Engineering Economics Ch 3

Engineering Economics

Dr. Karim Kobeissi

Chapter 3: Engineering Costs & Cost Estimating

Keyword: Consumer Price Index

The Consumer Price Index (CPI) is a statistical measure

of change, over time, of the prices of a market basket

of consumer products (such as transportation, food

and medical care) purchased by households.

The CPI gives an overall image about the evolution of

prices over time in a specific country.

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, training, material handling,

fixtures and tooling, data management, technical

support, as well as general support costs

(overhead).

Types of Costs• Fixed, Variable and Total Costs• Marginal Costs & Average Costs• Sunk Costs • Incremental Costs • Opportunity Costs• Recurring & Non-recurring Costs• Cash Costs & Book Costs• Life-Cycle Costs

Fixed, Variable and Total Costs• Fixed Costs: constant, independent of the output or activity level.

– Property taxes, insurance– Management and administrative salaries– License fees, and interest costs on borrowed capital– Rental or lease

• Variable Costs: Proportional to the output or activity level. – Direct labor cost– Direct materials

• Total Variable Cost = Unit Variable Cost * Quantity Produced

• Total Cost = Fixed Cost + Total Variable Cost

Fixed, Variable and Total Costs

• 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 those who signed up.

• He gathered data and categorized these expenses as either

fixed or variable:

Bus Rental 80.00$ Event Tickets 12.50$ Gas Expense 75.00$ Refreshments 7.50$ Other Fuels 20.00$ Bus Driver 50.00$ Total FC 225.00$ Total VC 20.00$

People Fixed cost Variable cost Total cost0 225.00$ -$ 225.00$ 5 225.00$ 100.00$ 325.00$

10 225.00$ 200.00$ 425.00$ 15 225.00$ 300.00$ 525.00$ 20 225.00$ 400.00$ 625.00$

Fixed Costs Variable Costs

Total costs

$-

$200.00

$400.00

$600.00

$800.00

0 5 10 15 20

Volume

Cost ($

)

Total cost

Fixed cost

Breakeven Analysis• Total Revenue = Unit Selling Price * Quantity Sold

• Profit = Total Revenue - Total Cost

• The break-even point (BEP) is the point at which total cost and total

revenue are equal: there is no net loss or gain, and one has "broken

even."

• Applications of Breakeven analysis:- Determining the minimum number of units to be sold in order to

cover total cost

BE volume = Total Fixe Cost / [Unit Price – Unit Variable Cost]

- Forecast production profit / loss

TR

TC

FC

Breakeven Analysis

Units of Outputs

$

Break-even Point

Fixed Costs

Variable Costs

Total Costs

Total Revenue

Loss

Profit

Applied Example

XUnits of Outputs15

Fixed Costs= $225

Variable Costs= 20X

Total Costs= $225 + 20X

Total Revenue= 35X

Loss

Profit

$1000

$800

$600

$400

$200

$0105 20 25

Breakeven Charts

• DK developed an overall total cost equation for his business expenses.

• Now DK wants to evaluate the potential to make money from this chartered bus trip.Total Cost = Total Fixed Cost + Total Variable Cost

= $225 + ($20)(the number of people on the trip) • Let x = number of people on the trip• Thus,

Total Cost = 225 + 20x

• Using this relationship, DK can calculate the total cost for any number of people - up to the capacity of the bus.

Breakeven Charts• What he lacks is a revenue equation to offset his costs. • DK's total revenue from this trip can be expressed as:

Total Revenue = = (Charter ticket price)(number of people on the trip) = (ticket price)(x)

• Profit or loss can now be calculated as:Total Profit =

= [Total Revenue] - [Total Costs] = [ticket price]x – [225 + 20x]

If he charged a charter ticket price of $35, then = [35x] - [225 + 20x] = 15x - 225

Ticket priceBus Rental 80.00$ Event Tickets 12.50$ 35.00$ Gas Expense 75.00$ Refreshments 7.50$ Other Fuels 20.00$ Bus Driver 50.00$ Total FC 225.00$ Total VC 20.00$

People Fixed cost Variable cost Total cost Revenue Profit Region0 225.00$ -$ 225.00$ -$ (225.00)$ Loss5 225.00$ 100.00$ 325.00$ 175.00$ (150.00)$ Loss

10 225.00$ 200.00$ 425.00$ 350.00$ (75.00)$ Loss15 225.00$ 300.00$ 525.00$ 525.00$ -$ Breakeven20 225.00$ 400.00$ 625.00$ 700.00$ 75.00$ Profit25 225.00$ 500.00$ 725.00$ 875.00$ 150.00$ Profit30 225.00$ 600.00$ 825.00$ 1,050.00$ 225.00$ Profit35 225.00$ 700.00$ 925.00$ 1,225.00$ 300.00$ Profit40 225.00$ 800.00$ 1,025.00$ 1,400.00$ 375.00$ Profit

Fixed Costs Variable Costs

Profit-loss breakeven chart

$-

$500.00

$1,000.00

$1,500.00

0 5 10 15 20 25 30 35 40

Volume

Co

st

($)

Total cost

Fixed cost

Revenue

Marginal Costs and Average Costs

• Average Cost: total cost divided by the total number of units

produced.

