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Engineering Economy H00K1A Prof dr ir Liliane Pintelon 2018-2019 Course material Course Classes: 1st semester, every Thursday afternoon Work shops: 1st semester, 4 sessions (on Thursday) Course Powerpoints and journal articles on Handbook (highly recommended) Blank, L. and Tarquin, A., Engineering Economy, 8th edition, McGraw Hill, Boston, 2018 Evaluation - Exam Closed book exam Written part (+/- 25%) Small questions (insight?!) – multiple choice (guess correction)/open questions Oral part with (written) preparation time (+/- 75%) Exercise(s) Knowing which method to use and drawing the right conclusions from the results Formularium available Case study – Journal article Proving “maturity” in the subject Being able to evaluate a chosen approach Note: the course is taught and examined in English https://eng.kuleuven.be/studenten/lijstRM.html Engineering Economy Setting the scene 1 Overview Typical EE problems ‐ EE ? ‐ Since when ? ‐ Why important to engineers ? Role of EE in decision making ‐ The decision making process ‐ Ethics and decision making What costs or (dis)benefits ? Course content Note on course material 2 Typical EE problems Professional examples A chemical installation needs a new pump. You can buy either a cheap one with a life time of 2 years or a more expensive one with a life time of 10 years. What is the best choice if the company intends to operate the installation for another 8 years? The CEO is preparing performance reports. He wants to relate the costs of spare parts consumption to the value of the installations in questions. How can he obtain this value without going into detailed and lenghty calculations? 3 Typical EE problems Professional examples (c’d) The production installation is generating rather high maintenance costs. Should this installation been replaced by a new one now or is it better to keep the old one (as is or maybe with a small investment in renovation) ? There is a new technology which could be used for a given production process. If the company does not go for it, but its competitors do, the company may loose a lot of market share. On the other hand, the new technology is maybe not very reliable yet. Would it be a good idea to spend some money to gain further insight in the technology and its possibilities and limitations? 4 Typical EE problems (Personal examples) You are buying a house. You have offers from different banks for your loan. Some banks recommend to pay back a fixed amount per month, others recommend to pay more interest in the beginning and less in the end. How does this work ? Do you – in the end ‐ pay the same amount of interest? You requested to borrow some money from different banks, but no bank was willing to lend you the amount needed. Now you saw some offers by small companies and private persons in the newspaper. The interest rates stated are given by week, by month or by year. How to compare these interest rates?? 5 Engineering economy (EE) ? Some definitions Engineering economics, previously known as engineering economy, is a subset of economics for application to engineering projects. Engineers seek solutions to problems, and the economic viability of each potential solution is normally considered along with the technical aspects. (Wikipedia, 2009) Engineering economy involves the systmatic evaluation of the economic merits of proposed solutions to engineering problems. (Sullivan et al., 2009) 6 Engineering economy (EE) ? Some definitions (c’d) Fundamentally, engineering economy involves formulating, estimating and evaluating the economic outcome when alternatives to accomplish a defined purpose are available. Another way to define engineering economy is as a collection of mathematical techniques that simplify economic comparison. (Blank & Tarquin, 2008) “… the art and science that involves formulating, estimating and evaluating economic outcomes … always concerned wiht the selection and possible execution of alternatives given the economic parameters associated with the project …” (Blank & Tarquin, 2008) 7 Since when ? Arthur M. Wellington Justification of railway projects 1870s E. Paul DeGarma Holger G. Thuesen textbooks on Engineering Economy 1942, 1950 Herny R. Towne The Engineer as Economist 1886 Industrial revolution °Management science Arthur Jr Lesser founding editor or “The Engineering Economist” 1955 8 Why important to Engineers ? Engineers “design” and create, often they must select between multiple alternatives. Designing involves economic decisions. Engineers must be able to incorporate economic analysis into their creative efforts, but are not responsible for detailed investment financing decisions. Multidisciplinary input needed: e.g. AOC (annual operating costs) is typically a technical input, revenues are a marketing input … A proper economic analysis for selection and execution is a fundamental aspect of engineering. Engineering economy models help here, but keep in mind: “People make decisions, tools don’t.” 9 Role of EE in decision making Decision making involves the estimation of future events/outcomes Engineering economy aids in quantifying past outcomes and forecasting future outcomes Engineering Economy provides a framework for modeling problems involving: Money – “cash flows” Time – times of occurrence of cash flows Interest rates – time value of money Measures of economic worth – for selecting an alternative 10 Role of EE in decision making Time value of money All firms make use of investment of funds; investments are supposed to earn a return. Money possesses a “time value” Interest = manifestation of the time value of money The “time value” of money is the most important concept in engineering economy 11 Illustration a) b) Calculate the amount deposited 1 year ago to have $1000 now at an interest rate of 5% per year. Calculate the amount of interest earned during this time period. Solution a) The total amount accrued ($1000) is the sum of the original deposit and the earned interests. If X is sthe original deposit: total accrued = original + original*interest rate $1000 = X+X*0.05 X = $952.38 = original amount b) The interest earned is $1000 ‐ $ 952.38 = $47.62 12 The Decision Making Process 1. 2. 3. 4. 5. 6. 7. Understand the problem – define objectives Collect relevant data Define the set of feasible alternatives Identify the criteria for decision making Evaluate the alternatives and apply sensitivity analysis Select the “best” alternative Implement the alternative and monitor results All monetary criteria ? Brainstorming Nominal group technique “do nothing” = alternative 15 16 Ethics and decision making Universal morals or ethics Fundamental beliefs: stealing, lying, harming or murdering another are wrong Personal morals or ethics Beliefs that an individual has and maintains over time; how a universal moral is interpreted and used by each person Professional or engineering ethics Formal standard or code that guides a person in work activities and decision making 1‐18 All professional disciplines have a formal code of ethics. National Society of Professional Engineers (NSPE) maintains a code specifically for engineers; many engineering professional societies have their own code 1‐19 Money‐related situations with ethical dimensions In the design stage While the system is operating Bhopal, 1984 Challenger, 1986 20 21 23 24 Extra: What costs or (dis)benefits ? This topic is a topic for a course on Management Accounting. Let’s go over what we need here … CAPEX vs OPEX Tangibles (€, …) vs intangibles Standard, historical, predicted costs Labor costs, material costs, etc. Direct/indirect costs Case‐(non)related costs Current/non‐recurrent costs Sunk costs Internal/external costs