Chapter 4
More Interest Formulas
EGN 3615
ENGINEERING ECONOMICS
WITH SOCIAL AND GLOBAL
IMPLICATIONS
1
Chapter Contents
 Uniform Series Compound Interest Formulas
 Uniform Series Compound Amount Factor
 Uniform Series Sinking Fund Factor
 Uniform Series Capital Recovery Factor
 Uniform Series Present Worth Factor
 Arithmetic Gradient
 Geometric Gradient
 Nominal Effective Interest
 Continuous Compounding
2
Uniform Series Compound Amount Factor
0
A
A
A
A
1
2
3
4
A
0
1
0
1
1
0
1
2
3
4
F1
2
3
A
0
F1+F2+F3+F4 =F
2
2
4
F1  A (1  i)3
F2
3
4
A
F3
3
4
F2  A (1  i)
2
F3  A (1  i )
A=F4
0
1
2
3
4
F4  A
3
Uniform Series Compound Amount Factor
That is, for 4 periods,
F = F1 + F2 + F 3 + F4
= A(1+i)3 + A(1+i)2 + A(1+i) + A
= A[(1+i)3 + (1+i)2 + (1+i) + 1]
4
Uniform Series Compound Amount Factor
For n periods with interest (per period),
0
A
A
A
A
A
1
2
3
4
n-1
F
A
n
F = F1 + F2 + F3 + … + Fn-1 + Fn
= A(1+i)n-1 + A(1+i)n-2 + A(1+i)n-3 + … + A(1+i) + A
= A[(1+i)n-1 + (1+i)n-1 + (1+i)n-3 + …+ (1+i) + 1]
5
Uniform Series Compound Amount Factor
 (1  i) n  1
F A
  A ( F / A, i, n)
i


Notation
Uniform Series
Compound Amount Factor
i = interest rate per period
n = total # of periods
6
Uniform Series Formulas (Compare to slide 25)
(1) Uniform series compound amount: Given A, i, & n, find F
F = A{[(1+i)n – 1]/i} = A(F/A, i, n)
(4-4)
(2) Uniform series sinking fund: Given F, i, & n, find A
A = F{i/[(1+i)n – 1]} = F(A/F, i, n)
(4-5)
(3) Given F, A, & i, find n
n = log(1+Fi/A)/log(1+i)
(4) Given F, A, & n, find i
There is no closed form formula to use.
But rate(nper, pmt, pv, fv, type, guess)
7
Uniform Series Compound Amount Factor
 Question: If starting at EOY1, five annual deposits of $100 each are
made in the bank account, how much money will be in the account
at EOY5, if interest rate is 5% per year?
F= 552.6
i=0.05
0
1
2
3
4
5
$100 $100 $100 $100 $100
 (1  i) n  1
 (1  0.05)5  1
F A
  100 
  100(5.526)
i
0.05




8
QUESTION CONTINUES (USING INTEREST TABLE)
F= 552.6
i=0.05
0
1
2
3
4
5
$100 $100 $100 $100 $100
 (1  i) n  1
F A
  A ( F / A, i%, n)  100 (5.526)
i


9
QUESTION CONTINUES(SPREADSHEET)
A= ($100)
i=
5%
n=
5
$552.6
Go to XL
--Chap 4 extended examples-A1
Use function: FV(rate, nper, pmt, pv, type)
10
Uniform Series Compound Amount Factor
 Question: Five annual deposits of $100 each are made into an
account starting today. If interest rate is 5%, how much money
will be in the account at EOY5?
F1
F2= 580.2
4
5
i=5%
0
1
2
3
$100 $100 $100 $100 $100
 (1  i ) n  1
 (1  0.05) 5  1
F1  A 
  100 
  100 (5.526)  552.6
i
0.05




F2  F1 (1  i ) n  552.6 (1  0.05)1
11
QUESTION CONTINUES(INTEREST TABLE)
F1
F2= 580.2
4
5
i=5%
0
1
2
3
$100 $100 $100 $100 $100
 (1  i) n  1
F1  A 
  A ( F / A, i%, n)
i


