Chapter 3COST ESTIMATION

TECHNIQUES

ENGINEERING ECONOMIC(BPK30902)

HJ ZUIKARNAIN DAUD

COST ESTIMATION TECHNIQUES

–Introduction–Selected Estimating Techniques

(Model)–Parametric Cost Estimating

Introduction

CET - Introduction

• Estimating future cash flows for feasible alternatives

• Comprehensive planning & design process (engineering designers/ marketing/ manufacturing/finance/ top management personnel)

• Result of cost estimating are used for variety of purpose

CET – Variety Of Purpose1. Providing information used in setting a selling

price for quoting, bidding, evaluating contracts

2. Determining whether a proposed product can be made & distributed at a profit

3. Evaluating how much capital can be justified for process changes or other improvements

4. Establishing benchmarks for productivity improving programs

CET - Approaches

1. Top-down approach– Uses past similar engineering projects to

estimate cost, revenue, & other data or current project by modifying the data

– Best used early in estimating process when alternatives being developed & refined

CET - Approaches

2. Bottom-up approach– More detailed method of CE– Break down a project to small, manageable units

& estimates their economic consequences– The smaller unit costs are added together with

other types of costs to obtain overall CE– Best works when detail concerning the desire

output (product/ service) has been defined & clarified

The ICE approach has three major components.

• Work breakdown structure (WBS)• Cost and revenue structure

(classification)• Estimating techniques (models)

Integrated Cost Estimation

Work Breakdown Structure (WBS)

• A basic tool in project management• A framework for defining all project work

elements and their relationships, collecting and organizing information, developing relevant cost and revenue data, and management activities.

• Each level of a WBS divides the work elements into increasing detail.

A WBS has other characteristics.• Both functional and physical work elements

are included.• The content and resource requirements for

a work element are the sum of the activities and resources of related subelements below it.

• A project WBS usually includes recurring and nonrecurring work elements.

Cost and Revenue Structure• Used to identify and categorize the costs

and revenues that need to be included in the analysis.

• The life-cycle concept and WBS are important aids in developing the cost and revenue structure for a project.

• Perhaps the most serious source of errors in developing cash flows is overlooking important categories of costs and revenues.

Estimating Techniques

• Order-of-magnitude estimates (±30%)• Semi detailed, or budget, estimates (±15%)• Definitive (detailed) estimates (±5%)

REMEMBER! The purpose of estimating is to develop cash-flow projections—not to produce exact data about the future, which is virtually impossible. Cost and revenue estimates can be classified according to detail, accuracy, and their intended use.

The level of detail and accuracy of estimates depends on

• time and effort available as justified by the importance of the study,

• difficulty of estimating the items in question,

• methods or techniques employed,• qualifications of the estimator(s), and• sensitivity of study results to particular

factor estimates.

Selected Estimating Techniques (Model)

SET (Models)

Applicable for order-of-magnitude estimates & many semi detailed or budget estimates

Useful in the initial selection of feasible alternatives for further analysis & in the conceptual/preliminary design phase of a project

1. Indexes2. Unit Technique3. Factor Technique

SET (Models) - Indexes

• Costs & prices vary with time for a number of reasons, (1) technological advances, (2) availability of labor & materials, (3) inflation

• Index is a dimensionless number that indicate how a cost/ prices has changed with time with respect to base year

• Indexes provide a convenient means for developing present & future cost & price estimates from past data

SET (Models) – Indexes (Single Item)

An estimate of the cost or selling price of an item in year n can be obtained by multiplying the cost/ price of the item at an earlier point in time (year k) by the ratio of the index value in year n (In) to the index value in year k (Ik)

Cn = Ck (In/Ik)k = reference year for which cost/price of item is knownn = year of which cost/price is to be estimated (n > k) Cn = estimated cost/price of item in year Ck = cost/price of item in reference year k

SET (Models) – Indexes (Single Item)Example 1Company XYZ installed a 50,000 kg/hour boiler for

RM525,000 in 2000 at index value of 468. The company must install another boiler of the same size in 2007 at index 542. What is the approximate cost of the new boiler?

