CVE 4000Engineering Economy and Planning

Summer 2015Lecture #3

Engineering Costs and EstimatingMay 27, 2015

Dr. Troy Nguyen, P.E.Link Building, Room 234tnguyen@fit.edu

Homework #1 Question #1 Ethics Issue

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Problem 1-34 in textbookYou are working for a private engineering company. You are being asked to sign a documents verifying information that you believe it is not true.You like your job and your co-workers in the company and your family depends on your income.What criteria can you use to guide your decision regarding this issue?

Is it legal? (legality criterion) Is it fair? (fairness and equity criterion) What harmful effects it might have on other people (human safety &

health criteria) What harmful effects it might have on the environment (long-term human

safety & health criteria) Will you be able to live with your decision for the rest of your life

(conscience criteria)

Types of Costs - Review

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Fixed Costs & Variable Costs Marginal Costs & Average Costs Sunk Costs & Opportunity Costs Recurring & Non-recurring Costs Incremental Costs Cash Costs & Book Costs Life-Cycle Costs

Cost Estimating Models

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Per-Unit Model (Unit Technique) Segmenting Model Cost Indexes Power-Sizing Model Triangulation Improvement and the Learning Curve

Per-Unit Model

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Construction cost per square foot (building) Capital cost of power plant per kW of capacity Revenue / Maintenance Cost per mile (of highway) Utility cost per square foot of floor space Fuel cost per kWh generated Revenue per customer served

Notes: Per-unit model does not take into account the economy of scale, i.e. no savings

for larger projects Per unit model intended to provide ball-park of costs and benefits

Example - Cost Estimating using Per-Unit Model (1)

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Evaluation of 4-acre waterfront property for rental condominium development Front lot: 2 acres Rear lot: 2 acres 32-unit condominium building on each parcel

Initial Costs: Lot purchase prices: $400,000/acre front lot, $200,000/acre rear lot Legal fees, applications, permits, etc.: $80,000 Site clearing and preparation: $3000/acre Paving roadways, parking, curbs, and sidewalks: 25% of total lot at

$40,000/acre Construction costs: $3,000,000 per building

Rear2 acres

Front2 acresWhat data do we have?

Example - Cost Estimating using Per-Unit Model (2)

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Recurring Costs Taxes and insurance: $5000/month per building Landscaping: 25% of lot at $1000/acre/month Security: $1000/building + $1500/month Other costs: $2000/month

Revenue (assume 90% annual occupancy) Front lot units: $2500/unit/month Rear lot units: $1750/unit/month Other revenue: $5000/month

Estimate1. The total initial cost, annual cost, and annual revenue 2. Estimate profitability over 10 years with following assumptions

Ignoring the time value of money 90% occupancy

Example - Cost Estimating using Per-Unit Model (3)

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Solution:

Total Initial Cost Purchase price: ($400,000/ac 2 ac)+($200,000/ac2 ac) = $1,000,000 Legal costs = $ 80,000 Site clearing & preparation: $3000/ac4 ac = $12,000 Roadways, etc.: (0.254 ac)$40,000/ac = $40,000 Construction: $3,000,000/ building2 buildings = $6,000,000

$7,132,000Annual Cost Taxes and insurance: $5000/m/bldg2 bldg x 12 m = $120,000 Landscaping: (0.254 ac)$1000/ac12 m = $12,000 Security: ($1000/bldg2 bldg)+ ($1500/m12 m) = $20,000 Other costs: $2000/m12 m = $24,000

$176,000

ac = acrebldg = buildingm = month

Example - Cost Estimating using Per-Unit Model (4)

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Solution:

Annual Revenue Front lot leases: (32 units$2500/unit/m12 m).90 = $864,000 Rear lot leases: (32 units$1750/unit/m12 m).90 = $604,800 Other revenue: $5000/m12 m = $60,000

$1,528,000/yr

Net Prot= [$1,528,000/yr10 yr][$7,132,000 + ($176,000/yr10 yr)]= $6,388,000

Cost Estimating Models Segmenting Model

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Divide and Conquer Approach Estimate is decomposed into individual components Estimates are made at component level Individual estimates are aggregated back together