 A ( F / A, 5%, 5)
 100 (5.526)  552.6
F2  F1 ( F / P, 5%, 1)  552.6 (1.050)
12
QUESTION CONTINUES(SPREADSHEET)
A= ($100)
i=
5%
n=
5
$552.6
$580.2
13
Uniform Series Compound Amount Factor
 Question: If starting at EOY1, five annual deposits of $100 each are
made in the bank account, how much money will be in the account
at EOY5, if interest rate is 6.5% per year?
F= ?
i=6.5%
0
1
2
3
4
5
$100 $100 $100 $100 $100
( F / A,6.5%,5)  ?
( F / A,6%,5)  5.6371
( F / A,7%,5)  5.7507
14
INTERPOLATION
6.0
0.5
1
6.5
( F / A,6%,5)  5.6371
( F / A,7%,5)  5.7507
5.6371
X
0.1136
7.0
5.7507
0.5
X

1
0.1136
X  0.0568
(F/A, 6.5%, 5)  5.6371  0.0568  5.6939
F  100(F/A, 6.5%, 5)  100 (5.6939)  569.39
Interpolation
15
Uniform Series Sinking Fund Factor
F=Given
i=Given
0
1
2
3
4
5
n=Given
A=?
A= Equal Annual Dollar Payments
F= Future Some of Money
i = Interest Rate Per Period
n= Number of Interest Periods
16
Uniform Series Sinking Fund Factor


i
AF
 F (A / F , i%, n)

n
 (1  i)  1
Uniform Series
Sinking Fund
Factor
Notation
17
Uniform Series Sinking Fund Factor
 Question: A family wishes to have $12,000 in a bank account
by the EOY 5. to accomplish this goal, five annual deposits
starting at the EOF year 1 are to be made into a bank account
paying 6% interest. what annual deposit must be made to
reach the stated goal?
F=$12,000
i=5%
0
1
2
3
4
5
n=5
A =$2172




i
0.05
AF
 12,000 
 12,000 (0.1810)


n
5
 (1  i)  1
 (1  0.05)  1
18
QUESTION CONTINUES(INTEREST TABLE)
F=$12,000
i=5%
0
1
2
3
4
5
n=5
A = $2172


i
AF
 F ( A/F, i%, n)

n
 (1  i )  1
 F ( A/F, 5%, 5)
 12,000 (0.1810)
19
Uniform Series Sinking Fund Factor
 Question: A family wishes to have $12,000 in a bank account by the
EOY 5. to accomplish this goal, six annual deposits starting today
are to be made into a bank account paying 5% interest. What
annual deposit must be made to reach the stated goal?
F=$12,000
i=5%
0
1
2
3
4
5
n=6
A = $1764




i
0.05
AF
 12,000 
 12,000 (0.1470)


n
6
 (1  i)  1
 (1  0.05)  1
20
$1764
QUESTION CONTINUES(INTEREST TABLE)
F=$12,000
i=5%
0
1
2
3
4
5
n=6
A = $1764


i
AF
 F ( A/F, i%, n)

n
 (1  i )  1
 F ( A/F, 5%, 6)
 12,000 (0.1470)
21
Uniform Series Sinking Fund Factor
 Example: the current balance of a bank account is $2,500. starting
EOY 1 six equal annual deposits are to be made into the account.
The goal is to have a balance of $9000 by the EOY 6. if interest rate
is 6%, what annual deposit must be made to reach the stated goal?
F=$9000
i=6%
0
P=$2,500
1
2
3
4
5
6
A = $796.23
FPvalue  2,500 (F/P, 6%, 6)  2,500 (1.419)  3547.50
FTotal  9000  3547.50
A  5452.50 (A/F, 6%, 6)
A  5452.50 (0.1434)
 5452.50
22
Uniform Series Capital Recovery Factor
P=Given
i=Given
0
1
2
3
4
5
n
A=?
P= Present Sum of Money
A= Equal Annual Dollar Payments
i = Interest Rate
n= Number of Interest Periods
23
Uniform Series Capital Recovery Factor



i
i
n
AF
 P(1  i ) 


