This document provides an overview of engineering cost estimation. It defines various types of engineering cost estimates such as rough, semi-detailed, and detailed estimates. It discusses common difficulties in making cost estimates such as one-of-a-kind estimates and limitations of time and resources. The document also describes several common mathematical models used for cost estimating, including the per unit model, segmenting model, cost indexes, power-sizing model, and triangulation. It provides examples of how to use these models to estimate costs. Finally, it discusses the impact of learning curves on cost estimates over time.

1. 3: Engineering Cost and Estimation
Dr. Mohsin Siddique
Assistant Professor
msiddique@sharjah.ac.ae
Ext: 29431
Date: 23/09/2014
Engineering Economics
University of Sharjah
Dept. of Civil and Env. Engg.

2. 2
Part I

3. Outcome of Today’s Lecture
3
After completing this lecture…
The students should be able to:
Define various cost concepts
Provide specific examples of how and why these engineering
cost concepts are important

4. Engineering Costs
4
Fixed Cost: These constant or unchanging regardless of the level
of output or activity.
Example: In production environment, costs for factory floor space and
equipment remains the same regardless of production quantity, number of
employee and level of work in progress.
Variable Costs: These are not constant and depends in level of out
or activity.
Example: Labor costs are variable cost because it depend on number of
employees and number of hours they work
Marginal Costs: It is the variable cost for one more unit.
Average Costs: It is total cost divided by number of units

5. Engineering Costs
5
Let’s take example of two student enrolled in engineering one is full time
while other is a part time. student A full time student can enroll in 12-18
credit hours for a fixed fee of $1800. Overload credits are charged at
$120/credit
Part-time student: It is case of variable cost. He has to pay according to
number of credits he choose.
Full-time student: It is the case of fixed cost, regardless of number of
credit hours he has to pay tuition fee.
Fixed cost: $1800 per semester for 12-18 credits
Avg. cost if student chooses 12 credits: 1800/12= $150/credit
Avg. cost if student chooses 18 credits: 1800/18= $ 100/credit
Variable cost: It will cost of addition credit beyond 18 and if the student
chooses to enroll in 21 credits.
Variable cost for addition credit beyond17: $0
Variable cost for addition credit beyond18: $120

6. Example 2-1
6

7. Example 2-1
7
Total Cost= Total Fixed cost + TotalVariable cost

8. Engineering Costs
8
Profit-loss breakeven chart. It
is plot of revenue against costs for
various level of outputs (activity).
It allows to understand breakpoint
and region of profit and loss.
Terminologies:
Break-even point: the level of
activity at which total cost of
product/goods/services equals to
revenue
Profit region: values of variable x
greater than break point where
total revenue is greater than total
cost
Loss region:Values of variable x
less than breakpoint where total
revenue is less than total cost
Profit loss breakeven
chart figure here

9. Engineering Costs (Example 2-1 continued)
9
Profit-loss breakeven chart. It
is plot of revenue against costs for
various level of outputs (activity).
It allows to understand breakpoint
and region of profit and loss.
Terminologies:
Breakeven point: 15 students
Profit region: for more than 15,
FSSe will be in profit
Loss region: For less than 15, FSSe
will be in loss

10. Engineering Costs
10
Linear and non-linear cost
relation
For example:
Employees are often paid 150% of
their hourly rate for overtime. In
this case fixed cost is €3000 while
variable cost
Up to 10 units is €200 per unit
and
For 5 more units €300 per unit
Avg. cost at 5 units = (3000+200x5)/5= € 800
Avg. cost at 10 units = (3000+200x )/10=
Avg. cost at 12 units = (3000+200x10+300x2)/12= € 467
Avg. cost at 15 units = (3000+200x10+300x5)/15=
Total Cost= Total Fixed cost +
TotalVariable cost

11. Engineering Costs
11
Problem:
A private owned summer camp for youngsters has the following data for a
12 week session:
Charges per camper: $120 per week
Fixed costs: $48000
Variable cost per camper: $80 per week
Capacity: 200 campers
(a). Develop the mathematical relationship for total cost and total revenue
(b).What is the total number of campers that will allow the camp to just
break even
(c).What is profit or loss for the 12 week session if the camp operates at
80% capacity.

