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MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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Page 1: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

MER439- Design of Thermal Fluid Systems

Engineering Economics Lecture 2- Using Factors

Professor AndersonSpring 2012

Page 2: 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

Page 3: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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

Page 4: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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

Page 5: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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%

Page 6: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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.

Page 7: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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.

Page 8: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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

Page 9: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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

Page 10: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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?

Page 11: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

Factors

Fortunately these factors are tabulated…

And Excel has nice built in functions to calculate them too….

Page 12: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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

Page 13: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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.

Page 14: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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

Page 15: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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.

Page 16: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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.

Page 17: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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%

Page 18: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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)

Page 19: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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 = ?

Page 20: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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?

Page 21: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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

Page 22: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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

Page 23: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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.

Page 24: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

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?

Page 25: MER439- Design of Thermal Fluid Systems Engineering Economics Lecture 2- Using Factors Professor Anderson Spring 2012

Eng Econ Practice Problems

Check Website for Practice Problems…

Remember you ALL have a quiz on

Engineering Econ on Monday, not just the

economists!