– Basis for normal pricing

• Marginal Cost: the change in total cost (or total variable cost) that

comes from producing one additional unit in the short run.

– The purpose of analyzing marginal cost is to determine at what point an

organization can achieve economies of scale (Capacity Planning). The

calculation is most often used among manufacturers as a means of

isolating an optimum production level.

– Basis for last-minute pricing

Capacity Planning

Sunk Costs

Sunk Costs are irreversible expenses incurred

previously. They are irrelevant to present

decisions. For example, if you decide to have

your employees work three shifts instead of two,

your rent should stay the same.

• In the 1970's Lockhead spent $ 1 billiondeveloping a new airplane (Tristar). Aftersinking the money, it was clear that the venture

was not going to be a success.• Lockhead went to its creditors, and askedfor more money, saying, “We have to keepgoing, or else the $ 1 billion will be totallywasted.”

Sunk Costs

Some of the creditors said, “Why put in

more money, since there is no way we can

recoup our investment?”

Sunk Costs

Who was right?

Answer: Both arguments are wrong. The billion

dollar initial investment is a sunk cost that is

irrelevant to the present decision.

We should compare the extra

revenue of continuing with the

extra cost.

Sunk Costs

“Don’t cry over spilt milk”!!

Incremental Costs

Incremental Costs: Difference in costs between two alternatives.

– Suppose that A and B are mutually exclusive alternatives. If A has an initial cost of $10,000 while B has an initial cost of $17,500, the incremental initial cost of (B - A) is $7,500.

Example

Choose between alternative models A and B. What incremental costs occur with model B?

Cost Items A BPurchase price 10,000.00$ 17,500.00$ 7,500.00$ Installation costs 3,500.00$ 5,000.00$ 1,500.00$ Annual maintenance costs 2,500.00$ 750.00$ (1,750.00)$ Annual utility expenses 1,200.00$ 2,000.00$ 800.00$ Disposal costs after useful life 700.00$ 500.00$ (200.00)$

Model Costs

Incremental

Opportunity Costs

In engineering economics, the opportunity cost concept (hidden cost) is useful

in decision involving a choice between different alternative courses of action.

Resources are scarce We cannot produce all the commodities For the

production of one commodity, we have to forego the production of another

commodity We cannot have everything we want. We are, therefore, forced

to make a choice. Opportunity cost of a decision represents the benefits or

revenue forgone by pursuing one course of action rather than another.

The economic significance of opportunity cost is as follows:

1. It helps in determining relative prices of different products.

2. It helps in determining normal remuneration to a factor of production.

3. It helps in proper allocation of factor resources.

Recurring Costs and Non-recurring Costs

• Recurring Costs: Repetitive and occur when a firm produces similar goods and services on a continuing basis– Office space rental– Salaries– Etc …

• Non-recurring Costs: Not repetitive, even though the total expenditure may be cumulative over a period of time– Typically involve developing or establishing a capability or capacity

to operate– Examples are purchase cost for real estate and the construction

costs of the plant

Cash Costs & Book Costs

• A cash cost requires the cash transaction of dollars "out of one

person's pocket" into "the pocket of someone else“. When you buy

dinner for your friends or make your monthly automobile payment

you are incurring a cash cost or cash flow. Cash costs and cash flows

are the basis for engineering economic analysis.

• 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. asset

depreciation, is a common example of book cost. Book costs do not

ordinarily represent cash flows and thus are not included in

engineering economic analysis.

Life-Cycle Costs

• Life-Cycle Costs: Summation (+) of All costs, both

recurring and nonrecurring, related to a product,

structure, system, or service during its life span.

• Life cycle begins with the identification of the

economic needs or wants (the requirements) and

ends with the retirement and disposal activities.

Cost Estimating

Engineering economic analysis involves present and

future economic factors; thus, it is critical to obtain

reliable estimates of future costs, benefits and

other economic parameters.

• Estimating is the foundation of economic analysis.

