N. ENGINEERING ECONOMY
N1. INTRODUCTION
Engineering Economy _ study of economic theories and their applications to engineering
problems with the concept of obtaining maximum benefit at the least cost.
Time value of money is central to this study.
But first of all, what is economics? Borrowing the words of Jodie Beggs, PhD (Harvard U)
“Economics_ is the study of scarcity. Resources are limited, and every society wants to figure
out how to allocate its resources for maximum benefit. “
The field of economics serves in large part to help answer this resource allocation
question.
Economists study topics such as:






How prices and quantities of items are determined in market economies
How much value markets create for society
How taxes and regulation affect economic value
Why some goods and services are under-supplied in a market economy
How firms compete and maximize profit
How households decide what to consume, how much to save, and how much to work (or,
more generally, how people respond to incentives)
 Why some economies grow faster than others
 What effect monetary and fiscal policy has on economic well-being
 How interest rates are determined
In order to fully understand what economics is, it’s important to also understand a bit about
what economics isn’t. For example, economics and finance are related but separate fields,
and it’s not an economist’s job to tell people what stocks and bonds they should be
investing in.
It is, on the other hand, an economist’s job to understand the relationship between interest
rates and bond prices. In a similar fashion, many of the topics discussed in The Economist
deal with politics and current events and are not specifically economic-related, despite the
title of the publication.
Understanding what economists do and don’t study is important, since sometimes
economists are called on to answer questions that they are not technically qualified to
analyze, and this can have unsatisfactory results.
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1.1
Principles of Engineering Economy
Reasons for Studying Engineering Economy:
1. Engineering designs and operations must be equated with costs for practical applications
2. Engineers evolve into managers of their own or other enterprises
Uses of Engineering Economy
1. Application on various fields of engineering
2. Determining of limiting factors
3. Tool in selection of alternates
4. Investment of capital
5. Tool in decision making
Course Objectives
After completing this course, the student must be able to:
1. Solve problems involving interest and the time value of money;
2. Evaluate project alternatives by applying engineering economic principles and methods and select
the most economically efficient one; and
3. Deal with risk and uncertainty in project outcomes by applying the basic economic decision making
concepts.
1.2 Engineering Economic Decisions
What do engineers play within a firm? What specific tasks are assigned to the engineering staff,
and what tools and techniques are available to it to improve a firm’s profits. Engineers are called upon
to participate in a variety of decision making processes, ranging from manufacturing, through
marketing decisions.
The term engineering economic decision refers to all investment decisions relating to engineering
projects. The facet of an economic decision of most interest from an engineer’s point of view is the
evaluation of costs and benefits associated with making a capital investment.
The five main types of engineering economic decisions are
1. Equipment and process selection
2. Equipment replacement
3. New product and product expansion
4. Cost reduction
5. Service improvement
The factors of time and uncertainty are the defining aspects of any investment project
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1.3
Engineering Economy and Design Process
An engineering economy study is accomplished using a structured procedure and
mathematical modeling techniques. The economic results are then used in a decision situation that
involves two or more alternatives and normally includes other engineering knowledge and input.
Engineering Economic Analysis Procedure
Engineering Design Process
1. Problem
recognition,
evaluation
1. Problem/ need definition
2. Problem/ need formulation and evaluation
definition
and
2. Development of feasible alternatives
3. Synthesis of possible solutions (alternatives)
3. Development of the outcomes and cash flows
for each alternative.
4. Analysis, optimization, and evaluation
4. Selection of criterion (or criteria)
5. Analysis and comparison of alternatives
6. Selection of the preferred alternatives.
5. Specification of preferred alternative
7. Performance monitoring and post-evaluation of
results
6. Communication
The design process
Needs definition
Recognition
of a problem
to be solved
Problem Formulation
Possible Solution
Analysis
Specification
(Preferred
Alternative)
Communication
Completely
specified
solution
Approaches to Design Problems
The idea of design and development is what distinguishes engineering from science, which
concerns itself principally with understanding the world as it is. Decisions made during engineering design
phase of a product’s development determine the majority of the costs of manufacturing that product. As
design and manufacturing processes become more complex, the engineer, increasingly, will be called upon
to make decisions that involve money.
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1.4 Cost Concepts for Decision Making
The term cost estimating is frequently used to describe the process by which the present
and future cost consequences of engineering designs are forecast. A primary difficulty in
estimating for economic analyses is that most prospective projects are relatively unique; that is,
substantially similar design efforts have not been previously undertaken to meet the same
functional requirements and economic constraints. Hence, accurate past data that can be used
in estimating costs and benefits directly, without substantial modification, often does not exists.
It may be possible, however, to develop data on certain past design outcomes that are related to
the outcomes being estimated and to adjust that data based on design requirements and expected
future conditions.
Following are some of the basic uses of cost estimation
1. Providing information used in setting selling price of quoting, bidding or evaluating
contracts,
2. Determining whether a proposed product can be made and distributed at a profit,
3. Evaluating how much capital can be justified for process changes or other
improvements, and
4. Establishing benchmarks for productivity improvement programs.