Smaller segments are simpler to estimate For example, it is easier to estimate the cost of

building a Boeing 787 by breaking the plane into to small subsystems

Example - Cost Estimating using Segmenting Model (1)Cost estimate of lawn mower

Cost Item EstimateB.1 Engine $38.50B.2 Starter assembly 5.90B.3 Transmission 5.45B.4 Drive disc assembly 10.00B.5 Clutch linkage 5.15B.6 Belt assemblies 7.70Subtotal $72.70

A. Chassis

B. Drive Train

Cost Item EstimateA.1 Deck $7.40A.2 Wheels 10.20A.3 Axles 4.85Subtotal $22.45

Example - Cost Estimating using Segmenting Model (2)

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Cost estimate of lawn mower (continued)

Cost Item EstimateC.1 Handle assembly $3.85C.2 Engine linkage 8.55C.3 Blade linkage 4.70C.4 Speed control linkage 21.50C.5 Drive control assembly 6.70C.6 Cutting height adjuster 7.40Subtotal $52.70

C. Controls D. Cutting/Collection system

Cost Item EstimateD.1 Blade assembly $10.80D.2 Side chute 7.05D.3 Grass bag &

adapter7.75

Subtotal $25.60

Total material cost = $22.45 + $72.70 + $52.70 + $25.60 = $173.45

Work Breakdown Structure (from cost segmenting model) Logical components (A.1, A.2, A.3,

B.1, etc.) from previous example

Known as Work Breakdown Structure (WBS)

WBS decomposes large work package into smaller, more manageable parts

Homework #1 Question #3 Costs

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Problem 2-34 in textbookA pump in your facility has failed. The facility will be completely replaced in 3 years. Here is the scenario A brass pump costs $6,000 and will last three years An used stainless steel has been in storage. If we were to use to this pump, it

should an extra 3 years This stainless steel pump cost $13,000 new Right now, this stainless steel pump is worth $7,000 according to your accountant It will require an extra cost of $500 to reconfigure for use You have a buyer who will buy the pump as is for $4,000

1. What is the book cost of the stainless steel pump?Answer: $7,000, why?

2. What is the opportunity cost of the stainless steel?Answer: $4,000, why?

3. How much cheaper or more expensive to use the stainless steel pump instead of the brass pump?Answer: Brass pump costs $6,000, Stainless steel pump costs $4,000 + $500 so stainless steel is $1,500 cheaper

4. Bonus question: What type of cost is the $13,000 that you paid for the stainless steel pump?Answer: Sunk cost

Cost Estimating Models Cost Indexes

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Example - Consumer price index (CPI) measures changes in the price level of a market of consumer goods and services purchased by households

(Eq. 2-2)

Cost indexes reflect historical change in cost Cost index could be individual cost items (labor, material, utilities), or group of

costs (consumer prices, producer prices) Indexes can be used to update historical costs

=

http://en.wikipedia.org/wiki/Consumer_price_index

Example - Cost Estimating using Cost Indexes

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= 10 10 = $575,500 188124 = $871,800 = 3 3 = $2,455,000 715544 = $3,227,000

You are asked to estimate the annual labor and material costs for a new production facility. Following data are available to youLabor costs

Labor cost index value was 124 ten years ago and is 188 today Annual labor costs for similar facility were $575,500 ten years ago

Material costs Material cost index value was 544 three years ago and is 715 today Annual material costs for similar facility were $2,455,000 three years ago

=

Cost Estimating Models Power Sizing Model

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Used to estimate costs of industrial plants & equipment Scales up or scales down known costs

(Eq. 2-3)

=

= Power-sizing exponent

Notes: Power-Sizing model takes into account the economy of scale, i.e. savings for

larger size or capacity

Power-Sizing Exponent Values Industrial Equipment

Equipment/Facility xBlower, centrifugal 0.59Compressor 0.32Crystallizer, vacuum 0.37Dryer, drum 0.40Fan, centrifugal 1.17

Equipment/Facility xFilter, vacuum 0.48Lagoon, aerated 1.13Motor 0.69Reactor 0.56Tank, Horizontal 0.57

Notes: X is usually less than 1the larger the size the more savings What if X > 1?