Solution n = 2007 and k = 2000. Approximate cost of the boiler

in 2007 isC2007 = RM525,000 (542/468)

= RM608,013

SET (Models) – Indexes (Multiple Items)

• A composite index is created by averaging the ratios of selected item costs in a particular year to the same items in a reference year

• the developer of an index can assign different weights to the item in the index according to their contribution to total cost

• The weights W1, W2, …..W3 can sum to any positive number, typically 1 to 100

SET (Models) – Indexes (Multiple Items)

W1(Cn1/Ck1)+ W2(Cn2/Cn2) + …Wm(Cnm/Ckm)In =

W1 + W2 + …Wm Where; M = total number items in the index (1 ≤ m ≤ M)Cnm = unit cost/price ot the mth item in year nCkm = unit cost/price of the mth item in year kWm = weight assigned to the mth item Ik = composite index value in year k

x Ik

SET (Models) – Indexes (Multiple Items)Example 2

• Based on the following data, develop a weighted index for the price of a gallon of gasoline in 2006, when 1992 is the reference year having index value of 99.2. The weight placed on regular unleaded gasoline is 3 times that of premium or unleaded plus because 3 times as much regular unleaded is solid compared with premium or unleaded plus.

Price (Sen/Liter) in Year

1992 1996 2006

Premium 114 138 240

Unleaded Plus 103 127 230

Regular unleaded 93 117 221

SET (Models) – Indexes (Multiple Items)Solution

K = 1992, n = 2006, the value of I2006 =

(1)(240/114) + (1)(230/103) + (3)(221/93) -------------------------------------------------------- x 99.2 = 227.5 1 + 1 + 3Now, index value in 2008 to be 253, determine the

corresponding 2008 prices of gasoline from I2006 = 227.5

Premium : 240 sen/liter (253/227.5) = 267 sen/literUnleaded plus: 230 sen/liter (253/227.5) = 256 sen/literRegular unleaded: 221 sen/liter (253/227.5) = 246 sen/liter

SET (Models) – Indexes (Multiple Items)Exercise 1

• Develop a weighted index for the price of a ton matrix of pump oil in 2009, when 2005 is the reference year having index value of 677. The weight placed on regular unleaded pump oil is 3 times and unleaded super is 2 times that of premium due to it’s solidness compared.

Price (RM/Ton Matrix in Year

2005 2006 2009

Premium 752 835 1150

Unleaded super 682 755 1008

Regular unleaded 598 692 973

SET (Models) – Indexes (Multiple Items)Solution

K = 2005, n = 2009, the value of I2009 =

(1)(1150/752) + (2)(1008/682) + (3)(973/598) ------------------------------------------------------------- x 677 = 1,056.8 1 + 2 + 3Now, index value in 2010 to be 845, determine the corresponding 2012

prices of gasoline from I2009 = 1,056.8

Premium : 1,150 M/T(845/1056.8) = 919.52M/TUnleaded plus: 1,008 M/T(845/1056.8) = 805.98M/TRegular unleaded: 973M/T(845/1056.8) = 777.99M/T

SET (Models) – Unit Technique

Involves using a per unit factor that can be estimated effectively. E.g.1. Capital Cost of plant per kilowatt of capacity 2. Revenue per kilometer3. Capital cost for installed telephone4. Temperature loss per 1,000 meter steam pipe5. Operating cost per kilometer6. Constructions cost per square meter

Such unit when multiplied by appropriate unit gave total estimates of cost , saving or revenue

SET (Models) – Unit Technique

• Suppose we need a preliminary estimate of cost of a particular house.

• Using RM95 per sq meter, and the house is 2,000 sq meter, the estimate cost is RM95 x 2,000= RM190,000

SET (Models) – Factor Technique • Extension of the unit method in which the product of several

quantities or components are sum and added these to any

components estimate directly.C = ∑ Cd + ∑ fm Um

C = cost being estimatedCd = cost of selected component d that is estimated directly

fm = cost per unit of component m

Um = number of units of component m

d m

SET (Models) – Factor Technique Example

• Suppose that we need a slight refined estimate of the cost of a house consisting of 2,000 sq meter, 2 porches, & a garage. Using unit factor of RM85, RM10,000 per porch, & RM8,000 per garage for 2 directly components; estimate as(RM10,000x 2) + RM8,000 +(RM85 x 2,000) = RM198,000

SET (Models) – Factor Technique Exercise 1

• EE team is considering to develop 2 alternative plants.

