Factors:

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Chapter 2
ENGINEERING ECONOMY Sixth Edition
Blank and Tarquin

• Factors
• How Time and Interest Affect Money

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Learning Objectives

• F/P and P/F Factors (Single Payment Factors)
• P/A and A/P Factors (Uniform Series Present Worth
  Factor and Capital Recovery Factor)
• F/A and A/F Factors (Sinking Fund Factor and
  Uniform-Series Compound Amount Factor)
• P/G and A/G Factors (Arithmetic Gradient Factors)
• Geometric Gradient
• Calculate i (unknown interest rate)
• Calculate n (number of years)

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Single Payment Factors(F/P and P/F)

• Objective
• Derive factors to determine the present or future
  worth of a cash flow
• Cash Flow Diagram basic format

Fn
i / period
0 1 2
3
n-1 n
P0
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Basic Derivation F/P factor
Fn P0(1i)n ?(F/P, i, n) factor Excel
FV(i, n, ,P) P0 Fn1/(1i)n ?(P/F, i, n)
factor Excel PV(i, n, , F)
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Derivation F/P factor

• Find F given P
• F1 P Pi P(1i)
• F2 F1 F1i F1(1i)..or
• F2 P(1i) P(1i)i P(1i)(1i) P(1i)2
• F3 F2 F2 i F2(1i) P(1i)2 (1i)
• P(1i)3
• In general
• FN P(1i)n
• FN P (F/P, i, n)

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Present Worth Factor from F/P

• Since FN P(1i)n
• We solve for P in terms of FN
• P F1/ (1i)n F(1i)-n
• P F(P/F,i,n) where
• (P / F, i, n) (1i)-n

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P/F factor discounting back in time

• Discounting back from the future

P/F factor brings a single future sum back to a
specific point in time.
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Example- F/P Analysis

• Example P 1,000n3i10
• What is the future value, F?

F3 1,000 F/P,10,3 1,0001.103
1,0001.3310 1,331.00
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Example P/F Analysis

• Assume F 100,000, 9 years from now. What is
  the present worth of this amount now if i 15?

i 15/yr
P0 100,000(P/F, 15,9) 100,000(1/(1.15)9)
100,000(0.2843) 28,426 at time t 0
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Uniform Series Present Worth and Capital Recovery
Factors
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Uniform Series Present Worth

• This expression will convert an annuity cash
  flow to an equivalent present worth amount one
  period to the left of the first annuity cash flow.

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Capital Recovery Factor (CRF) A/P factor
CRF calculates the equivalent uniform annual
worth A over n years for a given P in year 0,
when the interest rate is i.
The present worth point of an annuity cash flow
is always one period to the left of the first A
amount.

• Given the P/A factor

Solve for A in terms of P
Yielding.
A/P, i, n
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Modified HW Problem 2.5

• A maker of microelectromechanical systems MEMS,
  believes it can reduce product recalls if it
  purchases new software for detecting faulty
  parts. The cost of the new software is 225,000.
• How much would the company have to save each year
  for 4 years to recover its investment if it uses
  a minimum attractive rate of return of 15 per
  year?
• (A/P, 15, 4) p. 745 Table 19 (interest rate
  i15) ? column A/P, row n4
• A/P factor 0.35027
• Company would have to save 225,000 x 0.35027
  78,810.75 each year.

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HW Problem 2.12

• Comparison of Table with equation
• V-Tek Systems is a manufacturer of vertical
  compactors, and it is examining its cash flow
  requirements for the next 5 years. The company
  expects to replace office machines and computer
  equipment at various times over the 5-year
  planning period. Specifically, the company
  expects to spend 9000 two years from now, 8000
  three years from now, and 5000 five years from
  now. What is the present worth of the planned
  expenditures at an interest rate of 10 per year?