Exponent Values for Different Types of Industrial Facilities and Equipment

Power-Sizing Model - Example

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A. First, consider power-sizing index change. Apply eq. 2-3

25002 = 10002 2500210002 0.55 = $50,000 2500210002 0.55 = $82,800B. Second, considering cost index change. Apply eq. 2-2

= 5 5 = $82,800 14871306 = $94,300

You are asked to estimate the cost today of a 2500-ft2 heat exchanger for a new facility. Following data are available to you

Your company paid $50,000 for a 1000-ft2 heat exchanger 5 years ago Heat exchangers within this capacity range have power-sizing exponent (x) of 0.55 Five years ago the Heat Exchanger Cost Index (HECI) was 1306; it is 1487 today

Cost Estimating Models Triangulation

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Triangulation techniques used in surveying Map points of interest by using three fixed points

and horizontal angular distance

Triangulation technique used in target identification Use signals from multiple sources to estimate

location of a target

Application in Economic Analysis Use multiple sources of data or multiple quantitative

models to estimate costs and benefits

www.directionsmag.com

Triangular Method in Target ID

Triangular Method in Surveying

www.e-education.psu.edu

Cost Estimating Models Improvement and Learning Curve (1)

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Learning Phenomenon: As the number of repetitions increase, performance of people becomes faster and more accurate.

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 This called the learning curve percentage or

learning curve ratewww.intropsych.com

S-Shaped Learning Curve

Example: It takes 60 minutes to produce the 5th unit in a production run. The production involves a task with 90% learning curve. How much time is required to produce the 10th unit?

In this case, the 10th unit doubles the output so it will take (60 minutes)*(0.90) = 54 minutes

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Let T1 = Time to perform the 1st unitTN = Time to perform the Nth unitb = Learning-curve exponentN = Number of completed units

= 1 = log( )log 2

Improvement and Learning Curve (2)

Estimating time for repetitive tasks

T1 = 32.0 min.N = 100Learning-curve rate = 80%

= log(0.80)log 2 = 0.3219100 = 1 100 = 32.0 1000.3219 = 7.27.

Improvement and Learning Curve (3)

Example: Calculate the time required to product 100th unit of a production run 1st unit took 32.0 minutes to produce Learning-curve rate for production is 80%

Solution: Use equations on previous slide: = 1 = log( )log 2

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Solution:Known T16 = 100 hrs.

Learning-curve rate = 85%

= log(0.85)log 2 = 0.234516 = 1 16 = 1001 = 100160.2345 = 191.6 . = 1 = 191.6 0.2345

Improvement and Learning Curve (4)Example: You are asked to estimate the manufacturing labor costs to produce twenty (20) 2-MW wind turbines for a new wind farm. Following data are available to you

Learning-curve rate for labor is 85% Steady-state manufacturing will be reached with the 16th unit Steady-state production rate per unit is 100 hours, with 15 workers per unit Average labor rate with benefits is $25 per hour

Plot time vs. number of production units using equation

= 1 = 191.6 0.2345N TN1 191.62 162.83 148.14 138.45 131.46 125.97 121.48 117.69 114.410 111.7

N TN11 109.212 107.013 105.014 103.215 101.516 100.017 100.018 100.019 100.020 100.0

Improvement and Learning Curve (5)

We will do the rest of the solution as a homework problem

Cash Flow Diagrams (CFD)

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CFD summarize costs & benefits occur over time

CFD illustrates the size, sign, and timing of individual cash flows

CFD in engineering economic analysis is analogous to Free Body Diagram in an engineering statics analysis

Components of CFD A segmented time-based horizontal line,

divided into time units A vertical arrow representing a cash flow is

added at the time it occurs Arrow pointing down for costs and up for

benefits

CFD

FBD

Cash Flow Diagrams (CFD) Example

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Timing of Cash Flow Size of Cash FlowAt time zero (now) Positive $1001 time period from today Negative $1002 time periods from today Positive $1003 time periods from today Negative $1504 time periods from today Negative $1505 time periods from today Positive $50

