1. Engineering Economic Analysis Canadian Edition
Chapter 3:
Interest and Equivalence

2. Chapter 3 … Defines the time value of money.
Distinguishes between simple and compound interest in engineering economic analysis.
Explains equivalence of cash flows.
Defines nominal, effective, and periodic interest rates.
Applies single payment compound interest formulas.

3. Chapter Assumptions in Solving Economic Analysis Problems
End-of-year convention
simplifies calculations
Viewpoint of the firm
Sunk costs
the past has no bearing on current decisions
Owner-provided capital
no debt capital
Stable prices
No depreciation and no income taxes

4. Economic Decision Components
Where economic decisions are immediate or short term, we need to consider:
amount of expenditure (costs) and benefits (income)
taxes (if after-tax analysis)
Where economic decisions occur over a considerable period of time, we also need to consider interest and inflation.
For the present discussion, we have assumed no inflation and no taxes.

5. Computing Cash Flows Cash flows have: Examples 3-1 & 3-2
Costs (disbursements) > a negative number
Benefits (receipts) > a positive number
Examples 3-1 & 3-2

6. Time Value of Money (TVM)
Money has value.
Money can be leased or rented.
The payment is called interest.
If you put $100 in a bank at 9% interest for one time period, you will receive your original $100 plus $9.
Original amount to be returned = $100
Interest to be returned = $100  9% = $9
Total amount received = $109

7. Importance of Time
Time is a critical factor in engineering economic studies (except for very short-term projects).
There is a preference for consuming goods and services sooner rather than later.
The stronger the preference for current consumption, the greater the importance of time in investment decisions.
The amount and timing of a project’s cash flows (operating expenses; revenues/ sales) are crucial to the value of a project’s worth.

8. TVM Illustration (Projects A & B)
Project Parameters
Project A
Project B
Initial investment
$10,000
Salvage value $0
Life
5 years
End-of-year cash flows
Year 1
$3,000
$5,000
Year 2
$4,000
Year 3
Year 4
$2,000
Year 5
$1,000

9. TVM Illustration (Projects A & B)…
Projects A and B have identical initial costs, duration, and cumulative cash flows over 5 years ($15,000) but different cash flow patterns.
If the discount rate i = 0%, then A and B are equivalent.
If i ≠ 0%, then A and B are not equivalent.

10. Interest Rate
The importance of time is inherent in the rate of interest.
The rate of interest is
the rate of return received by a lender for lending money (foregoing or trading off current consumption for future consumption), or
the rate of return paid by a borrower for the use of a lender’s funds.

11. Interest Rate For discrete cash flows, we have:
Simple interest: interest calculated once only and paid at the end of the term.
Compound interest: finite number (m) of within-year compounding periods, e.g.
12% compounded monthly (m = 12)
12% compounded quarterly (m = 4)
12% compounded annually (m = 1)
Continuous compounding (infinite number of compounding periods per year) is possible in theory; but, continuous cash flows don’t exist, so we will ignore continuous compounding.

12. I = $100  9%/period  ½ period = $4.50
Simple Interest
Interest that is computed only on the original sum or principal.
The term is usually up to one year.
Total interest earned: I = Pin
where
P – principal
i – interest rate (%) per period
n – number of periods (usually years)
For example: P = $100; i = 9% per year; 6 month term
I = $100  9%/period  ½ period = $4.50

13. Future Value of a Loan with Simple Interest
Amount of money due at the end of a loan
F = P + Pin; or F = P(1 + in), where F = future value.
Would you accept payment with simple interest terms? Would a bank?
F = $100 (1 + 9%  ½) = $104.50

14. Simple Interest Simple annual interest rate = 14%
End-of-year
Amount Borrowed
Interest for Period
Amount Owed at EOY
$1,000 1
$140
$1,140 2
$1,280 3
$1,420
$1,280 = $1,000+2(14%)$1,000

15. Compound Interest
Interest that is computed on the original unpaid principal and on the unpaid interest.
Total interest earned: In = P(1 + i)n – P
where
P = present sum of money
i = interest rate per period
n = number of periods
For example: P = $100; i = 9% compounded annually; 5-year term
I5 = $100  (1+9%)5  $100 = $53.86

16. Compound Interest Compound annual interest rate = 14%
End-of-year
Amount Borrowed
Interest for Period
Amount Owed at EOY
$1,000 1
$140.00
$1,140.00 2
$159.60
$1,299.60 3
$181.94
$1,481.54
$1, = $1,000(1+14%)2

17. Future Value of a Loan with Compound Interest
Amount of money due at the end of a loan
F = P(1+i)(1+i)  (1+i) = P(1+i)n; where F = future value
Would you be more likely to accept payment with compound interest terms? Would a bank?
For example: P = $100; i = 9%; 5-year term
F = $100(1 + 9%)5 = $153.86

18. Comparison of Simple and Compound Interest Over Time
The difference is negligible over a short period of time, but over a long period of time it may a considerable amount.
Check the table to see the difference over time.
Short or long? When is the difference significant? You choose the time period.