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Sinking Fund and Series Compound Amount Factors
(A/F and F/A)

• Take advantage of what we already have
• Recall
• Also

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Modern context of a Sinking Fund

• In modern finance, a sinking fund is a method
  enabling an organization to set aside money over
  time to retire its indebtedness. More
  specifically, it is a fund into which money can
  be deposited, so that over time its preferred
  stock, debentures or stocks can be retired. For
  the organization that is retiring debt, it has
  the benefit that the principal of the debt or at
  least part of it, will be available when due. For
  the creditors, the fund reduces the risk the
  organization will default when the principal is
  due.
• In some US states, Michigan for example, school
  districts may ask the voters to approve a
  taxation for the purpose of establishing a
  Sinking Fund. The State Treasury Department has
  strict guidelines for expenditure of fund dollars
  with the penalty for misuse being an eternal ban
  on ever seeking the tax levy again.
• Historical Context A Sinking Fund was a device
  used in Great Britain in the 18th century to
  reduce national debt.

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HW Problem 2.23

• Southwestern Moving and Storage wants to have
  enough money to purchase a new tractor-trailer in
  3 years. If the unit will cost 250,000, how much
  should the company set aside each year if the
  account earns 9 per year?
• (A/F, 9, 3) or n 3, F 250,000, i 9
• Using Table 14 (pg 740), the A/F 0.30505
• A 250,000 x 0.30505 76262.50

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Interpolation (Estimation Process)

• At times, a set of interest tables may not have
  the exact interest factor needed for an analysis
• One may be forced to interpolate between two
  tabulated values
• Linear Interpolation is not exact because
• The functional relationships of the interest
  factors are non-linear functions
• From 2-5 error may be present with
  interpolation.

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Example 2.7

• Assume you need the value of the A/P factor for i
  7.3 and n 10 years.
• 7.3 is likely not a tabulated value in most
  interest tables
• So, one must work with i 7 and i 8 for n
  fixed at 10
• Proceed as follows

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Basic Setup for Interpolation

• Work with the following basic relationships

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Arithmetic Gradient Factors

• An arithmetic (linear) Gradient is a cash flow
  series that either increases or decreases by a
  constant amount over n time periods.
• A linear gradient always has TWO components
• The gradient component
• The base annuity component
• The objective is to find a closed form expression
  for the Present Worth of an arithmetic gradient

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Linear Gradient Example

• Assume the following

This represents a positive, increasing arithmetic
gradient
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Example Linear Gradient
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Arithmetic Gradient Factors

• The G amount is the constant arithmetic change
  from one time period to the next.
• The G amount may be positive or negative.
• The present worth point is always one time period
  to the left of the first cash flow in the series
  or,
• Two periods to the left of the first gradient
  cash (G) flow.

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Present Worth Point
700
600
500
400
300
200
100
X
The Present Worth Point of the Gradient
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Present Worth Linear Gradient

• The present worth of a linear gradient is the
  present worth of the two components
• 1. The Present Worth of the Gradient Component
  and,
• 2. The Present Worth of the Base Annuity flow
• Requires 2 separate calculations.

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Present Worth Gradient Component(Example 2.10)

• Three contiguous counties in Florida have agreed
  to pool tax resources already designated for
  county-maintained bridge refurbishment. At a
  recent meeting, the county engineers estimated
  that a total of 500,000 will be deposited at the
  end of next year into an account for the repair
  of old and safety-questionable bridges throughout
  the three-county area. Further, they estimate
  that the deposits will increase by 100,000 per
  year for only 9 years thereafter, then cease.
• Determine the equivalent (a) present worth and
  (b) annual series amounts if county funds earn
  interest at a rate of 5 per year.