Six-tenths-factor rule_ If a cost of given unit at one capacity is known, the cost of a similar
unit with X times the capacity of the first is approximately (X)0.6 times the cost of the initial
unit.
0.6

 CapA 

CostA  Cost B 
 CapB 
Consumer price index (CPI)_ measures changes in the price level of
consumer goods and services purchased by households. The CPI is a
statistical estimate constructed using the prices of a sample of representative
items whose prices are collected periodically
 Index A 

CostA  Cost B 
Index
B


 Combination of Price Cost Index and Sixteenth Rule:
 Index A  CapA 


CostA  Cost B 
 Index B  CapB 
0.6
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N.2 SELECTION AND EQUIVALENCE IN PRESENT ECONOMY
Present economy involves the analysis of problems for manufacturing a product or rendering a service upon
the basis of present or immediate costs. It is highlighted when the effects of time such as interest and depreciation are
negligible. Present economy is employed when the alternatives to be compared will provide the same result and the
period involved in the study is relatively short.
When alternatives for accomplishing a specific task are being compared over one year or least-and influence
of time on money can be ignored, engineering economic analyses are referred to as present economy studies.
Present economy studies occur in the following situations:
a. Selection of material_ In many cases, economic selection among materials cannot be based solely on the costs of
materials. Frequently, a change in materials will affect the design and processing costs, and shipping costs may
also be altered.
b. Selection of method to be used_ In mechanical or chemical operations a product may be made by two or more
methods giving equivalent results. Some goods may be delivered by various methods such as using different
capacity trucks, and the results would still be the same regardless of the truck used. These are but a few of the
examples that may be cited to show that certain operations are capable of being done by two or more methods.
c. Selection of design_ In the design of a machine to produce a certain product, the engineer responsible for the work
will usually make as many designs as possible and from which, by a process of elimination, he will select the
design best fitted for the work to be done with particular care being given to the one which will do the work with
most economy.
d. Selection location or site for a project_ In the choice of a factory site many factors are often considered such as the
cost of the land, the cost of construction in the different sites, and the difference in transportation cost, and many
other factors.
e. Comparison of proficiency among workers_ In industrial operations where the efficiency of the workers is a factor
affecting costs, it is usually observed that workers have varying efficiencies. In some occupations only those with
better average proficiencies are acceptable. In teaching for example, other factors being equal, preference should
be given to those teachers who did better than average in their studies.
f. Economy of tool and equipment maintenance_ In many activities, tools have to be sharpened from time to time,
and equipment have to be kept in good operating condition all the time. In certain cases, experience will show the
best time to perform certain operations to maintain equipment at the optimum operating efficiency.
g. Economy of number of laborers_ In certain industrial operations it is observed that a certain number of workers
cooperating on a certain phase of work will lead to the highest efficiency. An increase beyond this number will
