Cost and Time Value of $$
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Cost and Time Value of $$
Prof. Eric Suuberg
ENGINEERING 90
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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
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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
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Have I Got a Deal for You Would you be interested in investing in
a company that has $1 million in annual sales?
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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)
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ProfitProfit = Sales (revenues) - expenses (costs) Basis for taxation - What goes into the
calculation is of great interest to Uncle Sam
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In Our Example Profit = $1,000,000/yr - $900,000/yr
= $100,000/yr Is this a good business?
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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.
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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
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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
Information
Required
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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
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Capital Investment- Manufacturing
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.
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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
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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
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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
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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
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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
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Take Home Message Not all dollars of profit are the same Those that come earlier are “worth”
more
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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?
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Cost ComparisonsWhat 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.
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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
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Unequal Lifetime Cost Comparisons
Repeatability Assumption
(to get to same time basis)
Annuity Comparison
Co-termination assumption
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First Some Useful Mathematical Machinery
Uniform periodic annual payments (annuities)
Projects frequently generate recurring income or cost streams on an annual basis
1 65432 (m-1) m0
x x x x x x x x
x = annuity
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Discounting a Series of Payments
m
j 2 mj 1
1 1 1 1P X X ...(1 i) (1 i) (1 i) (1 i)
1(1 i) a
2 3 mP X a a a ... a 2 3 4 m 1Pa X a a a ... a
m 1P X(1 a) x(a a )
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Discounting a Series of Payments con’t
m 1 ma a 1 aP X Xa1 a 1 a
m
m
m
11 1 i 11 1 iX X11 i i(1 i)1
1 i
m 1 m
m
(1 i) 1 (1 i)P X Xi(1 i) i
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Capital Recovery Factor
m
i captial recovery factor1 (1 i)
m
iX P1 (1 i)
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Future Equivalents of Annuities
mm
m
(1 i) 1P S(1 i) Xi(1 i)
m(1 i) 1S Xi
m
iX S(1 i) 1
Link to summary of useful formulae
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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?
What set of annual revenues for N years will entice me to invest $P, given the same alternative as above?
NS P(1 i)
N
iX P1 (1 i)
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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?
What annual interest rate (bank, etc.) would be required to make an investment returning $S in N years on a present investment of $P?
Examples
N1 (1 i)P Xi
NSi 1P
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A Simple Replacement Problem
Process to be operated for 4 years and then junked
Do you buy a new low-maintenance machine now or not???
DATA (neglect tax effects)
Options Stick w/old Buy newPurchase Price ($) 0 4000Operating Cost ($/yr) 2000 500Lifetime (yrs) 4 4
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Cash Flow Time Lines1 4320
$2000 $2000 $2000$2000
OLD
1 4320
$500 $500 $500$500
NEW
4
old1 (1 i)P $2000
i
4
new1 (1 i)P $4000 500
i
$4000
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The Key Role of Interest Rates
4
old1 (1 i)P $2000
i
4
new1 (1 i)P $4000 500
i
If management demands i = 10 %/yrPold=$6340, Pnew=$5585 new is better choice
If management demands i = 20 %/yrPold=$5180, Pnew=$5295 old is better choice
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Note In a replacement problem like this you
could have added revenues to the analysis, but no need to do so if they are the same for both options.
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Financial Comparisons with Unequal Lifetimes
Simple Example: Choose between 2 pieces of equipment, one of which is better built and has a longer lifetime
N is not the same for both
Not a fair comparison with N=2 unless process is to be shut down and both options have no residual value
Well Built
Poorly Built
20 year life
2 year life
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What to Do? Option 1 - Repeatability
Well Built 20 year life
Poorly Built 2 year life
(Buy 1)
(Buy 10)
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Option 2 - Annualized Costs
Convert the investment and maintenance for both options into a single annual payment
Alternative 1 Alternative 2Purchase Price ($)
Annual Op. Cost ($/yr)
Salvage Value ($)Service Life (yrs)
10,000 20,000
1500 1000
500 1000
2 3
i = 0.15 / yr
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Annualized Cost of Alternative 1
50 500
$1500$10,000 $15001 2
0
$7418 $74181 2=
2
1500 1500 500P 10,000 $12,0601 .15 (1 .15)
Now 2
iX P1 (1 i)
2 2
i .15X 12060 12060 $74181 (1 i) 1 (1.15)
0 500
150010,000 15001 2
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Annualized Cost of Alternative 2
01000
100020,000 1000
1
1000
0
9472 9472
1
9472
32=
In this case, choose alternative 1 because yearly cost is lower.
2 3