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 ID: 755435
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Slide1
Chapter 4More Interest FormulasEGN 3615ENGINEERING ECONOMICS WITH SOCIAL AND GLOBAL IMPLICATIONS
1Slide2
Chapter ContentsUniform Series Compound Interest FormulasUniform Series Compound Amount FactorUniform Series Sinking Fund FactorUniform Series Capital Recovery FactorUniform Series Present Worth FactorArithmetic GradientGeometric GradientNominal Effective Interest Continuous Compounding2Slide3
Uniform Series Compound Amount Factor 0
F1+F2+F3+F4 =
F
1
2
3
4
0
A
A
A
A
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
3Slide4
Uniform Series Compound Amount Factor4That is, for 4 periods,F = F1 + F2 + F3 + F4 = A(1+i)3 + A(1+i)2 + A(1+i) + A = A[(1+i)3 + (
1+i)
2
+
(
1+i) +
1] Slide5
Uniform Series Compound Amount Factor5For n periods with interest (per period),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]
0
A
A
A
A
1
2
3
4
A
n
A
n-1
FSlide6
Uniform Series Compound Amount Factor
Uniform Series
Compound Amount Factor
Notation
6
i
= interest rate per period
n
= total # of periodsSlide7
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.But
rate(nper, pmt, pv, fv, type, guess)Slide8
Uniform Series Compound Amount FactorQuestion: 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
$100
$100
$100
$100
F
= 552.6
i
=0.05
8Slide9
QUESTION CONTINUES (USING INTEREST TABLE)
0
1
2
3
4
5
$100
$100
$100
$100
$100
F
= 552.6
i
=0.05
9Slide10
QUESTION CONTINUES(SPREADSHEET)Go to XL
--Chap 4 extended examples-A1
Use function: FV(rate,
nper
, pmt,
pv
, type)
10Slide11
Uniform Series Compound Amount FactorQuestion: 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
= 580.2
F1
0
1
2
3
4
5
$100
$100
$100
$100
$100
i=5%
11Slide12
QUESTION CONTINUES(INTEREST TABLE)F2= 580.2
F1
0
1
2
3
4
5
$100
$100
$100
$100
$100
i=5%
12Slide13
QUESTION CONTINUES(SPREADSHEET)13Slide14
Uniform Series Compound Amount FactorQuestion: 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
$100
$100
$100
$100
F
= ?
i=6.5%
14Slide15
INTERPOLATION6.0
7.0
6.5
5.6371
5.7507
0.5
X
1
0.1136
Interpolation
15Slide16
Uniform Series Sinking Fund FactorA= Equal Annual Dollar PaymentsF= Future Some of Moneyi = Interest Rate Per Periodn= Number of Interest Periods
0
1
2
3
4
5
F=Given
i=Given
A=?
n=Given
16Slide17
Uniform Series Sinking Fund Factor
Uniform Series
Sinking Fund Factor
Notation
17Slide18
Uniform Series Sinking Fund FactorQuestion: 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
18Slide19
QUESTION CONTINUES(INTEREST TABLE)F=$12,000
0
1
2
3
4
5
i=5%
A = $2172
n=5
19Slide20
Uniform Series Sinking Fund FactorQuestion: 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
20Slide21
QUESTION CONTINUES(INTEREST TABLE)
0
1
2
3
4
5
F=$12,000
i=5%
A = $1764
n=6
21
$1764Slide22
Uniform Series Sinking Fund FactorExample: 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
22Slide23
Uniform Series Capital Recovery FactorP= Present Sum of MoneyA= Equal Annual Dollar Paymentsi = Interest Raten= Number of Interest Periods
0
1
2
3
4
5
P=Given
i=Given
A=?
n
23Slide24
Uniform Series Capital Recovery Factor
Uniform Series
Capital Recovery Factor
Notation
24Slide25
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.But
rate(nper
, pmt, pv, fv, type, guess)Slide26
Uniform Series Capital Recovery FactorExample: 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%
26Slide27
Example CONTINUES(INTEREST TABLE)
0
1
2
3
4
5
A =26,380
P
= $100,000
i=10%
27Slide28
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
28Slide29
Uniform Series Capital Recovery FactorExample: 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? 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
29Slide30
EXAMPLE CONTINUES 0
1
2
3
21
22
F
= 245,868
i=10%
23
24
A=$2500
0
1
2
3
12
13
i=10%
14
15
P
= $245,868
A =$32,332
30Slide31
Uniform Series Present Worth FactorA= Equal Annual Dollar PaymentsP= Present Sum of Money (at Time 0)i = Interest Rate/Periodn= Number of Interest Periods
0
1
2
3
4
5
P=?
