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:
Figure 1 : Physical and Economic Components of an Engineering System
Figure 1 shows how engineering is composed of physical and economic components.
system output(s) Physical (efficiency ) = ------------------- system input(s)
System output(s) Economic (efficiency ) = ----------------- System input(s)
system worth Economic (efficiency ) = ----------------- system cost
Economics deals with the behavior of people, and as such, economic concepts are usually qualitative in nature, and not universal in application.
ECONOMY OF ORGANIZATION
CLASSIFICATION OF COST
A key objective in engineering applications is the satisfaction of human needs, which will nearly always imply a cost.
INTEREST RATE
TIME VALUE OF MONEY
The time-value of money is the relationship between interest and time. i.e.
Figure 2 : Time-Value of Money
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
Economy of Material Selection
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.
Figure 3 : Typical Cash Flow Diagrams
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.
P = Principal i = Interest rate n = Number of years (or periods) I = Interestt.
Interest is due at the end of the time period. For fractions of a time period, multiply the interest by the fraction.
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
Example : A loan of $1,000 is made at an interest of 12% for 5 years. The interest is due at the end of each year with the principal is due at the end of the fifth year. The following table shows the resulting payment schedule:
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
Example : A loan of $1,000 is made at an interest of 12% for 5 years. The principal and interest are due at the end of the fifth year. The following table shows the resulting payment schedule:
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)
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
Figure 4 : Cash Flow for Single Payment Compound Amount
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.
Figure 5 : Schematic of Equal Payment-Series Compount Amounts
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.
Figure 6 : Schematic of Equal-Payment-Series Capital Recovery
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
================================================================ 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)
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
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
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 | *- -* *- -*
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)]
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
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
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
In case of multiple IRR, other methods can be used.
t = n Sum [ F_t > 0 ] t = o
t = n Sum F_t * [1 + i]^[-t] >= 0 t = o
Written by Mark Austin, David Lovell and Bilal Ayyub
Last Modified February 1, 1998
Copyright © 1996-1998, Department of Civil Engineering, University of Maryland