ENGINEERING ECONOMICS TUTORIAL
ENGINEERING ECONOMICS TUTORIAL
WHY DO ENGINEERS NEED TO LEARN ABOUT ECONOMICS?
Ages ago, the most significant barriers to engineers were technological. The things that engineers wanted to do, they simply did not yet know how to do, or hadn't yet developed the tools to do. There are certainly many more challenges like this which face present-day engineers. However, we have reached the point in engineering where it is no longer possible, in most cases, simply to design and build things for the sake simply of designing and building them. Natural resources (from which we must build things) are becoming more scarce and more expensive. We are much more aware of negative side-effects of engineering innovations (such as air pollution from automobiles) than ever before.
For these reasons, engineers are tasked more and more to place their project ideas within the larger framework of the environment within a specific planet, country, or region. Engineers must ask themselves if a particular project will offer some net benefit to the people who will be affected by the project, after considering its inherent benefits, plus any negative side-effects (externalities), plus the cost of consuming natural resources, both in the price that must be paid for them and the realization that once they are used for that project, they will no longer be available for any other project(s).
Simply put, engineers must decide if the benefits of a project exceed its costs, and must make this comparison in a unified framework. The framework within which to make this comparison is the field of engineering economics, which strives to answer exactly these questions, and perhaps more. The Accreditation Board for Engineering and Technology (ABET) states that engineering "is the profession in which a knowledge of the mathematical and natural sciences gained by study, experience, and practice is applied with judgment to develop ways to utilize, economically, the materials and forces of nature for the benefit of mankind".1
It should be clear from this discussion that consideration of economic factors is as important as regard for the physical laws and science that determine what can be accomplished with engineering. The following figure shows how engineering is composed of physical and economic components:
![[Eng. Economics]](economics-fig1.gif)
Figure 1 shows how engineering is composed of physical and economic components.
Physical efficiency takes the form:
system output(s)
Physical (efficiency ) = -------------------
system input(s)
Satisfaction of the physical and economic environments is linked through production and construction processes. Engineers need to manipulate systems to achieve a balance in attributes in both the physical and economic environments, and within the bounds of limited resources. Following are some examples where engineering economy plays a crucial role:
With items 1 and 2 in particular, note that coursework in engineering should provide sufficient means to determine a good design for a furnace, or a suitable robot for an assembly line, but it is the economic evaluation that allows the further definition of a best design or the most suitable robot.
In item 1 of the list above, what is meant by " high-efficiency"? There are two kinds of efficiency that engineers must be concerned with. The first is physical efficiency, which takes the form:
System output(s)
Economic (efficiency ) = -----------------
System input(s)
For the furnace, the system outputs might be measured in units of heat energy, and the inputs in units of electrical energy, and if these units are consistent, then physical efficiency is measured as a ratio between zero and one. Certain laws of physics (e.g., conservation of energy) dictate that the output from a system can never exceed the input to a system, if these are measured in consistent units. All a particular system can do is change from one form of energy (e.g. electrical) to another (e.g., heat). There are losses incurred along the way, due to electrical resistance, friction, etc., which always yield efficiencies less than one. In an automobile, for example, 10-15% of the energy supplied by the fuel might be consumed simply overcoming the internal friction of the engine. A perfectly efficient system would be the theoretically impossible Perpetual Motion Machine!
The other form of efficiency of interest to engineers is economic efficiency, which takes the form:
system worth
Economic (efficiency ) = -----------------
system cost
You might have heard economic efficiency referred to as "benefit-cost ratio". Both terms of this ratio are assumed to be of monetary units, such as dollars. In contrast to physical efficiency, economic efficiency can exceed unity, and in fact should, if a project is to be deemed economically feasible. The most difficult part of determining economic efficiency is accounting for all the factors which might be considered benefits or costs of a particular project, and converting these benefits or costs into a monetary equivalent. Consider for example a transportation construction project which promises to reduce everyone's travel time to work. How do we place a value on that travel time savings? This is one of the fundamental questions of engineering economics.
In the final evaluation of most ventures, economic efficiency takes precedence over physical efficiency because projects cannot be approved, regardless of their physical efficiency, if there is no conceived demand for them amongst the public, if they are economically infeasible, or if they do not constitute the "wisest" use of those resources which they require.