Hidden costs Overhead costs Out‐of‐pocket/opportunity cost Fixed/variable costs Life cycle costs (etc) PS: the term “cost” is used to describe costs and (dis)benefits 27 About: Tangibles and intangibles tangible • increase in sales or profit • decrease in information processing cost • decrease in operating cost • decrease in required investment • increased operational ability and efficiency Illustration intangible • new or improved information availability • improved abilities in computation and analysis • improved customer service • improved employee morale • improved managerial decisions • improved competitive position • improved image Remark: some tangibles are #, but no $ (e.g; accidents) 28 How to compute/estimate costs and benefits ? About: Standard, historical, predicted costs Point of view: Cost calculation method About: Labor costs, material costs, … Point of view: Cost nature About: Case (non)related costs Be careful to include “all” costs, but beware NOT to take into account non‐relevant costs, also DON’T take the same costs into account twice … 29 About: Direct/indirect costs ABC (activity based costing) Direct: easy to relate “consumption” to “product” e.g. raw material, labor Indirect: difficult (or too expensive) to relate “consumption to “product” e.g. energy, warehouse About: Overhead costs Be careful, can be tricky ! Example: Supervision hours + global percentage for management About: Recurrent/non‐recurrent costs Point of view: Occurence frequency About: Internal/external costs Point of view: “who pays ?” Example: Environmental fee: charge/km Insurance & risk May same straighforward at first, but … 30 About: Fixed/variable costs output level output level cost cost output level cost cost cost Beware: • Variable costs are not necessarily linear in function of the output level • Fixed costs are not necessarily constant for any output level 31 output level output level Exercise You are working for a professional organization for warehouse managers. The organization regularly organizes short courses for their members. For the upcoming short course there are two options: regular or plus. Data Fee for participants: € 200 – Costs for development course material: € 200 Speaker cost: € 300 – Handouts for participant: € 35 – Mailing to announce course € 2100 Option 1: “plus”: Renting facility (hotel ): € 2500 – Lunch participant: € 60 Option 2: “regular”: Use in-house room: € 500 – Lunch participant: € 15 Questions: How many participants do you need to at least cover costs (both options) ? If 40 participants are expected for option 1 and 25 participants for option 2. How much profit can you expect (both options) ? 32 • • • • break even: p*q=f+v*q q = number participants p = prce (income) f = fixed cost v = variable cost short course financial overview 14000 12000 • • • • p = € 200 f1 = € (200 + 300 + 2500 + 2100) = € 5100 f2 = € (200 + 300 + 500 + 2100) = € 3100 v1 = € (35 + 60) = € 95 v2 = € (35 + 15) = € 50 • q1 = 49 en q2 = 21 participants • euro 10000 8000 income 6000 cost plus 4000 cost regular 2000 0 0 10 20 30 40 50 60 70 number of participants (q) Sensitvity analysis • • profit1 = € [200 * 40 – 5100 – 95 * 40] = - € 900 profit2 = € [200 * 25 – 3100 – 50 * 25] = € 1650 33 About: Out‐of‐pocket/opportunity costs An explosion in the chemical plant caused severe damages to the equipment. The installations had to close down for more than a month in order to repair all damages. The costs of materials and labor for these repairs were considerable. This downtime caused production problems. Penalties were to be paid to some major customers because of late deliveries. Sales also felt that they missed some sales because some potential customers preferred to go to their competitors because of the explosion and its bad impact on the company’s image. Illustration 34 About: Hidden costs A hospital is planning to change from a manual pharmacy picking system to a fully automated picking system, including a state‐of‐the‐art distribution robot. A consultant helped to gather all necessary data for an economic evalaution: investment cost in both the hardware and the software for the automated system (data from supplier), savings in labor costs, maintenance costs and energy costs to be expected and some material costs (packaging). Some costs for training were also included. Illustration Example “hidden cost”: technical and infrastructural investments 35 About: Sunk costs A company plans to build a new production unit. A detailed study has been carried out by an engineering firm and some basic construction works have been carried out (flattening the site). The engineering department is challenging this decision and claims that renovating the old production unit would – economically – be a better option. The manufacturing manager on the other hand considers outsourcing the production in question. The CEO decides to go ahead with the new production unit seen the fact that so much money already has been spent. Illustration “No use crying over spilled milk” 36 About: CAPEX vs OPEX CAPEX (capital expenditure) Costs made for obtaining or upgrading non‐consumable, physical assets, such as installations and buildings E.g. purchasing cost, commissioning cost OPEX (operational expenditure) Costs made to keep asset “usable” E.g. energy, spare parts, personnel Seems pretty obvious, but … Example: Quid major renovation of a distillation plant in CPI ? Investment or cost ? Will make a difference in financial results !! Note: EBITDA: earnings before interest, taxes, depreciation and amortization 37 Typical trade‐off (Fertilizer Focus, 2004) 38 LCC, TCO: related, but not synonyms About: Life cycle costs LCC, TCO = {CAPEX, OPEX} disposal replacement specifications decommissioning inception design maintenance O&M modifications operations build/buy commissioning construction acquisition installation/construction 39 40 http://www.assda.asn.au/lifecycle1.html (source: Pintelon & Van Puyvelde, 2013) 41 (source: Barringer & Associate, 2008) 42 Careful consideration needed! CAPEX vs OPEX Tangibles (€, …) vs intangibles Standard, historical, predicted costs Labor costs, material costs, etc. Direct/indirect costs Case‐(non)related costs Current/non‐recurrent costs Sunk costs Internal/external costs Hidden costs Overhead costs Out‐of‐pocket/opportunity cost Fixed/variable costs Life cycle costs (etc) PS: the term “cost” is used to describe costs and (dis)benefits 44 Course content ② Classes ①: this class ⑦ ③ ④ ⑫ ⑤ ⑥: Nov 2, holiday, no class ⑧: Nov 17, Athens, no class ⑨ ⑪ ⑩ 45 Note on course material Toledo Powerpoints: for each class on Toledo (beforehand) Some topics will be elaborated on, background material will be provided Handbook Well explained, many exercises and cases, examples, … Computational help: Excel (not in classes) 49 Wrap‐up Ch. 1 ‐ Extra Setting the scene What is EE ? What type of decisions ? Why important for engineers Cost classification Course content 50 Engineering Economy Basics 1 Overview Time value of money explored Factors: How time and interest affect money Combining factors Nominal and effective interest rates 2 Time Value of Money: Why ? All firms make use of investment of funds, these investments are expected to earn a return. In engineering economy computations it is important to taken into account the “time value” of money. Interest is the manifestation of the time value of money. Time value of money 3 Interest rates and returns Point of view on interests borrower’s point of view interest owed = amount owed now – original amount lender’s point of view interest earned = total amount now – original amount Return on investment interest accrued per time unit * 100% original amount interest accrued per time unit * 100% rate