usually cause the taking into effect of the law of diminishing returns. In certain cases, the excess of laborers will
result in some laborers not working at certain times while waiting for the work of other laborers to be finished.
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SAMPLE PROBLEMS:
1.
The ore of a gold mine contains an average of 14 grams of gold per ton. One method of processing costs
P1,000 per ton and recovers 93% of the gold, while another method costs only P900 per ton and recovers 81%
of the gold. If gold can be sold at P90/gram, which method is better and by how much?
Solution: Basis 1-ton ore
First Method
Second Method
Recovered Gold
14 (0.93) = 13.02 gm
14 (0.81) = 11.34 gm
Sales from gold
(13.02 gm) (P90/gm) = P 1,171.80
(11.34 gm) (P90/gm) = P1,020.60
Cost
P 1000
P 900
Net receipt
P 171.80
P120.60
Answer: First method of processing is better by P51.20/ton ore! (171.80-120.60)
2.
A certain masonry dam requires 200,000 m3 gravel for its construction. The contractor found two possible
sources for the gravel with following data: Determine which of the two sites will give lesser cost
Data:
Source A
Source B
Distance, gravel pit to dam
3.0 km
1.2 km
Gravel Cost @ pit, P/m3
None
P100.00
Purchase price of pit
P 8,000,000
None
Cost of Road construction
P 4,500,000
None
Overburden costs
P3,700,000
Hauling Costs, P/km/m3
P40.00
P40.00
Solution:
Purchase price of pit
Gravel Cost @ pit, P/m3
Cost of Road
construction
Overburden costs
Hauling Costs, P/km/m3
Source A
P 8,000,000
P 24,000,000
(1.2) (40) (200,000)
None
(3)(40)(200,000)
Total Costs
Source B is cheaper by P 3,200,000
3.
P 4,500,000
Source B
None
P100.00 (200,000 m3) P20,000,000
None
P 36,500,000
(36,500,000-33,300,000)
P 3,700,000
P 9,600,000
P 33,300,000
Two workers, A and B, produce the same product on identical machines. Worker A receives P50/h and he
produces 100 units/h. Worker B is able to produce 120 units/h. The machine cost of operation used by them is P
250/h.
a. Determine the cost per piece for worker A
b. Determine the hourly wage of worker B in order that his cost per piece will equal that of A.
Solution:
a. Cost per piece for Worker A
P 50  P 250 P3.00
P / piece 

100 units
unit
b.
Cost per piece for Worker B = Cost per piece for Worker A
Let X = hourly wage of worker B
X  P 250 P3.00

120 units
unit
X 
P 110.00
h
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4.
The monthly demand for ice cream cans being manufactured by a company is 3,200 pieces. With a manually
operated guillotine, the unit cutting cost is P25.00. An electrically operated hydraulic guillotine was offered to
the company at a price of P 300,000 which will cut by 30% the unit cutting cost. Disregarding the cost of money,
how many months will the company be able to recover the cost of the machine if he decides to buy now?
Solution:
Manually operated machine
Monthly Cutting Cost 253,200  P80,000
Electrically Operated Hydraulic Guillotine
Monthly Cutting Cost  253,2001  0.3  P56,000
Savings  80,000  56,000  P24,000 / month
P 300,000
 12.5 months
P 24,000/month
A company has been selling a soap containing 30% by weight water at a price of P150 per kilo f.o.b. (i.e., freight
on board, which means the laundry pays the freight charges). The company offers an equally effective soap
containing 5% water. The water content is to no important to the laundry, and it is willing to accept the soap
containing 5% water if the delivered costs are equivalent. If the freight rate is P0.70 per kilo, how much should
the company charge the laundry per kilo f.o.b. for the soap containing 5% water?
Number of months to recover 
5.
Solution:
For Soap containing 30% water, the Cost delivered per kg of 100% dry soap:
P 150  0.70 P 215.29

