ENGINEERING COSTS & ESTIMATION

1. ENGINEERING COSTS
Evaluating a set of feasible alternatives requires that many costs be analyzed. Examples include costs
for initial investment, new construction, facility modification, general labor, parts and materials,
inspection and quality, contractor and subcontractor labor, training, material handling, fixtures and
tooling, data management, and technical support, as well as general support costs (overhead).
In this section we discuss the several concepts for classifying and understanding these costs.

1.1. Fixed, Variable, Marginal, and Average Costs

Fixed costs are constant or unchanging regardless of the level of output or activity. In contrast,
variable costs depend on the level of output or activity. A marginal cost is the variable cost for one
more unit, while the average cost is the total cost divided by the number of units. Examples of the
different costs are discussed below.
Fixed costs-In a production environment, fixed costs include the costs for factory floor space and
equipment, which remain the same even though production quantity, number of employees, and
level of work-in-process may vary.
Variable costs-Labor costs are classified as a variable cost because they depend on the number of
employees in the factory.
Thus fixed costs are level or constant regardless of output or activity, and variable costs are changing
and related to the level of output or activity.
Decision makers use average cost to attain an overall cost picture of the investment on a per unit
basis. Marginal cost is used to decide whether the additional unit should be made, purchased, or
enrolled in.
From the above analysis we can write the single total cost equation as follows:
= + ……………1.1
The relationship between total cost, fixed cost and variable cost are shown in figure 1.1.
The fixed-cost portion of $3000 is the same across the entire range of the output variable x.
Often, the variable costs are linear (y equals a constant times x); however, the variable costs can be
nonlinear.
For example, employees are often paid at 150% of their hourly rate for overtime hours, so that
production levels requiring overtime have higher variable costs.
Total cost in Figure 1.1 is a fixed cost of Rs 3000 plus a variable cost of Rs 200 per unit for straight-
time production of up to 10 units and Rs 300 per unit for overtime production of up to 5 more units.
Figure 1.1 can also be used to illustrate marginal and average costs.
At a volume of 5 units the marginal cost is Rs 200 per unit, while at a volume of 12units the marginal
cost is Rs 300 per unit. The respective average costs are Rs 800 per unit, or (3000 + 200 x 5)/5, and Rs
467 per unit, or (3000 + 200 x 10 + 300 x 2)/12.
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Figure 1.1 − Fixed, variable and total cost

The profit equation of the firm can be expressed as follows.

= − …………………..1.2
The equations 1.1 and 1.2 allow us to illustrate a breakeven point (in terms of costs and revenue) and
regions of profit and loss for some business activity. These terms can be defined as follows.

Breakeven point: The level of business activity at which the total costs to provide the product, good,
or service are equal to the revenue (or savings) generated by providing the service. This is the level at
which one "just breaks even."
Profit region: The output level of the variable x greater than the breakeven point, where total
revenue is greater than total costs.
Loss region: The output level of the variable x less than the breakeven point, where total costs are
greater than total revenue.

Example 1.1- An entrepreneur named DK was considering the money-making potential of chartering
a bus to take people from his hometown to an event in a larger city. DK planned to provide
transportation, tickets to the event and refreshments on the bus for his customers. He gathered data
and categorized the predicted expenses as either fixed or variable.

DK's Fixed Costs DK's Variable Costs

Bus rental Rs 80 Event ticket Rs 12.50 per person

Gas expense 75 Refreshments 7.50 per person
Other fuels 20
Bus driver 50
Ticket price = Rs 35 per person.

Determine the breakeven point.

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Solution: From the above problem we can find the expression for total cost function.
Total fixed cost = 80 + 75 + 20 + 50 = Rs 225
Total Variable cost = 12.50 + 7.50 = Rs 20 per person
If we consider the no. of persons on the trip to be x, total variable cost = 20x
Therefore Total cost = 225 + 20x….. (From 1.1)
On the other hand, the total revenue of the entrepreneur is the product of the price of the ticket per
person and no. of persons on the trip.
Thus total revenue = 35x
Thus the total profit = 35x – (225 + 20x) = 15x – 225…… (From 1.2)
We can represent the total revenue and total cost functions in the following figure.

Figure 1.2 – Profit-loss breakeven chart

In the figure, the breakeven point for the number of persons on the charter trip is 15 people. For
more than 15people, DK will make a profit. If fewer than 15 sign up there will be a net loss. At the
breakeven level the total cost to provide the charter equals the revenue received from the
15passengers. We can solve for the breakeven point by setting the total costs and total revenue
expressions equal to each other and solving for the unknown value of x.