of return original amount interest rate Time value of money 4 Example 1.3 (p 12) You borrow $10,000 for one full year and must pay back $10,700 at the end of one year Interest Amount (I) = $10,700 ‐ $10,000=$700 for the year The $700 represents the return to the lender for this use of his/her funds for one year Interest rate (i) = 700/$10,000 = 7%/Yr The interest rate (i) is 7% per year 7% is the interest rate charged to the borrower; 7% is the return rate earned by the lender Time value of money 5 Equivalence Different sums of money at different times may be equal in economic value $106 one year from now 0 1 Interest rate = 6% per year $100 now $100 now is said to be equivalent to $106 one year from now, if the $100 is invested at the interest rate of 6% per year. Time value of money 6 Simple and compound interest Simple interest (one period) = (principal)*(interest rate) Simple interest (more periods) = (principal)*(number of periods)*(interest rate) Compound interest (one period) = (principal+all accrued interest)*(interest rate) , 1.8 and 1.9 Time value of money 7 Exercise I make a deposit of €10000 in a savings account for 5 years. The interest rate is 5 %. How much interest do I earn ? deposit interest rate Every year I take up the amount of interest earned. Time value of money 10000euro 5% single interest money interest in bank earned Every year I also deposit the interest in my account. compound interest money interest in bank earned after 1 year 10000 500 10000 500 after 2 years 10000 500 10500 525 after 3 years 10000 500 11025 551 after 4 years 10000 500 11576 579 after 3 years 10000 total 500 2500 12155 total 608 2763 8 “Trick” The time required for an initial amount to double in size with compound interest (rule of 72): 72 estimated n i The time required for an initial amount to double in size with simple interest (rule of 100): 100 estimated n i 2X=X+X*n*i/100 2=1+n*i/100 1=n*i/100 n=100/i Time value of money 9 Terminology and symbols t represents time P = a present sum of money at a time designated as t = 0 (now) F = a future amount of money at some point in time later than t = 0 A = a series of equal, end‐of‐period (i.e. “annual”) cash flows n = the number of interest periods (study period, horizon) i = the interest rate or rate of return per time period, in percent Time value of money 10 Computational help Tables See end of book Time value of money 11 Excel financial functions Present Value P: =PV(i%,n,A,F) Future Value F: =FV(i%,n,A,P) Equal, periodic value: =PMT(i%,n,P,F) No. of periods: =NPER((i%,A,P,F) Compound interest rate: =RATE(n,A,P,F) Compound interest rate: =IRR(first_cell:last_cell) Present value of a series: =NPV(i%,second_cell:last_cell) + first_cell Example =FV(5%,5,,10000) → (€12767) Time value of money 12 MARR (minimun attractive rate of return) Investors expect to earn a return on their investment (commitment of funds) over time → in order to be interes ng (profitable) an investment should return funds in excess of the investment amount → MARR, also called “hurdle” rate for an investment The MARR (%) is established by the financial managers of the firm; it is based on the cost of all types of capital and the allowance for risk, thus always higher than the return for a safe investment ! (see also Ch 10) Time value of money 13 ROR≥MARR>cost of capital Time value of money 14 15 Types of financing Topics in italics will be elaborated on later in the course. Equity Financing the firm uses funds either from retained earnings, new stock issues, or owner’s infusion of money Debt Financing the firm borrows funds from outside sources the cost of debt financing = the interest rate charged on the debt (loan) amounts Time value of money 16 Combination of debt‐equity financing is possible, it results in a WACC, a weighted average cost of capital Example A music system will be financed for 40 % with a credit card at 18% per year and for 60% with a savings account funds earning of 5% per year, then WACC=0.4*18+0.6*5=12.2% per year 17 Inflation Topics in italics will be elaborated on later in the course. A social‐economic occurrence in which there is more currency competing for constrained goods and services Where a country’s currency becomes worth less over time thus requiring more of the currency to purchase the same amount of goods or services in a time period Inflation impacts: Purchasing Power (reduces) Operating Costs (increases) Rate of Returns on Investments (reduces) Time value of money 18 Taxes Topics in italics will be elaborated on later in the course. Taxes represent a significant negative cash flow to the for‐ profit firm. A realistic economic analysis must assess the impact of taxes, called and AFTER‐TAX cash flow analysis. Not considering taxes is called a BEFORE‐TAX cash flow analysis A Before‐Tax cash flow analysis (while not as accurate) is often performed as a preliminary analysis A final, more complete analysis should be performed using an After‐Tax analysis Both are valuable analysis approaches Time value of money 19 Cash flows Cash inflows vs Cash outflows Cash inflows Money flowing INTO the firm from outside Revenues, savings, salvage values, etc Cash outflows Disbursements, money flows OUT First costs of assets, labor, salaries, taxes paid, utilities, rents, interest, etc Time value of money 20 Cash flow computation For many practical engineering economy problems the cash flows must be either: Assumed known with certainty Estimated A range of possible realistic values provided Generated from an assumed distribution and simulated Assumption: All cash flows are assumed to occur at the end of an interest period Tip: Visualizing cash flows in a cash flow diagram helps your analysis. Time value of money 21 Beware of conventions Time value of money Cash flow diagram Interest Factors : ANSI Standard Notation Consists of two cash flow symbols, the interest rate, and the number of time periods General form: (X/Y,i%,n) X represents what is unknown ‐ Y represents what is known i and n represent input parameters; can be known or unknown depending upon the problem Example: (F/P,6%,20) is read as: the factor needed to find F, given P when the interest rate is 6% and the number of time periods equals 20. In problem formulation, the standard notation is often used instead of the closed‐form equivalent relations (factor) Factors 24 29 30 Future value Present value Factors F/P and P/F Single‐payment factors Objective Derive the present of future worth of a cash flow Fn Cash flow diagram i% / period 0 1 2 3 n‐1 n P0 Formulas F P (1 i ) n 1 and P F n (1 ) i !!! “THE” FORMULA !!! Factors 32 Computation: straightforward F1 P iP P (1 i ) F2 P (1 i ) iP (1 i ) P (1 i i i 2 ) P (1 i )2 F3 F2 iF2 (1 i )F2 P (1 i )3 etc. after n periods : Fn F P (1 i )n Factors 33 Example Known: P=$12000; n=24 years; i=8% per year Unknown: F Finding F: (a) F=P*(1+i)n=$12000*(1+0.08)24=$76094 (b) FV(i%,n,A,P)=FV(8%,24,,12000)=($76094) Factors 34 Example How long will it take for $1000 to double if the interest rate is 5% per year ? The n value can be obtained using the P/F or F/P factor P=F(P/F,i,n) → 1000=2000*(P/F,5%,n) → (P/F,5%,n)=0.5 From the interest tables we see that n is between 14 and 15 years, by interpolation we obtain: n=14.2 years. Alternatively, we can use the spreadsheet function NPER(5%,0,‐1000,2000) to obtain n=14.2 years. It is not so much a matter of studying all kinds of different formulas, but rather of knowing the basics and of working carefully. Excel can be a powerful help, as are the tables; use them correctly. Factors 35 36 Present worth factor (uniform series) Captal recovery factor (uniform series) Factors P/A and A/P Uniform series factors Cash flow diagram i% per interest period . . . . 