Engineering Economic Analysis Canadian Edition

1
Engineering Economic AnalysisCanadian Edition

• Chapter 4 More Interest Formulas

2
Chapter 4. . .

• examines uniform series compound interest
  formulas.
• applies nominal and effective interest rates.
• uses arithmetic and geometric gradients to solve
  problems.
• discusses discrete and continuous compounding.
• uses spreadsheets and functional functions to
  solve problems.

3
Chapter Assumptions in Solving Economic Analysis
Problems

• End-of-year convention
• simplifies calculations
• Viewpoint of the firm
• Sunk costs
• past has no bearing on current decisions
• Owner provided capital
• no debt capital

4

• Stable prices
• No depreciation, income taxes
• Annual compounding and cash flows

5
Components of Engineering Economic Analysis

• Calculation of P, A, and F are fundamental.
• Some problems are more complex and require an
  understanding of added components
• uniform series
• arithmetic or geometric gradients
• nominal and effective interest rates
• continuous compounding

6
Uniform Series

• Equidistant and equal-valued cash flows during a
  period of time
• all inflows or outflows
• Matching frequencies cash flows and interest
  compounding
• Conversion of series to single sum equivalents
• present worth or future worth

7
Economic Criteria

• Projects are judged against an economic criterion.

8
Transformation of Uniform Series
SERIES DISCOUNT AMOUNT (PW)
SERIES COMPOUND AMOUNT (FW)
Transformations
UNIFORM SERIES
9
Uniform Series Future Worth
FW ??
1,000
1,000
1,000
1,000
1000
EQUIVALENCE

1
2
3
4
5
0
0
5
What is the value in five years of five
end-of-year 1,000 deposits beginning one year
from today?
10
Financial Table
F P(F/A,i,N) 1,000(F/A,10,5)
1,000(6.1051) 6,105.10
11
Uniform Series Future Worth

• From the financial table
• (F/A,10,5) 6.1051
• (F/P,10,5-1) (F/P,10,5-2) (F/P, 10,5-3)
  (F/P,10,5-4) (F/P,10,5-5)
• 1.464 1.331 1.210 1.1 1
• 6.1051

12
(F/A,i,N)

• Ten annual deposits 800 in a fund that pays
  10 interest annually. What is the fund balance
  after 10 years?
• F A(F/A,i,N) 800(F/A,10,10)
• 800(15.9374) 12,750
• All things being equal, you would be indifferent
  between 800 at the end of each year of 10 years
  12,750 in 10 years from today

13
(F/A,10,15)
14
Increasing Importance of Interest Income
15
Annuities Present Worth
PW ??
1000
1000
1000
1000
1000

1
2
3
4
5
0
0
What is the Present Worth of five end-of-year
1,000 deposits beginning one year from now?
16
Financial Table
P A(P/A,i,N) 1,000(P/A,10,5)
1,000(3.7908) 3,791
17
Uniform Series Present Worth

• From the financial table
• (P/A,10,5) 3.7908
• (P/F,10,1) (P/F,10,2) (P/F,10,3)
• (P/F,10,4) (P/F,10,5)
• .9091 .8264 .7513 .6830 .6209
• 3.7908

18
Example 4-5

• A 140/month
• i 1/month
• n 30 months
• Is the above equivalent to 6,800 now?

19
Example (P/A,i,N)

• How much should you spend (i.e., the purchase
  price) for an energy-saving device with a
  five-year life (and no salvage value) if the
  device is to result in annual savings of 200?
• Assume end-of-year savings and interest rate of
  10.

20

• P A(P/A,i,N)
• 200(P/A,10,5)
• 200(3.7908)
• 758

21
Financial Table
22

• The present value of a series of uniform future
  payments
• P A(P/A,i,n)

23
Uniform Payment SeriesCompound Amount Factor F

• The future value of an investment based on
  periodic, constant payments and a constant
  interest rate.
• F A(F/A,i,n)

24
Example 4-1
25

• At 5/year, F 500(F/A, 5,5)
• 500(5.526)
• 2763

26
Example 4-6
27

• F1 100(F/A,15,3)
• 347.25
• F2 347.25(F/P,15,2)
• 459.24
• F2 100(P/A,15,3) x (F/P,15,5)
• 459.24

or
28
(No Transcript)
29

• PW 20(P/F,15,2) 30(P/F,15,3)
• 20(P/F,15,4)
• 46.28
• FW 46.28(F/P,15,4)
• 80.94

30
Uniform Payment Series Sinking Fund Factor

• The constant periodic amount, at a constant
  interest rate, that must be deposited to
  accumulate a future value.
• A F(A/F,i,n)

31
Uniform Payment Series Capital Recovery Factor

• The series of uniform payments that will recover
  an initial investment.
• A P(A/P,i,n)

32
Financial Table
33
Example (A/F,i,N)

• Your dream is to purchase a 50,000 sports car
  following university graduation in four years.
• How much should you deposit in a bank account at
  the end of each of the 4 years in order to pay
  cash for the car?
• Assume the rate of interest on your deposits is
  10 compounded annually.

34

• A F(A/F,i,N)
• 30,000(A/F,10,4)
• 30,000(0.2155)
• 6,465

35
Example (A/P,i,N)

• Five years ago, a couple purchased a fifth wheel
  for 100,000. After touring the Americas, the
  fifth wheel had a negligible market value.
• If the prevailing interest rate was 10
  compounded annually during the last 5 years, what
  was the annual equivalent cost of the fifth
  wheel?