There are numerous examples of engineering systems that have physical design but little economic worth (i.e it may simply be too expensive !!). Consider a proposal to purify all of the water used by a large city by boiling it and collecting it again through condensation. This type of experiment is done in junior physical science labs every day, but at the scale required by a large city, is simply too costly.
ROLE OF UNCERTAINTY IN ENGINEERING
When conducting engineering economic analyses, it will be assumed at first, for simplicity, that benefits, costs, and physical quantities will be known with a high degree of confidence. This degree of confidence is sometimes called assumed certainty. In virtually all situations, however, there is some doubt as to the ultimate values of various quantities. Both risk and uncertainty in decision-making activities are caused by a lack of precise knowledge regarding future conditions, technological developments, synergies among funded projects, etc. Decisions under risk are decisions in which the analyst models the decision problem in terms of assumed possible future outcomes, or scenarios, whose probabilities of occurrence can be estimated. Of course, this type of analysis requires an understanding of the field of probability. Decisions under uncertainty, by contrast, are decision problems characterized by several unknown futures for which probabilities of occurrence cannot be estimated. Other less objective means exist for the analysis of such problems.
For the purposes of this brief tutorial, we cannot delve further into the analytical extensions required to accommodate risk or uncertainty in the decision process. We must recognize that these things exist, however, and be careful about reaching strong conclusions based on data which might be susceptible to these. Because engineering is concerned with actions to be taken in the future, an important part of the engineering process is improving the certainty of decisions with respect to satisfying the objectives of engineering applications.
THE ENGINEERING PROCESS
Engineering activities dealing with elements of the physical environment take place to meet human needs that arise in an economic setting. The engineering process employed from the time a particular need is recognized until it is satisfied may be divided into a number of phases:
This step involves finding out what people need and want that can be supplied by engineering. People's wants may arise from logical considerations, emotional drives, or a combination of the two.
The factors that stand in the way of attaining objectives are known as limiting factors. Once the limiting factors have been identified, they are examined to locate strategic factors -- those factors which can be altered to remove limitations restricting the success of an undertaking. A woman who wants to empty the water from her swimming pool might be faced with the limiting factor that she only has a bucket to do the job with, and this would require far greater time and physical exertion than she has at her disposal. A strategic factor developed in response to this limitation would be the procurement of some sort of pumping device which could do the job much more quickly, with almost no physical effort on the part of the woman.
This step involves discovering what means exist to alter strategic factors in order to overcome limiting factors. In the previous example, one means was to buy (or rent) a pump. Of course, if the woman had a garden hose, she might have been able to siphon the water out of the pump. In other engineering applications, it may be necessary to fabricate the means to solve problems from scratch.
It is usually possible to accomplish the same result with a variety of means. Once these means have been described fully, in the form of project proposals, economic analysis can be employed to determine which among them, if any, is the best means for solving the problem at hand.
It is commonplace for the final decision-making responsibility to fall on the head(s) of someone other than the engineer(s). The person(s) so charged, however, may not be sufficiently knowledgeable about the technical aspects of a proposal to determine its relevant worth compared to other means. The engineer can help to bridge this gap.
ENGINEERING ECONOMIC STUDIES
The four key steps in planning an economic study are :
Future cash flows are assigned to each alternative, consisting of the time-value of money.
Decisions among system alternatives should be made on the basis of their differences.
For a small number of real world systems there will be complete knowledge. Dll facts/information and their relationships, judgements and predictive behavior become a certainty.
For most systems, however, even after all of the data that can be bought to bear on it has been considered, some areas of uncertainty are likely to remain. If a decision must be made, these areas of uncertainty must be bridged by consideration of non-quantitative data/information, such as common sense, judgement and so forth.
Examples :
Economics deals with the behavior of people, and as such, economic concepts are usually qualitative in nature, and not universal in application.
UTILITY
VALUE
CONSUMER AND PRODUCER GOODS
UTILITY OF GOODS
ECONOMY OF EXCHANGE
ECONOMY OF ORGANIZATION
Through organizations, ends can be attained or attained more economically by:
CLASSIFICATION OF COST
A key objective in engineering applications is the satisfaction of human needs, which will nearly always imply a cost.