Plant Land Warehouse Production Line

A 2500 sq. meter 2 units 2 units

Cost RM75 (per sq. meter)

RM2,000 (per unit)

RM3,250(per unit)

B 2,000 sq. meter 1 unit 2 units

Cost RM90(per sq. meter)

RM3,000(per unit)

RM3,000 (per unit)

SET (Models) – Factor Technique Solution

Plant Land Warehouse Production Line Total

A 2,500 sq. meter 2 units 2 units

Cost RM75 (per sq. meter)

RM2,000 (per unit)

RM3,250(per unit)

Estimation 75 x 2,500 = 187,500

2 x 2,000= 4,000

2 x 3,250= 6,500

198,000

B 2,000 sq. meter 1 unit 2 units

Cost RM90(per sq. meter)

RM3,000(per unit)

RM3,000 (per unit)

Estimation 90 x 2,000= 180,000

1 x 3,000 = 3,000

2 x 3,000 = 6,000

189,000

Parametric Cost Estimating

PCE

• PCE is the use of historical cost data & statistical technique to predict future costs

• Statistical technique are used to develop cost estimating relationships (CERs) that tie the cost/price of an item(e.g. a product/ service/activity) to one or more independent variables (i.e. drivers)

• Para metric models are in early design stage to get the idea of project/ product cost used to gauge the impact of design decision on the total cost

• CERs – the power sizing technique & learning curve to overview of the procedure

PCE – Power Sizing Technique

• Sometimes referred to as an exponential model is frequently used for developing capital investment estimates for industrial plant & equipment

CA/CB = (SA/SB)X

CA = CB(SA/SB)X

CA = cost of plant A

CB =cost for plant B

SA = size of plant A

SB = size of plant B

X = cost-capacity factor to reflect economies of scale

(Both in RM as of the point in time for which the estimate is desired)

(both in same physical units

PCE – Power Sizing TechniqueExample

• An aircraft manufacture desires to make a preliminary estimate of the cost of building a 600-MW fossil-fuel plant for the assembly of its new long-distance aircraft . A 200-MW cost RM100 million 20 year ago, cost index 400, cost index now 1,200.the cost-capacity factor is 0.79

PCE – Power Sizing TechniqueSolution

1. Use the cost index information to update the known cost of the 200-MW plant 20 years ago to the current cost

CB = RM100 million (1200/400) = RM300 million

2. Use the power-sizing model to estimate the cost of the 600-MW plant (CA)

CA = RM300 million (600-MW/200-MW)0.79

CA = RM300 million x 2.38 = RM714 million

PCE – Learning & Improvement

• A learning curve is a mathematical model that explains the phenomenon of increase worker efficiency & improved organizational performance with respective production or service

• Also called as experiences curve or manufacturing progress function

• Basic concept – input resources decrease (energy cost, labor hours, materials costs, engineering hours) or on a per output in basis as produced increase

PCE – Learning & Improvement

• E.g. 100 hours required to produce output unit 90%then 100(.90) = 90 hours would be required to produce for second unit

• Similarly, 100(0.9)2 = 81 labor hours needed to produces fourth unit,100(0.9)3 = 72.9 hours to produce the eight unit and so on

• Resource requirement a assuming a consistent percentage reduction in resources each time the output quantity is double

Zu = K(un)

u = output unit numberZu = number of input resource units needed to produce output unit u

K= number of input resource units needed to produce the first output units = the learning curve slope parametric expressed as a decimal

(s = 0.0 for a 90% learning curve n = log s / log2 = the learning curve exponent

PCE – Learning & ImprovementExample

• The EE team is designing a formula car for national competitions and has 100 hours to complete. The improvement / learning rate) is 0.8 which means that output is doubled, assemble time reduced by 20%. Determine:a. The time it will take the team to assemble the 10th carb. The total time required to assemble the first 10 carsc. The estimated cumulative average assembly time for the

first 10 cars

PCE – Learning & ImprovementSolution

a. Assuming a proportional decrease in assembly time for output units between doubled quantities;

