Engineering Economics

1
Engineering Economics

• Introduction

2
Why Engineering Economics?

• Accreditation requirement in Ontario
• Engineers have to understand financial
  implications
• Communicate with the bean counters

3
Method of Instruction

• 1 Lecture per week
• Lecture PLUS (Participation, Learning,
  Understanding, Success)
• Participation - in class
• Learning - together
• Understanding - individual
• Success - individual

A B C
4
Demonstration - Lecture PLUSQuick Checks

• You are buying a new car and have three choices
• Choice A B C
• Auto 57 Chevy 87 Honda 82 Mercedes
• Price () 12,000 7,000 20,000
• Operation() 200/mo. 50/mo. 150/mo.
• Resale() 13,000 6,000
  20,000
• Which would you choose?

5
Module 4 Outline

• Intro to Engineering Economic Decisions
• Read Chapter 1 of the text
• Study the Chapter Summary (pp. 36)
• The time value of money
• Equivalence and Interest Formulas ( Chapter 2)
• Real world examples (Chapter 3)
• Analysis of independent Investments (Chapter 4)

6
Why Pay Interest?

• The borrower pays the lender for
• administrative costs associated with the loan
• compensation for the risk of default
• opportunity cost associated with not investing
  the money elsewhere

7
Key Notation

• Interest I
• nominal interest rate r
• effective interest rate (per period) i
• Present Value (or Worth) P
• Total Number of interest Periods N
• Future Value.. F
• Annuity amount .. A
• A discrete payment or receipt at
• the end of an interest period An

8
Interest Rate Example(Quick Check)

• If you borrow 1000 from me and agree to pay the
  1000 plus 125 extra at the end of one year,
  what is the interest rate?
• A. 8
• B. 12.5
• C. .125

9
Simple Interest

• Interest is paid only on the original principal
  amount, not on any accumulated interest
• In Pin
• Fn P In P(1in)
• Rarely used

10
Simple InterestExample (QC)

• How much interest is due for a loan of
  10,000 at 10 simple interest
• 1) after 1 yr?
• I Pin 10000(.10)(1) 1000
• 2) after 2 yrs?
• A) 1,000 B) 1,100 C) 2,000

11
Compound Interest

• The interest accumulates interest
• F1 P(1i) where F1 future value an the end of
  period 1
• F2 F1 (1i) P(1i) (1i) P(1i)2
• F3 P(1i)3
• Fn P(1i)n
• where, (1i)n (F/P,i,n),is called the compound
  amount factor

12
Compound Interest (QC)

• How much borrowed today will have to be repaid by
  10,000 in three periods(i 10) ?
• A) 9 000
• B) 9 091
• C) 7 513

13
Compound Interest Factors

• There are three categories of factor
• those that convert a single payment to a present
  or future amount
• those that convert a series of payments to a
  present or future amount
• those that convert a series of payments that
  increase or decrease by some constant amount G
  (for Gradient series) every period to an
  equivalent uniform series

14
Single Payment Factors
15
Single Payment
F
i
Yrs
1 2 3 4 5
P
Two amounts P, F
16
Example (QC)
F?
i 6
yrs
1 2 3 4 5
P200
A. 267.65 B. 267.64 C. 268.00
17
Uniform Series Payment Factors
Factor Formula Name Description
Converts a series of uniform payments to a
single future amount
Uniform series compound amount factor
(F/A,i,n)
Converts a single future amount into a series
of equal payments that would be necessary to
accumulate F in n periods
Sinking fund factor
(A/F,i,n)
18
More Uniform Series Factors
Factor Formula Name Description
Converts a series of equal payments to a
present amount
Series present worth factor
(P/A,i,n)
Finds the series of payments necessary to
recover (repay) a present amount in a fixed
number of periods.
Capital recovery factor
(A/P,i,n)
19
Uniform Series (Annuity)
A A A A A
yrs
5
0 1 2 3 4
i
F
Example, RRSP
20
Uniform Series
A A A A A
yrs
5
1 2 3 4
i
P
Example - bank loan/mortgage
21
Example - Loan repayment
You take out a mortgage (150 000) for a term of
20 yrs, with i5.25 per yr. What are your
annual payments?
A A A A A
1 2 3 19 20
i 5.25
P 150 000
22
Solution by formula
A A A A A
1 2 3 19 20
i 5.25
A P (A/P,i,n)
P 150 000
A 12 292,83
23
Solution by interpolation
A P (A/P,i,n)
A 150000 (A/P,5.25,20)
(A/P, 5, 20) 0.0802 (A/P, 6, 20) 0.0872
0.0872
? (A/P, 5.25,20)
0.0802
5.25
5
6
(A/P, 5.25,20)
0.08195
A 150000(0.08195) 12 292,50
24
Gradient present worth factor
Factor Formula Description
Converts an increasing or decreasing series of
payments by a constant amount G to a
single present value.
(P/G,i,n)
Arithmetic gradient to annuity conversion factor
(A/G,i,n)
25
Arithmetic Gradient
4G
3G
2G
G
Yrs
5
1 2 3 4
i
P
26
Example (QC)
Find the Present value
P?
i 10
7
500
150
500/yr
2000
A. 2 018 B. 3 804 C. 3 208
27
Formula Summary