Economic analyses may be based on a number of cost classifications:
SUPPLY AND DEMAND (not covered)
INTEREST RATE
Interest is a rental amount charged by financial institutions for the use of money.
TIME VALUE OF MONEY
The time-value of money is the relationship between interest and time. i.e.
![[Time-Value of Money]](economics-fig2.gif)
Money has time-value because the purchasing power of a dollar changes with time.
EARNING POWER OF MONEY
The earning power of money represents funds borrowed for the prospect of gain.
Often these funds will be exchanges for goods, services, or production tools, which in turn can be employed to generate and economic gain.
PURCHASING POWER OF MONEY
The prices of goods and services can go upward or downward, and therefore, the purchasing power of money can change with time.
Economy of Material Selection
Standardization and Simplification
Economy of Resource Input for Organizations
Knowledge and Information
Cash flow diagrams are a means of visualizing (and simplifying) the flow of receipts and disbursements (for the acquisition and operation of items in an enterprise).
The diagram convention is as follows:
All disbursements and receipts (i.e. cash flows) are assumed to take place at the end of the year in which they occur. This is known as the "end-of-year" convention.
Arrow lengths are approximately proportional to the magnitude of the cash flow.
Expenses incurred before time = 0 are sunk costs, and are not relevant to the problem.
Since there are two parties to every transaction, it is important to not that cash flow directions in cash flow diagrams depend upon the point of view taken.
Example :
![[Cash Flows]](economics-fig3.gif)
Figure 3 shows cash flow diagrams for a transaction spanning five years.
The transaction begins with a $1000.00 loan.
For years two, three and four, the borrower pays the lender $120.00 interest.
At year five, the borrower pays the lender $120.00 interest
plus the $1000.00 principal.
Interest formulae play a central role in the economic evaluation of engineering alternatives.
TYPES OF INTEREST
P = Principal
i = Interest rate
n = Number of years (or periods)
I = Interestt.
Example : Suppose that $50,000 is borrowed at a simple interest rate of 8% per annum. At the end of two years the interest owed would be:
I = $ 50,000 * 0.08 * 2
= $ 8,000
Principal P = $1000.00
Interest Rate i = 0.12.
Number of years (or periods) n = 5.
================================================================
Amount at Interest at Owed amount at
Year start of year end of year end of year Payment
================================================================
1 $1000.00 $120.00 $1120.00 $120.00
2 $1000.00 $120.00 $1120.00 $120.00
3 $1000.00 $120.00 $1120.00 $120.00
4 $1000.00 $120.00 $1120.00 $120.00
5 $1000.00 $120.00 $1120.00 $120.00
Principal P = $1000.00
Interest Rate i = 0.12.
Number of years (or periods) n = 5.
================================================================
Amount at Interest at Owed amount at
Year start of year end of year end of year Payment
================================================================
1 $1000.00 $120.00 $1120.00 $0.00
2 $1120.00 $134.40 $1254.40 $0.00
3 $1254.40 $150.53 $1404.93 $0.00
4 $1404.93 $168.59 $1573.52 $0.00
5 $1573.52 $188.82 $1762.34 $1762.34
INTEREST FORMULAE (Discrete Compounding and Discrete Payments)
Notations:
i = The annual interest rate
n = The number of annual interest periods
P = A present principal sum
A = A single payment, in a series of n equal payments, made at the end
of each annual interest period
F = A future sum, n annual interest periods hence
F = P.[1 + i]^n
Example 1 : Let the principal P = $1000, the interest rate i = 12%, and the number of periods n = 4 years. The future sum is:
F = $1000 [1 + 0.12] ^ 4
= $1,573.5
![[Single Payment]](economics-fig6.gif)
Figure 4 shows the cash flow for the single present amount (i.e. P = 1000) and the single future amount (i.e. F = $1,573.5).
F
P = =========
[1 + i]^n
The factor 1.0/[ 1 + i ]^n is known as the single-payment present-worth factor, and may be used to find the present worth P of a future amount F.
Example 1 : Let the future sum F = $1000, interest rate i = 12%, and number of periods n = 4 years. The single payment present-worth factor is:
F $1000.00
P = ========= = ============== = $635.50.
[1 + i]^n [ 1 + 0.12 ]^4
The present worth P = $635.50.
