8/10/2019 Engineering Economy Factors
1/55
Chapter 2
Factors:
How Time and Interest Affect Money
8/10/2019 Engineering Economy Factors
2/55
Learning Objectives
12/21/2014 2
1. F/P and P/F Factors (Single Payment Factors)
2. P/A and A/P Factors (Uniform Series Present WorthFactor and Capital Recovery Factor)
3. F/A and A/F Factors (Sinking Fund Factor and Uniform-Series Compound Amount Factor)
4. P/G and A/G Factors (Arithmetic Gradient Factors)
5. Geometric Gradient
6. Calculate i (unknown interest rate)
7. Calculate n (number of years)
8/10/2019 Engineering Economy Factors
3/55
Single Payment Factors(F/P and P/F)
Objective: Derive factors to determine the present or future
worth of a cash flow
Cash Flow Diagrambasic format
0 1 2 3 n-1 n
P0
Fn
i% / period
8/10/2019 Engineering Economy Factors
4/55
Basic Derivation: F/P factor
12/21/2014 4
Fn= P0(1+i)n(F/P, i%, n) factor:
P0= Fn1/(1+i)n(P/F, i%, n) factor:
8/10/2019 Engineering Economy Factors
5/55
Derivation: F/P factor
Find F given P
F1= P + Pi = P(1+i)
F2= F1+ F1i= F1(1+i)..or
F2= P(1+i) + P(1+i)i= P(1+i)(1+i) = P(1+i)2
F3= F2+ F2 i = F2(1+i) =P(1+i)2 (1+i)
= P(1+i)3
In general:
FN= P(1+i)n
FN= P (F/P, i%, n)
12/21/2014 5
8/10/2019 Engineering Economy Factors
6/55
Present Worth Factor from F/P
Since FN= P(1+i)n
We solve for P in terms of FN
P = F{1/ (1+i)
n}
= F(1+i)
-n
P = F(P/F,i%,n) where
(P / F, i%, n) = (1+i)-n
12/21/2014 6
8/10/2019 Engineering Economy Factors
7/55
P/F factordiscounting back in time
Discounting back from the future
12/21/2014 7
P
Fn
N
.
P/F factor brings a single futuresum back to a specific point intime.
8/10/2019 Engineering Economy Factors
8/55
Example- F/P Analysis
Example: P= $1,000;n=3;i=10%
What is the future value, F?
12/21/2014 8
0 1 2 3
P=$1,000
F = ??
i=10%/year
F3= $1,000 [F/P,10%,3] = $1,000[1.10]3
= $1,000[1.3310] = $1,331.00
8/10/2019 Engineering Economy Factors
9/55
ExampleP/F Analysis
Assume F = $100,000, 9 years from now. What is the presentworth of this amount now if i =15%?
12/21/2014 9
0 1 2 3 8 9
F9= $100,000
P= ??
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
8/10/2019 Engineering Economy Factors
10/55
12/21/2014 10
Uniform Series Present Worth and Capital Recovery Factors
8/10/2019 Engineering Economy Factors
11/55
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.
(1 ) 1 0
(1 )
n
n
iP A for i
i i
/ %,P A i n factor
12/21/2014 11
8/10/2019 Engineering Economy Factors
12/55
Capital Recovery Factor (CRF) = A/P factor
Given the P/A factor
(1 ) 1 0
(1 )
n
n
iP A for i
i i
(1 )
(1 ) 1
n
n
i iA P
i
12/21/2014 12
Solve for A in terms of P
Yielding.
A/P, i%, n
The present worth point of an annuity cash flow is always one period to the left of the firstA amount.
CRF calculates the equivalent uniform annual worthA over n years for a given Pinyear 0, when the interest rate is i.
8/10/2019 Engineering Economy Factors
13/55
12/21/2014 13
8/10/2019 Engineering Economy Factors
14/55
Example
A maker of micro-electromechanical systems [MEMS], believes it canreduce product recalls if it purchases new software for detectingfaulty parts. The cost of the new software is $225,000.
