Transcript A Compressed Air System

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
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
Equivalence
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.
Cash Flow Diagrams:
An Important Tool
time
Salvage
“Costs”
Operating &
Initial Capital Maintenance
Cost
Costs
Replacement
Costs
Income
- 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
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
t=0
n = 12, i = 15%
$4500
P =Present Value (in dollars)
F = Future Value (in dollars)
F
t=12
Factors
Single Payment Compound Amount
Factor (future worth)
(F/P, i%, n) :
F
 (1  ieff ) n
P
Single Payment Present Worth Factor
(P/F, i%, n):
P
1

F (1  ieff ) n
n is in years if the ieff is used.
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.
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
Uniform Series
Present Worth
Factor
(1  ieff ) n  1
P

A ieff (1  ieff ) n
Capital
Recovery
Factor
n
i
(
1

i
)
A eff
eff

P (1  ieff ) n  1
Annuities
Can Relate an Annuity to a future value:
Uniform Series
Compound
Amount Factor
F

A
(1  ieff ) n  1
ieff
Uniform Series
Sinking Fund
Factor
ieff
A

F (1  ieff ) n  1
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?
Factors
Fortunately these factors are
tabulated…
And Excel has nice built in functions
to calculate them too….
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 endof-period cash payments, Type = 1
means beginning-of-period payments,
guess is a guess value of the interest
rate
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.
Gradient Factors
The amount in the
initial year is called
a base
amount, and it
doesn’t need to
equal the gradient
amount
The Uniform
amount of increase
each period is the
gradient amount
Present Value @ time zero
Gradient Factors
To get the Gradient Factors we subtract off the base
amount, and start things in year (period) 2:
PT (total) = PG+PA
PA comes from using the P/A
factor on an annuity equal to the
base amount.
PG = Present worth of the
gradient starting in year 2…
This is what is calculated by
P/G factor.
PG/G and AG/G
(i  1)  in  1
( P / G, i, n) 
2
n
i (1  i)
n
P/G = factor to convert a gradient series to a
present worth.
1
n
( A / G, i, n )  
n
i (1  i )  1
A/G = factor to convert a gradient series to an
equivalent uniform annual series.
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%
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)
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 = ?
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?
Use of Multiple Factors
Many cash flow situations do not fit the single
factor equations.
It is often necessary to combine equations
Example? 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
Use of Multiple Factors
Several Methods:
1. Use P/F of each payment
2. F/P of each and then multiply by P/F
3. 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
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.
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?
Eng Econ Practice Problems
Check Website for Practice
Problems…Remember you ALL have a quiz on
Engineering Econ on Monday, not just the
economists!