Hence, Total cost = Total revenue

Or, Rs 225 + 20x = 35x
Or, x = 15 people.

1.2. Sunk Costs

A sunk cost is money already spent as a result of a past decision. Sunk costs should be disregarded in
our engineering economic analysis because current decisions cannot change the past. For example,
dollars spent last year to purchase new production machinery is money that is sunk: the money
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allocated to purchase the production machinery has already been spent-there is nothing that can be
done now to change that action. As engineering economists we deal with present and future
opportunities.

1.3. Opportunity Costs

An opportunity cost is associated with using a resource in one activity instead of another.
Every time we use a business resource (equipment, dollars, manpower, etc.) in one activity, we give
up the opportunity to use the same resources at that time in some other activity.
Each resource that a firm owns can feasibly be used in several alternative ways.
For instance, the assembly line could produce a different product, and the parking lot could be
rented out, used as a building site, or converted into a small airstrip. Each of these alternative uses
would provide some benefit to the company.
A firm that chooses to use the resource in one way is giving up the benefits that would be derived
from using it in those other ways. The benefit that would be derived by using the resource in this
"other activity" is the opportunity cost for using it in the chosen activity. Opportunity cost may also
be considered a forgone opportunity cost because we are forgoing the benefit that could have been
realized. A formal definition of opportunity cost might be:
An opportunity cost is the benefit that is forgone by engaging a business resource in a chosen activity
instead of engaging that same resource in the forgone activity.

1.4. Recurring and Nonrecurring Costs

Recurring costs refer to any expense that is known, anticipated, and occurs at regular intervals.
Nonrecurring costs are one-of-a-kind expenses that occur at irregular intervals and thus are
sometimes difficult to plan for or anticipate from a budgeting perspective.
Annual expenses for maintenance and operation are examples of recurring expenses.
Examples of nonrecurring costs include the cost of installing a new machine, the cost of augmenting
equipment based on older technology to restore its usefulness, emergency maintenance expenses,
etc.
In engineering economic analyses recurring costs are modeled as cash flows that occur at regular
intervals. Their magnitude can be estimated, and they can be included in the overall analysis.
Nonrecurring costs can be handled easily in our analysis if we are able to anticipate their timing and
size.

1.5. Incremental Costs

One of the fundamental principles in engineering economic analysis is that in making a choice among
a set of competing alternatives, focus should be placed on the differences between those
alternatives. This is the concept of incremental costs. Thus it is the difference in costs of two
alternatives.
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1.6. Cash Costs versus Book Costs

A cash cost requires the cash transaction of dollars "out of one person's pocket" into "the pocket of
someone else."
Book costs do not require the transaction of dollars "from one pocket to another."
Rather, book costs are cost effects from past decisions that are recorded "in the books" (accounting
books) of a firm. In one common book cost, asset depreciation, the expense paid for a particular
business asset is "written off" on a company's accounting system over a number of periods.

1.7. Life-Cycle Costs

The products, goods, and services designed by engineers all progress through a life cycle very much
like the human life cycle. Like human beings, goods and services are also designed; go through a
growth phase, reach their peak during maturity, and then gradually decline and expire. The duration
of the different phases, the height of the peak at maturity, and the time of the onset of decline and
termination all vary depending on the individual product, good, or service.
Figure 1-3 illustrates the typical phases that a product, good or service progresses through over its
life cycle.

NEEDS, CONCEPTUAL DETAILED PRODUCTION OPERATIONAL DECLINE AND

ASSESSMENT OR DESIGN PHASE OR USE PHASE RETIREMENT
AND PRELIMINARY CONSTRUCTION PHASE
JUSTIFICATION DESIGN PHASE •ALLOCATION OF PHASE •OPERATIONAL USE
PHASE RESOURCES •USE BY ULTIMATE •PHASE OUT
• IMPACT ANALYSIS •DETAILED •PRODUCT, GOODS CUSTOMER •RETIREMENT
SPECIFICATIONS OR SERVICES BUILT •MAINTENANCE •RESPONSIBLE
•REQUIREMENTS •PROOF OF
CONCEPT •COMPONENT AND •ALL SUPPORTING AND SUPPORT DISPOSAL
•OVERALL SUPPLIER
FEASIBILITY •PROTOTYPE FACILITIES BUILT •PROCESS,
SELECTION MATERIALS AND
•CONCEPTUAL •DEVELOPMENT •OPERATIONAL USE
•PRODUCTION OR PLANING METHOD USE
DESIGN PLANNING AND TESTING
CONSTRUCTION •DECLINE AND
•DETAILED DESIGN PLANNING RETIREMENT
PLANNING
PLANNING

