Introduction to Engineering Economics
From the organization’s point of view, efficient and effective
functioning of the organization would certainly help it to
provide goods/services at a lower cost which in turn will
enable it to fix a lower price for its goods or services.
Engineering economics deals with the methods that enable
one to take economic decisions towards minimizing costs
and/or maximizing benefits to business organizations.
It deals with the concepts and techniques of analysis useful in
evaluating the worth of systems, products, and services in
relation to their costs.
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Role of Engineers in the Industry
Why do engineers care about engineering economics?
Engineering designs are intended to produce good results.
They are accompanied by undesirables (costs).
If outcomes are evaluated in dollars, and “good” is defined as profit,
then decisions will be guided by engineering economics.
This process maximizes goodness only if all outcomes are
anticipated and can be monetized.
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Engineering Decision Making
Role of Engineers:
Manufacturing Profit
− Estimating a
Required investment. Planning
− Forecasting a
product demand.
− Estimating a selling
price.
− Estimating a
manufacturing cost.
− Estimating a product
life. Investment Marketing
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Accounting Vs Engineering Economy
Evaluating past performance Evaluating and predicting future events
Accounting Engineering Economy
Past Present Future
Types of Strategic Engineering Economic Decisions in
Manufacturing Sector:
Service Improvement.
Equipment and Process Selection.
Equipment Replacement.
New Product and Product Expansion.
Cost Reduction.
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Time Value of Money
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 back your original $100 plus $9.
Original amount to be returned = $100
Interest to be returned = $100 x .09 = $9
The “value” of money depends on the amount and when it is received
or spent.
Example: What amount must be paid to settle a current debt of $1000 in two
years at an interest rate of 8% ?
• Solution: $1000 (1 + 0.08) (1 + 0.08) = $1166
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Types of Costs
• There are usually two types of costs/Benefits associated with an
engineering project, one-time costs, which include first costs
and salvage costs, and annual costs (or benefits) that occur
every year or several years of the project.
One time costs:
First Costs or Initial Costs are the costs necessary to implement a
project, including: Costs of new equipment, Costs of shipping and
installation, Costs of renovations needed to install equipment, Cost
of engineering, Cost of permits, licenses, etc.
Salvage value is the money that can be obtained at the end of the
project by selling equipment. Salvage value is a benefit rather than
a cost.
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Types of Costs
Annual Costs/Benefits:
Direct operating costs such as labor, supervision,
supplies, maintenance, material, electricity, fuel, etc.
Indirect operating costs sometimes included, such as a
portion of building rent, a portion of secretarial
expenses, etc.
Depreciation of equipment.
Finally, need to include savings or profits from the
project and tax.
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Cash Flow Diagram
The graphic presentation of the costs and benefits over the
time is called the cash flow diagram.
It is a presentation of what costs have to be incurred and
what benefits are received at all points in time.
The costs and benefits of engineering projects over time are
summarized on a cash flow diagram (CFD).
Specifically, CFD illustrates the size, sign, and timing of
individual cash flows, and forms the basis for engineering
economic analysis.
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Cash Flow Diagram
A CFD is created by first drawing a segmented time-based horizontal line,
divided into appropriate time unit. Each time when there is a cash flow, a
vertical arrow is added, pointing down for costs and up for revenues or
benefits. The cost flows are drawn to relative scale.
Horizontal axis = time;
vertical axis = costs and benefits.
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Cash Flow Diagram
P-Pattern “Present”
1 2 3 n
F-Pattern “Future”
1 2 3 n
A-Pattern “Annual”
1 2 3 n
G-Pattern “Gradient”
1 2 3 n
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Example 1: A man borrowed $1,000 from a bank at 8% interest. Two end-
of-year payments
– At the end of the first year, he will repay half of the $1000 principal
plus the interest that is due.
– At the end of the second year, he will repay the remaining half plus the
interest for the second year. Cash flow for this problem is:
End of year Cash flow
0 +$1000
1 -$580 (-$500 - $80)
2 -$540 (-$500 - $40)
• q1
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Interest and their applications
Interest rate:
Is the rental value of money. It represents the growth of capital per
unit period. The period may be a month, a quarter, semiannual or
a year.