n
n
 (1  i )  1
 (1  i )  1
 i (1  i ) n 
AP
  P (A / P, i %, n)
n
 (1  i )  1
Uniform Series
Capital
Recovery Factor
Notation
24
Uniform Series Formulas (Compare to slide 7)
(1) Uniform series present worth: Given A, i, & n, find P
P = A{[(1+i)n – 1]/[i(1+i)n]} = A(P/A, i, n)
(4-7)
(2) Uniform series capital recovery: Given P, i, & n, find A
A = P{[i(1+i)n]/[(1+i)n – 1]} = P(A/P, i, n)
(4-6)
(3) Given P, A, & i, find n
n = log[A/(A-Pi)]/log(1+i)
(4) Given P, A, & n, find i (interest/period)
There is no closed form formula to use.
But rate(nper, pmt, pv, fv, type, guess)
25
Uniform Series Capital Recovery Factor
 Example: A person borrows $100,000 from a commercial bank. The
loan is to be repaid with five equal annual payments. If interest rate
is 10%, what should the annual payments be?
P= $100,000
i=10%
0
1
2
3
4
5
A =26,380
 i(1  i ) n 
 0.1(1  0.1)5 
AP
  100,000 

n
5
 (1  i)  1
 (1  0.1)  1
 100,000 (0.2638)
26
Example CONTINUES(INTEREST TABLE)
P= $100,000
i=10%
0
1
2
3
4
5
A =26,380
A  P ( A/P, i%, n)
 100,000 ( A/P, 10%, 5)
 100,000 (0.2638)
27
Uniform Series Capital Recovery (MS EXCEL)
 Use function: PMT(rate, nper, pv, fv, type)
rate = interest rate/period
nper = # of periods
pv = present worth
fv = balance at end of period n (blank means 0).
type = 1 (payment at beginning of each period) or
0 (payment at end of a period)(blank means 0)
 See spreadsheet
28
Uniform Series Capital Recovery Factor
 Example: At age 30, a person begins putting $2,500 a
year into account paying 10% interest. The last deposit
is made on the man’s 54th birthday (25 deposits).
Starting at age 55, 15 equal annual withdrawals are
made. How much should each withdrawal be?
Solution
 Step 1: First A will be converted into F.
 Step2: F will be considered as P.
 Step3: P will be converted into Second A
29
EXAMPLE CONTINUES
F = 245,868
i=10%
0
1
2
3
21
22
23
24
A=$2500
F  2500 ( F/A, 10%, 25)  2500 (98.347)
i=10%
0
1
2
3
12
13
14
15
A =$32,332
P= $245,868
A  245,868 ( A/P, 10%, 15)  245,868 (0.1315)
30
Uniform Series Present Worth Factor
A=Given
0
1
2
3
4
5
n=Given
i=Given
P=?
A= Equal Annual Dollar Payments
P= Present Sum of Money (at Time 0)
i = Interest Rate/Period
n= Number of Interest Periods
31
Uniform Series Present Worth Factor
 (1  i) n  1
PA
 A ( P / A, i, n)
n 
 i(1  i) 
Uniform Series
Present Worth
Factor
Notation
32
Uniform Series Present Worth Factor
 Example: A special bank account is to be set up. Each year, starting
at EOY 1, a $26,380 withdrawal is to be made. After five withdrawals
the account is to be depleted. if interest rate is 10%, how much
money should be deposited today?
A=26,380
0
1
2
3
4
5
i=10%
P= 100,001
 (1  i) n  1
 (1  0.1)5  1
PA
 26,380 
 26,380(3.791)
n 
5
 i(1  i) 
 0.1(1  0.1) 
33
EXAMPLE CONTINUES (USING INTEREST TABLE)
A=26,380
0
1
2
3
4
5
i=10%
P= 100,001
P  A ( P/A, i%, n)
 26,380 ( P/A, 10%, 5)
 26,380 (3.791)
34
Uniform Series Present Worth (Using MS EXCEL)
 Use function: PV(rate, nper, pmt, fv, type)
rate = interest rate/period
nper = total # of periods (payments)
pmt = constant payment/period
fv = balance at end of period n (blank means 0)
type = 1 or 0
PV(0.1, 5, -26380) = $100,000.95
 See spreadsheet
35
Uniform Series Present Worth Factor
 Example: Eight annual deposits of $500 each are made into a
bank account beginning today. Up to EOY 4, the interest rate
is 5%. After that, the interest rate is 8%. What is the present
worth of these deposits?
PTotal  ?
i=8%
i=5%
0
P3
1
2
P2
3
A=500
4
5
6
7
P1
36
EXAMPLE CONTINUES
PTotal
i=8%
i=5%
0
1
2
3
4
5
6
7
A=500
P1
P2
P2  500(P/A, 5%, 4)
P1  500 (P/A, 8%, 3)
 500(3.546)  1773
 (P/F, 5%, 4)
 500(2.577)(0.8227)
 P1  P2  P3  3333.05
 1060.05
P3
P3  500
PTotal
37
EXAMPLE CONTINUES (Using MS EXCEL)
P1 = PV(0.08, 3, -500)(1+0.05)–4
= (1288.55)(0.8227)
= $1,060.09
P2 = PV(0.05, 4, -500)
= $1,772.98
PTotal  P1  P2  P3  $3,333.07
38
Arithmetic Gradient
 Arithmetic Gradient series (G): each annual amount differs
from the previous one by a fixed amount G.
A+3G
A+G
A+4G
A+2G
A
0
1
2
3
4
P  A ( P / A, i, n)
 G ( P / G , i, n)
5
=
0
A
A
A
A
A
1
2
3
4
5
3G
4G
4
5
+
G
2G
0
0
1
2
3
39
Arithmetic Gradient Present Worth Factor
 Given G, i, & n, find P
(4-19)
 (1  i) n  in  1
PG 2
  G ( P / G, i%, n)
n
 i (1  i)

