible
Frequent
Probable
Occasional
Improbable
Remote
High
High
High
Low
High
High
Medium
Minimal
High
Medium
Low
Minimal
Medium
Low
Minimal
Minimal
Low
Minimal
Minimal
Minimal
The door (with a 1-hour-rated fire
door with self-closing and positivelatching hardware) between the laboratory with the solvent still and a
room filled with flammable and combustible liquids was blocked open.
Automatic sprinkler system protection was not installed.
The fact that this should not have
happened during the life of the facility,
but did, gives it likelihood rating of
‘‘Occasional.’’ Similar accidents have
occurred at the University of California
at Berkeley in 1996 and the University
of Texas at Austin in 1997. These
observations lend further validity to
the ‘‘Occasional’’ likelihood rating.6
Using the matrix, the residual risk
(the risk remaining after controls are in
place, with consideration of reliability
and certainty of the controls and risk of
control failure) associated with a solvent still is high. Whether operation
with high residual risk should be
authorized (acceptable) is the responsibility of line management. While this
judgment is determined in a case-bycase study, UCI hopes to replace 30
chemical distillation devices with columns.7 The column method lowers the
likelihood to a remote level during the
purification. Hence, the residual risk
associated with the column method is
low (acceptable).
The second issue that implementing a process improvement raises is
whether the cost is acceptable. Many
emerging technologies may be technically sound and lower the risk associated with the process they seek to
improve and yet may be economically
unacceptable. From a business viewpoint, the acceptable level may be
achieved when the costs of decreasing
a given risk further are greater than the
costs realized from the occupational
exposure to hazardous chemicals.
Tangible costs for implementing technologies that lower risk in a chemical
operation have not been published.
On the other hand, criteria for establishing cost-effective dose-reduction
measure are well established for
the nuclear industry.8 Cost–benefit
analyses typically apply monetary
equivalents of $1,000 to $10,000
per person-rem with the recommended nominal value being $2,000
per person-rem. Optimization analysis are performed whenever the cost
of these measures exceeds $50,000 or
the collective dose to be avoided is
greater that 5 person-rem. In addition, the cost incurred from radiological exposures have been compared
to other non-radiological accidents
including property losses, as shown
in Table 3.9
Using the UCI accident as an example, the following approach is taken to
obtain a reasonable cost–benefit analysis between the two methods of solvent purification:
Table 3. Criteria that Trigger an Unusual Occurrence
Groups of Unusual Occurrences
Criteria
Facility condition: loss of control of radioactive
>100,000 (dpm)/100 cm2
material or spread of radioactive contamination*
Personnel radiological protection: radiation exposure >5 rem**
Value Base Reporting: cost-based occurrences
¼$1,000,000
*
dpm ¼ disintegrations per minute.
rem ¼ Roentgen equivalent man.
**
Chemical Health & Safety, July/August 2003
17
1. An accident resulting from a solvent
still operation is classified as a costbased occurrence.
2. The UCI accident is rated as an
‘‘Unusual Occurrence’’ in the Value Base Reporting category because over 1 million dollars in
property damage occurred.
3. This type of occurrence is categorized at the same level as a radiation
exposure of greater than 5 Roentgen
equivalent man (rem), as shown in
Table 3.
4. Recommended nominal value of
$2,000 per person-rem typically is
applied.
Using this recommended nominal
value, over $10,000 could be spent
on process improvements to reduce
the likelihood of fire and explosion
hazards and still be cost-effective.
Furthermore, a study conducted by
the University Wisconsin, Madison
(UWM), concluded that over the long
term the column method cost less.10
Results of cost comparison of the solvent still and column methods are
compiled in Table 4. While the column
method has a higher set up cost, the
annual operating are lower.
The quality of the distilled product
needs to be considered. For most purposes (e.g., Grignard reactions) the use
of common inorganic drying reagents
such as anhydrous magnesium sulfate,
MgSO4 (stirred overnight and filtered)
are sufficient. The UWM study also
concluded that the column method
was an ineffective choice for purifying
tetrahydrofuran.
Another alternative is the purchase
of pure anhydrous solvent. In the past,
if only small quantities of pure anhydrous solvent were needed, it was
sometimes considered more cost
effective to buy these solvents directly
from a supplier. Cost increase varies
per solvent, but in general the cost
Table 4. Solvent Purification Cost
Comparison
Method
Solvent still
Column
18
Set Up
Costs ($)
Annual
Operating
Costs ($)
4800
6300
4000
100
of an anhydrous solvent is two to
three times that of a certified American Chemical Society (ACS) solvent
and costs of standard high purity solvents are 20–50% more expensive
than that of certified ACS solvents.11
Depending on volumes, required costs
vary. Once operating costs of the
solvent stills, the capital costs of the
column, and the liability of both
the purification methods are weighed;
the additional costs seem more reasonable.