How much would the company have to save each year for 4 years torecover its investment if it uses a minimum attractive rate of return of15% per year?
(A/P, 15%, 4)Using Tables (interest rate i=15%)
column: A/P, row: n=4 A/P factor = 0.35027
Company would have to save $225,000 x 0.35027 = $78,810.75 eachyear.
12/21/2014 14
8/10/2019 Engineering Economy Factors
15/55
Example
V-Tek Systems is a manufacturer of vertical compactors, and it is examining itscash flow requirements for the next 5 years. The company expects to replace officemachines and computer equipment at various times over the 5-year planningperiod. 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?
12/21/2014 15
8/10/2019 Engineering Economy Factors
16/55
Sinking Fund and Series Compound Amount Factors(A/F and F/A)
Take advantage of what we already have
Recall:
Also:
1
(1 )nP F
i
(1 )
(1 ) 1
n
n
i iA P
i
12/21/2014 16
8/10/2019 Engineering Economy Factors
17/55
12/21/2014 17
8/10/2019 Engineering Economy Factors
18/55
Modern context of a Sinking Fund
In modern finance, a sinking fund is a method enabling an organization to set asidemoney over time to retire its indebtedness. More specifically, it is a fund into whichmoney can be deposited, so that over time its preferred stock, debentures or stockscan be retired. For the organization that is retiring debt, it has the benefit that theprincipal 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 isdue.
In some US states, Michigan for example, school districts may ask the voters toapprove a taxation for the purpose of establishing a Sinking Fund. The StateTreasury 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 Fundwas a device used in Great Britain in the 18thcentury to reduce national debt.
12/21/2014 18
8/10/2019 Engineering Economy Factors
19/55
Example
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
12/21/2014 19
8/10/2019 Engineering Economy Factors
20/55
Arithmetic Gradient Factors
12/21/2014 20
An arithmetic (linear) Gradient is a cash flow series that eitherincreases 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
8/10/2019 Engineering Economy Factors
21/55
Linear Gradient Example
Assume the following:
12/21/2014 21
0 1 2 3 n-1 N
A1+G
A1+2G
A1+n-2G
A1+n-1G
This represents a positive, increasing arithmetic gradient
8/10/2019 Engineering Economy Factors
22/55
Example: Linear Gradient
12/21/2014 22
8/10/2019 Engineering Economy Factors
23/55
Arithmetic Gradient Factors
12/21/2014 23
The G amount is the constant arithmetic change fromone time period to the next.
The G amount may be positive or negative.
The present worth point is always one time period to theleft of the first cash flow in the series or,
Two periods to the left of the first gradient cash (G) flow.
8/10/2019 Engineering Economy Factors
24/55
Present Worth Point
12/21/2014 24
0 1 2 3 4 5 6 7
$100
$200
$300
$400
$500
$600
$700
X
The Present Worth Point of theGradient
8/10/2019 Engineering Economy Factors
25/55
Present Worth: Linear Gradient
The present worth of a linear gradient is the presentworth of the two components:
1. The Present Worth of the GradientComponentand,
2. The Present Worth of the Base Annuity flow
Requires 2 separate calculations.
12/21/2014 25
8/10/2019 Engineering Economy Factors
26/55
Present Worth: Gradient Component(Example 2.10*)
Three contiguous counties in Florida have agreed to pool taxresources already designated for county-maintained bridgerefurbishment. At a recent meeting, the county engineers estimatedthat a total of $500,000 will be deposited at the beginning of next
year into an account for the repair of old and safety-questionablebridges throughout the three-county area. Further, they estimate thatthe deposits will increase by $100,000 per year for only 9 yearsthereafter, then cease.
Determine the equivalent (a) present worth and (b) annual seriesamounts if county funds earn interest at a rate of 5% per year.