Notation
Arithmetic Gradient
Present Worth Factor
40
Arithmetic Gradient Present Worth Factor
 Question: You has purchased a new car. the following maintenance
costs starting at EOY 2 will occur to pay the maintenance of your car
for the 5 years. EOY2 $30, EOY3 $60, EOY4 $90, EOY5 $120. If
interest rate is 5%, how much money you should deposit into a
bank account today?
$120
$30
$60 $90
0
0
1
2
3
4
5
i=5%
G=$30
P= $247.11
 (1  i) n  in  1
 (1  0.05)5  0.05(5)  1
PG 2
  30
  30(8.237)
n
2
5
 i (1  i)

 0.05 (1  0.05)

41
QUESTION CONTINUES (INTEREST TABLE)
$30
$120
$90
$60
0
0
1
2
3
4
5
i=5%
G=$30
P= $247.11
P  G (P/G, i%, n)
 30 (P/G, 5%, 5)
 30(8.237)
42
Arithmetic Gradient Present Worth Factor
 Question: If interest rate is 8%, what is the present worth of
the following sums?
550
600
500
450
400
400
1
2
400
400
400
4
5
150
200
4
5
400
0
1
2
3
4
5
=
0
3
+
50
100
0
0
1
2
3
43
QUESTION CONTINUES
0
400
400
1
2
400
3
400
400
4
5
150
200
4
5
ATotal  A1  A2
A1  400
50
100
0
0
1
2
3
 400  92
 492
P  492 ( P / A, 8%, 5)
 492 (3.9927)
 1,964.41
A2  50 ( A / G, 8%, 5)  50 (1.8465)
 92.33  92
44
Arithmetic Gradient Uniform Series Factor
Convert an arithmetic gradient series into a uniform series
Given G, i, & n, find A
(4-20)
 (1  i)  in  1
AG
  G ( A / G, i%, n)
n
 i(1  i)  i 
n
Notation
Arithmetic Gradient
Uniform Series Factor
45
Arithmetic Gradient Uniform Series Factor
 Question: Demand for a new product will decrease as competitors
enter the market. What is the equivalent annual amount of the
revenue cash flows shown below? (interest 12%)
3000
3000 3000 3000 3000 3000
2500
2000
1500
1000
0
1
2
3
4
5
=
0
AEQIV  3000  500 ( A / G , 12%, 5) 0
 3000  500(1.775)
 2112.50
1
2
3
4
5
4
5
+
1
0
2
500
3
1000
1500
2000
46
Geometric Series Present Worth Factor
 Geometric series: Each annual amount is a fixed percentage
different from the last. In this case, the change is 10%.
 We will look at this problem in a few slides.
g=10%
$100 $110 $121
0
1
2
3
$133
4
?
5
?
6
?
7
?
8
?
9
?
10
i=5%
P=?
47
Geometric Gradient
 Unlike the Arithmetic Gradient where the amount of period-by-
period change is a constant, for the Geometric Gradient, the periodby-period change is a uniform growth rate (g) or percentage rate.
First year maintenance cost
Cash Flow
1
2
Uniform growth rate (g)
100
100  10%(100) 
100(1  0.1)1
 100  A1
 110  A 2
3
.
110  10%(110) 
.
100(1  0.1) 2
.
 121  A 3
.
.
.
.
.
.
.
.
.
n
A n -1  10%( A n -1 ) 
100(1  g ) n 1
 An
Year
48
Geometric Series Present Worth Factor
Given A1, g, i, & n, find P
(4-29) & (4-30)
1  (1  g ) n (1  i)  n 
P  A1 
 where i  g
ig