• See pp. 94
• Complete Assignment 1(from Chapter 2)

28
Chapter 2 Review Problems

• The following questions from Chapter 2 are
  recommended
• Level 1 2.1 to 2.8 (answers on pp. 905), 2.20 to
  2.22, 2.33 to 2.35
• Level 2 2.40 to 2.42, 2.50 to 2.51
• Level 3 2.65, 2.67 a only

29
Nominal and Effective Interest

• Generally interest rates are quoted on an annual
  basis (annual percentage rate - APR) but the
  contract may specify that compounding occurs more
  frequently
• monthly
• quarterly (every 3 months)
• daily
• etc

30
Nominal Vs. Effective

• What is the effect of compounding more frequently
  than once per yr?
• Given
• r nominal interest per yr (APR - annual
  percentage rate)
• m number of times per year (sub-periods) that
  interest is compounded
• is r/m effective interest rate per period

31
Example

• 10 000 is deposited in the bank at a nominal
  rate of 10 per year, compounding is quarterly..
• Given
• r 10
• m 4 periods per yr.
• Therefore
• is r/m 10/4 2.5 per quarter

32
Quick Check

• Do you think the corresponding effective annual
  interest rate will be
• A. Less
• B. Equal
• C. More
• than the nominal rate?

33
Example Contd

• We have P10,000, is2.5
• Therefore
• F3moP(1 is) 10 000(1.025) 10 250
• F6mo F3mo(1 is) 10250(1.025) 10 506
• F9mo F6mo(1 is) 10 506(1.025) 10 769
• F12mo F9mo(1 is) 11 038
• Therefore the accumulated interest at the end of
  the yr 1,038 which is more than the 1,000
  That we would receive with simple interest
• The Effective annual rate
• ie (F-P)/P (11038-10000)/10000 10.38

34
Formula for calculating ia for 1 yr
ia (1 r/m)m -1
where
ia effective annual rate r nominal interest
rate per year m number of compounding periods
per yr
Note 1. This formula applies when compounding is
more frequent than once per yr. 2. If m1, ia
r 3. The factor (1r/m)m increases with
m,therefore the more frequent the compounding,
the more interest will accumulate.
35
Quick Check

• Choose between
• A) borrowing at 12 compounded monthly
• B) borrowing at 13 compounded semi- annually
• C) borrowing at 11.5 compounded daily.

36
Example

• Would you prefer to receive
• 200 after 1 yr (case 1)
• 150 in 1 yr and another 100 after 2 yrs (case
  2)?
• The interest rate at the bank is 15 compounded
  monthly.

