Cost and Time Value of $$
Prof. Eric Suuberg
ENGINEERING 90
Cost and Time Value Lecture


What is our goal?
» To gain an understanding
of what is and what is not
a good project to undertake
from a financial point of view.
What are our tools?
» Material presented by
Prof. Crawford
» Discounting / Time Value
of Money
» Tax Savings through
Depreciation
So, what are we starting today?







Go through some of the “fun”
math for Present Value
Calculations
Do a teaching example of
purchasing a machine for a
manufacturing plant
Talk about costs – both the
obvious kind as well as the
non-obvious types
Time value of money
calculations
Cost Comparisons
Depreciation
Put it all together – inc.
continuous discounting and
after-tax cost comparisons
Have I Got a Deal for You

Would you be interested in investing in
a company that has $1 million in annual
sales?
What More Would
You Like to Know?
Annual operating expenses (salaries,
raw materials, etc.)
 Suppose these were $900,000/yr
 Are you interested? (Come on - I’ve got
to know now. There are a lot of people
interested)

Profit
Profit = Sales (revenues) - expenses (costs)
 Basis for taxation - What goes into the
calculation is of great interest to Uncle Sam
In Our Example
= $1,000,000/yr - $900,000/yr
= $100,000/yr
 Is this a good business?

Profit
What Would You be Willing
to Pay Me for this Business?
$1 million?
$2 million?
 How do you decide?
 This is one of the questions that we will
answer in this part of the course.

Present Value
Calculations
Essential element of evaluating a
business opportunity
 Different variants

» Simple discounting
» Replacement and abandonment
» Venture Worth, Present Value, Discounted
Cash Flow Rate of Return
What information
do we need?



Investment (Capital assets,
working capital)
Lifetime and Salvage Values
Operating Costs
» Fixed
» Variable




Interest Rate
Tax Rate
Depreciation Method
Revenues
Capital Investment Facility









Purchased Process Equipment
Field Constructed Equipment
Wiring, Piping, Instrumentation
Construction, Installation Costs
Site Preparation, Buildings
Storage Areas
Utilities
Services (Cafeterias, Parking lots, etc.)
Contingency
Capital InvestmentManufacturing
Costs of process equipment may
represent only 25% of actual
investment!
 Costs of process equipment scale
according to the “six-tenths rule”
» C2/C1 = (Q2/Q1)0.6
 See, for example:
» “Cost and Optimization Engineering”
by F.C. Jelen and J.H. Black,
McGraw-Hill, 1983.

Other Items

Working Capital
» Raw materials and supplies inventory
» Finished goods in stock and Work in
Progress
» Accounts Receivable, Taxes payable

Operating Costs
» Labor and Raw Materials
» Utilities and Maintenance
» Royalties

Fixed Costs
» Insurance, rent, debt service, some taxes
Time Value of Money

$1 today is more valuable than
the promise of $1 tomorrow
» Has nothing to do with inflation


“Discounting” is the term used
to describe the process of
correcting for the reduced
value of future payments
Discount rate is the return that
can be earned on capital invested today
Future Worth of an
Investment
P = Principal
 i = Annual Interest Rate
 S = Future value of investment

Compound Interest Law
S1 = P (1+i)
at the end of one year
S2 = S1(1+i) = P(1+i)2
at end of year 2
Sn = P (1+i)n
at end of year n
Present Value of a
Future Amount
P = Sn / (1 + i)n
= Sn (1 + i)-n
(1 + i)-n = Present Value Factor or
Discount Factor
The promise of $1 million at a time 50 years in
the future @ i = 15%/yr
P = $1,000,000(1+0.15)-50 = $923
Simple Example
What is the PV of $10.00
today if I promise to give it to
you in fifteen years, given a
discount rate of 20%?
 PV = 10(1.20)-15
 = $.65
 Not enough to buy a soda
these days

Take Home Message
Not all dollars of profit are the same
 Those that come earlier are “worth”
more

Start with Simple Example
from Everyday Life

Do you buy the better made equipment
with the higher price tag? or the low first
cost equipment that has high
maintenance?
Cost Comparisons
What are we doing here?
 Comparing one project to another
 Deciding to buy the expensive computer
that has free maintenance versus the
cheap one that makes you pay for service
vs.
Simple Cost
Comparisons

Strategy
» Reduce costs (and/or revenues) to a
common instant, usually the present time
» Work on full year periods
» approximate costs or revenues which
occur over the year as single year-end
amounts

Basic Rule: All comparisons must be
performed on an equal time period
basis
Unequal Lifetime Cost
Comparisons