Example 2 : Let the future sum F = $1,573.5, the interest rate i = 12%, and the number of periods n = 4 years. The single payment present-worth factor is:
F $1573.50
P = ========= = ============== = $1000.00.
[1 + i]^n [ 1 + 0.12 ]^4
The present worth P = $1000.00.
Some economic studies require the computation of a single factor value that would accumulate from a series of payments occurring at the end of succeeding interest periods.
![[Compound Amounts]](economics-fig4.gif)
Figure 5 represents this scenario in graphical terms. At the end of Year 1 a payment of $ A begins the accumulation of interest at rate i% for (n-1) years. At the end of Year 2 a payment of $ A begins the accumulation of interest at rate i% for (n-2) years. End of year payments of $ A continue until Year N (or n as written below).
The total accumulatio of funds at Year N is simply the sum of $A payments multiplied by the appropriate single-payment present-worth factors. In tabular format we have:
End of Compound Amount at Total
Year the end of n Years Compound Amount
==========================================================
1 $ A . [ 1 + i ] ^ (n - 1)
2 $ A . [ 1 + i ] ^ (n - 2)
3 $ A . [ 1 + i ] ^ (n - 3)
n - 1 $ A . [ 1 + i ]
n $ A
==========================================================
$ F
==========================================================
The total compound amount is simply the sum of the compound amounts for years 1 though n. This sumation is a geometric series:
F = A + A.[1 + i] + A.[1+i]^2 + ..... + A.[1+i]^(n-1)
With a little bit of mathematical manipulation, it can be shown;
[ 1 + i ]^n - 1
F = A * ---------------
i
Example 1: Let A = 100, i = 12%, and n = 4 years.
[ 1 + 0.12 ]^4 - 1
F = 100 * ------------------ = 477.9
0.12
Given a future amount F, the equal payments compound-amount relationship is:
i
A = F * ---------------
[ 1 + i ]^n - 1
A = required end-of-year payments to accumulate a future amount F.
Example 1: Let F = 1000, i = 12%, and n = 4 years.
0.12
A = 1000 * ------------------ = 209.2
[ 1 + 0.12 ]^4 - 1
A deposit of amount P is made now at an interest rate i. The depositor wishes to withdraw the principal plus earned interest in a series of year-end equal payments over N years, such that when the last withdrawl is made there should be no funds left in the account.
![[Capital Recovery]](economics-fig5.gif)
Figure 6 summarizes the flow of disbursements and receipts (from the depositors point of view) for this scenario.
Equating the principle $P (plus accumulated interest) with the accumulation of equal payments $A (plus appropriate interest) gives:
[ [ 1 + i ]^n - 1 ]
P [ 1 + i ]^n = A -------------------
i
which can be rearranging to give:
i * [ 1 + i ]^n
A = P * ---------------
[ 1 + i ]^n - 1
This is the case of loans (mortgages).
Example 1: Let P = 1000, i = 12%, and n = 4 years
0.12 * [ 1 + 0.12 ]^4
A = $1000 * --------------------- = $329.2
[ 1 + 0.12 ]^4 - 1
This can be described as
[ 1 + i ]^n - 1
P = A * ---------------
i * [ 1 + i ]^n
Example 1: Let A = 100, i = 12%, and n = 4 years.
[ 1 + 0.12 ]^4 - 1
P = 100 * --------------------- = 303.7
0.12 * [ 1 + 0.12 ]^4
Often periodic payments do not occur in equal amounts, and may increase or decrease by constant amounts (e.g. $100, $120, $140, $160 .... $200).
The gradient (G) is a value in the cash flow that starts with 0 at the end of year 1, G at the end of year 2, 2G at the end of year 3, and so on to (n-1)G at the end of year n.
This can be described as
*- -*
| 1 n |
A = G . | - - --------------- |
| i [ 1 + i ]^n - 1 |
*- -*
Example 1 : Let G = 100, i = 12%, and n = 4 years.
*- -*
| 1 4 |
A = $100 . | ---- - ------------------ | = $ 135.9
| 0.12 [ 1 + 0.12 ]^4 - 1 |
*- -*
DISCRETE AND CONTINUOUS COMPOUNDING (compounding frequency)
Nominal interest rate : is expressed on an annual basis. Financial institutions refer to this rate as annual percentage rate (APR).