12. Engineering Costs
12
x = number of campers
(a) Total Cost = Fixed Cost +Variable Cost
= $48,000 + $80 (12) x
Total Revenue = $120 (12) x
(b) Break-even when Total Cost = Total Revenue
$48,000 + $960 x = $1,440 x
$4,800 = $480 x
x = 100 campers to break-even
(c) capacity is 200 campers
80% of capacity is 160 campers
@ 160 campers x = 160
Total Cost = $48,000 + $80 (12) (160) = $201,600
Total Revenue = $120 (12) (160) = $230,400
Profit = Revenue – Cost = $230,400 - $201,600 = $28,800

13. Engineering Costs
13
Sunk Cost:
A sunk cost is money already spent as a
result of past decision.
e.g., (i) If 5 students signed up for the
engineering course than in example 2-1,
the advertising cost would be sunk cost
(ii). Price of two years old pc purchase
at $2000 is sunk cost which has no
influence on current market value of
$400 of pc
(iii). _________________________
____________________________
____________________________

14. Engineering Costs
14
Opportunity cost: An opportunity cost is associated with using a
resource in one activity instead of another.
Every time we use business resource in one activity we, we give up
the opportunity to use the same resource at that time in some
other activity.
“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”
Example

15. Engineering Costs
15
Opportunity cost:
Examples:
Suppose that friends invite a college student to travel through
Europe over the summer break.
Cost analysis: In considering the offer, student computes the total
cost of travel as $ 3000 for 10 weeks and after checking his bank
account he agrees to go with them.
However, true cost to the student includes not only his out-of pocket
cash costs but also opportunity costs. By taking the trip, the student
is giving up the opportunity to earn $ 5000 as a summer intern at a
local business.
Thus the student total cost will comprise the $3000 cash cost as
well as $5000 opportunity cost.

16. Example 2-2
16

17. Example 2-2
17

18. Engineering Costs
18
Recurring and Nonrecurring costs
Recurring costs refers to any expense that is know and anticipated
and that occurs at regular interval.
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.
________________________________________________
________________________________________________
________________________________________________
In engineering economics analysis, recurring costs are modeled as
cash flow that occur at regular interval. Nonrecurring costs can also
be handled if we are able to anticipate their timing and size.

19. Engineering Costs
19
Incremental costs
One of fundamental principle in engineering economics analysis is that in
choosing between competing alternatives, the focus is one difference
between those alternative.
For example:You may be interested in comparing two car lease
options. In both of the cases (say case A and case B) total car prices
after lease time are same.
What else would you like analyze to make a decision ??
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________
_________________________________________________

20. Example 2-3
20

21. Engineering Costs
21
A cash cost is a cash transaction, or cash flow. If a company
purchases an asset, it realizes a cash cost.
A book cost is not a cash flow, but it is an accounting entry
that represents some change in value.
When a company records a depreciation charge of $4 million
in a tax year, no money changes hands. However, the company
is saying in effect that the market value of its physical,
depreciable assets has decreased by $4 million during the year.
Life-cycle costs refer to costs that occur over the various
phases of a product or service life cycle, from needs
assessment through design, production, and operation to
decline and retirement.

22. 22
Part II
Date: 25/09/2014

23. Outcome of Today’s Lecture
23
After completing this lecture…
The students should be able to:
Define engineering estimating
Explain the three types of engineering estimates, as well as common
difficulties encountered in making engineering cost estimates
Use several common mathematical estimating models in cost estimating
Discuss the impact of learning curve on cost estimates
State the relationship between cost estimating and estimating project
benefits
Draw cash flow diagrams to show project costs and benefits

24. Cost Estimating
24
Engineering economic analysis focuses on future consequences of
present decisions and therefore one must estimate all cost related
variables
Estimates for engineering economic analysis include purchase costs,
annual revenue, yearly maintenance, interest rates for investments,
annual labor and insurance costs and tax rates etc
Types of Estimates
1. Rough Estimate
2. Semi-detailed estimate
3. Detailed estimate

25. Cost Estimating
25
Types of Estimates
1. Rough Estimates: These are order of magnitude estimate used for
high level planning for determining project feasibility and in a project
initial planning and evaluation phase.
These estimates require minimum resources to develop and their
accuracy is -30 to 60%.
2. Semi-detailed estimates: These are used for budgeting purpose at
a project’s conceptual or preliminary design stage.These are more
detailed than rough but still require time and resources.
Their accuracy is generally -15 to 20%
3. Detailed estimates: These are used during project detailed design
and contract bidding phases.These involve most time and resources and
thus are much more accurate than rough estimates.
The accuracy is general -3 to 5%

26. Cost Estimating
26
Difficulties in Estimation
One of a kind estimates: The first time something is done, it is difficult
to estimate costs required to design, produce, and maintain a product over
its life cycle.
E.g., NASA first space mission, ___________________________
__________________________________________________
Time and effort available: Our ability to develop engineering estimates is
constrained by the time and man-power availability
Estimator Expertise:The past is our greatest teacher and knowledge is
power.
The more experienced and knowledgeable this estimator is, the more
accurate the estimate will be.