Repeatability Assumption
(to get to same time basis)

Annuity Comparison

Co-termination assumption
First Some Useful
Mathematical Machinery
Uniform periodic annual payments
(annuities)
 Projects frequently generate recurring
income or cost streams on an annual
basis

x
1
x
2
x
3
0
x = annuity
x
4
x
5
x
6
x
(m-1)
x
m
Discounting a Series
of Payments
 1
1
1
1 
P  X
 X

 ... 
j
2
m
(1

i)
(1

i)
(1

i)
(1

i)
j1


m
1
(1  i)  a
2
3
m

P  X a  a  a  ...  a 
2
3
4
m 1

Pa  X a  a  a  ...  a 
m1
P  X(1  a)  x(a  a
)
Discounting a Series
of Payments con’t
 a  am1 
 1  am 
P  X
  Xa 

 1 a 
 1 a 
  1 m 
m
 1 


1  i  1

 1    1 i  
 X
X

m


1
1

i
i(1

i)

 1


1

i


 (1  i) 
1  (1  i)
 P  X
 X
m 
i
 i(1  i) 

m1
m



Capital Recovery Factor


i
1  (1  i)m   captial recovery factor




i
X
P
m 
1  (1  i) 
Future Equivalents
of Annuities
P  S(1  i)m
(1  i)m  1
X
i(1  i)m
(1  i)m  1
SX
i


i
X  S

m
(1

i)

1


Link to summary of useful formulae
Examples

What future payment N years from now shall I
accept in return for an investment of $P now,
given I could instead invest my money
elsewhere (e.g. a bank) and earn i %/yr?
S  P(1  i)
N

What set of annual revenues for N years will
entice me to invest $P, given the same
alternative as above?


i
X  P
N 
1

(1

i)


Examples

What price should I pay for an investment
which returns $X/yr for N years, if i %/yr is
available to me in a bank?
1  (1  i)N 
P  X

i



What annual interest rate (bank, etc.) would
be required to make an investment returning
$S in N years on a present investment of $P?
S
i
1
P
N
A Simple Replacement Problem
Process to be operated for 4 years and
then junked
 Do you buy a new low-maintenance
machine now or not???

DATA (neglect tax effects)
Options
Purchase Price ($)
Operating Cost ($/yr)
Lifetime (yrs)
Stick w/old
0
2000
4
Buy new
4000
500
4
Cash Flow Time Lines
OLD
0
1
$2000
NEW
2
3
4
$2000 $2000 $2000
0
1
2
3
$4000
$500
$500
$500
4
$500
1  (1  i)4 
Pold  $2000 

i


Pnew
1  (1  i)4 
 $4000  500 

i


The Key Role of Interest Rates
1  (1  i)4 
Pold  $2000 

i


Pnew
1  (1  i)4 
 $4000  500 

i



If management demands i = 10 %/yr
Pold=$6340, Pnew=$5585
 new is better choice

If management demands i = 20 %/yr
Pold=$5180, Pnew=$5295 
 old is better choice
Note

In a replacement problem like this you
could have added revenues to the
analysis, but no need to do so if they
are the same for both options.
Financial Comparisons with
Unequal Lifetimes


Simple Example: Choose between 2 pieces
of equipment, one of which is better built and
has a longer lifetime
N is not the same for both
20 year life
Well Built
Poorly Built

2 year life
Not a fair comparison with N=2 unless
process is to be shut down and both options
have no residual value
What to Do?

Option 1 - Repeatability
Well Built
20 year life
Poorly Built
2 year life
(Buy 1)
(Buy 10)
Option 2 - Annualized Costs

Convert the investment and
maintenance for both options into a
single annual payment
Purchase Price ($)
Alternative 1 Alternative 2
10,000
20,000
Annual Op. Cost ($/yr)
1500
Salvage Value ($)
Service Life (yrs)
500
2
i = 0.15 / yr
1000
1000
3
Annualized Cost of
Alternative 1
0
500
2
1
10,000
1500
1500
1500
1500  500
P  10,000 

 $12,060
2
1 .15  (1 .15)
Now


i
X  P
2 
1

(1

i)






i
.15
X  12060 

12060
 $7418

2 
2 
1  (1  i) 
1  (1.15) 
50
1
$10,000
0
500
$1500
2
$1500
=
1
$7418
2
$7418
Annualized Cost of
Alternative 2
1000
0
1
20,000
1000
2
1000
3
1000
=
0
1
9472
2
9472
In this case, choose alternative 1
because yearly cost is lower.
3
9472