Effective interest rate : is an interest rate that is compounded using a time period less than a year. The nominal interest rate in this case is the effective rate times the number of compounding periods in a year. Then, it is referred to as nominal rate compounded at the period less than a year.
Example : If the effective rate is 1% per month, it follows that the nominal rate is 12% compounded monthly.
Relationship between the two rates. Let's define:
r = nominal interest rate per year
i = effective interest rate in the time interval
l = length of the time interval (in years)
m = reciprocal of the length of the compounding period (in years)
Therefore, the effective interest rate for any time interval is given by:
*- -* ^ l*m
| r |
i = | 1 + --- | - 1.0
| m |
*- -*
Clearly if l*m = 1, then i is simply r/m.
The product l*m is called c = the number of compounding periods in the time interval l. Note that c should be > 1.
Continuous compounding: The limiting case for the effective rate is when compounding is performed an infinite times in a year, that is continuously. Using l = 1, the following limit produces the continuously compounded interest rate
*- -* ^ m
| r |
i_a = Limit | 1 + --- | - 1.0
m->infinity | m |
*- -*
resulting into and effective interest rate
i_a = e^r - 1
Example :
================================================================
Compounding Number of Effective interest Effective annual
frequency Periods rate per period interest rate
================================================================
Annually 1.0 18% 18.00%
Semiannually 2.0 9% 18.81$
Quarterly 4.0 4.5% 19.25%
Monthly 12.0 1.5% 19.56%
Weekly 52.0 0.3642% 19.68%
Daily 365.0 0.0493% 19.74%
Continuously infinity 0.0000% 19.72%
Interest Formulas (Continuous Compounding and Discrete Payments)
Interest Formulas (Continuous Compounding and Continuous Payments)
Handout. (Table)
Why? It is common in engineering to compare alternatives.
THE MEANING OF EQUIVALENCE
In engineering economy two things are said to be equivalent when they have the same effect. Unlike most individual involved with personal finance, industrial decision makers using engineering economics are not so much concerned with the timing of a project's cash flows as with the profitability of that project. This means that mechanisms are needed to compare projects involving receipts and disbursements occurring at different times, with the goal of identifying an alternative having the largest eventual profitability [Lindeberg82].
EQUIVALENCE CALCULATIONS INVOLVING A SINGLE FACTOR
The interest equations are affected by three factors: (a) amounts, (b) times of occurrence of amounts, and (c) rate of interest.
F = P.[1 + i]^n
Example 1 : Let P = $1000, i = ??, n = 4 years, and F = $1200. The interest rate is
F = $1200.00 = $1000.00 [1 + i ]^4
Rearranging terms in this equation gives i = 1.2^0.25 - 1 = 0.046635.
Example 2 : Let P = $1000, i = 10%, n = ?? years, and F = $1200.
F = $1200.00 = $1000.00 [1 + 0.10 ]^n
Rearranging terms in this equation gives n = 1.91 years.
F
P = ---------
[1 + i]^n
Example 1 : Let F = $1000, i = 12%, n = 4 years, and P = ?
$1000.00
P = ------------ = $635.5
[1 + 0.12]^4
[ 1 + i ]^n - 1
F = A * ---------------
i
The derivation of this formula can be found on page 46 of the economics text.
Example 1 : Let A = $100.00, i = 10%, and F = $1000.00. How many years n are needed ?
[ 1 + 0.12 ]^n - 1
$1000.00 = $100.00 * ------------------
0.12
Rearranging the terms in this equation gives n = 7.27 years.
i
A = F * ---------------
[ 1 + i ]^n - 1
Example 1 : Paying towards a future amount. Let F = $1000.00, i = 12%, and n = 4 years. What is A ?
0.12
A = $1000.00 * ------------------ = $209.20
[ 1 + 0.12 ]^4 - 1
i * [ 1 + i ]^n
A = P * ---------------
[ 1 + i ]^n - 1
Example 1 : Paying back a loan. Let P = $1000, i = 12%, n = 4 years, and A = ?