12/21/2014 26
8/10/2019 Engineering Economy Factors
27/55
12/21/2014 27
8/10/2019 Engineering Economy Factors
28/55
Example 2.10 (b)*
12/21/2014 28
Determine the equivalent annual series amounts ifcounty funds earn interest at a rate of 5% per year.
8/10/2019 Engineering Economy Factors
29/55
Equations for P/G and A/G
12/21/2014 29
8/10/2019 Engineering Economy Factors
30/55
Geometric Gradients
12/21/2014 30
An arithmetic (linear) gradient changes by a fixed dollaramount each time period.
A GEOMETRIC gradient changes by a fixed percentage
each time period.We define a UNIFORM RATE OF CHANGE (%) for eachtime period
Define g as the constant rate of change in decimal
form by which amounts increase or decrease from oneperiod to the next
8/10/2019 Engineering Economy Factors
31/55
cash flow diagrams for geometric gradient series
12/21/2014 31
8/10/2019 Engineering Economy Factors
32/55
Geometric Gradients: Increasing
12/21/2014 32
Typical Geometric Gradient Profile
LetA1= the first cash flow in the series
0 1 2 3 4 .. n-1 n
A1
A1(1+g)A1(1+g)
2A1(1+g)
3
A1(1+g)n-1
The series starts in year 1 at an initial amount A1, not considered a base
amount as in the arithmetic gradient.
8/10/2019 Engineering Economy Factors
33/55
Geometric Gradients
12/21/2014 33
For a Geometric Gradient the following parameters are required:
The interest rate per periodi
The constant rate of changeg
No. of time periodsn
The starting cash flowA1
8/10/2019 Engineering Economy Factors
34/55
Pg /A Equation
12/21/2014 34
In summary, the engineering economy relation and factorformulas to calculate Pgin period t = 0 for a geometricgradient series starting in period 1 in the amount A1 and
increasing by a constant rate of g each period are
8/10/2019 Engineering Economy Factors
35/55
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 behigh 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.
12/21/2014 35
8/10/2019 Engineering Economy Factors
36/55
12/21/2014 36
$1700$1700(1.11)1
$1700(1.11)2
$1700(1.11)3
$1700(1.11)5
0 1 2 3 4 5 6
PW(8%) = ??
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 worthof the future expenses at time t = 0.
First, draw a cash flow diagram to represent the model.
continued
8/10/2019 Engineering Economy Factors
37/55
Cash flow diagram
12/21/2014 37
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 PTis
continued
8/10/2019 Engineering Economy Factors
38/55
12/21/2014 38
continued *
8/10/2019 Engineering Economy Factors
39/55
Payment Periods (PP), Interest Periods (IP) and
Compounding Periods (CP)
PPhow often cash flows occur IPhow often the interest is incurred CPhow often interest in compounded If PP = CP, no problem concerning effective i rate
PP IP or PP CP or IP CPThen there is a problem
8/10/2019 Engineering Economy Factors
40/55
Payment Periods (PP), Interest Periods (IP) and
Compounding Periods (CP)
If MARR of the cash flow below is 6% per monthcompounded semiannually then:
PP IP or PP CP or IP CPThen there is a problem
PP=1 month, IP=1 month, CP=6 months
8/10/2019 Engineering Economy Factors
41/55
Nominal and Effective interest Rates
PP IP or PP CP or IP CPThen there is a problem
In an effective interest rate: IP = CP.
All factors are using effective interest rate.
If IP CP then this is a nominal interest rate.
8/10/2019 Engineering Economy Factors
42/55
Nominal and Effective interest Rates
PP IP or PP CP or IP CPThen there is a problem
Important rules:
If nominal interest rate is divided or multiplied by anynumber the value of the interest and the IP period are
multiplied or divided. Example:
6% per month compounded semiannually x 6 =
36 per semi annual compounded semi annually.
