Chapter 4 More Interest Formulas EGN 3615 ENGINEERING ECONOMICS
WITH SOCIAL AND GLOBAL IMPLICATIONS 1 Slide 2 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 Slide 3 Uniform Series Compound Amount
Factor 0 F1+F2+F3+F4 = F 1 2 3 4 0 AAAA 1 2 3 4 0 A 1 2 3 4 F1 0 A
1 2 3 4 F2 0 A 1 2 3 4 F3 0 A=F4 1 2 3 4 3 Slide 4 Uniform Series
Compound Amount Factor 4 That is, for 4 periods, F = F 1 + F 2 + F
3 + F 4 = A(1+i) 3 + A(1+i) 2 + A(1+i) + A = A[(1+i) 3 + (1+i) 2 +
(1+i) + 1] Slide 5 Uniform Series Compound Amount Factor 5 For n
periods with interest (per period), F = F 1 + F 2 + F 3 + + F n-1 +
F n = 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] 0 AAAA 1 2 3 4 A n A n-1
F Slide 6 Uniform Series Compound Amount Factor Uniform Series
Compound Amount Factor Notation 6 i = interest rate per period n =
total # of periods Slide 7 Uniform Series Formulas (Compare to
slide 25) 7 (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. Butrate(nper, pmt, pv, fv, type, guess)
Slide 8 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? 0 1 2 3 4 5 $100 F= 552.6 i=0.05 8 Slide 9
QUESTION CONTINUES (USING INTEREST TABLE) 0 1 2 3 4 5 $100 F= 552.6
i=0.05 9 Slide 10 QUESTION CONTINUES(SPREADSHEET) Go to XL Go to XL
--Chap 4 extended examples-A1 Use function: FV(rate, nper, pmt, pv,
type) 10 Slide 11 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? F2= 580.2F1 0 1 2 3 4 5 $100 i=5% 11 Slide 12
QUESTION CONTINUES(INTEREST TABLE) F2= 580.2F1 0 1 2 3 4 5 $100
i=5% 12 Slide 13 QUESTION CONTINUES(SPREADSHEET) 13 Slide 14
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? 0 1 2 3 4 5 $100 F= ? i=6.5% 14 Slide 15
INTERPOLATION 6.0 7.0 6.5 5.6371 5.7507 0.5 X 10.1136 Interpolation
15 Slide 16 Uniform Series Sinking Fund Factor A= Equal Annual
Dollar Payments F= Future Some of Money i = Interest Rate Per
Period n= Number of Interest Periods 0 1 2 3 4 5 F=Given i=Given
A=? n=Given 16 Slide 17 Uniform Series Sinking Fund Factor Uniform
Series Sinking Fund Factor Notation 17 Slide 18 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 0 1 2 3 4 5 i=5% A =$2172 n=5 18
Slide 19 QUESTION CONTINUES(INTEREST TABLE) F=$12,000 0 1 2 3 4 5
i=5% A = $2172 n=5 19 Slide 20 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? 0 1 2 3 4 5
F=$12,000 i=5% A = $1764 n=6 20 Slide 21 QUESTION
CONTINUES(INTEREST TABLE) 0 1 2 3 4 5 F=$12,000 i=5% A = $1764 n=6
21 $1764 Slide 22 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? 0 1 2 3
4 5 F=$9000 i=6% A = $796.23 6 P=$2,500 22 Slide 23 Uniform Series
Capital Recovery Factor P= Present Sum of Money A= Equal Annual
Dollar Payments i = Interest Rate n= Number of Interest Periods 0 1
2 3 4 5 P=Given i=Given A=? n 23 Slide 24 Uniform Series Capital
Recovery Factor Uniform Series Capital Recovery Factor Notation 24
Slide 25 Uniform Series Formulas (Compare to slide 7) 25 (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.
Butrate(nper, pmt, pv, fv, type, guess) Slide 26 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? 0 1 2 3 4 5 A =26,380 P= $100,000 i=10% 26 Slide 27 Example
CONTINUES(INTEREST TABLE) 0 1 2 3 4 5 A =26,380 P= $100,000 i=10%
27 Slide 28 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 Slide 29 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 mans
54 th birthday (25 deposits). Starting at age 55, 15 equal annual
withdrawals are made. How much should each withdrawal be? Step 1:
First A will be converted into F. Step2: F will be considered as P.