37
Solution
200
Case 1
i 15/12 1.25
12
0
P 200 (P/F,1.25,12) 172,30
150
100
i 15/12 1.25
Case 2
12
24
0
P 150 (P/F,1.25,12) 100(P/F,1.25,24) 203.45
By calculating a single present amount that is
equivalent to each cash flow we can compare the
2 cases directly.
38
Quick Check
A student borrowed 10,000 from the bank to buy a
new car. The bank charges 10 compounded
annually but give the student three repayment
options. Which would you choose?
End Yr Plan A Plan B Plan C 1 2638
- 1800 2 2638 - 1800 3 2638
- 1800 4 2638 - 1800 5 2638
16105 6916 Total 13190 16105 14116
39
Effective Interest rate any payment period
Effective interest are usually based on the
payment period, but can Be calculated for any
desired period. For example, if cash flow
payments Occur quarterly, but interest is
compounded monthly we may wish To calculate an
effective quarterly interest rate.
i (1 r/m)C -1
(1 r/CK)C -1
Where MThe number of compounding periods per
year C The number of compounding periods per
payment period KThe number of payment periods
per year Note that MCK
40
Example effective rate per period
Suppose you make quarterly deposits in a savings
account which earns 9 interest compounded
monthly. Calculate the effective interest rate
per quarter.
r9, C 3 compounding periods per quarter K4
quarterly payments per year, and M12 compounding
periods per year. Or, MCK 3x412
i (1 r/m)C -1
i (1 0.09/12)3 1 2.27
41
Example explained
r 9 compounded monthly, i9/12 0.75
i (1 0.09/12)3 1 2.27 Effective
interest rate per quarter
ia (1 r/m)m 1 (1.0075)12-1
(1.0227)4-1 9.38
42
Equivalence and demand loans

• A demand loan is a loan that either the lender or
  borrower can decide to have terminated by
  immediate payment of the outstanding balance. In
  practice, borrowers often choose to pay off a
  loan in order to save on interest charges. Using
  the principal of equivalence, the remaining
  balance of a loan can be determined in two ways.
• Example Julie borrowed 1 000 from the bank at
  9 compounded monthly. She agreed to repay the
  loan in 6 monthly payments but since this is a
  demand loan, she can clear the balance at any
  time. Immediately following her second payment,
  Julie wins big in the lottery and she decides to
  clear off her debt. How much must she pay to
  clear the loan?

43
Solution - Method 1

• Payments A 1000(A/P,9/12,6)
  
  1000(.1711) 171.10/mo
• We find an equivalent value of all preceding
  transactions just after the second payment
• B2 1000(F/P,.75,2) - 171.1(F/A,.75,2)
• 1000(1.015)-1.71(2.008) 671.76
• is the remaining balance after the second payment

44
Solution - Method 2

• We simply discount the remaining payments to the
  time at which the loan is to be paid off. I.e.,
  we calculate the PW of the remaining payments.
• B2 171.1(P/A,.75,4)
  171.1(3.9261) 671.76

45
Loan Tables
capital recovered 163.60 164.83 166.06 167.30 16
8.56 169.80
A balance 1 000.00 836.40 671.57 505.51 338.20 16
9.63 0.00
B payments 171.10 171.10 171.10 171.10 171.10 17
1.10
A(.0075) interest 7.50 6.27 5.04 3.79 2.54 1.27
End Yr 0 1 2 3 4 5 6
46
Calculations involving compound interest

• The frequency of payments is not always the same
  as the compounding period
• Usually we have 3 of P,A,F,i,n and we have to
  find the forth
• We generally assume that P occurs at the
  beginning of the first period (or at the end of
  period 0) and that F and A occur at the end of
  the period (end of period convention)

47
Example - Single Transaction
You have 1000 available to invest. You also
know that 6 years from now you have a requirement
for 1600. What rate of return i is necessary
achieve 1600?
48
Solution - single transaction
F1600
i?
n5
P1000
F P(F/P,i,n) therefore, (F/P,i,6) F/P 1.6
(1i)6
? (1i) 1.6(1/6) 1.08148 therefore, i
0.08148 8.15/yr
Therefore, we require an internal rate of return
of 8.15/ yr. To meet the 1600 requirement.
49
Multiple TransactionQuick Check

• In general, most projects have a combination of
  individual transactions and annuities.
• Example An investment pays 10,000 immediately
  and 1,000 at the end of each yr for a period of
  5 yrs and also has an individual payment of
  2,000 at the end of the 5th year. If i7,
  what is the present worth of the transactions?

50
Special Cases

• There are 3 special cases that prevent the direct
  application of the compound interest factors
• Compounding is more frequent than payments
• The interest rate is not constant for the whole
  period
• Annuities occur at the start of the period

51
Compounding more frequent than payments

• Example three deposits of 2,500 are made every
  2 years starting in 2 years. i 10/yr, how
  much will be in the bank at the end of 6 yrs?
• The series compound amount factor (F/A, i, n)
  cannot be used directly because the payment
  interval does not match the compounding period.

