Session 6B – Evaluation & Development Phases
Engineering Economy In Brief
In industrial engineering, it is often necessary to compare among alternative designs,
procedures, plans and methods. Since the available alternative courses of action involve
different amounts of investment and different operating costs and benefits, the inevitable
question is “Will it pay?” The method of solution requires us to express each alternative in
some common form and then choose the best, taking both the monetary and intangible factors
It answers one of 2 questions:
(1) Which of the choices considered is best from a financial point of view? (i.e., which
equipment offers the required service at lowest cost?), or
(2) What is the expected return on investment (ROI) in using this equipment? (i.e., if I
were to purchase this equipment, would I make a satisfactory return for the money
In most computations, a minimum attractive rate of return (MARR) is used. This is the
smallest interest rate at which one is willing to invest money. It represents the interest rate a
company expects to make on its investments.
The interest rate is the rent charged on the money lent for a defined period of time.
Under simple interest, the interest owed upon repayment of a loan is proportional to the length
of time the principal sum has been borrowed.
Whenever the interest charged for any interest period is based on the remaining principal
amount plus any accumulated interest charged up to the beginning of that period, the interest is
said to be compounded. [Put simply, the interest earned is added to the principal, and the total
(principal + interest) is carried forward to the following period to continue to earn interest.]
Compound interest is much more common in practice than simple interest, and the interest
obtained due to the effect of compounding is higher than that obtained from simple interest
calculation. The difference would be much greater for a larger amount of money, a higher
interest rate, or a greater number of interest periods.
IED VE Notes Session 6B – Evaluation & Development Phases [WCH Dec/11] Page 1
Consider the following variables in an engineering economy problem:
P = present value (or principal sum);
F = future value;
I = interest owed;
i = interest rate (in decimals) per year;
n= no. of interest periods
To calculate simple interest,
I = Pin, or F = P (1 + in).
To calculate compound interest,
F = P (1+ i) n, or P = F / (1+ i) n.
Another variable commonly used is the uniform annual amount A, which represents a constant
value, A, occurring at the end of each interest period over a continuous number of interest
Time Value Of Money
As money can earn a certain interest through its investment for a period of time, a dollar
received at some future date is not worth as much as a dollar in hand at present. The
relationship between interest and time leads to the concept of the time value of money.
[The concept of time value of money is related directly to the concept of opportunity cost. If
instead of receiving a certain sum right now, and you had to wait until the end of a whole year
for it, you would have lost the interest that you could have earned.]
The Future Value of Present Money
F = P (1+ i) n,
The Present Value of Future Money
P = F / (1+ i) n or F (1+ i) – n.
IED VE Notes Session 6B – Evaluation & Development Phases [WCH Dec/11] Page 2
(1+ i)n is also given the factor notation: (F/P, i %, n) and 1 / (1+ i)n given the notation
(P/F, i %, n), where the top letter of the ratio is the sum of money you want to find (i.e.,
P), given the bottom sum (F), i is the interest rate, and n the number of periods.
F = P (F/P, i %, n) means “the calculated amount, F, at the point in time it occurs (i.e., n
periods from when P occurs), is equivalent to the known value of P at the point in time it
occurs, for the given interest rate, i %.” Tables of different values of i and n are available
for engineering economy calculations.
In engineering economy, 2 things are said to be equivalent when they have the same
effect. When 2 or more alternatives are to be compared, their relative merits are often not
directly apparent from a simple statement of their future receipts and disbursements. To
compare, these amounts must first be placed on an equivalent basis, for example, at the
same point in time. When interest is earned, monetary amounts can be directly added
only if they occur at the same point in time.
The factors involved in the equivalence of sums of money are:
- the amounts of the sums;
- the times of occurrence of the sums; and
- the interest rate.
In the interest formulas, P occurs at the start of an interest period, and F and A payments
occur at the end of interest periods. Tables of interest factors are available for practically
Cash Flow Diagrams
A cash flow diagram is a means of visualizing and simplifying the flow of receipts and
disbursements for the acquisition and operation of items in a company.