Step3: P will be converted into Second A Solution 29 Slide 30
EXAMPLE CONTINUES 0 1 2 3 21 22 F = 245,868 i=10% 2324 A=$2500 0 1
2 3 12 13 i=10% 1415 P= $245,868 A =$32,332 30 Slide 31 Uniform
Series Present Worth Factor A= Equal Annual Dollar Payments P=
Present Sum of Money (at Time 0) i = Interest Rate/Period n= Number
of Interest Periods 0 1 2 3 4 5 P=? i=Given A=Given n=Given 31
Slide 32 Uniform Series Present Worth Factor Uniform Series Present
Worth Factor Notation 32 Slide 33 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? 0 1 2 3 4 5 A=26,380 P=
100,001 i=10% 33 Slide 34 EXAMPLE CONTINUES (USING INTEREST TABLE)
0 1 2 3 4 5 A=26,380 P= 100,001 i=10% 34 Slide 35 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 Slide 36 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? 0 1 2 3 4 5 A=500 6 7 i=5% i=8% 36 Slide
37 EXAMPLE CONTINUES 0 1 2 3 4 5 A=500 6 7 i=5% i=8% 37 Slide 38
EXAMPLE CONTINUES (Using MS EXCEL) 38 P 1 = PV(0.08, 3,
-500)(1+0.05) 4 = (1288.55)(0.8227) = $1,060.09 P 2 = PV(0.05, 4,
-500) = $1,772.98 Slide 39 Arithmetic Gradient Arithmetic Gradient
series (G): each annual amount differs from the previous one by a
fixed amount G. + A 0 1 2 3 4 5 A A A A 0 0 1 2 3 4 5 G 2G 3G 4G 0
1 2 3 4 5 A A+G A+2G A+3G A+4G = 39 Slide 40 Arithmetic Gradient
Present Worth Factor Given G, i, & n, find P(4-19) Arithmetic
Gradient Present Worth Factor Notation 40 Slide 41 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? 0 1 2 3 4 5 G=$30 P=
$247.11 i=5% 0 $30 $60 $90 $120 41 Slide 42 QUESTION CONTINUES
(INTEREST TABLE) 0 1 2 3 4 5 G=$30 P= $247.11 i=5% 0 $30 $60 $90
$120 42 Slide 43 Arithmetic Gradient Present Worth Factor Question:
If interest rate is 8%, what is the present worth of the following
sums? + 400 0 1 2 3 4 5 0 0 1 2 3 4 5 50 100 150 200 0 1 2 3 4 5
400 450 500 550 600 = 400 43 Slide 44 QUESTION CONTINUES 0 0 1 2 3
4 5 50 100 150 200 400 0 1 2 3 4 5 44 Slide 45 Arithmetic Gradient
Uniform Series Factor Convert an arithmetic gradient series into a
uniform series Given G, i, & n, find A(4-20) Arithmetic
Gradient Uniform Series Factor Notation 45 Slide 46 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%) + 0 1 2 3 4 5 0 0 1 2 3 4 5 500 1000 1500 2000 0 1 2
3 4 5 1000 1500 2000 2500 3 000 = 46 Slide 47 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. 0 1 2 3 4 5
P=? i=5% $100 $110 $121 $133 6 7 8 9 10 g=10% ? ? ? ? ? ? 47 Slide
48 Geometric Gradient Unlike the Arithmetic Gradient where the
amount of period-by- period change is a constant, for the Geometric
Gradient, the period- by-period change is a uniform growth rate (g)
or percentage rate. Uniform growth rate (g) First year maintenance
cost 48 Slide 49 Geometric Series Present Worth Factor Geometric
Series Present Worth Factor When Interest rate equals the growth
rate, 49 Given A 1, g, i, & n, find P (4-29) & (4-30) Slide
50 Geometric Series Present Worth Factor Question: What is the
present value (P) of a geometric series with $100 at EOY1 (A 1 ),
5% interest rate (i), 10% growth rate (g), and 10 interest periods
(n)? 0 1 2 3 4 5 P=? i=5% $100 $110 $121 $133 6 7 8 9 10 g=10% ? ?
? ? ? ? 50 Slide 51 Geometric Series Present Worth Factor 0 1 2 3 4
5 P= $1184.67 i=5% $100 $110 $121 $133 6 7 8 9 10 g=10% $146 $161
$195 $177 $214 $236 51 Slide 52 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
Slide 53 Problem 4-7 53 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. Slide 54 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 Slide 55 Problem 4-9 Solution (a)Let x =
the toll/vehicle. Then F = $25,000,000i = 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 Slide 56
Problem 4-9 Solution (b)Let x = the toll/vehicle. Then F = $25Mi =
(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 56 Slide 57 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 Slide 58
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 Slide 59 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]}(4-5) capital recovery (F=0) A = P{[i(1+i) n ]/[(1+i) n 1]}
(4-6) or combined (P 0, and F 0) rate(n, A, P, F, type, guess)
returns i, given n, A, P, and F 59 Slide 60 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)F =
P(1+i) n (3-3) series compound amount (P=0)F = A{[(1+i) n 1]/i}
(4-4) or combined (A 0, and P 0) 60 Slide 61 Use of MS EXCEL
nper(i, A, P, F, type) returns n, given i, A, P, and F. If A = 0,n
= log(F/P)/log(1+i) single payment If P = 0,n =
log(1+Fi/A)/log(1+i)uniform series If F = 0,n =
log[A/(A-Pi)]/log(1+i)uniform series effect(r, m) returns i a,
given r and m. effective annual interest rate i a = (1+r/m) m 1
(3-7) nominal(i a, m) returns r, given i a and m. nominal annual
interest rate r = m[(i a 1) 1/m + 1] 61 Slide 62 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 i a = (F/P) 1/n 1 = (300/1.5) 1/20 1 = 30.332% i a =
rate(20, 0, 1.5, 300) = 30.332% i a = rate(20, 0, -1.5, 300) =
30.332% 62 Slide 63 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
Slide 64 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 Slide 65
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 Slide 66
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
