Ch2 (part1)econ factors_rev2

1. 1
Factors: HowTime and MoneyFactors: HowTime and Money
Affect Interest (CH2)Affect Interest (CH2)

2. 2
ObjectivesObjectives
To understand and use factors to account
for TMV.
◦ Single payment compound amount (F/P)
◦ Uniform Series Present Worth (P/A)
◦ Uniform Series Sinking Fund (A/F)
◦ Interpolation in interest tables
◦ Arithmetic Gradient Factors (P/G , A/G)
◦ Geometric Gradient Factors

3. 3
Single Payment factors (F/P & P/F)Single Payment factors (F/P & P/F)
F/P : Determines the amount of money F
accumulated after n periods from a single
present worth P with interest
compounded one time per period
Only for one payment
Can you define P/F factor?
n
iPF )1(/ +=
n
iFP −
+= )1(/

4. 4
Solving Factor ProblemsSolving Factor Problems
Factor problems can be solved in several ways
◦ a) Using Equations
◦ Using equations: F = P(1+i)n
= 12,000(1+8%)24
= $76,
094.17

5. 5
Solving Factor ProblemsSolving Factor Problems
b) Using tables:
F=P(F/P,i,n)=12,000(F/P,8%,24)=12,000 x
(6.3412)=$76,094.40
c) Using Excel
◦ (F/P,i, n) FV(i%, n,P)
◦ (P/F,i,n) PV(i%, n,P)

6. 6
Uniform Series Present Worth (P/A) &Uniform Series Present Worth (P/A) &
Capital Recovery (A/P) FactorsCapital Recovery (A/P) Factors
P/A is calculated as
◦ Uniform Series Present Worth Factor
Derive A/P equation
◦ Capital Recovery Factor
How P/A is related to P/F?
0
)1(
1)1(
≠





+
−+
= i
ii
i
AP n
n

7. 7
P/A Problem
How much you should be willing to pay for
a project that brings $600 for the next 9
years starting next year, the rate of return is
16% per year.
P =600(P/A,16%,9) = 600(4,606.5)= 2763.90

8. 8
Sinking Fund Factor & Uniform –Series
Compound Amount Factor (A/F &
F/A)
Sinking Fund Factor (A/F) determines the
uniform annual series that is equivalent to a
given future worth F.
Uniform Series Compound Amount Factor (F/A)
gives the future worth of a uniform series.






−+
=
1)1( n
i
i
FA





 −+
=
i
i
AF
n
1)1(

9. 9
F/A Problem
Plant x wants to know the equivalent
future worth of a $1 million capital
investment each year for 8 yrs, starting 1
yr from now. Capital earns at a rate of
14% per yr.
In $1000 units, F=1000(F/A,14%,8) =
$13,232.80

10. 10
Linear Interpolation & Interest Tables
For unlisted values in the tables we can
use
Factors equations
Linear interpolation
%
7 0.14238
7.3 x
8 0.14903

11. 11
Arithmetic Gradient Factors (P/G and
G/P)
Arithmetic gradient is a cash flow series that
increases or decreases by a constant amount.
Different from previous factors
Gradient: constant arithmetic change in the
receipts or disbursements from one time period to
the next, G can be positive or negative

12. 12
Arithmetic Gradient Factors (P/G and
G/P)
Cash Flow in yr n (CFn) = base amount+(n-1)G
The relation to convert an arithmetic gradient
G (not including base amount) for n years into
a present worth at yr 0
The equivalent uniform annual series (A value)
for an arithmetic gradient G






+
−
+
−+
= nn
n
i
n
ii
i
i
G
P
)1()1(
1)1(






−+
−=
1)1(
1
n
i
n
i
GA

13. 13
Arithmetic Gradient Factors
 F/G factor (future worth for arithmetic gradient) is
calculated as
 Base amount (A) and gradient (G) are considered
separately
 Total present worth
◦ Increasing PT=PA + PG
◦ Decreasing PT=PA - PG
 Equivalent total annual series
◦ Increasing AT=AA + AG
◦ Decreasing AT=AA - AG












−
−+






= n
i
i
i
GF
n
1)1(1

14. 14
Arithmetic Gradient Problem
Gradient problems are solved for 1) base &
2) G, then add two amounts.

15. 15
Arithmetic Gradient Problem
Total present worth = PA+PG =
500(P/A,5%,10) +100(P/G,5%,10)
Total annual series = AA+AG = 500
+100(A/G,5%,10)

16. 16
Geometric Gradient