An interest rate 15% compounded annually means that for every
hundred dollar invested now, an amount of $15 will be added to
the account at the end of the first year. So, the total amount at the
end of the first year will be $ 115.
At the end of the second year, again 15 % of $ 115, i.e. $ 17.25
will be added to the account. Hence the total amount at the end
of the second year will be $ 132.25. The process will continue
thus till the specified number of years.
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Interest formulas:
Simple interest: the interest is calculated, based on the initial
deposit for every interest period. In this case, calculation of interest on
interest is not applicable.
Compound interest: the interest for the current period is computed
based on the amount (principal plus interest up to the end of the
previous period) at the beginning of the current period.
The notations which are used in various interest formulae are as
follows:
P = principal amount (Initial amount).
n = No. of interest periods.
i = interest rate
F= future amount at the end of year n.
A = equal amount deposited at the end of every interest period.
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Single-Payment Compound Amount:
Here, the objective is to find the single future sum (F) of the initial
payment (P) made at time 0 after n periods at an interest rate i
compounded every period. The formula to obtain the single-payment compound
amount is:
F= P(1 + i)n = P(F| P, i, n)
Where:
P = principal amount (Initial amount).
n = No. of interest periods.
i = interest rate
F= future amount at the end of year n.
A = equal amount deposited at the end of every interest period.
(F|P, i, n) is called as single-payment compound amount factor.
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Example: A person deposits a sum of $ 20,000 at the interest
rate of 18% compounded annually for 10 years. Find the
maturity value after 10 years.
Solution: P= $20,000, i = 18% compounded annually, n= 10
years
F= P(1 + i)n = P(F|P, i, n)
= 20,000 (F|P, 18%, 10)
= 20,000 * 5.234 = $ 104,680
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Single-Payment Present Worth Amount:
Here, the objective is to find the present worth amount (P)
of a single future sum (F) which will be received after n
periods at an interest rate of i compounded at the end of
every interest period.
Example:
A person wishes to have a future sum of $100,000 for his son’s
education after 10 years from now. What is the single-payment that he
should deposit now so that he gets the desired amount after 10 years?
The bank gives 15% interest rate compounded annually.
Ans: $24,720
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Equal-payment series future worth amount:
The objective is to find the future worth of n equal payments which are
made at the end of every interest period till the end of the n-th interest
period.
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Example:
A person who is now 36 years old is planning for his retired
life. He plans to invest an equal sum of $10,000 at the end
of every year for the next 24 years starting from the end of
the next year. The bank gives 20% interest rate,
compounded annually. Find the maturity value of his
account when he is 60 years old.
Ans: $ 3,262,368
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Equal-payment series sinking fund:
In this type of investment mode, the objective is to find the equivalent
amount (A) that should be deposited at the end of every interest period
for n interest periods to realize a future sum (F) at the end of the nth
interest period at an interest rate of i.
Example: A company has to replace a present facility after 15 years at an outlay of
$500,000. It plans to deposit an equal amount at the end of every year for the next 15
years at an interest rate of 18% compounded annually. Find the equivalent amount that
must be deposited at the end of every year for the next 15 years.
Ans: $ 8,200, The annual equal amount which must be deposited for 15 years is $ 8,200.
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Equal-payment series present worth amount :
The objective of this mode of investment is to find the present worth of
an equal payment made at the end of every interest period for n interest
periods at an interest rate of i compounded at the end of every interest
period.
Example: A company wants to set up a reserve which will help the
company to have an annual equivalent amount of $1,000,000 for the next
20 years towards its employees welfare measures. The reserve is assumed
to grow at the rate of 15% annually. Find the single-payment that must be
made now as the reserve amount. Ans: $ 6,259,300
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Equal-Payment Series Capital Recovery Amount:
The objective of this mode of investment is to find the annual equivalent
amount (A) which is to be recovered at the end of every interest period
for n interest periods for a loan (P) which is sanctioned now at an interest
rate of i compounded at the end of every interest period.
Example: A bank gives a loan to a company to purchase an equipment worth
$ 1,000,000 at an interest rate of 18% compounded annually. This amount
should be repaid in 15 yearly equal installments. Find the installment
amount that the company has to pay to the bank. Ans: $196,400
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