0 1 2 3 n‐2 n‐1 $A per interest period n Find P Required: To find P given A Cash flows are equal, uninterrupted and flow at the end of each interest period Factors 37 Computing … Factors Typical computational procedure for EE formulas 38 Example How much money should you be willing to pay now for a guaranteed $600 per year for 9 years starting next year, at a rate of return of 16 % per year ? Cash flow diagram Computation with Factors or … (1 i ) n 1 P A 600 n i (1 i ) (1 0 .1 6 ) 9 1 $2764 9 0 .1 6(1 0 .1 6 ) 41 Typically more solution approaches possible present future year amount worth (P) worth n A 1/(1+0.16)^n amount (1+0.16)^(9‐n) 0 1 600 0.862 517.241 3.278 2 600 0.743 445.898 2.826 3 600 0.641 384.395 2.436 4 600 0.552 331.375 2.100 5 600 0.476 285.668 1.811 6 600 0.410 246.265 1.561 7 600 0.354 212.298 1.346 8 600 0.305 183.015 1.160 9 600 0.263 157.772 1.000 sum 2763.926 (F) amount 1967.049 1695.732 1461.838 1260.205 1086.384 936.538 807.360 696.000 600.000 10511.105 future * 1/(1+0.16)^9 2763.926 present 42 Sinking fund factor Uniform‐series compound amount factor Factors A/F and F/A Cash flow diagram F = given i% per interest period . . . . 0 1 2 3 n‐2 n‐1 A=? per interest period Find A, given F Formula n P 1 i (1 i ) n AF n n (1 i ) (1 i ) 1 i A F n (1 ) 1 i n (1 i ) 1 and if A is given and F is to be found F A i Factors 43 Exercises a) Formasa Plastics has major fabrication plants in Texas and Hong Kong. The president wants to know the equivalent future worth of a $1 million capital investment each year for 8 years, starting 1 year from now. Formosa capital earns at a rate of 14 % per year. Answer: $ 13 233 b) How much money must Carol deposit every year from now at 5.5 % per year in order to accumulate $ 6000 seven years from now ? Answer: $ 726 per year Factors 45 (a) F = 1000(F/A,14%,8) = 1000*13.2328 = $ 13233 46 (b) A = 6000 (A/F, 5.5 %, 7) = 6000 * 0.12098 = $ 725.76 per year (0.12282+0.11914)/2 =0.12098 49 Gradient factors ‐ 1 Arithmetic gradient series Definition: Cash flow series that either increases or decreases by a constant amount; the amount of the increase or the decrease is the (arithmetic) gradient G; G can be positive or negative Cash flow profile CFn=base amount+(n‐1)G Formula Factors G (1 i ) n 1 n P= i i (1 i ) n (1 i ) n 52 53 54 55 Note that … A1+(n-1)G A1+(n-2)G Find P, given gradient cash flow G A1+2G A1+G “Conventional gradient” Base amount = A1 0 1 2 3 n-1 n CFn = A1 ± (n‐1)G Factors 56 Also Arithmetic‐gradient‐uniform‐series factor (A/G,i,n) 1 n AG n i (1 i ) 1 Arithmetic‐gradient future worth factor (F/G,i,n) 1 (1 i ) n -1 F G - n i i 59 some algebra 60 Example (ex. 2.10) (a) P ? (b) AT ? Cash flow diagram 61 62 63 Gradient factors ‐ 2 Geometric gradient series Definition: Cash flow series that either increases or decreases by a constant percentage from period to period; the uniform rate of change is called the geometric gradient Cash flow diagram Formula Factors 1 g n 1 1 i gi Pg A1 ig 64 Start with: A1 A1 (1 g ) A1 (1 g ) 2 A1 (1 g ) n 1 Pg ... (1 i )1 (1 i ) 2 (1 i )3 (1 i ) n (1) Factor out A1 out and re‐write 1 (1 g )1 (1 g ) 2 (1 g ) n 1 ... Pg A1 2 3 (1 i ) (1 i ) (1 i ) (1 i ) n (2) Multiply by (1+g)/(1+i) to obtain Eq. (3 ) Pg (1 + g ) (1 + g ) 1 (1 g ) 1 (1 g ) 2 (1 g ) n 1 . .. A1 (1 + i) ( 1 + i ) (1 i ) (1 i ) 2 (1 i ) 3 (1 i ) n (3) Subtract Eq. (2 ) from Eq. (3 ) to yield (1 g ) n 1 1+g Pg 1 A1 n 1 1 i 1+i (1 i ) Solve for Pg and simplify to yield…. In case (g=i), go back to (1), replace g by i and simplify the equation Factors 1 g n 1 1 i gi Pg A1 ig nA1 Pg (1 i) 65 Example 2.11 66 1 g n 1 1 i Pg A1 gi ig 67 Combining factors Most real life cash flow series do not fit exactly the series for which the basic formulas above were developed. Therefore it is almost always necessary to make combinations. Combining factors 69 Example Combining factors 71 Taking closer look … Cash flow diagram Total present worth at time 0 Present worth of uniform annual series at time other than 0 Present worth of uniform annual series at time 0 Combining factors 72 PT P0 PA with P0 5000 In year 0 and PA P * (P / F ,8%,2) ' A and P 500 * (P / A,8%,6) ' A (1 i )n 1 also (P / A, i %, n ) i (1 i )n thus Of the shifted series PA’=present worth of the A series (numbers below the line) but also a future worth for the current time (numbers on above the line) PT 5000 500 * (P / A,8%,6) * (P / F ,8%,2) with P / F 1/ (1 i )n PT 5000 500 * (4.6229) * (0.8573) PT $ 6981 Combining factors 73 Example Combining factors Example (Ex. 3.8) 74 Type Base amount ($) Gradient ($) Period (years) Start PI investment 2000 500 5 Year 1 P2 withdrawal 5000 ‐1000 5 Year 6 P3 withdrawal 2000 ‐‐‐ 2 Year 11 75 Nominal and effective interest rates Nominal interest rate r An interest rate that does not include any consideration of compounding(*) Thus r = interest rate per period * number of periods Examples “8 % per year” “1.5 % per month” Nominal and effective interest rates 80 (*) interest earned on interest Effective interest rate ia The effective interest rate is the actual rate that applies for a stated period of time; it accounts for the compounding of interest during the time period of the corresponding nominal rate. It is is commonly expressed on an annual basis denoted as “ia” This is what really matters to properly account for the time value of money Examples “4 % per year compounded monthly” Nominal and effective interest rates 81 In previous examples all interest rates had t and CP = 1 year, so m=1; the interest rates were effective rates Interest rates – three time based units Time Period t – The period over which the interest is expressed (always stated). Ex: “1% per month” Compounding Period (CP) – The shortest time unit over which interest is charged or earned. Ex: “8% per year, compounded monthly” Compounding Frequency – The number of times m that compounding occurs within time period t. Ex: “10% per year, compounded monthly” has m = 12 Ex: “1 % per month, compounded monthly” has m=1 and r % per time period t r effective rate per CP m compounding periods per t m Nominal and effective interest rates 82 Nominal and effective interest rates 83 Effective annual interest rates Nominal and effective interest rates 84 Let r = nominal interest rate per year m = number of compounding periods per year i = effective interest rate per compounding period (CP) = r/m ia = effective interest rate per year Then F=P+iaP=P(1+ia) and also F=P(1+i)m Which leads to i a (1 i ) 1 m 85 Example r i=r/m F=P*(1+ia) Nominal and effective interest rates ia=(1+i)m‐1 87 Nominal and effective interest rates 88 Effective interest rates for any time period Compounding period (CP) and payment period (PP) do not necessarily coincide. Example A company deposits money each month into an account that pays a nominal interest of 14 % per year, compounded semiannualy, then compounding period CP=6 months, payment period PP=1 month Formula m r effective i 1 1 m where r nominal interest rate per payment period (PP ) m number of compounding periods per payment period (CP per PP ) Nominal and effective interest rates 89 90 Example (p 10) 91 Example Example (Ex. 4.4) Nominal and effective interest rates 92 Equivalence relations: PP vs CP Equivalence concept decision support EE PP<CP or PP>CP Straightforward application of formulas Different ways of computing Special case: continuous compounding 98 Interest rates that vary over time In practice, interest rates do not stay the same over time unless by contractual obligation. There can exist “variation” of interest rates over time. If required, it is best to build a spreadsheet model. It can be a cumbersome task to perform Nominal and effective interest rates 115 Example Example (Ex. 4.13) Nominal and effective