i=Given
A=Given
n=Given
31Slide32
Uniform Series Present Worth FactorUniform Series Present Worth Factor
Notation
32Slide33
Uniform Series Present Worth FactorExample: 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%
33Slide34
EXAMPLE CONTINUES (USING INTEREST TABLE)
0
1
2
3
4
5
A=26,380
P
= 100,001
i=10%
34Slide35
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) = $100,000.95See spreadsheet35Slide36
Uniform Series Present Worth FactorExample: 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%
36Slide37
EXAMPLE CONTINUES
0
1
2
3
4
5
A=500
6
7
i=5%
i=8%
37Slide38
EXAMPLE CONTINUES (Using MS EXCEL)38P1 =
PV(0.08, 3, -500)
(1+0.05)
–4
= (1288.55)(0.8227)
= $1,060.09
P
2
=
PV(0.05, 4, -500)
= $1,772.98Slide39
Arithmetic GradientArithmetic 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
=
39Slide40
Arithmetic Gradient Present Worth FactorGiven G, i, & n, find P (4-19)
Arithmetic Gradient
Present Worth Factor
Notation
40Slide41
Arithmetic Gradient Present Worth FactorQuestion: 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
41Slide42
QUESTION CONTINUES (INTEREST TABLE)
0
1
2
3
4
5
G=$30
P
= $247.11
i=5%
0
$30
$60
$90
$120
42Slide43
Arithmetic Gradient Present Worth FactorQuestion: 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
400
400
400
43Slide44
QUESTION CONTINUES
0
0
1
2
3
4
5
50
100
150
200
400
0
1
2
3
4
5
400
400
400
400
44Slide45
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
45Slide46
Arithmetic Gradient Uniform Series FactorQuestion: 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
=
3
000
3
000
3
000
3
000
3
000
46Slide47
Geometric Series Present Worth FactorGeometric 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%
?
?
?
?
?
?
47Slide48
Geometric GradientUnlike 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
48Slide49
Geometric Series Present Worth FactorGeometric Series Present Worth Factor
When Interest rate equals the growth rate,
49
Given A
1
, g, i, & n, find P
(4-29) & (4-30)Slide50
Geometric Series Present Worth FactorQuestion: 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)?
0
1
2
3
4
5
P
=?
i=5%
$100
$110
$121
$133
6
7
8
9
10
g=10%
?
?
?
?
?
?
50Slide51
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
51Slide52
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 52Slide53
Problem 4-753
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.Slide54
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.54Slide55
Problem 4-9Solution(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 vehicle55Slide56
Problem 4-9Solution(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 vehicle56Slide57
Problem 4-32If i = 12%, for what value of B is the PW = 0?SolutionConsider now = time 1. ThenPW = B+800(P/A, 0.12, 3) – B(P/A, 0.12, 2) – B(P/F, 0.12, 3) = 1921.6 – 1.758BLetting PW = 0 yields B = $1,093.06For any cash flow diagram,if PW = 0, then its worth at anytime = 0!57Slide58
Problem 4-46SolutionFW = FW[1000(F/A, i, 10)](F/P, i, 4) = 28000By try and error:At i = 12%, LHS = [1000(17.549)](1.574) = $27,622 too lowAt i = 15%, LHS = [1000(20.304)](1.749) = $35,512 too highUsing linear interpolation:i = 12% + 3%[(28000 – 27622)/(35512 – 27622)] = 12.14%
58Slide59
Use of MS EXCELpmt(i, n, P, F, type) returns A, given i, n, P, and Fsinking 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
59Slide60
Use of MS EXCELpv(i, n, A, F, type) returns P, given i, n, A, and Fpresent 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)
60Slide61
Use of MS EXCELnper(i, A, P, F, type) returns n, given i, A, P, and F.If A = 0, n = log(F/P)/log(1+i) single paymentIf P = 0, n = log(1+Fi/A)/log(1+i) uniform seriesIf F = 0, n = log[A/(A-Pi)]/log(1+i) uniform serieseffect(r, m) returns ia, given r and m.effective annual interest rate ia
= (1+r/m)
m
– 1 (3-7)
nominal(
i
a
, m)
returns r, given
i
a
and m.
nominal annual interest rate r = m[(
ia – 1)
1/m + 1]
61Slide62
A Real Life CaseMr. 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?SolutionP = $1.5M, F = $300M, n = 20 yearsia = (F/P)1/n – 1 = (300/1.5)1/20 – 1 = 30.332%ia = rate(20, 0, 1.5, –300) = 30.332%i
a
=
rate(20, 0, -1.5, 300)
=
30.332%
62Slide63
End of Chapter 4Uniform Series Compound Interest FormulasUniform Series Compound Amount Factor: F/AUniform Series Sinking Fund Factor: A/FUniform Series Capital Recovery Factor: A/PUniform Series Present Worth Factor: P/AArithmetic GradientGeometric GradientSpreadsheet Solutions63Slide64
Interpolation-1Given: 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 soF(X1) = mX1 + bF(X2) = mX2 + bSubtract 2 from 1:F(X1)-F(X2) = m (X1-X2) m = (F(X1)-F(X2))/(X1-X2)
From 1 we get
b
= (F(X1) -
m
X1)
64Slide65
Interpolation-2F(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)
65Slide66
Interpolation-3F(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), thenF(i3) = F(i1) + (F(i1)-F(i2)) (i3 - i1) /(i1-i2)Return66