8/10/2019 Engineering Economy Factors
43/55
Effective Interest Rate Formula
i = effective rate per some stated period, e.g., quarterly,annually
r = nominal rate for same time period
m = frequency of compounding per same time period
8/10/2019 Engineering Economy Factors
44/55
Effective Interest Rate
Example:Find i per year, if m = 4 for quarterlycompounding, and r = 12% per year
Stated period for i is YEAR
i = (1 + 0.12/4)4 - 1 = 12.55%
rEffective i = (1+ ) 1
m
m
8/10/2019 Engineering Economy Factors
45/55
Nominal and Effective RatesNominal
r = rate/period periods
Example:Rate is 1.5% per month.Determine nominal rate perquarter, year, and over 2 years
Qtr: r = 1.5 3 mth = 4.5%
Year: r = 1.5 12 mth = 18%
= 4.5 4 qtr = 18%
2 yrs: r =1.5
24 mth = 36%= 18 2 yrs = 36%
Effective
Example:Credit card rate is 1.5% per monthcompounded monthly. Determine effectiverate per quarter and per year
Period is quarter:
r = 1.5 3 mth = 4.5%
m = 3
i = (1 + 0.045/3)31 = 4.57%per quarter
Period is year: r = 18% m = 12
i = (1 + 0.18/12)12- 1) = 19.6%per year
rEffective i = (1+ ) 1m
m
8/10/2019 Engineering Economy Factors
46/55
Effective Interest Rate
As m , continuous compounding is approached
effective i = (r1)
Example:r = 14% per year compoundedcontinuously
i = ( 0.14- 1) = 15.03% per year
8/10/2019 Engineering Economy Factors
47/55
Interpolation (Estimation Process)
12/21/2014 47
At times, a set of interest tables may not have the exactinterest factor needed for an analysis
One may be forced to interpolate between two tabulatedvalues
Linear Interpolation is not exact because:
The functional relationships of the interest factors arenon-linear functions
From 2-5% error may be present with interpolation.
8/10/2019 Engineering Economy Factors
48/55
Example 2.7
12/21/2014 48
Assume you need the value of the A/P factor for i = 7.3% andn = 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:
8/10/2019 Engineering Economy Factors
49/55
Basic Setup for Interpolation
12/21/2014 49
Work with the following basic relationships
8/10/2019 Engineering Economy Factors
50/55
12/21/2014 50
8/10/2019 Engineering Economy Factors
51/55
irate is unknown
12/21/2014 51
A class of problems may deal with all of the parametersknowexcept 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
8/10/2019 Engineering Economy Factors
52/55
Example: i unknown
12/21/2014 52
Assume one can invest $3000 now in a venture in anticipation ofgaining $5,000 in five (5) years.
If these amounts are accurate, what interest rate equates these twocash flows?
0 1 2 3 4 5
$3,000
$5,000
F = P(1+i)n
(1+i)5= 5,000/3000 = 1.6667
(1+i) = 1.66670.20
i = 1.10761 = 0.1076 = 10.76%
8/10/2019 Engineering Economy Factors
53/55
Unknown Number of Years
12/21/2014 53
Some problems require knowing the number of time periodsrequired given the other parameters
Example:
How long will it take for $1,000 to double in value if the discount rateis 5% per year?
Draw the cash flow diagram as.
0 1 2 . . . . . . . n
P = $1,000
Fn= $2000
i = 5%/year; n is unknown!
8/10/2019 Engineering Economy Factors
54/55
Unknown Number of Years
12/21/2014 54
Solving we have..
0 1 2 . . . . . . . n
P = $1,000
Fn= $2000
(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
8/10/2019 Engineering Economy Factors
55/55
WHAT A DIFFERENCE THE YEARS AND COMPOUNDINTEREST CAN MAKE
Real World Situation - Manhattan Island purchase.
It is reported that Manhattan Island in New York waspurchased for the equivalent of $24 in the year
1626. In the year 2001, the 375th anniversary of thepurchase of Manhattan was recognized.
F = P (1+i)n = 24 (1+ 0.06)382= 111,443,000,000 (2008)
F = P + Pin = 24 + 24(0.06)382= $550.08 (simple interest)
Top Related