36

• A P(A/P,i,N)
• 100,000(A/P,10,5)
• 100,000(0.2638)
• 26,380

37
Uniform Series
38
Transformation of Arithmetic Series
SERIES DISCOUNT AMOUNT (PW)
UNIFORM SERIES
ARITHMETIC SERIES
SERIES COMPOUND AMOUNT (FW)
INTERNAL RATE of RETURN
39
Increasing Arithmetic Gradient Series

• Composite series (two series in one)
• EUW A1 G(A/G,i,n), where G gradient and
  A1 first element of series
• (A/G,i,N) (1/i) -N/(1i)N 1
• PW A1(P/A1,i,n) G(P/G,i,n)

40

• (P/G,i,N) (1i)N - iN -1/ i2(1i)N
• FW A1(F/A1,i,n) G(F/G,i,n)
• (F/G,i,N) (1i)N 1/i - N

For a decreasing series, change the sign to a
sign.
41
Arithmetic Gradient Series
PW
1000
PW
1000
750
750
500
500
250
250


2
3
4
0
1
2
3
4
0
1
Decreasing or decaying series
Increasing or growing series
42
Increasing Arithmetic Series
Equivalent to
Actual Series
300
300
300
300
450
400
0
1
2
3
4
350

150
300
100
50
1
2
3
4
0
0
1
2
3
4
0
43
Arithmetic Gradient

• A uniform increasing amount.
• The first cash flow is always equal to zero.
• G the difference between each cash amount.

G 10
44
Arithmetic Gradient Present Worth Factor

• The equivalent present value of a uniformly
  increasing amount.
• PG G(P/G,i,n)

Example 4-8
45
Arithmetic Series Present Worth

• Present Worth
• 120(P/F,5,1) 150(P/F,5,2) 180(P/F,5,3)
  210(P/F,5,4) 240(P/F,5,5)
• 766.64

Example 4-8, no shortcut
46
Arithmetic Gradient Uniform Series Factor

• The equivalent present value of a uniformly
  increasing amount.
• AG G(A/G,i,n)

Example 4-9
47
Decreasing Arithmetic Series

• Equivalent to

• Actual Series

450
450
400
0
1
2
3
4
350
-
300
150
100
1
2
3
4
0
50
0
1
2
3
4
0
48
Arithmetic Gradient Series
49
Transformation of a Geometric Series
SERIES DISCOUNT AMOUNT (PW)
UNIFORM SERIES
GEOMETRIC SERIES
SERIES COMPOUND AMOUNT (FW)
RATE of RETURN (ROR)
50
Geometric Series

• Types of series
• decay
• exponential
• General Formula
• PW C/(1k)?(1k)/(1i)N, N 1 to n
• (P/C,i,k,N) C/(i-k)1-(1k)/(1i)N if k
  ? i
• (P/C,i,k,N) CxN/(1i)
• CN/(1k) if k i

51

• Where,
• C first term of series
• i discount factor
• k rate of growth () or decay (-)

52
Geometric Series Present Worth Factor

• The equivalent present value of a geometrically
  increasing amount.
• P A(P/A,g,i,n)

Example 4-12
53
Nominal and Effective Interest

• Nominal interest rate/year the annual interest
  rate w/o considering the effect of any
  compounding.
• 12/year
• Interest rate/period the nominal interest
  rate/year divided by the number of interest
  compounding periods.
• 12/year/12 months/year 1/period

54

• Effective interest rate/year the annual interest
  rate taking into account the effect of the
  compounding periods in the year.
• 12/year compounded monthly is equivalent to
  12.68/year compounded annually

55
Example Discrete Interest Rate

• Given an interest rate of 12 compounded
  quarterly
• Nominal rate 12
• Effective rate 1001(0.12/4)4 12.55
• Actual rate 12/4 3 per quarter
• Investing 1 at 3 per quarter is equivalent to
  investing 1 at 12.55 annually

56
Example Increasing Arithmetic Gradient Series

• The used car that you bought to attend university
  is projected to have the following maintenance
  record during your 4 years at university
• 300 (year 1)
• 350 (year 2)
• 400 (year 3)
• 450 (year 4)

57

• Find the equivalent annual value of the
  maintenance charges for an interest rate of 10
  compounded annually?
• A A1 G(A/G,i,N)
• 300 50(A/G,10,4)
• 50(1.381)
• 369.05

58
Annuity

• Actual Series

• Equivalent to

300
0
1
2
3
4

150
100
50
0
2
3
4
1
59
Financial Table
60
Decreasing Arithmetic Gradient Series

• Annual Equivalent A1 - G(A/G,i,n), where G
  gradient and A1 first element of series
• Present Worth A1(P/A1,i,n) - G(P/G,i,n)
• Future Worth A1(F/A1,i,n) - G(F/G,i,n)

61
Nominal vs. Effective Rates A Comparison - A
Deposit
62
Nominal vs. Effective Rates A Comparison - A
Loan Repaid Monthly
63
Nominal and Effective Interest Rates

• Credit card interest rate 18, compounded
  monthly
• Nominal rate 18
• Effective rate 1(0.18/12)12 1
• 0.231 or 23.1
• Actual rate per quarter 18/12 1.5
• Investing 1 at 1.5 per month is equivalent to
  investing 1 at 23.1 annually.