Geometric Series
Present Worth Factor
When Interest rate equals the growth rate,
P  A1n(1  i)
1
where i  g
49
Geometric Series Present Worth Factor
 Question: What is the present value (P) of a geometric
series with $100 at EOY1 (A1), 5% interest rate (i), 10%
growth rate (g), and 10 interest periods (n)?
g=10%
$100 $110 $121
0
1
2
3
$133
4
?
5
?
6
?
7
?
8
?
9
?
10
i=5%
P=?
50
Geometric Series Present Worth Factor
$100
0
1
$133
$121
$110
2
3
4
g=10%
$195
$177
$146 $161
5
6
7
8
$214
9
$236
10
i=5%
P= $1184.67
1  (1  g ) n (1  i)  n 
1  (1  0.1)10 (1  0.05) 10 
P  A1 
  100

ig
0.05  0.1




51
 100(11.8467)
Time for a Joke!
What is Recession?
Recession is when your neighbor loses his or her job.
What is Depression?
Depression is when you lose yours.
By Ronald Reagan
52
Problem 4-7
Purchase a car:
$3,000 down payment
$480 payment for 60 months
If interest rate is 12% compounded monthly, at what
purchase price P of a car can one buy?
Solution
i = 12.0%/12 = 1.o% per month, n = 60, and A = $480
P = 3000 + 480(P/A, 0.01, 60)
= 3000 + 480(44.955)
= $24,578
Important: P = $3000 + $480(60) = $31,800, if i = 0.
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Problem 4-9
$25 million is needed in three years.
Traffic is estimated at 20 million vehicles per year.
At 10% interest, what should be the toll per vehicle?
(a)
Toll receipts at end of each year in a lump sum.
(b)
Traffic distributed evenly over 12 months, and
toll receipts at end of each month in a lump sum.
54
Problem 4-9
Solution
(a)
Let x = the toll/vehicle. Then
F = $25,000,000
i = 10%/year, n = 3 years
Find A (=20,000,000x).
A = F(A/F, 0.1, 3)
20,000,000x = 25,000,000(0.3021)
x = $0.3776 = $0.38 per vehicle
55
Problem 4-9
Solution
(b) Let x = the toll/vehicle. Then
F = $25M
i = (1/12)10%/month, n = 36 months
Find A (=20,000,000x/12).
A = F{i/[(1+i)n–1]}
20,000,000x/12 = 25,000,000{(0.1/12)/(1+0.1/12)36–1}
x = $0.359 = $0.36 per vehicle
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Problem 4-32
If i = 12%, for what value of B is the PW = 0?
Solution
Consider now = time 1. Then
PW = B+800(P/A, 0.12, 3) – B(P/A, 0.12, 2) – B(P/F, 0.12, 3)
= 1921.6 – 1.758B
Letting PW = 0 yields B = $1,093.06
For any cash flow diagram,
if PW = 0, then its worth at anytime = 0!
57
Problem 4-46
Solution
FW = FW
[1000(F/A, i, 10)](F/P, i, 4) = 28000
By try and error:
At i = 12%, LHS = [1000(17.549)](1.574) = $27,622 too low
At i = 15%, LHS = [1000(20.304)](1.749) = $35,512 too high
Using linear interpolation:
i = 12% + 3%[(28000 – 27622)/(35512 – 27622)]