1  0.3 kg
kg
For Soap containing 5% water, the Cost per kg FOB = X
 X  0.70  P215.29
1  0.05 kg
kg
X
6.
P 203.82
kg
Powdered coal having a heating value of 10,000 BTU per pound is to be compared with bunker fuel priced at
P26.80/liter having a heating value of 18,600 BTU/lb. If the efficiency of conversion of fuel to power is 83%
for coal and 88% for bunker, with all other costs being equal, what is the allowable selling price for coal per
metric ton? Specific gravity of bunker is 0.96.
Solution:
Using bunker, the cost per BTU net of losses:
 1  P 773.50
lb
 P 26.80  l  kg 





 6
 l
 0.96 kg  2.205 lb  18.600 BTU  0.88  10 BTU
Solving for equivalent cost of coal per metric ton = Z
 mt 
lb

Z 
 2,205 lb  10,000 BTU
P 14,156
Z
mt
 1  P 773.50

 6
 0.83  10 BTU
7. Worker A can clean the garage in 3 hours, but it takes Worker B 4 hours to do the same job.
How long would it take them to clean the garage if they work together?
Solution: let t= time required to finish cleaning if A and B work together.
1 1 1
 
A B t
1 1 1
7
  ; t h
3 4 t
12
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N.2 PROBLEMS
1. A group from Bacolod is planning a six-day trip to Dumaguete. For transportation, the group will rent a car from
either the school motor pool or a local car dealer. The school charges P26 per kilometer, has no daily fee, and the
motor pool pays for the gas. The car dealer charges P2500 per day and P14 per km, but the group must pay for the
gas. The car’s fuel rating is 10 km per liter, and the price of the gas is estimated to be P28 per liter.
(a) At what point, km is the cost of both options equal? (nearest value)
a. 717 km
b. 1200 km
c. 1448 km
d. 1,630 km
e. 1,913 km
(b) The car dealer has offered a special student discount and will give the students 100 free km per day.
What is the new break-even point?
a. 717 km
b. 1200 km
c. 1448 km
d. 1,630 km
e.1,913 km
2. One method for developing a mine containing an estimated 100,000 tons of ore will result in the recovery of 62%
of the available ore deposit and will cost USD 23 per ton of material removed. A second method of development
will recover only 50% of the ore deposit, but it will cost only USD 15 per ton of material removed. Subsequent
processing of the removed ore recovers 300 pounds of metal from each ton and costs USD 40 per ton of ore
processed. The recovered metal can be sold for USD 0.80 per pound. What is maximum profit that could be
expected if method A is used to develop the mine?
a. USD10.974 M b. USD9.25 M
c. USD7.50M
d. USD6.75M
e. USD2.68M
3. What is the increase in profit if instead of Method A, Method B is used? (Pls refer to the preceding problem)
a. $1.724 M
b.(-) $1.724. M
c. (-)$1.55M
d. $1.55M
e. USD2.68M
4. Two currently owned machines are being considered for the production of a part. The capital investment
associated with the machine is about the same and can be ignored. The important differences between the
machines are their production capacities (production rate x available production hours) and their rejects
(percentage of parts produced that cannot be sold) Consider the following table:
Machine A
Machine B
Production Rate
100 parts/h
130parts/h
Hours Available for Production
7 h/day
6 h/day
Percent Parts Rejected
3%
10%
The material cost is P6 per part and all defect-free parts produced can be sold for P12 each (rejected parts have
negligible scrap value). For either machine, the operator is P25 per hour and the variable overhead rate for traceable
costs is cost P5 per hour. Assume that the daily demand for this part is large enough that all defect-free parts can be
sold. What is the maximum profit per day using machine A?
a. P3,738
b. P6,763
c. P5,670
d. P4,520
e. P2,350
5. What is the increase in profit if Machine B is used instead of Machine A?
a. P338
b. P174
c. (-P174)
d. P420
e. (-P420)