19. Four Ways to Repay a Debt
Plan
Repay Principal
Repay Interest
Interest Paid 1
Equal annual amounts
Interest on unpaid balance
Declines 2
End of loan
Constant 3
Declines at increasing rate 4
Compound and pay at end of loan
Compounds at increasing rate until end of loan

20. Loan Repayment — Four Options
Complete the calculator in this spreadsheet.

21. Equivalence
When an organization is indifferent as to whether it has an amount of money now (present value) or another amount of money in the future (future value), we say that the present amount of money is equivalent to the future amount of money.
Each of the plans on the previous slide is equivalent because each repays an amount of money that has the same value today at the same interest rate.

22. Equivalence …
Given the choice of these two plans which would you choose?
Year
Plan 1
Plan 2 1
$1400
$400 2
$1320 3
$1240 4
$1160 5
$1080
$5400
Total
$6200
$7000
To make a choice the cash flows must be altered (discounted) so a comparison can be made.

23. Technique of Equivalence
Select a common point in time.
Determine a single equivalent value at that point in time for plan 1.
Determine a single equivalent value at that point in time for plan 2.
Judge the relative attractiveness of the two alternatives from the comparable equivalent values.

24. Repayment Plans Establish the Interest Rate
Principal outstanding over time
Amount repaid over time
For example,
if F = P(1 + i)n,
then i = (F/P)1/n  1

25. Comparing Alternatives …
Compare the investment alternatives for various interest rates. Is there a rate at which you are indifferent among all alternatives?

26. Interest Formulas
To understand equivalence, the underlying interest formulas must be analyzed.
Notation:
i = Interest rate per interest period
n = Number of interest periods
P = Present sum of money (Present value)
F = Future sum of money (Future value)

27. Single Payment Compound Interest
Year
Beginning balance
Interest for period
Ending balance 1 P iP
P(1+i) 2
iP(1+i)
P(1+i)2 3
iP(1+i)2
P(1+i)3 n
P(1+i)n-1
iP(1+i)n-1
P(1+i)n
P at time 0 becomes P(1+i)n at the end of period n. Stated otherwise, Future value = Present value  (1+i)n.

28. Notation for Calculating Future and Present Values
Formula: F = P(1+i)n is the single payment future value factor.
Find the value in five years of $1000 today if the rate of interest is 7½%. (Ans: $ )
Formula: P = F(1/(1+i)n) = F(1+i)n is the single payment present value factor.
Find the value today of $1000 in five years if the rate of interest is 7½%. (Ans: $696.56)
Note: never use financial tables.

29. Nominal, Effective and Periodic Interest Rates
Interest rate = 8% compounded quarterly
Nominal rate = 8% (per year) compounded quarterly, i.e. 4 times per year.
Effective rate = [1+(8%/4)]4 – 1 = %.
Periodic rate (per quarter) = 8%/4 = 2%.
Investing $1 at 2% per quarter is equivalent to investing $1 at % annually.

30. Examples
If you borrow $5000 today, then $10,000 in eight months, calculate the two payments you must make in six months and one year, with the second payment being twice the size of the first, to repay the debt if interest is 6¾% per year.
$ in six months and $10, in one year, using one year from now as the focal date
$ in six months and $10, in one year, using the sequential payment technique

31. Examples …
Determine which of the following you would rather have: A. $10,000 today; B. $5000 today and $7000 in four years; or C. $3000 at the end of each of the next four years. Your opportunity cost of capital (interest rate) is 8½% compounded annually.
B is best with a value of $10, today.
A is second with a value of $10,000 today.
C is worst with a value $ today.

32. Examples …
Find how long it takes for $10,000 invested today to reach a value of $11, if the interest rate is 4¼% compounded monthly.
42 months or 3½ years
You promise to pay $10,000 in 30 months to a lender who gives you $ today. Find the interest rate the lender is charging. Express it as a nominal rate compounded semiannually and as an effective rate.
5½% compounded semiannually
% effective

33. Suggested Problems
3-2, 3-6, 3-9, 3-10, 3-13, 3-15, 3-16, 3-17, 3-19, 3-20.