40 1 2 3 5

$100

$100

$100

$150 $150

+

-

$50

Categories of Cash Flows

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First cost: expenses 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 Revenues: annual receipts due to sale of products or services Overhaul: major capital expenditure that occurs during the

assets life

Drawing a Cash Flow Diagram

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CFD shows when all cash flows occur In a CFD, the end of period t is the same

time as the beginning of period t+1 Rent, lease, and insurance payments are

usually treated as beginning-of-period cash flows

O&M, salvage, revenues, and overhauls are assumed to be end-of-period cash flows

The choice of time 0 is arbitrary, i.e. we can draw a CFD for a project that takes place in the future

t t+1

Drawing Cash Flow Diagrams with Spreadsheet

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YearCapital Costs O&M Overhaul

0 -$80,000

1 $(12,000)

2 $(12,000)

3 $(12,000) $(25,000)

4 $(12,000)

5 $(12,000)

6 $ 10,000 $(12,000) $(90,000)$(80,000)$(70,000)$(60,000)$(50,000)$(40,000)$(30,000)$(20,000)$(10,000)

$-$10,000$20,000

0 1 2 3 4 5 6

Year

Cash

Flo

ws

Capital Costs O&M Overhaul

Interest & Equivalent (Chapter 3 in Textbook)

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Outline Time Value of Money Interest Calculations Cash Flow Equivalence Single Payment Compound Interest Formulas Nominal and Effective Interest Rate

Objectives Understand the concept of time value of money Distinguish between simple and compound interest Understand the concept of equivalence of cash flows Solve problems using Single Payment Compound Interest Formulas

Computing Cash Flows

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Question: Would you rather Receive $1000 today; or Receive $1000 10 years from today?

Answer: Of course today! Why?

I could invest $1000 today to make more money I could buy a lot of stuff today with $1000 Who knows what will happen in 10 years

Because money is more valuable today than in the future, we need to describe cash receipts and disbursements at time they occur.

Example - Cash Flows of 2 Payment Options

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You want to purchase a new $30,000 Honda Accord. Following are your payment options Pay the full price now minus a 3% discount; or Pay $5000 now; $8000 at the end of year 1; and $6000 at

the end of each of the next 4 years

Example - Cash Flows of 2 Payment Options

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Cash flow table and diagram

End of Year Cash Flow

0 (now) -$29,1001 02 03 04 05 0

Pay in full

40 1 2 3 5

Pay in 5 years

End of Year Cash Flow

0 (now) -$5,0001 -8,0002 -6,0003 -6,0004 -6,0005 -6,000

40 1 2 3 5

$29,100

$5,000$8,000

CVE 4000Homework #1 Question #1 Ethics IssueTypes of Costs - ReviewCost Estimating Models Per-Unit ModelExample - Cost Estimating using Per-Unit Model (1)Example - Cost Estimating using Per-Unit Model (2)Example - Cost Estimating using Per-Unit Model (3)Example - Cost Estimating using Per-Unit Model (4)Cost Estimating Models Segmenting ModelExample - Cost Estimating using Segmenting Model (1)Example - Cost Estimating using Segmenting Model (2)Work Breakdown Structure (from cost segmenting model)Homework #1 Question #3 CostsCost Estimating Models Cost IndexesExample - Cost Estimating using Cost IndexesCost Estimating Models Power Sizing ModelPower-Sizing Exponent Values Industrial EquipmentPower-Sizing Model - ExampleCost Estimating Models TriangulationCost Estimating Models Improvement and Learning Curve (1)Improvement and Learning Curve (2)Improvement and Learning Curve (3)Improvement and Learning Curve (4)Improvement and Learning Curve (5)Cash Flow Diagrams (CFD)Cash Flow Diagrams (CFD) ExampleCategories of Cash FlowsDrawing a Cash Flow DiagramDrawing Cash Flow Diagrams with SpreadsheetInterest & Equivalent (Chapter 3 in Textbook) Computing Cash FlowsExample - Cash Flows of 2 Payment OptionsExample - Cash Flows of 2 Payment Options