interest rates 116 Note: About the handbook Every chapter is nicely structured and situated in the course material (course book). Every chapter has many worked‐out examples. (additional to ones discussed in class). Every chapter has Chapter summary Problems Additional problems and FE exam review questions Case study 121 Exercises: rather straightforward computations & somewhat harder ‐ Insights, links … important see workshops 122 Multiple choice format (FE=fundamentals of engineering) 125 More/less computational work, more insight and critical managemt thinking required, “case study” See workshops & classes SEE WORKSHOP 1 128 130 131 Ch. 1‐4 Wrap‐up Time value of money: equivalence, simple and compounded interest, MARR, cash flow diagram, WACC Factors: F/P, P/F, P/A, A/P, A/F, F/A, gradient factors – Combining factors Interest rates: nominal, effective, compounding period, payment period, continuous compounding, varying over time Symbols P, A, F, G, … and Notation (X/Y,i%,n) Excel built‐in functions – Tables p 581 132 Engineering Economy Present worth and annual worth analysis 1 Overview Formulating alternatives Analysis – “Criteria” Present worth (PW) analysis Annual worth (AW) analysis 2 Formulating alternatives Process (fig. 5.1) Formulating alternatives 3 First “filter” Example A company is expecting a change in regulations within a few years. This will mean that their current machines will have to be replaced by new ones (solvents, Pb). They can replace the machines now or wait for the new legislation (Do nothing). Example Your company needs to boost production in order to cope with an increased market demand. Alternatives Expand capacity of current production line Build a similar second line Build a new line using new technology Buy a new line and customize Example A hospital wants to invest in new pediatric beds. After collecting information in and outside the hospital and some market research 12 different suppliers were identified. After a first screening 2 suppliers were discarded because of their location and 1 because of non‐compliance with patient safety guidelines. 5 Categories Mutual exclusive Every alternative is a viable project. Only one viable project can be selected. E.g. Pediatric bed example: only one supplier out of 9 potential suppliers will be selected by the economic analysis. Independent More than one viable project may be selected. E.g. The safety engineer has come up with a list of several projects to enhance occupational safety beyond what is legally compulsory. More than one project can be executed if selected by the economic analysis. Often however there is a budget limitation. 6 Type of alternatives Revenue Each alternative generates costs (disbursement) and revenue (or receipt) cash flows and/or savings. E.g. A new production line generates costs: initial investment, O&M costs, … and savings: personnel cost (automated line). Service Each alternative has only cost cash flows. Revenues or savings are not dependent upon the alternative selected. Sometimes these are also not estimable. E.g. Public projects (tunnel for pedestrians, highway). 7 Present worth analysis Objective One alternative: Selection (interesting yes/no ?) More than one alternative: Mutual exclusive: classification (most interesting ?) Independent: selection (all intereresting ones) What does “interesting” mean ? An alternative is “interesting” if the PW at MARR is ≥ 0. financially interesting Careful When comparing alternatives, the period considered should be “the same” for all alternatives. equivalence 8 Equal‐life alternatives Present worth analysis 9 =3.7908 (table 15) Present worth analysis =0.6209 (table 15) 11 Different life‐alternatives The PW of the alternaties must be compared over the same number of years and end at the same time, i.e. “equal service requirement”. Possible approaches Compare alternatives over a period of time equal to the least common multiple (LCM) of their lives. Compare the alternatives using a study period of length n years (planning horizon), which does not necessariy take into consideration the useful life of the alternatives. Present worth analysis 12 Example 13 Present worth analysis Consider 18 years Present worth analysis 14 15 16 Future worth analysis Example Present worth analysis 17 today Present worth analysis 18 Present worth analysis 19 Capitalized cost calculation and analysis Definition Capitalized cost (CC) is the present worth of an alternative that will last “forever”. Used Public projects such as bridges, dams, irrigation systems, railroads, highways, tunnels, … Permanent and charitable organization endowments Formula A CC i Present worth analysis 20 Formula derivation Start from On the right hand side, divide both numerator and denominator by (1+i)n If “n” approaches the above reduces to: Present worth analysis (1 i)n 1 P A n i (1 i ) 1 1 (1 i)n P A i A P i A CC i 21 Approach (p 139) Draw a cash flow diagram showing all non‐recurring cash flows and at least two cycles of all recurring cash flows Find the present worth (P) of all nonrecurring amounts: CC component Find the equivalent uniform annual worth (A) through one life cycle of all recurring amounts Add this to all other uniform amounts occurring in years 1 to infinity: uniform annual worth (AW) Divide this AW by i to obtain the corresponding CC component Add the CC of the non‐recurring amounts to the CC of the recurring amounts to obtain the overall CC. Present worth analysis 22 Example Present worth analysis 25 Suspension bridge: Initial cost Right‐of‐way Inspection and maintenance cost Concrete resurfacing Truss bridge: Initial cost Right‐of‐way Inspection and maintenance cost Sandblasting Painting Note: (1) Cost categories are similar but not identical 26 (2) Only costs – nothing on time needed, effectiveness, impact on citizens/businesses, …. Suspension Bridge Alternative: 0 1 2 3 4 . . . . . 10 11 …….. $50 Million (initial) Annual inspection costs = $35,000/yr=A1 $2 Million (right‐of‐way) Resurfacing=$100,000 A2=‐100000(A/F,6%,10)=‐7587 =‐50‐2=‐52 (106) = =‐709783 . =‐52 106‐709783=‐52.71 106 Truss Design: 0 1 2 3 4 5 6 7 8 9 10 11 ….. $25 Million (initial) Annual Maintenance costs = $20,000/yr=A1 Sandblast: $‐190,000 A3=‐190000(A/F,6%,10)=‐14415 Paint: $‐40,000A2=‐40000(A/F,6%,3)=‐12564 $15 Million (right‐of‐way) =‐25‐15=‐40 (106) = . =‐782983 =‐40 106‐782983=‐40.78 106 Payback period analysis The payback period estimates the time np to recover the initial investment in a project; there are two forms 1. 2. With no interest ‐‐ i = 0% (no‐return payback) With an assumed interest rate ‐‐ i > 0% (discounted payback analysis) Formulas for np t n p 0 P NCFt t 1 P For a uniform NCF: n p NCF t n p 0 P NCFt ( P / F , i, t ) t 1 (Discounted PBP) (Undiscounted/no return PBP) 37 Common managerial philosophy is that a shorter payback is preferred to a longer payback period; this is not a good approach from engineering economic point of view: PBP ignores all cash flows after the payback time period; may not use all of the cash flows in the cash flow sequence – myopic approach ! Never use payback analysis as the primary means of making an accept/reject decision on an alternative! Not a preferred method for final decision making – better as a screening tool Present worth analysis 38 Example 40 Excel or tables 41 Use predefined formula NPER =NPER(i,A,P,F)=NPER(0.15,3,‐18,3) Try to determine n through trial‐and‐error 42 Example 43 44 Life cycle cost Life cycle costing (LCC) Considers the whole life cycle of the equipment, project, etc. Cost elements: many different ones, not always easy to estimate (esp. if far in future) LCC <> TCO (total cost of ownership) Present worth analysis 45 47 Present worth analysis It is not unusual to have 75% to 85% of the entire life span LCC committed (locked in) during the preliminary and detail design stages. Watch out with ad hoc “cost‐savings” (eg use of inferior components) Actual costs (maybe higher than B if design flaws) Improved design 50 Example (ex. 5.9) Present worth analysis 52 Challenge is not to forget any cost element and to make accurate estimations ! Present worth analysis 53 PW= (0.5+1187+3067)+(6512+20144+13142)=$44822 million FW=PW(F/P, 18%,10)=$234.6 million 54 “Real life” example 55 Strictly speaking: TCO ! 