Types of Estimates

An engineer should ask himself “How accurate do I want my cost

estimation to be ?”. There are three general types of estimates:

1. Rough – order of magnitude, used for high level planning,

inaccurate, range from -30% to +60% of actual values.

2. Semi-detailed - based on historical records, reasonably

sophisticated and accurate, -15% to +20% of actual values.

3. Detailed - based on detailed specifications and cost models, very

accurate, within -3% to +5% of actual.

N.B. We must balance the needed accuracy with the cost to perform

the cost estimation.

C o s t E s ti m a ti n g M o d e l s

The per-unit model uses a "per unit" factor, such as cost

per square meter (m2), to develop the estimate desired.

This is a very simplistic yet useful technique, especially for

developing estimates of the rough type. The per unit

model is commonly used in the construction industry.

Per Unit

ExplanationModel

• Per-Unit Model (Unit Technique)– Construction cost per square foot (building)– Capital cost of power plant per kW of capacity– Revenue / Maintenance Cost per mile (hwy)– Utility cost per square foot of floor space– Fuel cost per kWh generated– Revenue per customer served

Cost Estimating using Per-Unit Model

We may be interested in a new home that is

constructed with a certain type of material and has

a specific construction style. Based on this

information a contractor may quote a cost of $65

per square meter for our home. If we are interested

in a 2000 square meter floor plan, our cost would

thus be: 2000 x 65 =$130,000.

Example of Cost Estimating using Per-Unit Model


The segmenting model breaks up the total estimation

task into segments (estimates are made at component

level). Each segment is estimated, then the segment

estimates are combined for the total cost estimate.

Segmenting

ExplanationModel

Example of Cost Estimating using Segmenting Model

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

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

A. Chassis B. Drive Train

Example of Cost Estimating using Segmenting Model

Cost estimate of lawn mower

Cost Item Estimate

C.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

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

adapter7.75

Subtotal $25.60

C. Controls D. Cutting/Collection system

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


Cost indexes can be used to reflect historical changes in costs. Cost

index could be individual cost items (labor, material, utilities), or group

of costs (Consumer Prices Index- CPI; Producer Prices Index - PPI).

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 Indexes

ExplanationModel

B

A

B

A

Index

Index

Cost

Cost

Example of Cost Estimating Using Cost Indexes

800,871$124

188500,575$

Index

IndexCostLaborCostLabor

yrs10

nowyrs10Now

000,227,3$544

715000,455,2$

Index

IndexCostMaterialCostMaterial

yrs3

nowyrs3Now

Project Cost now = Project Cost 5 years [CPI now /CPI 5 years ] = $420000 [116/103]= $ 473000

C o s t E s ti m a ti n g M o d e l s ( c o n )

The power-sizing model accounts explicitly for economies of scale. For

example, the cost of constructing a six floor building will typically be less

than double the construction cost of a comparable three floor building. To

estimate the cost of B based on the cost of comparable item A, we 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 including industry reference books, research reports, and technical

journals. 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.

Power Sizing

ExplanationModel

E x a m p l e o f X = P o w e r S i z i n g E x p o n e n tEquipment/Facility X

Blower, centrifugal 0.59

Compressor 0.32

Crystallizer, vacuum 0.37

Dryer, drum 0.40

Fan, centrifugal 1.17

Equipment/Facility X

Filter, vacuum 0.48

Lagoon, aerated 1.13

Motor 0.69

Reactor 0.56

Tank, horizontal 0.57

Example of Cost Estimating Using Power-Sizing Model

Based on her work ,Miriam has been asked to estimate the cost today of a 2500

ft^2 heat exchange system for the new plant being analyzed. She has the

following data:

- Her company paid $50,000.0for a 1000 ft^2 heat exchanger 5 years ago.

- Heat exchangers within this range of capacity have a power sizing exponent

(x) of .0.55 (< 1 economy of scale ).

- Five years ago the Heat Exchanger Cost Index (HECI) was 1306; it is 1487

today.

800,82$1000

2500000,50$

ft1000

ft2500CostCost

55.0

55.0

2

2

ft1000ft2500 22

SOLUTION

Miriam will first use the power sizing equation to scale up the cost of

the 1000 ft^2 exchanger to one that is 2500 ft^2 using the 0.55

power-sizing exponent :


Miriam knows that the $82,8000 reflects only the scaling up of the cost

of the 1000 ft^2 model to a 2500 ft^2 model.