52
Solution
F?
i 10
1
2
3
4
5
6
2500
2500
2500
Method 1
F 2500 2500(F/P, 10,2) (F/P, 10,4)
2500 2500(1.21) (1.4641) 9,185
Method 2 we calculate an effective interest rate
to match the compounding period. I.e. an
effective 2 yr interest rate.
ie (1.10)2-1 0.21 21 F 2500(F/A,21,3)
2500(3.6741) 9185
53
Variable interest rates (QC)
If the interest rate changes for different
periods, the present worth has to be calculated
in steps.
Ex Given the following CFD, calculate the PW
A. 428 B. 89 C. 484
54
Annuities at the start of the period

• For annuities where the first payment is made
  today instead of at the end of the first period
• P A A(P/A,i,n-1)
• Ex Jaqueline needs to rent a machine for 5 yrs.
  She has to choose between paying 10,000 up front
  or paying 2,100 at the start of each of the 5
  yrs. Use i 8.
• P 2100 2100(P/A,8,4) 2100(13.3121)
  9,055
• Conclusion it is less expensive to pay monthly.
• The difference between the up front cost and the
  PW of the annuity is 945. How would you
  interpret this amount?

55
Interpretation - Present Worth
Yr 0 1 2 3 4 5
A Balance at the start of the yr. (C) - (D)
7900.00 6432.00 4846.56 3134.28 1285.03
B Interest (_at_8) during the yr.
632.00 514.56 387.72 250.74
C Balance at the end of yr (A)
(B) 10000.00 8532.00 6946.56 5234.28 3385.03
D Payments at start of yr 2100 2100 2100 2100
2100
56
Quick Check

• The PW of the remaining balance after the last
  payment is
• A) P 1285.03(P/A,8,5)
• B) P 1285.03(P/F,8,4)
• C) P 1285.03(P/F,8,5)

57
Chapter 3 Review Problems

• The following questions from Chapter 3 are
  recommended
• Level 1 3.2 to 3.9, 3.11 to 3.17 (answers on pp.
  905)
• Level 2 3.22, 3.26, 3.27 a and b only, 3.31,
  3.38, 3.40, 3.43, 3.45, 3.54, 3.58, 3.59, 3.91
• Level 3 3.95. For part b, use Excels IRR
  function to find the effective interest rate.

58
Analysis of Independent Investments
59
Outline

• Cash flow representation of projects
• Payback Period
• without interest
• with interest
• Decision Criteria
• NPW
• AW
• FW
• IRR
• Capitalized Value

60
The Payback Method

• Justification
• important to know when a project starts to make a
  profit
• traditional payback approach ignores the time
  value of money
• do not require a MARR for calculations
• Useful for project screening

61
Payback Method
Given Ft the net sum of all payments up to period
t, the payback period is the smallest value of n
that satisfies
62
Example (QC)
1 000 invested that returns 200 per yr for 10
yrs (i8)
The payback period without interest 5
ans with interest 1000200(P/A,8,n)? n ?
A. n5 B. n6 C. n7
63
Disadvantages of payback period

• Suppose there were a second option in the
  previous example that returned the 1 000 after
  only 1 yr but had no subsequent receipts. The
  payback period would be 1 yr, but the investment
  would not earn any interest!!!
• Even though the first option earned
• 1000200(P/A,i,10) ? i15.1 (IRR)
• Because payback ignores transactions that occur
  after the payback has occurred it should only be
  used as supplemental information and never as the
  main decision.

64
Payback period Ex. 2i15
With interest 5 ans
Payback without interest 4 ans
65
Typical Project Cash Flow

• Initial investment
• Followed by a series of expected profits

66
Project Example

• New Factory
• Initial investment
• construction costs
• investment in production equipment
• training
• Expected profits
• decision required
• accept/reject the project
• need a decision criteria

67
Example
The company Fevolard of Kingston is considering
the purchase of a new packaging machine. One
model has an initial cost of 120 000 and a
salvage value of 5 000 after its 10 yr service
life. Given an anticipated increase in sales
revenue the machine is anticipated to result in a
cost savings of 15 000 the first yr.
Increasing by 5 000 each of the following 9 yrs.
Supplemental operating costs associated with
this piece of equipment are 10 000 per yr. The
company uses a required rate of return of 12.
Should they invest in the machine?
i 12
68
PW Criteria
i 12
5.6502
20.2541
0.3220
Conclusion Accept the project because PW(_at_12)
gt 0
69
Quick Check