Figure 1.3 Typical life cycle for products, goods and services

Life-cycle costing refers to the concept of designing products, goods, and services with a full and
explicit recognition of the associated costs over the various phases of their life cycles. Two key
concepts in life-cycle costing are that the later design changes are made, the higher the costs, and
that decisions made early in the life cycle tend to "lock in" costs that are incurred later.
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Figure 1-4 illustrates how costs are committed early in the product life cycle-nearly 70-90% of all
costs are set during the design phases. At the same time, as the figure shows, only 10-30% of
cumulative life-cycle costs have been spent.
Figure 1-5 reinforces these concepts by illustrating that downstream product changes are more
costly and that upstream changes are easier (and less costly) to make. When planners try to save
money at an early design stage, the result is often a poor design, calling for change orders during
construction and prototype development. These changes, in turn, are more costly than working out a
better design would have been.
The figures are illustrated as follows.

Figure 1.4 − Cumulative life-cycle costs committed and money spent

Figure 1.5 − Life-cycle design change costs and ease of change.

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2. COST ESTIMATION

Engineering economic analysis focuses on the future consequences of current decisions.

Because these consequences are in the future, usually they must be estimated and cannot be known
with certainty. Examples of the estimates that may be needed in engineering economic analysis
include purchase costs, annual revenue, yearly maintenance, interest rates for investments, annual
labor and insurance costs, equipment salvage values, and tax rates.
To ensure that an analysis is a reasonable evaluation of future events, it is very important to make
careful estimates.

2.1. Types of Estimates

There are three general types of estimates with different purposes, accuracies, and underlying
methods.

a) Rough estimates

Rough estimates tend to involve little detail or accuracy. The intent is to quantify and consider the
order of magnitude of the numbers involved. These estimates require minimum resources to
develop, and their accuracy is generally -30% to +60%. They are used in a project's initial planning
and evaluation phases.

b) Semi detailed estimates

These estimates are more detailed, and they require additional time and resources to develop.
Greater sophistication is used in developing semi detailed estimates than the rough-order type, and
their accuracy is generally -15 to +20%. They are used for budgeting purposes at a project's
conceptual or preliminary design stages.

c) Detailed estimates
Detailed estimates involve the most time and resources to develop and thus are much more accurate
than rough or semi detailed estimates. The accuracy of these estimates is generally -3 to +5%. They
are used during a project's detailed design and contract bidding phases.

In considering the three types of estimate it is important to recognize that each has its unique
purpose, place, and function in a project's life. As one moves from rough to detailed design, one
moves from less to much more accurate estimates.
However, this increased accuracy requires added time and resources. Figure 1.6 illustrates the trade-
off between accuracy and cost. In engineering economic analysis, the resources spent must be
justified by the need for detail in the estimate. As an illustration, during the project feasibility stages
we would not want to use our resources (people, time, and money) to develop detailed estimates for
unfeasible alternatives that will be quickly eliminated from further consideration.
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Figure 2.1 - Accuracy versus cost trade-off in estimating.

2.2. Estimating Models

This section develops several estimating models that can be used at the rough; semi detailed, or
detailed design levels.

a) Per-unit mode l

The per-unit model uses a "per unit" factor, such as cost per square foot, to develop the estimate
desired. This is a very simplistic yet useful technique, especially for developing estimates of the
rough or order-of-magnitude type. Per unit model is commonly used in the construction industry.
Examples where per unit factors are utilized include,

Service cost per customer

Safety cost per employee
Gasoline cost per mile
Cost of defects per batch
Maintenance cost per window
Mileage cost per vehicle
Utility cost per square foot of floor space
Housing cost per student

Example 2.1- Estimate the total transportation cost for transporting 24 foreign students from the city
to the camp. The data is given as follows.
Van rental from the city to the camp (one way) is Rs 50 per 15 person van plus gas.
Camp is 50 miles away, the van gets 10 miles per gallon, and gas is Re 1 per gallon.

Solution: Two vans would be required for 24 students (as each van carries 15 passengers).
Two trips of the two vans would be required for the to and fro journey.
Therefore the total van rentals = 2 x Rs 50 x 2 = Rs 200.
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Since the camp is 50 miles away, gas required for 2 vans for the total journey would be 5 x 2 x 2 = 20
gallons.
Therefore cost of gas = 20 x Re 1 = Rs 20.
Hence the total transportation cost for the to and fro journey = Rs 200 + Rs 20 = Rs 220.

b) Segmenting Model
In this case an estimate is decomposed into its individual components, estimates are made at those
lower levels, and then the estimates are aggregated (added) back together. It is much easier to
estimate at the lower levels because they are more readily understood. This approach is common in
engineering estimating in many applications and for any level of accuracy needed.

