3. eng. cost and estimation

3: Engineering Cost and Estimation

Dr. Mohsin Siddique

Assistant Professor

[email protected]

Ext: 29431

Date: 23/09/2014

Engineering Economics

University of SharjahDept. of Civil and Env. Engg.

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Part I

Outcome of Today’s Lecture

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� 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

Engineering Costs

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� 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

Engineering Costs

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� 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

Example 2-1

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Example 2-1

7� Total Cost= Total Fixed cost + Total Variable cost

Engineering Costs

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� 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

Engineering Costs (Example 2-1 continued)

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� 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, FSSewill be in loss

Engineering Costs

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� 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 +

Total Variable cost

Engineering Costs

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� 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.

Engineering Costs

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� 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

Engineering Costs

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� 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). _________________________

� ____________________________

� ____________________________

Engineering Costs

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� 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

Engineering Costs

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� 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.

Example 2-2

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Example 2-2

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Engineering Costs

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� 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.

Engineering Costs

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� 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 ??

� _________________________________________________

� _________________________________________________

� _________________________________________________

� _________________________________________________

� _________________________________________________

Example 2-3

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Engineering Costs

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� 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.

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Part II

Date: 25/09/2014

Outcome of Today’s Lecture

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� 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

Cost Estimating

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� 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

Cost Estimating

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� 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%

Cost Estimating

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� 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.

Estimating Models

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� 1. Per unit model

� 2. Segmenting model

� 3. Cost indexes

� 4. Power-sizing model

� 5. Triangulation

� 6. Improvement and learning curve

Estimating Models

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� 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

� ________________________________________________

� ________________________________________________

� ________________________________________________

Estimating Models

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� 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.

Estimating Models

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� 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)

Example 2-7

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� 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

Example 2-7

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� Solution: Miriam will use cost index equation for cost estimates for annual labor and material cost

871800$124

188755005cost today Annual

ago years 10 eIndex valu

todayeIndex valuago years 10cost Annualcost today Annual

ago years 10 eIndex valu

todayeIndex valu

ago years 10cost Annual

cost today Annual

==

=

=

3227000$544

7152455000cost today Annual

ago years 3 eIndex valu

todayeIndex valu

ago years 3cost Annual

cost today Annual

==

=

Annual Labor Cost

Annual Material Cost

Estimating Models

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� 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.

Example 2-8

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� 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

Example 2-8

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� 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

148782800cost today Annual

ago years 5 eIndex valu

todayeIndex valu

ago years 5cost Equipment

cost todayEquipment

==

=

Problem

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� 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 the VMIC machine and wants to list its PSE value in his next issue. Given the following data calculate the PSE that Padre should publish.

�

� Cost of VMIC-200 today= $100,000

� Cost of VMIC-100 10 years ago=$45,000

� VMIC equipment index today= 856

� VMIC equipment index 10 years ago=604

Problem

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� Solution:

� Using Cost Index Model

� Cost of VMIC – 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

Estimating Models

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� 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.

Estimating Models

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� 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.

Estimating Models

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� 6. Improvement and the learning curve

� The following expression can be used for time estimating in repetitive tasks

Estimating Models

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� 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

timeSteady state

Example

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� 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%.

Problem

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� 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%

Estimating benefits

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� Same concepts of cost can be applied to estimate benefits

Cash Flow: Estimation and Diagramming

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� 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.

Cash Flow Diagram (CFD)

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� 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.

Cash Flow Diagram (CFD)

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� 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

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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

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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

Cash Flow Diagram

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� 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.

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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

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Thank You

Feel Free to Contact

[email protected]

Tel. +971 6 5050943 (Ext. 2943)