• If PW(_at_ MARR) 0 we should
• A. Accept the project
• B. reject the project

70
Annual Equivalent (AE)
We could solve the same problem by calculating an
single annuity amount that is equivalent to
entire project cash flow as follows
i 12
0.0570
0.1770
3.5847
AE -120000(A/P, 12,10)150005000(A/G,12,10)-100
005000(A/F,12,10) 1 968
71
Annual Equivalent (AE)

• Of course, if we already have the PW, we can
  calculate the AE directly
• AEPW(A/P,12,10) 11 132(0.1770)
  1 970

72
Quick Check

• Do you believe that the PW criteria and the AE
  criteria will always yield the same decision?
• A. Yes
• B. No
• C. It depends

73
Future Equivalent

• The same reasoning applies if we chose to use FE
  as our decision criteria. The FW, PW and AE can
  be calculated directly from each other by
  multiplying by a positive constant amount.
  Therefore, all three methods will yield
  consistent results.

PW1 FE1 AE1


Note
PW2 FE2 AE2
74
Present Worth Profile

• To help understand the relationship between the
  PW and the IRR, examine the PW as a function of
  the interest rate i.

75
Comparison Example - Equipment Purchase
76
NPW for Machine 1 (check)
S 3,000
i10
1
5
2
3
4
A 1,000
10,000
A. P -10000 - 1000(P/F,10,5)
3000(P/F,10,5) B. P -10000 - 1000(P/A,10,5)
3000(P/F,10,5) C. P -10000 -
1000(A/P,10,5) 3000(P/F,10,5)
77
NPW - Machine 1
NPW -10000 - 1000P/A,10,5 3000P/F,10,5
-11,928
78
AEC (Check)
A. AEC -10000(A/P,10,5) - 1000
3000A/F,10,5
B. AEC -10000(P/A,10,5) - 1000
3000A/F,10,5
C. AEC -11,928(A/P,10,5) equivalent NPW
79
Solution - Machine 1
NPW -10000 - 1000P/A,10,5 3000P/F,10,5
-11,928
AEC -10000(A/P,10,5) - 1000 3000A/F,10,5
-3,146.50
AEC 11,928(A/P,10,5) -3,146.50
80
Solution - Machine 2
NPW -6000 - 2000P/A,10,5 1500P/F,10,5
-12,650
AEC -6000(A/P,10,5) - 2000 1500A/F,10,5
-3,337.10
AEC NPW (A/P,10,5)
-12,650(A/P,10,5) -3,337.10
81
Solution Summary
Machine 1 NPW -11,928 AEC
-3,146.50
Machine 2 NPW -12,650 AEC
-3,337.10
With equal lives, the two methods yield
consistent results.
82
The Net Present Worth Function
PW(i)
0
5 10 15 20
i
83
PW Function Check
Suppose your MARR 6, would you a. Select
project A b. Select project B c. Select project
A and B
84
Internal Rate of Return IRRrendering revenue
equivalent to expenses

• Ex Your company can buy a machine for 10,000
  and then rent it out for 2500 per yr for its
  service life of 5 yrs. What is the is the
  interest rate (IRR) for this investment?
• P 10 000 A 2 500 n 5 i ?
• Therefore, 100002500(P/A,i,5), (P/A,i,5 )
  10000/2500 4
• From the interest tables (P/A,8,5) 3.993 and
  (P/A,7,5) 4.1001
• By interpolation i .07.01(4.1001-4)/(4.1001-3
  .993)
• .07935 or IRR 7.9
• Using Excel 7.93 DEMO

85
Symbol Convention

• i represents the interest rate that makes the
  NPV of the project equal to zero
• IRR represents the internal rate of return of the
  investment, for simple investments IRRi, and
  this is frequently referred to as the rate of
  return ROR

86
Chapter 4 Review Problems

• The following questions from Chapter 4 are
  recommended
• Level 1 4.1, 4.7 a and c only, 4.8,4.10, 4.18,
  4.23, 4.28, 4.35
• Level 2 4.38, 4.43, 4.46, 4.62
• Level 3 4.67 use excel to find i