56 Annual worth analysis Popular analysis technique, because rather “intuitive” and easily understood Results reported in $ per time period, e.g. $/yr Eliminates the least common multiple problem associated with the present worth method: Only one life cycle to evaluate Applicable to a variety of engineering economy studies such as: Asset replacement; Breakeven analysis; Make‐or‐Buy decisions; Studies dealing with manufacturing costs; Economic value added (EVA) analysis Formula AW = PW(A/P,i%,n) or AW = FW(A/F,i%,n) Annual worth analysis 57 Repeatability assumption Given alternatives with unequal lives, the assumptions are 1. The services provided are needed forever 2. The first cycle of cash flows is repeated for all successive cycles in the same manner 3. All cash flows will have exactly the same estimated values in every life cycle. Similar cycles Not very reasonable in an industrial context Annual worth analysis At least for the least common multiple (LCM) of the lives of the alternatives Use study period approach 58 Example (5.2 – 6.1) Recall: Comparison of two alternatives. T(A)=6 years, T(B)=9 years LCM of 6 and 9 = 18 years; outcome: PWA=$‐45036 and PWB=$‐41384 Annual worth analysis 59 5.2 – 6.1) Recall: Comparison of two alternatives. T(A)=6 years, T(B)=9 years LCM of 6 and 9 = 18 years; outcome: PWA=$‐45036** and PWB=$‐41384 Question: Would an AW analysis yield the same result ? Is already annual worth Test: Here in first life cycle: AW=‐15000(A/P,15%,6)+1000(A/F,15%,6)‐3500=$‐7349 Retake **: AW=‐45036(A/P,15%,18)=$‐7349 Answer: Yes Annual worth analysis 60 Capital recovery and AW values An economic alternative should have the following cash flow estimates made Initial investment ‐‐ P Estimated future salvage value ‐‐ S Beware: S>0 if there is a market or trade‐in value and S<0 if it will cost money for disposal, S=0 is also possible Estimated annual operating costs – AOC Costs for service alternatives. Costs and receipts for revenue alternatives and Estimated life ‐‐ n Interest rate ‐‐ i% (this is usually the MARR) Annual worth analysis 62 Annual worth (AW) Is composed of two factors: capital recovery (CR) for the initial investment and the equivalent annual amount (AOC), annual operating or maintenance & operating cost) AW=‐CR‐AOC Capital recovery (CR) Is the equivalent annual amount that the asset (or process) must earn every year to just recover the initial investment plus a stated rate of return over its expected life; any expected salvage value is to be taken into account CR=‐[P(A/P,i,n)‐S(A/F,i,n)] Annual worth analysis 63 Example (ex. 6.2) 65 Present worth (PW) of the two separate investment amounts (P0 and P1): 8+5(P/F,12%,1) Capital recovery CR=‐[P(A/P,i,n)‐S(A/F,i,n)]=capital recovery = ‐ {[8+5(P/F,12%,1)](A/P,12%,8) ‐ 0.5(A/F,12%,8)} =$‐2.47 Interpretation: Lockheed Martin should have an equivalent total revenue from the tracker of at least $‐2.47 106 just to recover the initial present worth investment plus the required return of 12% per year. This does not include the annual operating costs AOC of $0.9 106 each year. Total annual worth AW=‐CR‐AOC=‐2.47‐0.9 or AW=$‐3.37 106 per year. This is the AW for all future life cycles (provided …) 66 Example (ex. 6.3) Annual worth analysis 67 68 Mutual exclusive alternatives and AW Calculate AW at the MARR for each alternative; If AW≥0, MARR is met or exceeded, the alternative in question is economically justifiable. Select one with lowest AW of costs (service) or highest AW of net incomes (revenue), i.e. select the numerically largest AW alternative. If AW < 0 at MARR, the (revenue) alternative is not economically justifiable, since initial investment P is not recovered over n years at the required rate of MARR = i% per year Annual worth analysis 70 Example 72 Initial cost Salvage value 73 Additional exercise Problem 6.8 74 FE review problem 76 Ch. 5‐6 Wrap‐up Formulating alternatives Process steps. Types. Present/future worth analysis PW of (non)equal life alternatives. FW analysis. Capitalized cost (CC). Payback period (PBP). Life cycle cost (LCC). Annual worth analysis Importance. AW calculation. Capital recovery (CR). Alternative selection by AW. 78 ! Cash flow diagrams ! Right formulas ! Management interpretation ! Engineering Economy Rate of return analysis: Single alternatives – Multiple alternatives 1 Overview ROR analysis for single alternatives Basis Multiple ROR ROR analysis for multiple alternatives (mutual exclusive) Two alternatives More than two alternatives 2 Single alternative 4 Interpretation of a rate of return (ROR) ROR (rate of return), also called IRR (Internal Rate of Return) method or marginal efficiency of capital method is one of the popular measures of investment worth. Definition: ROR is either the interest rate paid on the unpaid balance of a loan, or the interest rate earned on the unrecovered investment balance of an investment such that the final payment or receipt brings the terminal value to exactly equal “0”. The ROR is found using a PW (present worth) or AW (annual worth) relation. The rate determined is called i* Single alternative 5 ROR (i*) is the interest rate earned/charged on the unrecovered balance of a loan or investment project. ROR is not the interest rate earned on the original loan amount or investment amount (P) The i* value is compared to the MARR (minimal attractive rate of return) If i* > MARR, investment is justified If i* = MARR, investment is justified (indifferent decision) If i* < MARR, investment is not justified Mathematically: ‐100%≤i*≤+∞ An i* = ‐100% signals total and complete loss of capital One can have a negative i* value (feasible) but not less than –100% All values above i* = 0 indicate a positive return on the investment Single alternative 6 Example Single alternative 7 Single alternative 8 Single alternative 9 ROR calculation using a PW or AW equation Solve for i* either PWcosts/disbursements=PWincomes/receipts or AWcosts/disbursements=AWincomes/receipts using either trial and error or Excel There is no closed formula for the RoR Trial and error procedure 1. 2. 3. 4. 