0.12 * [ 1 + 0.12 ]^4
A = 1000 * --------------------- = 329.2
[ 1 + 0.12 ]^4 - 1
[ 1 + i ]^n - 1
P = A * ---------------
i * [ 1 + i ]^n
Example 1 : Let A = 100, i = 10%, and n = 8 years.
[ 1 + 0.10 ]^8 - 1
P = 100 * ---------------------- = 533.49
0.10 * [ 1 + 0.10 ]^8
As explained above, the gradient (G) is a value in the cash flow that starts with 0 at the end of year 1, G at the end of year 2, 2G at the end of year 3, and so on to (n-1)G at the end of year n.
This can be described as
*- -*
| 1 n |
A = G . | - - --------------- |
| i [ 1 + i ]^n - 1 |
*- -*
Example 1 : Let G = 100, i = 12%, n = 4 years, and A = ??
*- -*
| 1 4 |
A = 100.| ---- - ------------------ | = 135.9.
| 0.12 [ 1 + 0.12 ]^4 - 1 |
*- -*
EQUIVALENCE CALCULATIONS INVOLVING CASH FLOW
Two cash flows need to be presented along the same time period using a similar format to facilitate comparison.
When interest is earned, monetary amounts can be directly added only if they occur at the same point in time.
EQUIVALENCE BETWEEN CASH FLOWS
Equivalent cash flows are those that have the same value.
Example : Two equivalent cash flows.
Cash Flow 1 Cash Flow 2
=====================================================
P = $1000.00 P = $0.00
i = 12% i = 12%
n = 4 years n = 4 years
F = $0.00 F = $$1000*[1+0.12]^4 = $1,573.50
The equivalence can be established at any point in time. Arbitrarily setting n = 8 years, for example, gives:
For cash flow 1 : F = $1000.0 * [1 + 0.12]^8 = $2475.96 For cash flow 2 : F = $1573.5 * [1 + 0.12]^4 = $2475.96
Note : Two or more distinct cash flows are equivalent if they are equivalent to the same cash flow.
EQUIVALENCE FOR DIFFERENT INTEREST RATES
Example 1 : Given the following cash flow:
===================================================
Interest rate applicable
from previous year (t-1)
Year End Amount to current year end (t)
===================================================
0 $0.00 NA
1 $200.00 12% compounded quarterly
2 $0.00 12% compounded quarterly
3 $100.00 7% compounded annually
4 $100.00 10% compounded annually
5 $100.00 10% compounded annually
The above cash flow can be converted to its present value as follows:
Assuming the following cash flow:
===================================================
Time (Year End) Receipts Disbursements
===================================================
0 $0.00 -$1000.00
1 $0.00 -$500.00
2 $482.00 $0.00
3 $482.00 $0.00
4 $482.00 $0.00
5 $0.00 -$250.00
6 $482.00 $0.00
7 $482.00 $0.00
In this case, equivalence states that the actual interest rate earned on an investment is the one that sets the equivalent receipts to the equivalent disbursements.
For the above table, the following equality can be set:
$1000 + $500.00 (P/F,i,1) + $250(P/F,i,5) =
$482(P/A,i,3)(P/F,I,1) + $482.00 (P/A,i,2)(P/F,i,5)
By trial and error i = 10% will make the above equation valid. The equivalence can be made at any point of reference in time, it does not need to be the origin (time = zero) to produce the same answer.
If the receipts and disbursement of cash flow are equivalent for some inter= est rate, the cash flows of any equivalent portion of the investment are eq= ual at that interest rate to the negative (-) of the equivalent amount of t= he cash flows that constitute the remaining portion on the investment.
For example, break-up the above cash flow between year 4 and 5. Perform th= e equivalence at the 4th year produces the following:
-1000(F/P,10,4)-500(F/P10,3)+482(F/A,10,3) = -(-250(P/F,10,1)+482(P/A,10,2)(P/F,10,1))
-1000(1.464)-500(1.331)+482(3.310) = -$(-250(0.9091)+482(1.7355)(0.9091))
-$534 = -$534
EQUIVALENCE CALCULATIONS WITH MORE FREQUENT COMPOUNDS
Assume for example that
Interest = 10% compounded semiannually => 5% per semiannual period Payments are done semiannually for three years => 3(2) = 6 periods
The calculations from here on are the same as before.