Now she will use the cost indexes equation and the HECI data to

estimate the cost of a 2500 ft^2 exchanger today. Miriam's cost

estimate would be:


C o s t E s ti m a ti n g M o d e l s ( c o n )

The theory of the learning curve is based on the simple idea that

the time required to perform a task decreases as a worker gains

experience. The basic concept is that the time, or cost, of

performing a task decreases at a constant rate as cumulative

output doubles. Learning curves are useful for preparing cost

estimates, bidding on special orders, setting labor standards,

scheduling labor requirements, evaluating labor performance,

and setting incentive wage rates. In general, as output doubles

the unit production time will be reduced to some fixed

percentage, the learning curve percentage or learning curve

rate.

Learning Curve

ExplanationModel

Cost Estimating Using Learning Curve Model

For example, a learning curve rate of 70% represents

much faster learning than a rate of 90%. If an

operator exhibits learning on a certain task at a rate

of 70%, the time required to complete production

unit 50, for example, is only 70% of the time

required to complete unit 25.

Learning Curve Rates in Different Industries

1) Aerospace 85%

2) Shipbuilding 80-85%

3) Complex machine tools for new models 75-85%

4) Repetitive electronics manufacturing 90-95%

5) Repetitive machining or punch-press operations 90-95%

6) repetitive electrical operations 75-85%

7) Repetitive welding operations 90%

8) Raw materials 93-96%

9) Purchased Parts 85-88%

Wright's Cumulati ve Average Model

In Wright's Model, the learning curve function is defined as follows:

Y = aXb

where:

Y = the cumulative average time (or cost) per unit.

X = the cumulative number of units produced.

a = time (or cost) required to produce the first unit.

b = slope of the function when plotted on log-log paper.

= Log of the learning rate in decimal form /Log 2

For an 80% learning curve b = Log .8/Log 2 = -.09691/.301 = -.32196If the first unit required 100 hours, the equation would be: Y = 100X-.322

Wright's Cumulati ve Average Model

Let T1 = Time to perform the 1st unit

TN = Time to perform the Nth unit

b = Learning curve exponent = Constant based on learning curve rate = log (learning curve rate in decimal form) / log 2N = Number of completed units

b1N NTT

2ln

%ln

2log

%logb

Cost Estimating Using Learning Curve Model

Example of Cost Estimating Using Learning Curve Model

Example 2-9 Cost Estimating using Learning Curve

N TN

1 9.602 8.163 7.424 6.945 6.586 6.317 6.088 5.909 5.73

10 5.59

N TN

11 5.4712 5.3613 5.2614 5.1715 5.0916 5.0017 5.0018 5.0019 5.0020 5.00

2345.0b1N N)6.9(NTT

0.00

2.00

4.00

6.00

8.00

10.00

12.00

1 3 5 7 9 11 13 15 17 19

N

TN

Example 2-9 Cost Estimating using Learning Curve

Example of Cost Estimating Using Learning Curve Model

The learning curve slope indicates "how fast"

learning occurs.

Estimating Benefits

• So far we have focused on cost terms and cost estimating.

• However, engineering economists must often also estimate

benefits. Benefits include sales of products, revenues from bridge

tolls and electric power sales, cost reductions from reduced

material or labor costs, reduced time spent in traffic jams, and

reduced risk of flooding.…

• These benefits are the reasons that many engineering projects

are undertaken.

• The cost concepts and cost estimating models can also be applied

to economic benefits.

Cash Flow Diagrams

• The costs and benefits of engineering projects occur over time and are

summarized on a Cash Flow Diagram (CFD).

• Specifically, a CFD illustrates the size, sign, and timing of individual cash

flows. In this way the CFD is the basis for engineering economic analysis.

• A Cash Flow Diagram is created by first drawing a segmented time-based

horizontal line, divided into appropriate time units.

• The time units on the CFD can be years, months, quarters or any other

consistent time unit.

• Then at each time when there is a cash flow, a vertical arrow is added -

pointing down for costs and up for revenues or benefits.

• These cash flows are drawn to relative scale.

Cash Flow Diagrams

Cash Flow Diagrams• Summarizes the flow of money over time• Can be represented using a spreadsheet

Year Capital costs O&M Overhaul Total0 (80,000.00)$ (80,000.00)$ 1 (12,000.00)$ (12,000.00)$ 2 (12,000.00)$ (12,000.00)$ 3 (12,000.00)$ (25,000.00)$ (37,000.00)$ 4 (12,000.00)$ (12,000.00)$ 5 (12,000.00)$ (12,000.00)$ 6 10,000.00$ (12,000.00)$ (2,000.00)$

Cash flow

$(100,000.00)

$(80,000.00)

$(60,000.00)

$(40,000.00)

$(20,000.00)

$-

$20,000.00

0 1 2 3 4 5 6

Year

Ca

sh

flo

w

Overhaul

O&M

Capital costs

Documents

Engineering Economics Ch 3