mer439- design of thermal fluid systems engineering economics lecture 2- using factors
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MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012. Some Definitions. Capital: Invested money and resources Interest: The return on capital - PowerPoint PPT PresentationTRANSCRIPT
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MER439- Design of Thermal Fluid Systems
Engineering Economics Lecture 2- Using Factors
Professor AndersonSpring 2012
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Some Definitions
Capital: Invested money and resources
Interest: The return on capital
Nominal IR: the interest rate per year without adjusting for the number of compounding periods
Effective IR: the interest rate per year adjusting for the number of compounding periods
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Different sums of money at different times can be equal in economic value.
i.e. $100 today with i = 6% is equivalent to $106 in one year.
Equivalence depends on the interest rate!
Equivalence occurs when different cash flows at different times are equal in economic value at a given interest rate.
Equivalence
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Cash Flow Diagrams: An Important Tool
Income
time
Initial Capital Cost
ReplacementCosts
Operating &Maintenance
Costs
Salvage“Costs”
- Arrows up represent “income” or “profits” or “payoffs”- Arrows down represent “costs” or “investments” or “loans”- The “x axis” represents time, most typically in years
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Time Value of Money
If $4500 is invested today for 12 years at 15% interest rate, determine the accumulated amount. Draw this.
F = P(1+i)n
P =Present Value (in dollars)
F = Future Value (in dollars)
$4500
Ft=0
t=12
n = 12, i = 15%
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Factors
Single Payment Compound Amount Factor (future worth)
(F/P, i%, n) :
Single Payment Present Worth Factor (P/F, i%, n):
neffi
PF
)1(
neffiF
P
)1(
1
n is in years if the ieff is used.
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Example - Factors
How much inheritance to be received 20 years from now is equivalent to receiving $10,000 now? The interest rate is 8% per year compounded each 6-months.
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Uniform Series (Annuity)
An Annuity is a series of equal amount money transactions occurring at equal time periods
Ordinary Annuity - one that occurs at the end of each time period
neffeff
neff
ii
i
AP
)1(
1)1(
1)1(
)1(
neff
neffeff
i
ii
PA
Uniform Series Present Worth
Factor
Capital Recovery
Factor
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Annuities
Can Relate an Annuity to a future value:
eff
neff
i
i
AF 1)1(
1)1(
n
eff
eff
i
i
FA
Uniform Series Compound
Amount Factor
Uniform Series Sinking Fund
Factor
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Annuity Example
How much money can you borrow now if you agree to repay the loan in 10 end of year payments of $3000, starting one year from now at an interest rate of 18% per year?
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Factors
Fortunately these factors are tabulated…
And Excel has nice built in functions to calculate them too….
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Spreadsheet Function
P = PV(i,N,A,F,Type)F = FV(i,N,A,P,Type)i = RATE(N,A,P,F,Type,guess)Where, i = interest rate, N = number of
interest periods, A = uniform amount, P = present sum of money, F = future sum of money, Type = 0 means end-of-period cash payments, Type = 1 means beginning-of-period payments, guess is a guess value of the interest rate
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Gradient Factors
Engineering Economic problems frequently involve disbursements or receipts that increase or decrease each year (i.e. equipment maintenance)
If the increase is the same every year this is called a uniform arithmetic gradient.
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Gradient Factors
Present Value @ time zero
The Uniformamount of increaseeach period is thegradient amount
The amount in the initial year is calleda baseamount, and itdoesn’t need toequal the gradientamount
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Gradient Factors
To get the Gradient Factors we subtract off the base amount, and start things in year (period) 2:
PG = Present worth of thegradient starting in year 2…This is what is calculated byP/G factor.
PT (total) = PG+PA
PA comes from using the P/Afactor on an annuity equal to the base amount.
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PG/G and AG/G
n
n
ii
ininiGP
)1(
1)1(),,/(
2
1)1(
1),,/(
ni
n
iniGA
P/G = factor to convert a gradient series to a present worth.
A/G = factor to convert a gradient series to an equivalent uniform annual series.
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Gradient Example
Find the PW of an income series with a cash flow in Year 1 of $1200 which increases by $300 per year through year 11. Use i = 15%
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Review of Factors
Using the tables..
Single Payment factors (P/F), (F/P)
Uniform Series factors (P/A), (F/A)
Gradients (A/G), (P/G)
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Unknown Interest Rates and Years
Unknown Interest rate:
-i.e. F = $20K, P = $10K, n = 9 i = ?
-Or A = $1770, n = 10, P = $10K i =?
Unknown Years – sometimes want to determine the number of years it will take for an investment to pay off ( n is unknown)
-A = $100, P = $2000, i = 2% n = ?
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Unknown interest example
If you would like to retire with $1million 30 years from now, and you plan to save $6000 per year every year until then, what interest rate must your savings earn in order to get you that million?
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Use of Multiple Factors
Many cash flow situations do not fit the single factor equations.
It is often necessary to combine equationsExample? What is P for a series of $100
payments starting 4 years from now?
1 2 3 4 5 6 7 8 9 10 11 12 13
$100
P = ?
years
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Use of Multiple Factors
Several Methods:1. Use P/F of each payment2. F/P of each and then multiply by P/F3. Get F =A (F/A, i,10), then P = F (F/P,i,13)4. Get P3 = A(P/A,I,10) and P0 = P3(P/F,i,3)
1 2 3 4 5 6 7 8 9 10 11 12 13
$100
P = ?
years
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Use of Multiple Factors
Step for solving problems like this:1. Draw Cash Flow Diagram.2. Locate P or F on the diagram.3. Determine n by renumbering if necessary.4. use factors to convert all cash flows to
equivalent values at P or F.
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Multiple Factors: Example
A woman deposited $700 per year for 8 years. Starting in the ninth year she increased her deposits to $1200 per year for 5 more years. How much money did she have in her account immediately after she made her last deposit?
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Eng Econ Practice Problems
Check Website for Practice Problems…
Remember you ALL have a quiz on
Engineering Econ on Monday, not just the
economists!