Example 2.2- Clean Lawn Corp. a manufacturer of yard equipment is planning to introduce a new
high-end industrial-use lawn mower called the Grass Grabber. The Grass Grabber is designed as a
walk behind self-propelled mower. Clean Lawn engineers have been asked by the accounting
department to estimate the material costs that will make up the new mower. The material cost
estimate will be used, along with estimates for labor and overhead to evaluate the potential of this
new model.

The engineers decide to decompose the design specifications for the Grass Grabber into its
subcomponents, estimate the material costs for each of the subcomponents, and then sum these
costs up to obtain their overall estimate. The engineers are using a segmenting approach to build up
their estimate. After careful consideration; the engineers have divided the mower into the following
major subsystems: chassis, drive train controls, and cutting/collection system. Each of these is farther
divided as appropriate, and unit material costs were estimated at this lowest of levels.
In Example 1.2 the engineers at Clean Lawn Corp. decomposed the cost estimation problem into
logical elements. The scheme they used of decomposing cost items and numbering the material
components is known as a work breakdown structure. This technique is commonly used in
engineering cost estimating and project management of large products, processes, or projects. A
work breakdown structure decomposes a large "work package" into its constituent parts which can
then be estimated or managed individually.

c) Cost Indexes
Cost indexes are numerical values that reflect historical change in engineering (and other) costs. The
cost index numbers are dimensionless, and reflect relative price change in either individual cost
items (labor, material, utilities) or groups of costs (consumer prices, producer prices). Indexes can be
used to update historical costs with the basic ratio relationship given in Equation 2.1

= … … … … … … … … .2.1

Equation 2.1 states that the ratio of the cost index numbers at two points in time (A and B) is
equivalent to the money cost ratio of the item at the same times.
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Example 2.3- Estimate the annual labour costs for a new production facility. The following is the
labour cost data:

Labour cost index value was at 124 ten years ago and is 188 today.
Annual labor costs for a similar facility were Rs 575,500 ten years ago.

Solution: Using equation 2.1, we can get the following relation.

Thus,

Annual labour cost today Index value today

=
Annual labour cost 10 years ago Index value 10 years ago

Annual labour cost today 188

or, =
575500 124

188
or, Annual labour cost today = X 575500 = Rs 872532.257
124

d) Power sizing model

The power-sizing model is used to estimate the costs of industrial plants and equipment.
The model "scales up" or "scales down" known costs, thereby accounting for economies or
diseconomies of scale that are common in industrial plant and equipment costs.
"Economies of scale" is a long run concept and refers to reductions in unit cost as the size or capacity
of a facility or a plant increase. Diseconomies of scale is the opposite of economies of scale.
The power-sizing model uses the exponent (x), called the power-sizing exponent, to reflect
economies of scale in the size or capacity.
The power-sizing exponent (x) can be 1.0 (indicating a linear cost-versus-size/capacity relationship)
or greater than 1.0 (indicating diseconomies of scale), but it is usually less than 1.0 (indicating
economies of scale).
The relation used in the case of Power-sizing model to estimate the costs is as follows.

( )
= … … … … .2.2
( )

Where, x is the power-sizing exponent, costs of A and B are at the same point in time and size or
capacity is in the same physical units for both A and B.
In Equation 2.2 equipment costs for both A and B occur at the same point in time. This equation is
useful for scaling equipment costs but not for updating those costs. When the time of the desired
cost estimate is different from the time in which the scaling occurs (per Equation 2.2) cost indexes
accomplish the time updating. Thus, in cases like the following example involving both scaling and
updating, we use the power sizing model together with cost indexes.
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Example 2.4- Estimate the cost today of a 2500 ft2 heat exchange system for the new plant being
analyzed. The following data has been given.
Company paid Rs 50,000 for a 1000 ft2 heat exchanger 5 years ago.
Heat exchangers within this range of capacity have a power sizing exponent (x) of 0.55.
Five years ago the Heat Exchanger Cost Index (HECI) was 1306; it is 1487 today.

Solution: We first use Equation 2.2 to scale up the cost of the 1000 ft2 exchanger to one that is
2500 ft2 using the 0.55 power-sizing exponent. However it is the cost which was 5years ago.
.
Cost of 2500 ft equipment Size of 2500 ft equipment
Thus, =
Cost of 1000 ft equipment Size of 1000 ft equipment
.
2500
or, Cost of 2500 ft (5yrs ago) = X 50,000 = Rs 82763.149
1000
Now we use equation 2.1 and the HECI data to estimate the cost of a 2500 ft2 exchanger today.