= 12.14%
58
Use of MS EXCEL
pmt(i, n, P, F, type) returns A, given i, n, P, and F
sinking fund (P=0)
A = F{i/[(1+i)n – 1]}
capital recovery (F=0) A = P{[i(1+i)n]/[(1+i)n – 1]}
or combined (P ≠ 0, and F ≠ 0)
(4-5)
(4-6)
rate(n, A, P, F, type, guess) returns i, given n, A, P, and F
59
Use of MS EXCEL
pv(i, n, A, F, type) returns P, given i, n, A, and F
present worth (A=0)
P = F/(1+i)n
(3-5)
series present worth (F=0) P = A{[(1+i)n – 1]/[i(1+i)n]} (4-7)
or combined (A ≠ 0, and F ≠ 0)
fv(i, n, A, P, type) returns F, given i, n, A, and P
compound amount (A=0)
series compound amount (P=0)
or combined (A ≠ 0, and P ≠ 0)
F = P(1+i)n
F = A{[(1+i)n – 1]/i}
(3-3)
(4-4)
60
Use of MS EXCEL
nper(i, A, P, F, type) returns n, given i, A, P, and F.
If A = 0,
If P = 0,
If F = 0,
n = log(F/P)/log(1+i)
n = log(1+Fi/A)/log(1+i)
n = log[A/(A-Pi)]/log(1+i)
single payment
uniform series
uniform series
effect(r, m) returns ia, given r and m.
effective annual interest rate ia = (1+r/m)m – 1
(3-7)
nominal(ia, m) returns r, given ia and m.
nominal annual interest rate r = m[(ia – 1)1/m + 1]
61
A Real Life Case
Mr. Goodman set up a trust fund of $1.5M for his 2
children in 1991. It is worth more than $300M today
(January 2012).
What is the effective annual interest rate?
Solution
P = $1.5M, F = $300M, n = 20 years
ia = (F/P)1/n – 1 = (300/1.5)1/20 – 1 = 30.332%
ia = rate(20, 0, 1.5, –300) = 30.332%
ia = rate(20, 0, -1.5, 300) = 30.332%
62
End of Chapter 4
 Uniform Series Compound Interest Formulas
 Uniform Series Compound Amount Factor: F/A
 Uniform Series Sinking Fund Factor:
A/F
 Uniform Series Capital Recovery Factor:
A/P
 Uniform Series Present Worth Factor:
P/A
 Arithmetic Gradient
 Geometric Gradient
 Spreadsheet Solutions
63
Interpolation-1
 Given: F(X1); F(X2)
 What is F(X3) where X1 < X3 < X2?
 Assuming linearity so that a linear equation will do:
 Basic equation: y = mx + b so
1. F(X1) = mX1 + b
2. F(X2) = mX2 + b
 Subtract 2 from 1:
 F(X1)-F(X2) = m (X1-X2)  m = (F(X1)-F(X2))/(X1-X2)
 From 1 we get b = (F(X1) - mX1)
64
Interpolation-2
 F(X1)-F(X2) = m (X1-X2)  m = (F(X1)-F(X2))/(X1-X2)
 From 1 we get b = (F(X1) - mX1)
 Now
 F(X3) = m X3 + b
= m X3 + F(X1) - mX1 = m (X3 - X1) + F(X1)
= (X3 - X1) (F(X1)-F(X2))/(X1-X2) + F(X1)
= F(X1) + (F(X1)-F(X2)) (X3 - X1) /(X1-X2)
65
Interpolation-3
 F(X3) = F(X1) + (F(X1)-F(X2)) (X3 - X1) /(X1-X2)
 Suppose that the Xs are interest rates, i, and the Fs are
the functions (F/A,i,n), then
 F(i3) = F(i1) + (F(i1)-F(i2)) (i3 - i1) /(i1-i2)
Return
66