The horizontal line is a time scale with progression of time moving from left to right, with the
interest periods or years written below the intervals of time. It is marked off in equal
increments, one per period, up to the duration of the project.
IED VE Notes Session 6B – Evaluation & Development Phases [WCH Dec/11] Page 3
The arrows represent cash flows. Arrows may go in opposing directions as cash flows may
represent either costs or incomes. Costs, expenditures, payments and disbursements (negative
cash flows or cash outflows) will be shown as downward pointing arrows. Benefits, revenues,
incomes or receipts (positive cash flows or cash inflows) are shown as upward pointing
arrows. Arrow lengths are approximately proportional to the magnitude of the cash flow. An
arrow head is normally added.
The “end-of-year” convention is used, where all disbursements and receipts (i.e., cash flows)
are assumed to take place at the end of the year in which they occur.
Expenses incurred before time = 0 are sunk costs and are not relevant to the problem.
However, salvage value is income, so it is drawn as positive.
The net cash flow is the arithmetic sum of receipts (+) and disbursements (–) that occur at the
same point in time.
It is important to draw the cash flow diagram for any proposal as it does not only give us a
pictorial view of the cash flows, it also allows us to write the cash flow equation (with
functional notations) more easily.
A = $14,667/- F = $44,560/-
1 2 3 4 5 years
P = $123,545/-
Converting everything to time 0 is called present worth (PW) or present value (PV) analysis:
Thus, PW or PV = – $123,545 + $14,667 (P/A, i %, 5) + $44,560 (P/F, i %, 5)
Once we know the interest rate, it is a matter of checking the interest factor tables, substitute
the factors into the equation, and calculate the PW.
IED VE Notes Session 6B – Evaluation & Development Phases [WCH Dec/11] Page 4
Basic Methods for Making Engineering Economy Studies
All engineering economy studies of capital projects should be made so as to include
consideration of the return that a given project will or should produce. As the patterns of
capital investment, revenue or saving flows and cost flows can be quite different in various
projects, there is no single method for making engineering economy studies that is ideal for all
There are 2 commonly used methods for comparing the relative financial merits of competing
alternatives in engineering economy problems:
(1) Payback period method;
(2) Present Value (PV) method.
Other methods are discussed in the IED module “Engineering Economy”.
The alternative selected will be that which offers the shortest payback period, the highest net
return, or is of the lowest net cost.
Payback Period Method
This method is concerned primarily with the period of time required to recover an investment.
This is useful when there are different rates of investment and annual cash flow spread over
the period. The payback period is obtained when the cumulative cash flow curve intersects the
x-axis. The alternative that returns the initial investment in the shortest time period is
Mathematically, payback period = net investment / net annual cash flow after taxes.
The merit of an investment may be judged by comparing the pay-back period with the
estimated life of the equipment.
This is the most popular method in general use. However, it will provide the wrong answer as
this method does not consider:
- the returns after payback (i.e., the economic life & the salvage value);
- the time value of money during the economic life of the equipment.
IED VE Notes Session 6B – Evaluation & Development Phases [WCH Dec/11] Page 5
The Present Value (PV) Method
The concept of PV discounts the future cash flows arising from an investment at a
predetermined standard rate of interest. This gives the present value which is then compared to
the amount being invested. To be realistically profitable, the PV must be greater than the sum
invested. The rate considered is usually the cost of capital.
The criterion is that as long as the present worth of the net cash flows is equal to or greater
than zero, the project is economically justified.
There must be a common analysis period when comparing alternatives. It is incorrect to
compare the PV of cost of Pump A, say, which is expected to last 6 years, with the PV of cost
of Pump B, expected to last 12 years. In this example, the assumption would be that Pump A
will be replaced by another identical Pump A at the end of 6 years. This gives a common
analysis period of 12 years.
In situations like this, restructure the problem so there is a common analysis period. This is
easy when the different lives of the alternatives have a practical least common multiple. When
this is not the case, some assumptions must be made to select a suitable common analysis
period. Or, just don’t use the PV method.
IED VE Notes Session 6B – Evaluation & Development Phases [WCH Dec/11] Page 6