Chapter 3 COST ESTIMATION TECHNIQUES ENGINEERING ECONOMIC (BPK30902) Chapter 3 COST ESTIMATION TECHNIQUES 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 Purpose Providing information used in setting a selling price for quoting, bidding, evaluating contracts Determining whether a proposed product can be made & distributed at a profit Evaluating how much capital can be justified for process changes or other improvements Establishing benchmarks for productivity improving programs

CET - Approaches 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 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. Integrated Cost Estimation The ICE approach has three major components. Work breakdown structure (WBS) Cost and revenue structure (classification) Estimating techniques (models)

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

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 Indexes Unit Technique 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 known n = 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 1 Company 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 is C2007 = 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 n Ckm = unit cost/price of the mth item in year k Wm = 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 + 3 Now, 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/liter Unleaded plus: 230 sen/liter (253/227.5) = 256 sen/liter Regular 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 + 3 Now, 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/T Unleaded plus: 1,008 M/T(845/1056.8) = 805.98M/T Regular 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. Capital Cost of plant per kilowatt of capacity Revenue per kilometer Capital cost for installed telephone Temperature loss per 1,000 meter steam pipe Operating cost per kilometer 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 estimated Cd = cost of selected component d that is estimated directly fm = cost per unit of component m Um = number of units of component m m d

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 Cost RM75 (per sq. meter) RM2,000 (per unit) RM3,250 B 2,000 sq. meter 1 unit RM90 RM3,000

SET (Models) – Factor Technique Solution Plant Land Warehouse Production Line Total A 2,500 sq. meter 2 units Cost RM75 (per sq. meter) RM2,000 (per unit) RM3,250 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 RM90 RM3,000 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 Technique Example 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 Technique Solution 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 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 number Zu = number of input resource units needed to produce output unit u K= number of input resource units needed to produce the first output unit s = 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 & Improvement Example 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: The time it will take the team to assemble the 10th car The total time required to assemble the first 10 cars The estimated cumulative average assembly time for the first 10 cars

PCE – Learning & Improvement Solution 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 hours 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 The cumulative average time for x units, Cx, is given by Cx = Tx /x C10 = T10 /10 = 631 /10 = 63.1 hours x x x u=1 10 u=1 u=1 u=1

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 steps Problem definition Data collection & normalization CER equation development 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

Price (RM/Ton Matrix in Year 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 2010 Civet Coffee Beans 2900 3250 3860 Robusta Coffee Beans 2580 3300 Arabica Coffee Beans 1980 2180 2890 If, index value in 2011 to be 4300, determine the corresponding 2012 prices of coffee beans from I2010

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 + 3 Now, 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/T Robusta Coffee Beans : 3300 M/T(4300/4806.50) = 2952.25M/T Arabica 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 unit b) 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) ……….Ck(In/Ik) = RM165,989 CA1 = CB(SA/SB)X = RM165,989 (120-KW/80-KW)0.6 = RM165,989 (1.2754) TC = RM211,702 + RM18,000 = RM229,702 CA2 = RM165,989 (40-KW/80-KW)0.6 = RM165,989 (0.6597) TC = RM109,503 +18,000 = RM127,503

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