Example 1 : Payments = 100 at year end for three years; Interest = 6% per year compounded quarterly.
i = 6/4 = 1.25%
F = $100 (F/P,1.25,8) + $100 (F/P,1.25,4) + $100
= $318.8
*- -*^l *- -*^4
| r | | 6 |
i = | 1 + --- | - 1 = | 1 + --- | - 1 = 6.14%
| m | | 4 |
*- -* *- -*
The solution of the previous example is
F = $100 (F/A,6.14,3) = $318.80
Example 2 : Assume that end of month payments = 100 with interest of 15% continuous compounding. What is the accumulated amount after 5 years ? The number of periods =12*5 = 60 years. The interest per month is 1/12= 1.25%. Then,
*- -* *- -*
| e^(rn) - 1 | | e^0.0125*60 - 1 |
F = A | ---------- | = $100.00 * | --------------- | = $8,865
| e^r -1 | | e^0.0125 - 1 |
*- -* *- -*
Usually no interest is paid for funds deposited during an interest period. In this case, funds earn interest after completing the next interest period.
BOND PRICES AND INTEREST
Suppose you can buy a bond for $900 that has a face value of $1000 with 6% annual interest that is paid semiannually. The bond matures in 7 years. The yield to maturity is defied as the rate of return on the investment for its duration. Using equivalence, the following expression can be developed:
$900 = $30 (P/A,i,14) + $1000 (P/F,i,14)
By trial and error, it can be determined that i = 3.94% per semiannual period.
The nominal rate is 2(3.94) = 7.88%.
The effective rate is 8.04%.
The bond market.
EQUIVALENCE CALCULATIONS FOR LOANS
The effective interest rate for a loan is defines as the rate that sets the= receipts equal to the disbursements on an equivalent basis.
REMAINING BALANCE OF A LOAN
Suppose a five-year loan of $10,000 (with interest of 16% compounded quarterly with quarterly payments) is to be paid off after the 13th payment. What is the balance?
The quarterly payment is
$10,000 (A/P,4,20) = $10.000 (0.0736) = $736.00
The balance can be based on the remaining payments as
$736 (P/A,4,7) = $736 * (6.0021) = $4418.
PRINCIPAL AND INTEREST PAYMENTS
Consider the case of a loan with fixed rate and constant payment A. Define the following:
I_t = Interest payment of A at time t.
B_t = Portion of payment of A to reduce balance at time t.
A = I_t + B_t for t = 1, 2, .... n
The balance at end of (t-1) is given by => A(P/A,i,n-(t-1)).
Therefore,
I_t = A(P/A,i,n-(t-1))(i)
B_t = A - I_t
= A[1-(P/A,i,n-(t-1))(i)]
Since
P/F,i,n = 1 - (P/A,i,n)(i) => B_t = A(P/F,i,n-t+1)
Example : For P = $1000.00 (loan), n = 4, and i = 15%, the following table can be constructed:
The payment is A = $1000(A/P,15,4) = $1000(0.3503) = $350.30
===============================================================
Year Loan Interest
End Payment Payment on Principal Payment
===============================================================
1 $350.30 $350.30(P/F,15,4) = $200.30 $150.00
2 $350.30 $230.32 $119.98
3 $350.30 $264.90 $85.40
4 $350.30 $304.62 $45.68
TOTAL $1401.12 $1000.14 $401.06
MEASURE OF INFLATION AND DEFLATION
The price index is the ratio between the current price of a commodity or
service to the price at some earlier reference time.
For example, the base year is 1967 (index =100),
and the commodity price is $1.46/lb.
The price in 1993 is $5.74/lb.
Therefore, the 1993 index is 5.74/1.4 = 393.2
Note : Actual Consumer price index (CPI) and annual inflation rates for 1965 are shown in Table 5.1 of the Economics Text.