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Example 2.10 (b)
Determine the equivalent annual series amounts if
county funds earn interest at a rate of 5 per
year.
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Equations for P/G and A/G
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Geometric Gradients

• An arithmetic (linear) gradient changes by a
  fixed dollar amount each time period.
• A GEOMETRIC gradient changes by a fixed
  percentage each time period.
• We define a UNIFORM RATE OF CHANGE () for each
  time period
• Define g as the constant rate of change in
  decimal form by which amounts increase or
  decrease from one period to the next

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cash flow diagrams for geometric gradient series
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Geometric Gradients Increasing

• Typical Geometric Gradient Profile
• Let A1 the first cash flow in the series

The series starts in year 1 at an initial amount
A1, not considered a base amount as in the
arithmetic gradient.
A1
A1(1g)
A1(1g)2
A1(1g)3
A1(1g)n-1
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Geometric Gradients

• For a Geometric Gradient the following parameters
  are required
• The interest rate per period i
• The constant rate of change g
• No. of time periods n
• The starting cash flow A1

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Pg /A Equation

• In summary, the engineering economy relation and
  factor formulas to calculate Pg in period t 0
  for a geometric gradient series starting in
  period 1 in the amount A1 and increasing by a
  constant rate of g each period are

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• Engineers at SeaWorld, a division of Busch
  Gardens, Inc., have completed an innovation on an
  existing water sports ride to make it more
  exciting. The modification costs only 8000 and
  is expected to last 6 years with a 1300 salvage
  value for the solenoid mechanisms. The
  maintenance cost is expected to be high at 1700
  the first year, increasing by 11 per year
  thereafter. Determine the equivalent present
  worth of the modification and maintenance cost.
  The interest rate is 8 per year.

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continued

• Assume maintenance costs will be 1700 one year
  from now.
• Assume an annual increase of 11 per year over a
  6-year time period.
• If the interest rate is 8 per year, determine
  the present worth of the future expenses at time
  t 0.
• First, draw a cash flow diagram to represent the
  model.

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Cash flow diagram
continued
Solution The cash flow diagram shows the salvage
value as a positive cash flow and all costs as
negative. Use Equation 2.24 for g ? i to
calculate Pg. The total PT is
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continued
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i rate is unknown

• A class of problems may deal with all of the
  parameters know except the interest rate.
• For many application-type problems, this can
  become a difficult task
• Termed, rate of return analysis
• In some cases
• i can easily be determined
• In others, trial and error must be used

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Example i unknown

• Assume one can invest 3000 now in a venture in
  anticipation of gaining 5,000 in five (5) years.
• If these amounts are accurate, what interest rate
  equates these two cash flows?

5,000

• F P(1i)n
• (1i)5 5,000/3000 1.6667
• (1i) 1.66670.20
• i 1.1076 1 0.1076 10.76

3,000
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Unknown Number of Years

• Some problems require knowing the number of time
  periods required given the other parameters
• Example
• How long will it take for 1,000 to double in
  value if the discount rate is 5 per year?
• Draw the cash flow diagram as.

i 5/year n is unknown!
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Unknown Number of Years

• Solving we have..

• (1.05)x 2000/1000
• X ln(1.05) ln(2.000)
• X ln(1.05)/ln(2.000)
• X 0.6931/0.0488 14.2057 yrs
• With discrete compounding it will take 15 years

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• Formulas and factors derived and applied in this
  chapter perform equivalence calculations for
  present, future, annual, and gradient cash flows.
  Capability in using these formulas and their
  standard notation manually and with spreadsheets
  is critical to complete an engineering economy
  study.
• Using these formulas and spreadsheet functions,
  you can convert single cash flows into uniform
  cash flows, gradients into present worths, and
  much more.
• You can solve for rate of return i or time n.
• A thorough understanding of how to manipulate
  cash flows using the material in this chapter
  will help you address financial questions in
  professional practice as well as in everyday
  living.

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WHAT A DIFFERENCE THE YEARS AND COMPOUND INTEREST
CAN MAKE

• Real World Situation - Manhattan Island purchase.
  
• It is reported that Manhattan Island in New York
  was purchased for the equivalent of 24 in the
  year 1626. In the year 2001, the 375th
  anniversary of the purchase of Manhattan was
  recognized.
• F P (1i)n 24 (1 0.06)382 111,443,000,000
  (2008)
• F P Pin 24 24(0.06)382 550.08 (simple
  interest)