27. Estimating Models
27
1. Per unit model
2. Segmenting model
3. Cost indexes
4. Power-sizing model
5. Triangulation
6. Improvement and learning curve

28. Estimating Models
28
1. Per-unit model: It is a simple but useful model in which a cost
estimate is made for a single unit, then the total cost estimate
results from multiplying the estimated cost per unit times the
number of units.
It is commonly used in construction industry.
Example:
Cost of $65 per square meter
Service cost per customer
Gasoline cost per km
________________________________________________
________________________________________________
________________________________________________

29. Estimating Models
29
2. Segmenting model: It can be described as Divide and Conquer
It partitions the total estimation task into segments. Each segment is
estimated, then the segment estimates are combined for the total
cost estimate.

30. Estimating Models
30
3. Cost indexes: It can be used to account for historical changes in
costs.The widely reported Consumer Price Index (CPI) is an
example.
Cost index data are available from a variety of sources.
Suppose A is a time point in the past and B is the current time. Let
IVA denote the index value at time A and IVB denote the current
index value for the cost estimate of interest.To estimate the current
cost based on the cost at time A, use the equation:
(Cost at time B)/ (Cost at time A) = (IVB / IVA)
Cost at time B = (Cost at time A) (IVB / IVA)

31. Example 2-7
31
Miriam is interested in estimating the annual labor and material
costs for a new production facility. She was able to obtain the
following labor and material cost data:
Labor costs:
Labor cost index value was at 124 ten years ago and is 188 today
Annual labor costs for a similar facility were $575,500 ten years ago
Material Costs
Material cost index value was at 544 three years ago and is 715 today
Annual material costs for a similar facility were $ 2,455,000 three year ago

32. Example 2-7
32
Solution: Miriam will use cost index equation for cost estimates for
annual labor and material cost
871800$
124
188
755005cost todayAnnual
agoyears10eIndex valu
todayeIndex valu
agoyears10costAnnualcost todayAnnual
agoyears10eIndex valu
todayeIndex valu
agoyears10costAnnual
cost todayAnnual
==
=
=
3227000$
544
715
2455000cost todayAnnual
agoyears3eIndex valu
todayeIndex valu
agoyears3costAnnual
cost todayAnnual
==
=
Annual
Labor
Cost
Annual
Material
Cost

33. Estimating Models
33
4. Power-sizing model: It accounts explicitly for economies of
scale.
For example, the cost of constructing a six-story building will
typically be less than double the construction cost of a comparable
three-story building.
To estimate the cost of B based on the cost of comparable item A,
use the equation
Cost of B = (Cost of A) [ ("Size" of B) / ("Size" of A) ] x
where x is the appropriate power-sizing exponent, available from a
variety of sources.
An economy of scale is indicated by an exponent less than 1.0.An
exponent of 1.0 indicates no economy of scale, and an exponent
greater than 1.0 indicates a diseconomy of scale.
"Size" is used here in a general sense to indicate physical size,
capacity, or some other appropriate comparison unit.

34. Example 2-8
34
Based on her work in example 2-7, Miriam has been asked to estimate the
today’s cost of a 2500ft2 heat exchange system for the new plant being
analyzed. She has the following data
Her company paid $50000 for a 1000ft2 heat exchanger 5 years ago
Heat exchangers within this range of capacity has a power sizing
exponent (x) of 0.55
Five years ago the Heat Exchanger Cost Index (HECI) was 1306; it is
1487 today
Solution:
Miriam will use the following equation of power sizing model;
Cost of 2500ft2 = (Cost of 1000ft2) [ ("Size" of 2500ft2) / ("Size" of 1000ft2) ] x
where x =0.55

35. Example 2-8
35
Cost of 2500ft2 equipment = (50000) [ (2500) / (1000) ] 0.55
Cost of 2500ft2 equipment =$82,800
Miriam knows that the $82,800 reflects only the scaling up of the
cost of the 1000 ft2 model to 2500 ft2 model. Now she will use
following equation of cost index model to determine the today’s
cost of equipment
300,94$
1306
1487
82800cost todayAnnual
agoyears5eIndex valu
todayeIndex valu
agoyears5costEquipment
cost todayEquipment
==
=

36. Problem
36
Padre works for a trade magazine that publishes lists of Power-Sizing
Exponent (PSE) that reflects economies of scale for developing engineering
estimates of various types of equipment. Padre has been unable to find any
published data on theVMIC machine and wants to list its PSE value in his
next issue. Given the following data calculate the PSE that Padre should
publish.
Cost ofVMIC-200 today= $100,000
Cost ofVMIC-100 10 years ago=$45,000
VMIC equipment index today= 856
VMIC equipment index 10 years ago=604