6. A machine part to be machined may be made either from an alloy of aluminum or steel. There is an order for 8000
units. Steel costs P380/kg, while aluminum costs P870/kg. If steel is used, the steel per unit weighs 110 grams; for
aluminum, 30 grams. When steel is used, 50 units can be produced per hour; for aluminum, 80 units per hour with the
aid of a tool costing P64,000, which will be useless after 8,000 units are finished. The cost of the machine and the
operator is P1080 per hour. If all other costs are identical, determine which material will be more economical.
a. Steel cheaper by P15.80 unit
b. Aluminum cheaper by P15.50
c. None of these
7. The ore of a gold mine contains an average of 14 grams of gold per ton. One method of processing costs P1,000 per
ton and recovers 93% of the gold, while another method costs only P900 per ton and recovers 81% of the gold. If gold
can be sold at P90/gram, which method is better and by how much?
a. 1st method is better by P51.20/ton ore
b. 2nd method is better by P51.20/ton ore
8
8. A certain masonry dam requires 200,000 m3 gravel for its construction. The contractor found two possible sources
for the gravel with following data: Determine which of the two sites will give lesser cost.
Data:
Source A
Source B
Distance, gravel pit to dam
3.0 km
1.2 km
Gravel Cost @ pit, P/m3
None
P100.00
Purchase price of pit
P 8,000,000
None
Cost of Road construction
P 4,500,000
None
Overburden costs
P3,700,000
Hauling Costs, P/km/m3
P40.00
P40.00
a. A is cheaper by P3,200,000
b. Source B is cheaper by P3,200,000
c. None of these
9. A company manufactures 1 million units of a product annually. A new design of the product will reduce material
cost by 12%, but will increase processing cost by 2%. If materials cost is P 12//unit and processing will cost P4/unit,
how much can the company afford to pay for the preparation of the new design and making changes in equipment?
a. P 1.36 M
b. P800,00
c. P600,000
d. P 505,000
10. In the design of an automobile radiator, an engineer has a choice of using either a brass-copper alloy casting (A) or
plastic molding (B). Either material provides the same service. However, the brass-copper alloy casting weighs 25
pounds, compared with 20 pounds for the plastic molding. Every pound of extra weight in the automobile has been
assigned a penalty of P600 to account for increased fuel consumption during the life cycle of the car. The brasscopper alloy costs P335 per pound, whereas the plastic molding costs P740 per pound. Machining costs per casting
are P600 for the brass-copper alloy. Which material should the engineer select, and what is the savings in unit costs.
a. A, P2,825
b. B, P2,825
c. A, P2,225
d. B, P2,225
11. A machine used for cutting materials in a factory has the following outputs per hour at various speeds and requires
periodic tool regrinding at intervals cited.
Speed
Output per hour
Tool regrinding
A
200 pieces
Every 8 h
B
250 pieces
Every 7 h
C
280 pieces
Every 5 h
A set of tools costs P4200 and can be ground 20 times. Each regrinding and change costs P180 and the
time needed to regrind and change tools is 1 h. The machine operator is paid P50/h, including the time the tool is
changed. The tool grinder who also sets the tools to the machine is paid P45/h. The hourly rate chargeable against
the machine is P150/h, regardless of speed.
Which speed should the engineer select for minimum costs?
a. A
12.
b. B
c. C
Workers A and B can finish a work in 2 days. Workers A and C can finish the same job in 3
days. Workers B and can finish the same job in 4 days. In how many days will the same job
be finished if workers A, B and C work together?
a. 0.5 d
b. 4.5 d
c. 30/24 d
d. 13/24 d
How many days will it take for each of worker A, B and C to finish the job if they work
individually?
a. 24/7 d
b. 13/24 d
c. 5/24 d
d. 24 d
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