5. Draw a cash flow diagram Set up the appropriate PW equivalence equation and set equal to 0 Select values of i and solve the PW equation Repeat for values of i until “0” is bracketed, i.e., the equation is balanced May have to interpolate to find the approximate i* value Excel functions Rate(n,A,P,F) or IRR(first_cell:last_cell, guess). Single alternative When an A series is given When the cash flows vary from period to period 10 Example Single alternative 11 Single alternative 12 Single alternative 14 RoR method vs PW, AW, FW When applied correctly, ROR method will always result in a good decision and should be consistent with PW, AW, or FW methods. However, for some types of cash flows the ROR method can be computationally difficult and/or lead to erroneous decisions Multiple RoRs Single alternative 15 Multiple ROR values A class of ROR problems exist that will possess multiple i* values. This is the case when there is an nonconventional or nonsimple cash flow series, i.e. the net cash flows switch from negative to positive from one year to another, so that there is more than one sign change. Single alternative 17 Tests for multiple i* values Descartes’ rule of signs: the total number of real values i* is always less than or equal to the number of sign changes in the original cash flow series Cumulative cash flow sign test Norstrom’s criterion – sufficient, not necessary condition: form the series of cumulative cash flows (CCF): S0, S1, S2, …; if S0 <0 and the sign changes only once in the CCF series, there is a single, real number positive i* Single alternative 18 Example Single alternative 19 Single alternative Descartes the total number of real values i* is always less than or equal to the number of sign changes in the original cash flow series Norstrom form the series of cumulative cash flows S0, S1, S2, …; if S0 <0 and the sign changes only once in the CCF series, there is a single, real number positive i* 20 21 Removing Multiple i* Values The result of follow‐up analysis to obtain a single ROR value when multiple, non‐useful i* values are present does not determine the internal rate of return (IROR) for nonconventional net cash flow series. The resulting rate is a function of the additional information provided to make the selected technique work and the accuracy is further dependent upon the reliability of this information. The resulting value will be referred to as external rate of return (EROR) as a reminder that it is different from the IROR discussed before. External = “external to the project “ 7‐26 Take the following view: You are the project manager and the project generates cash flows each year. Some years produce positive net cash flow (NCF) and you want to invest the excess money at a good rate of return, in some source external to the project. We will call this the investment rate or reinvestment rate ii. Other years, the NCF will be negative and you must borrow funds from some external source to continue the project. The interest rate you pay should be as small as possible, we will call this the borrowing rate or finance rate ib. 7‐27 Two approaches to determine External ROR (EROR) (1) Modified ROR (MIRR) (2) Return on Invested Capital (ROIC) Note that the results of the two methods for rectifying the multiple ROR situation will not be the same, because slightly different additional information is necessary and the cash flows are treated in slightly different fashions from the time value of money view point (1) Modified ROR Approach (MIRR) Four step procedure: Think of WACC (weighted cost of capital) Determine PW in year 0 of all negative CF at ib Determine FW in year n of all positive CF at ii Calculate EROR = i’ by FW = PW(F/P,i’,n) If i’ ≥ MARR, project is economically justified Think of MARR (minimum attractive rate of return) © 2012 by McGraw‐Hill All Rights Reserved 7‐30 31 MIRR example For the NCF shown below, find the EROR by the MIRR method if MARR = 9%, ib = 8.5%, and ii = 12% Year 0 1 2 3 NCF +2000 ‐500 ‐8100 +6800 Solution: PW0 = ‐500(P/F,8.5%,1) ‐ 8100(P/F,8.5%,2) = $‐7342 FW3 = 2000(F/P,12%,3) + 6800 = $9610 PW0 (F/P,i’,3) + FW3 = 0 ‐7342(1 + i’)3 + 9610 = 0 i’ = 0.939 (9.39%) Since i’ > MARR of 9%, project is justified© 2012 by McGraw‐Hill All Rights Reserved 7‐32 (2) Return on Invested Capital Approach (ROIC) Measure of how effectively project uses funds that remain internal to project ROIC rate, i’’, is determined using net‐investment procedure Extra funds available Three step Procedure (1) Develop series of FW relations for each year t using: Project uses all available funds Ft = Ft‐1(1 + k) + NCFt where: k = ii if Ft‐1 > 0 and k = i’’ if Ft‐1 < 0 (2) Set future worth relation for last year n equal to 0 (i.e., Fn= 0); solve for i’’ © 2012 by McGraw‐Hill All Rights (3) If i’’ ≥ MARR, project is justified; otherwise, reject Reserved 7‐33 ROIC Example For the NCF shown below, find the EROR by the ROIC method if MARR = 9% and ii = 12% Year 0 1 2 3 NCF +2000 ‐500 ‐8100 +6800 Solution: Year 0: F0 = $+2000 F0 > 0; invest in year 1 at ii = 12% Year 1: F1 = 2000(1.12) ‐ 500 = $+1740 F1 > 0; invest in year 2 at ii = 12% Year 2: F2 = 1740(1.12) ‐ 8100 = $‐6151 F2 < 0; use i’’ for year 3 Year 3: F3 = ‐6151(1 + i’’) + 6800 Set F3 = 0 and solve for i’’ ‐6151(1 + i’’) + 6800 = 0 i’’= 10.55% Since i’’ > MARR of 9%, project is justified © 2012 by McGraw‐Hill All Rights Reserved 7‐34 Important Points for EROR About the computation of an EROR value EROR values are dependent upon the selected investment and/or borrowing rates Commonly, multiple i* rates, i’ from MIRR and i’’ from ROIC have different values About the method used to decide For a definitive economic decision, set the MARR value and use the PW or AW method to determine economic viability of the project © 2012 by McGraw‐Hill All Rights 7‐35 Reserved Multiple alternatives 42 Two alternatives 43 Why incremental analysis is needed Illustration Assume MARR=16% ‐ Available budget for investment: $ 90000 Alternative A: investment: $ 50000 – IROR i*=35 % Alternative B: investment: $ 85000 – IROR i*=29% At first sight Alternative A is to be preferred But What to do with the excess investment capital ? Well, General assumption: excess funds are invested at MARR Overall RORA=[50000(0.35)+40000(0.16)]/90000=26.6 % Overall RORB=[85000(0.29)+5000(0.16)]/90000=28.3% Thus Alternative B is the most interesting one. Multiple alternatives 44 Under circumstances, project ROR values do not provide the same ranking of alternatives as do PW, AW and FW analyses. This situation does not occur if we conduct an incremental cash flow ROR. Equal service life assumption Multiple alternatives 45 Interpretation of ROR on extra investment Guidelines (mutual exclusive projects) For multiple revenue alternatives, calculate the IROR i* for each alternative, and eliminate all alternatives that have an i* <MARR. Compare the remaining alternatives incrementally. If the ROR available through the incremental cash flow equals or exceeds the MARR, the alternative with the extra investment should be selected. Note For independent projects, no comparison on the extra investment. The ROR is used to accept all projects with i* >MARR. Multiple alternatives 46 ROR evaluation using PW: incremental and breakeven Procedure for incremental ROR analysis Multiple alternatives 48 Example Multiple alternatives Is the more expensive B justified, after all it has lower annual costs and a higher salvage value ... ? 50 Multiple alternatives 51 year A B incremental cumulative incremental 0 ‐8000 ‐13000 ‐5000 ‐5000 1 ‐3500 ‐1600 +1900 ‐3100 2 ‐3500 ‐1600 +1900 ‐1200 3 ‐3500 ‐1600 +1900 700 4 ‐3500 ‐1600 +1900 2600 5 ‐3500 ‐1600+2000‐13000 +1900‐11000 ‐6500 6 ‐3500 ‐1600 +1900 ‐4600 7 ‐3500 ‐1600 +1900 ‐2700 8 ‐3500 ‐1600 +1900 ‐800 9 ‐3500 ‐1600 +1900 1100 10 ‐3500 ‐1600+2000 +1900+2000 5000 52 Multiple alternatives 53 Procedure with Breakeven Breakeven Interest Rate (Fisherian Intersection Rate) i*(B‐A) is the interest rate at which the two alternatives are economically equivalent. More on breakeven in Chapter 13. Multiple alternatives 54 Example 55 56 ROR using AW Two methods Solve AW relation on incremental cash flow series over LCM Solve equation of respective AW relations – one cycle 60 Example Multiple alternatives 61 (see higher) AWA=AWB (1) (2) Multiple alternatives 63 Many alternatives 64 Incremental ROR analysis of multiple, mutually exclusive alternatives Multiple alternatives 66 Example Multiple alternatives 67 C=challenger DN=defender Multiple alternatives 68 Some exercises … 74 81 intermezzo Single value IROR: (internal) rate of return; i* Multiple values EROR: external rate of return borrowing rate or finance rate; ib investment rate or reinvestment rate; ii MIRR: modified rate of return; i’ ROIC: return on invested capital, i’’ Two alternatives incremental ROR; ∆i* 