NOTES ON ENGINEERING ECONOMICS

Just as we could add 1 kg of apples to 1 kg of pears to get 2 kg of fruit, we can add $1 worth of apples to $1 worth of pears to get $2 worth of fruit. If steak is $2 per pound and a person earns $10 per hour then we are saying one hour of his time is worth 5 pounds of steak. Just as a unit for measuring mass is the kg, our unit for measuring economic worth is the dollar or the yen, peso, crown, mark or ruble.

An entity or system that has a mass of so many kg and an internal energy of so many joules, if kept in the same physical condition, will have the same mass and same internal energy tomorrow, next week, or next year. However, its economic worth in dollars, may change. Just as a change in physical state can change internal energy, it can also change economic worth. Economic worth may change in relation to economic state not involving change in physical state. Unfortunately, while the meaning or definition of kg doesn't change, the meaning of dollar does. We call this inflation or deflation.

Just as we are assuming ideal gas behavior in all problems in this course, we will assume no inflation. That doesn't mean inflation can be ignored, just that we can explore areas of economic analysis without bringing in the messy complications of inflation.

To the accountant, there is the basic principle that debts = credits. This is, in fact, a statement of the conservation principle applied to money as a conserved quantity.

Engineering economics is concerned with economic evaluations of alternatives. Suppose an urban community needs to augment its transportation system and is considering a subway system as alternative (1) or a bus system as alternative (2). Likely, the subway system will cost more to install than busses but will cost less to operate. How do we determine which is cheaper?

Consider a very simple case. You sell an acre of land for $1,000. Do you care whether the buyers pay you now or later? Even assuming the buyer is totally reliable, you should care. If you get the $1,000 now, you can invest it and have more than $1,000. Thus, in evaluating different alternatives with different costs at different points in time, we must include into our evaluation as part of the alternatives the differences in opportunity to invest or need to borrow.

If you have $1,000 now and the opportunity to invest it at 5% interest, one year from now you have $1,050. Then we might say that selling your acre for $1,000 now is economically equivalent to selling it for $1,050 one year from now. Another way of stating the same thing is to say that the present worth of $1,050 one year from now is $1,000.

A. COMPOUND INTEREST

If an amount of money P were invested today at an interest rate of 1% per annum, the amount of money available at the end of one year would be P(1 + i). After two years, it would be P(1 + i)2, etc. so that in general:
S = P (1 + i)n
where	S	=	Sum of money after N interest periods
	P	=	Principal or present amount of money
	i	=	Interest rate per interest period
	n	=	Number of interest periods

EXAMPLE

If $10,000 were invested at 4% per annum compounded quarterly (or 1% each quarter of a year), the amount of money after one quarter would be s1 = ($10,000) (1.01) = $10,100, after one year it would be S4 = ($10,000) (1.01)4 = $10,406.04 and after 10 years it would be S40 = ($10,.000) (1.01)40 = $14,888.64.

B. UNIFORM END OF YEAR PAYMENTS

If an amount of money A were invested at the end of each year for n years at an interest rate of 1 per annum, the total amount of money available could be found by adding the last investment which has drawn no interest, the next to last which drew interest for 1 year, the second to last investment which drew interest for two years, etc., on down to the initial investment which drew interest for n-1 years. Therefore, the total amount available can be written as:
S = A[1 + (1 + i) + (1 + i)2 + . . . + (1 + i)n-1]
multiply both sides by (1 + i):
S (1 + i) + A[(1 + i) + (1 + i)2 + (1 + i)3 + . . . + (1 + i)n]
Subtracting the first from the second yields
S (1 + i) - S = iS = A[(1 + i)n - 1]
or S i = A [ (1 + i)n - 1 ]
where	S	=	Sum of money after n interest periods

	A	=	Uniform investment at the end of each interest period

	i	=	Interest rate per interest period

	n 	=	Number of interest periods

EXAMPLE

If $1,000 were invested at the end of each year for 20 years with a constant interest rate of 6% per annum, the amount of money accumulated after 20 years is
S20 = $1000 [ (1.06)20 - 1 ] 0.06 = $36,786

Present Worth

If someone promises you S dollars n years from now, the present worth of this future income would be the amount of money P that would have to be invested today so that it would accumulate to S in n years. From the compound interest formula, it follows that
			P = S / (1 + i)n

		S	=	Future income after n interest periods

		P	=	Present worth

		i	=	Interest rate per interest period

		n	=	Number of interest periods
If someone promises to give you A dollars at the end of each year for n years, the present worth of this future income would be the amount of money P that would have to be invested today so that A dollars could be withdrawn at the end of each year for n years. From the uniform end of your payments formula, combined with the compound interest formula, it follows that
			P  =  A [ (1 + i)n - 1 ] / [ (1 + i)n i ]

		A	=	Uniform future payments at the end of each interest period

		P	=	Present worth

		i	=	Interest rate per interest period

		n	=	Number of interest periods

EXAMPLE

The present worth of $1,000 at the end of each year for 20 years if the interest rate stays constant at 6% per annum is
P = $1000 [ (1.06)20 - 1 ] / [ (1.06)20 (0.06) ] = $11,469.92

The author of these old notes is unknown.

Back to home page.