Equipment cost today Index value today

Therefore, =
Equipment cost 5 years ago Index value 5 years ago
1487
or, Equipment cost today = X 82763.149 = Rs 94233.386
1306

e) Improvement and the Learning Curve

One common phenomenon observed, regardless of the task being performed, is that as the number
of repetitions increases, performance becomes faster and more accurate. This is the concept of
learning and improvement in the activities that people perform. From our own experience we all
know that our fiftieth repetition is completed in much less time than we needed to accomplish the
task the first time.
The learning curve captures the relationship between task performance and task repetition.
In general, as output doubles the-unit production time will be reduced to some fixed percentage -
the learning curve percentage or learning curve rate.
For example, it may take 300 minutes to produce the third unit in a production run involving a task
with a 95% learning time curve. In this case the sixth (2 x 3) unit will take 300(0.95) = 285 minutes to
produce.
Sometimes the learning curve is also known as the progress curve, improvement curve, experience
curve or manufacturing progress function.
The following equation gives an expression that can be used for time estimating in repetitive tasks.

= … … … . .2.3

Where, TN = time requirement for the Nth unit of production,

Tinitial = time requirement for the first (initial) unit of production,
N = number of completed units (cumulative production),
b = learning curve exponent.
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b is calculated by the following formula.

( )
= … … … 2.4
.

Example 2.5- Calculate the time required to produce the hundredth unit of a production run if the first
unit took 32.0 minutes to produce and the learning curve rate for production is 80%.

.
Solution: In this example we have N=100, T1=32.0 and b = .
= −0.3219

Using equation 2.3,

.
T = T X 100
.
or, T = 32.0 X 100
or, T = 7.27 minutes

In the context of learning curve model, we get the idea of steady state.
Steady state is the time at which the physical constraints of performing the task prevent the
achievement of any more learning or improvement.
We consider the estimation of costs by this model taking into account the concept of steady state in
the following example.

Example 2.6- Estimate the overall labor cost portion due to a task that has a learning-curve rate of
85% and reaches a steady state value after 16 units of 5.0 minutes per unit. Labor and benefits are
$22 per hour, and the task requires two skilled workers. The overall production run is 20 units.

Solution: We can calculate the time required to produce the first unit by using the time required for
.
the 16th unit. We use equation 2.3. Here we have N=16 and b = = −0.2345
.
Thus by equation 2.3,
.
T = T X 16
.
or, 5 = T X 16
or, T = 9.6 minutes
Now we use Equation 2.3 again to calculate the time requirements for each unit in the production
run as well as the total production time required.
Thus,
T = 9.6 X N .
We use this equation and calculate the time required for each unit up to 20 units.
UNIT NUMBER(N) TIME(MIN) TO PRODUCE Nth UNIT CUMULATIVE TIME FROM 1TO N
1 9.6 9.6
2 8.2 17.8
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3 7.4 24.2
4 6.9 32.1
5 6.6 38.7
6 6.3 45.0
7 6.1 51.1
8 5.9 57.0
9 5.7 62.7
10 5.6 68.3
11 5.5 74.0
12 5.4 79.2
13 5.3 84.5
14 5.2 89.7
15 5.1 94.8
16 5.0 99.8
17 5.0 104.8
18 5.0 109.8
19 5.0 114.8
20 5.0 119.8
The Total cumulative time of the production run is 119.8 minutes (2.0 hours). Thus the total labour
cost estimate would be:
2.0 hours x 22/hour per worker x 2 workers = Rs 88

3. ESTIMATING BENEFITS
Engineering economists must often also estimate benefits. Examples of benefits include sales of
products, revenues from bridge tolls and electric power sales, etc.
The cost concepts and cost estimating models can also be applied to economic benefits.
Fixed and variable benefits, recurring and nonrecurring benefits, incremental benefits, and life-cycle
benefits all have meaning. Also, issues regarding the type of estimate (rough, semi detailed, and
detailed) apply directly to estimating benefits. Per unit, segmented, and indexed models are used to
estimate benefits.
The uncertainty in benefit estimates is also typically asymmetric, with a broader limit for negative
outcomes. Benefits are more likely to be overestimated than underestimated.
One difference between cost and benefit estimation is that many costs of engineering projects occur
in the near future (for design and construction), but the benefits are further in the future. Because
benefits are often further in the future, they are more difficult to estimate accurately, and more
uncertainty is typical.