Z10 = 100(10) log 0.8/ log 2

= 100(10)-0.322

= 100/2.099 = 47.6 hoursb. The total time to produce x units, Tx, is given by

Tx = ∑ Zu = ∑ K (un) = K ∑ un

T10 = 100 ∑ u-0.322 = 100(1-0.322 + 2-0.322 + ..... 10-0.322) = 631 hours

c. The cumulative average time for x units, Cx, is given by Cx = Tx /x

C10 = T10 /10 = 631 /10 = 63.1 hours

u=1

x x x

u=1 u=1

u=1

10

PCE – Developing a CERs

• CER is a mathematical model that describes the cost of an engineering project as a function of one or more design variables

• Four basic steps1. Problem definition2. Data collection & normalization3. CER equation development4. Model validation & documentation

PCE – Developing a CERs

1. Problem Definition A well defined problem is much easier to solve For the purpose of cost estimating, developing a work

breakdown structure (WBS) to describing the elements of the problem

A review of completed WBS can help potential cost drives for development of CERs

2. Data Collection & Normalization The most critical step in the development of CER WBS also help in collection phase Data can be obtained from internet & external sources Data collected must be normalized to account for differences

due to inflation, geographical location, labor rate & etc

PCE – Developing a CERs

3. CER Equation Development• Captures the relationship between the selected cost drive(s)

& project cost• Basic equation: Plot data on regular graph paper, if straight

line, an linear relationship is suggested; If curve, plot using semi log paper (straight line – relationship is logarithmic or exponential)or log-log paper (straight line – power curve)

• Value of coefficients: method of least squares seeks to determine a straight line through the data that minimizes the total deviation of the actual data from the predicted values

PCE – Developing a CERs

4. Model Validation & Document• Statistical “goodness of fit” measures as standard error & the

correlation coefficient• To infer how well the CER predicts cost as a function of

selected cost driver(s) by documenting for future use

Quiz 2 No.1

• Develop a weighted index for the price of a ton matrix of coffee beans in 2010, when 2006 is the reference year having index value of 3600. The weight placed on Civets Coffee Beans is 3 times and Robusta Coffee Beans is 2 times while Arabica Coffee Beans is 1 time to it’s quality compared.

Price (RM/Ton Matrix in Year

2006 2007 2010Civet Coffee Beans 2900 3250 3860Robusta Coffee Beans 2580 2900 3300Arabica Coffee Beans 1980 2180 2890

Solution Quiz 2 No 1

K = 2006, n = 2010, the value of I2010 =

(1)(2890/1980) + (2)(3300/2580) + (3)(3860/2900) ------------------------------------------------------------- x 3600 = 4806.50 1 + 2 + 3Now, index value in 2011 to be 4300, determine the corresponding 2012

prices of coffee beans from I2010 = 4806.50

Civets Coffee Beans : 3860 M/T(4300/4806.50) = 3453.24M/TRobusta Coffee Beans : 3300 M/T(4300/4806.50) = 2952.25M/TArabica Coffee Beans : 2890 M/T(4300/4806.50) = 2585.45M/T

Quiz 2 No. 2

• Six years ago, an 80-KW diesel electric set cost RM160,000. The cost index for this class of equipment six years ago was 187 and is now 194 the cost capacity factor 0.6 a) The EE team is considering a 120-KW unit of the same general design to power a small isolated plant. Assume we want to add a pre-compressor, which currently costs RM18,000. Determine the total cost of the 120-KW unitb) Estimate the cost of a 40-KW unit of the same general design, Include the cost of the RM18,000 pre-compressor

Solution Quiz 2 No 2

CB= RM160,000 (194/187)= RM165,989

CA1 = RM165,989 (120-KW/80-KW)0.6= RM165,989 (1.2754)= RM211,702 + RM18,000= RM229,702

CA2 = RM165,989 (40-KW/80-KW)0.6= RM165,989 (0.6597)= RM109,503 +18,000= RM127,503

Thank You