37. Problem
37
Solution:
Using Cost Index Model
Cost ofVMIC – 50 today = 45,000 (856/604) = $63,775
Using Power Sizing Model:
(63,775/100,000) = (50/100)x
log (0.63775) = x log (0.50)
x = 0.65

38. Estimating Models
38
5.Triangulation is used in engineering surveying. In this technique
surveyor is able to map points by using three fixed points and
horizontal distances.
Triangulation in cost estimating might involve using different sources
of data or using different quantitative models to arrive at the value
being estimated.
As decision makers we should approach our economic estimate
from different perspectives because such varied perspectives add
richness, confidence, and quality to the estimate.

39. Estimating Models
39
6. 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.
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 first
production run involving a task with a 95% learning time curve. In this case
the sixth (2 x 3) unit during 2nd run will take 300(0.95) = 285 minutes to
produce.

40. Estimating Models
40
6. Improvement and the learning curve
The following expression can be used for time estimating in
repetitive tasks

41. Estimating Models
41
6. Improvement and the learning curve
When thousands or even millions
of units are being produced, the
learning curve effect is
ignored/vanished at a time/stage
called 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.
Number of units
time
Steady state

42. Example
42
6. Calculate the time to required to produce the hundredth unit of a
production run if the first unit took 32 minute to produce and the
learning curve rate for production is 80%.

43. Problem
43
If 200 labor hours were required to produce the 1st unit in a production
run and 60 labor hours were required to produce the 7th unit, what was
the learning curve rate during production.
SOLUTION
TN=Tintial Nb ==> T(7) = T(1) x 7b
60 = (200) x 7b
0.300 = 7b
log 0.30= b log (7)
b = log (0.30)/log (7) = -0.62
b is defined as log (learning curve rate)/ log 20
b = [log (learning curve rate)/lob 2.0] = -0.62
log (learning curve rate) = -0.187
learning curve rate = 10(-0.187) = .650 = 65%

44. Estimating benefits
44
Same concepts of cost can be applied to estimate benefits

45. Cash Flow: Estimation and Diagramming
45
Cash flow
Cash inflows are the receipts,
revenues, incomes and saving
generated by project and business
activity. A plus sign indicates a cash
inflow
Cash outflows are costs, operation
and maintenance costs, disbursements,
expenses and taxes caused by projects
and business activity. A negative or
minus sign indicates a cash outflow.

46. Cash Flow Diagram (CFD)
46
Cash flow diagrams visually represent income and expenses over some
time interval.
It is graphical representation of cash flows drawn on the y-axis and a time
scale along x-axis.

47. Cash Flow Diagram (CFD)
47
Categories of Cash Flows
The expenses and receipts due to engineering projects usually fall into one
of the following categories:
First cost: expense to build or to buy and install
Operations and maintenance (O&M): annual expense, such as electricity,
labor, and minor repairs
Salvage value: receipt at project termination for sale or transfer of the
equipment (can be a salvage cost)
Revenues: annual receipts due to sale of products or services
Overhaul: major capital expenditure that occurs during the asset’s life
Revenue

48. 48
Drawing a 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

49. 49
Drawing a Cash Flow Diagram
In a cash flow diagram (CFD) the end of period t is the same as
the beginning of period (t+1)
Beginning of period cash flows are: rent, lease, and insurance
payments
End-of-period cash flows are: O&M, salvages, revenues, overhauls
The choice of time 0 is arbitrary. It can be when a project is
analyzed, when funding is approved, or when construction begins
One person’s cash outflow (represented as a negative value) is
another person’s inflow (represented as a positive value)
It is better to show two or more cash flows occurring in the
same year individually so that there is a clear connection from
the problem statement to each cash flow in the diagram

50. Cash Flow Diagram
50
For example, consider a truck that is going to be purchased for $55,000. It
will cost $9,500 each year to operate including fuel and maintenance. It will
need to have its engine rebuilt in 6 years for a cost of $22,000 and it will be
sold at year 9 for $6,000. Here is the cash flow diagram:
Note that it is customary to take cash flows during a year at the end of the
year, or EOY (end-of-year).There are certain cash flows for which this is
not appropriate and must be handled differently.

51. 51
Cash Flow Diagram
Example: 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)
$1,000
0
1 2
$580
$540

52. 52
ThankYou
Feel Free to Contact
msiddique@sharjah.ac.ae
Tel. +971 6 5050943 (Ext. 2943)