ANNUAL INFLATION RATE
The annual inflation rate at t+1 can be computed as
CPI(t+1) - CPI(t)
-----------------
CPI(t)
Assume the average inflation rate = f. The average rate can be computed as
CPI(t) [ 1 + f ] ^ n = CPI(t+n)
Example : Let the CPI(1966) = 97.2 and the CPI(1980) = 246.80. The average rate of inflation over the 14 year interval is:
*- -* ^ (1/14)
| 246.80 |
f = | -------- | - 1.0 = 6.88% per year.
| 97.2 |
*- -*
PURCHASING POWER OF MONEY
Purchasing power at time t in reference to time t-n is
CPI(t-n)
--------
CPI(t)
Now let's define k = annual rate of loss in purchasing power. Therefore the average rate of loss of purchasing power is
CPI(t+n) [ 1 - k] ^ n = CPI(t)
It follows that
1
[ 1 + f ]^n = -----------
[ 1 - k ]^n
This equation relates the average f inflation rate to k, the annumal rate of loss in purchasing power.
CONSTANT DOLLARS
By definition
1
Constant dollars = ------------ (actual dollars)
[ 1 + f] ^ n
When using actual dollars, use the market interest rate (i).
When using constant dollars, use the inflation-free interest rate (i*),
defined as
For one year
1 + i
i* = ----- - 1
1 + f
For several years
1 + i
i* = ----------- - 1
[ 1 + f ]^n
CURRENCY EXCHANGE
Add notes later ......
NET CASH FLOW OF INVESTMENT OPPORTUNITIES
Payments and disbursements need to be determined. Then a net cash flow can be developed.
PRESENT-WORTH AMOUNT
It is the difference between the equivalent receipts and disbursements at the present.
Assume F_t is a cash flow at time t, the present worth (PW) is
t = n
PW(i) = sum F(t) * [ 1 + i ] ^ -t
t = 0
for any interest -1 < i < infinity.
ANNUAL EQUIVALENT AMOUNT
The annual equivalent amount is the annual equivalent receipts minus the annual equivalent disbursements of a cash flow. It is used for repeated cash flows per year.
AE (i) = PW (i) * (A/P,i,n)
*- -* *- -*
| t = n | | i(1+i)^n |
= | sum F(t) * [ 1 + i ] ^ -t | * | ------------- |
| t = 0 | | (1+i)^n - 1 |
*- -* *- -*
Example : Given the following cash flow:
=======================================
Year end Receipts Disbursements
=======================================
0 $0.00 -$1000.00
1 $400.00 $0.00
2 $900.00 -$1000.00
3 $400.00 ....
4 $900.00 -$1000.00
... ....... ....
n-2 $900.00 -$1000.00
n-1 $400.00 $0.00
n $900.00 $0.00
Therefore,
AE(10) = [-1000+400(P/F,10,1)+900(P/F,10,2)](A/P,10,2)
for 10%, or
AE(10) = [-$1000+400(0.9091)+900(0.8265)](0.5762)
= $61.93
FUTURE WORTH AMOUNT
It is the difference between the equivalent receipts and disbursements at s ome common point in the future.
t = n
FW (i) = Sum F_t * [ 1 + i ]^[ n - t ]
t = o
for any interest rate -1 < i < infinity.
PW, AE, and FW differ in the point of time used to compare the equivalent amounts.
INTERNAL RATE OF RETURN
The internal rate of return (IRR) is the interest rate that causes the equi= valent receipts of a cash flow to be equal to the equivalent disbursements = of the cash flow. Solve for i* such that
t = n
0 = PW(i*) = sum F(t) * [ 1 + i* ] ^ -t
t = 0
Example : Given the following cash flow:
=======================================
Year end Receipts Disbursements
=======================================
0 $0.00 -$1000.00
1 $0.00 -$800.00
2 $500.00 $0.00
3 $500.00 $0.00
4 $500.00 $0.00
5 $1200.00 $0.00
By trial and error i = 12.8 %.
MEANING OF IRR
It represents the rate of return on the unrecovered balance of an investment (or loan). The following equation can be developed for loans:
U_t = U_(t-1) * [ 1 + i^* ] + F_t
where
U_o = Initial amount of loan or first cost of asset (F_o),
F_t = Amount received at the end of period t.
i^* = IRR.
It should be noted that the basic equation for i^* requires the solution of the roots of a nonlinear (polynomial) function. Therefore, more than one root might exist. The following conditions results in one root (single i*):
In case of multiple IRR, other methods can be used.
PAYBACK PERIOD
t = n
Sum [ F_t > 0 ]
t = o
t = n
Sum F_t * [1 + i]^[-t] >= 0
t = o