Last, the larger quantities of solvent
and peroxides on the used adsorbent
associated with the column method
must be addressed. The larger quantities of solvent are contained in
metal cans, which are safer than glass
bottles. Since the peroxides remain
unchanged on the alumina, the used
adsorbent should not be heated.
Stirring the adsorbent into an iron
sulfate solution can mitigate the peroxide hazard or spent columns can be
sent back to the manufacturer for
regeneration.
CONCLUSIONS
The results of this study demonstrate
that distillation and purification of
flammable solvents, using the solvent
still method, will continue to be an
integral part of chemical experimentation, if only on a limited basis as alternative methods become available.
Active metals increase the likelihood
of an explosion and fire with flammable liquids. In the past the solvent
still method of purifying organic solvents had an acceptable level of risk
because there was no alternative way
of obtaining a moisture- and oxygenfree product. Column methods of
aprotic solvent purification processes,
as well, as the purchasing of ultra-dry
organic synthesis solvents are now
cost effective alternative way of
obtaining a moisture- and oxygen-free
product.
Acknowledgements
The authors would like to acknowledge the Department of Energy and Los
Alamos National Laboratory Weapons
Engineering & Manufacturing Directorate for support of this work.
References
1. University of California, Irvine Independent Accident Investigation, ‘‘Injury and Fire Resulting from Benzene
Vapor Explosion in a Chemistry
Laboratory Frederick Reines Hall, University of California, Irvine, July 23,
2001,’’ Final Report, January 24, 2002.
2. Robert H. Grubbs, et al., ‘‘Safe and
Convenient Procedure for Solvent
Purification,’’ Organometallics 1996,
15, 1518. Robert K. Rosen, et al., ‘‘Safe
and Convenient Procedure for Solvent
Purification,’’ Journal of Chemical
Education January, 2001, 78 (1), 64.
3. James Farina and Paul Bouis, ‘‘Use of
Ether Solvents in Organometallics
Synthesis: The Good, Bad and the
Ugly,’’ 210th American Chemical Society Meeting, Las Vegas, NV, 1997.
4. System specification, SolventPurification.com & Innovative Technology,
Inc., 2 New Pasture Road, Newburyport, MA 01950.
5. Jeff Dare and Michael E. Cournoyer,
Ph.D., ‘‘A Risk Determining Model for
Hazardous Material Operations,’’
LA-UR 02-0378, Proceedings from
Probabilistic Safety Assessment and
Management 6 (PSAM 6), San Juan,
Puerto Rico, USA, June 22–29, 2002.
6. Kurt W. Dreger and Steven F. Pedersen, ‘‘Solvent Purification in Academic
Research Laboratories: A Comparison
of Old and New Technologies,’’ 2002
AIHCE Laboratory Health & Safety
Committee Presentations, http://
www2 . umdnj . edu / eohssweb / aiha /
administrative / presentations . htm #
Decommissioning.
7. Phil Willon, ‘‘Blame in UCI Fire Severity: No Sprinklers,’’ http://www.physsci.uci.edu/news/entries/2002022.html.
8. NUREG-1530 ‘‘Assessment of NRC’s
Dollar Per Person-Rem Conversion
Factor Policy,’’ 1995.
9. Steven Lee, Robert F. Grundemann,
Ph.D., and Michael E. Cournoyer, ‘‘An
Independent Analysis of a Glovebox
Glove Failure Incident,’’ LA-UR 012233, Journal of the American Society
of Mechanical Engineers, Proceeding
from ICEM’01, the 8th International
Conference on Radioactive Waste Management and Environmental Remediation, September 30–October 4, 2001.
10. ‘‘Research Laboratory Waste Reduction Project: Solvent Purification, Interim Report to DNR,’’ University of
Wisconsin, Madison, Submitted May 1,
2001.
11. Rich Motyka, Personal Communication, Mallinckrodt Baker, Phillipsburg,
NJ, July 17, 2002.
Chemical Health & Safety, July/August 2003
Tools for Making Acute Risk Decisions
with Chemical Process Safety Applications
CCPS – AIChE; 1995
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