48 Will be covered in the work shops 95 Wrap‐up Chapters 7‐8 ROR analysis: Single alternative Interpretation ROR using PW or AW equation Multiple ROR values ROR analysis: Multiple alternatives Two alternatives Why incremental analysis is necessary – Interpretaionof ROR on the extra investment ROR using PW: incremental and breakeven ROR using AW Many alternatives ROR analysis of multiple, mutual exclusive alternatives 97 Engineering Economy Making choices: Selecting the basic analysis tool 1 Overview What is the best method to use ? If no method is preselected … If method is predetermined … 2 What is the best method to use ? Comparing Mutually Exclusive Alternatives by Different Evaluation Methods Different problem types lend themselves to different engineering economy methods Different information is available from different evaluation methods Primary criteria for what method to apply Speed Ease of performing the analysis Note: “Best” method ≠ Only method (e.g. PW and AW) 3 If no method is preselected Aspects to consider Evaluation period (→ es mated life) Type of alternatives (private (income, cost) vs service‐based (cost)) Recommended method (≠ only possible method) Series to evaluate (CFs for 1 alternative, incremental CFs for two alternatives) 4 5 If method is predetermined Aspects to consider Equivalence relationship (basic: PW, AW – CC, FW, … see chapter 6) Lives of alternatives – Time period for analysis PW: LCM of all alternatives Incremental ROR and B/C: LCM of the two alternatives being compared AW: analysis over respective alternative lives Series to evaluate Estimated cash flows or incremental series Rate of return (interest rate) MARR Decision guideline Final choice 6 7 Examples (see handbook) 8 Wrap‐up Epilogue What is the best method to use ? If no method is preselected If method is predetermined INSIGHT IN METHODS! THINK! 17 Engineering Economy Benefit/cost analysis and public sector economic Benefit/cost analysis in a company context Overview Public (and private) sector Type of projects – B/C ratios ‐ Single and multiple projects Ethical considerations Private (and public) sector Type of projects – C/E analysis – C/B issues Public/private sector C/E analysis (CEA) ‐ multiple projects C/B: MADM ‐ MCDM Public sector projects Examples Public sector Characteristic Public sector Private sector Size of investment Larger Some large, more medium to small Life estimates Larger (30‐50+ years) Shorter (2‐25 years) Annual cash flow estimates No profit; costs, benefits and Revenues contribute to disbenefits are estimated profits; costs are estimated Funding Taxes, fees, bonds, private funds Stocks, bonds, loans, individual owners Interest rate Lower Higher, based on market cost of capital Alternative selection criteria Multiple criteria Primarily based on rate of return Environment of the evaluation Politically inclined Primarily economic Public sector Note‐1 Costs: estimated expenditures to the government entity for construction, operation and maintenance of the project Benefits: advantages to be experienced by the owners, the public Disbenefits: expected undesirable or negative consequences to the owners if the alternative is implemented, e.g. indirect economic disadvantage It is difficult to estimate and agree upon the economic impact of (dis)benefits for a public sector alternative. Note‐2 The viewpoint of the public sector analysis must be determined before cost, benefit an disbenefit estimates are made and before the evaluation is formulated and performed. There are several viewpoints for any situation and the different perspectives may alter how a cash flow estimate is classified. Public sector Note‐3 Contractor does not share project risk Fixed price ‐ lump‐sum payment Cost reimbursable ‐ Cost plus, as negotiated Contractor shares in project risk Public‐private partnerships (PPP), such as: Design‐build projects ‐ Contractor responsible from design stage to operations stage Design‐build‐operate‐maintain‐finance (DBOMF) projects ‐ Turnkey project with contractor managing financing (manage cash flow); government obtains funding for project Example costs A flood control project will have a first cost of $1.4 million with an annual maintenance cost of $40,000 and a 10 year life. Reduced flood damage is expected to amount to $175,000 per year. Lost income to farmers is estimated to be $25,000 per year. At an interest rate of 6% per year, should the project be undertaken? benefits disbenefits Note‐3: private contractors in public projects Contractors does not share project risk Fixed price ‐ lump‐sum payment Cost reimbursable ‐ cost plus, as negotiated Contractor shares in project risk Public‐private partnerships (PPP), such as: Design‐build projects ‐ Contractor responsible from design stage to operations stage Design‐build‐operate‐maintain‐finance (DBOMF) projects ‐ Turnkey project with contractor managing financing (manage cash flow); government obtains funding for project (out‐of‐scope) Example (ex. 9.1) Public sector Public sector Citizen of the city City budget Economic development C B D C B D C B D cost benefit disbenefit cost benefit disbenefit cost benefit disbenefit 1 Annual cost bonds X X X 2 Annual maintenance X X X 3 Annual parks development X X X 4 Annual loss commercial X 5 Loss of sales tax rebates X X X X X 6 Annual income park X X X 7 Savings flood control projects X X X 8 Property damage avoided X X X Citizen: “maximize quality and wellness of citizens, family and neighborhoods” City budget: “ensure budget is balanced and sufficient to fund rapidly growing city services” Economic development: “promote industrial and commercial development (jobs)” Benefit/cost analysis of a single project Basically: B/C ratio<1: not economically acceptable, B/C ratio ≥ 1: project acceptable PW of benefits AW of benefits FW of benefits B/C PW of cos ts AW of cos ts FW of cos ts (1) The conventional B/C approach benefits disbenefits B D conventional B / C cos ts C (2) The modified B/C approach mod ified B / C Public sector benefits disbenefits M & O cos ts initial investment Note: salvage value = negative cost (denominator) Benefit (savings) Cost (initial investment) Example (ex. 9.2) Cost (“M&O”) Public sector Disbenefit (negative consequence) Public sector Alternative selection using incremental B/C analysis for two alternatives Public sector Example (ex. 9.4) Public sector Public sector All costs in $ Alternative A Alternative B ∆ (B‐A) Costs 685000 1030750 345000 Benefits 200000 450000 250000 Disbenefits 500000 400000 ‐100000 Incremental B/C=250000/345250=0.72 A Incremental (B‐D)/C={250000‐(‐100000)}/345250=1.01 B Incremental B/C analysis of multiple, mutually exclusive alternatives Given three or more mutually exclusive alternatives; one must be selected Conduct the pair‐wise ∆B/C analysis (cfr ROR chapter) The selection rule is: Choose the largest‐cost alternative that is justified with an incremental B/C ≥ 1 when this alternative has been compared with another justified alternative Benefits Direct benefits Usage cost estimates Public sector Example (ex. 9.5) Public sector Public sector Public sector Public sector Ethical considerations Target group Time impact Community Money only ? Scope (disbenefits) Private sector “C/B” projects Examples Environment Occupational safety Different stakeholders Automation DSS‐BI Chemical company has to choose between alternatives for cleaning polluted site before selling it. Manufacturing company wants to invest in oxygen masks for the paint spraying department. Food company wants to buy new dust masks for their production workers. Hospital wants to replace all the current cots by new ones. Midsized company wants to implement ERP software. A company is considering to invest in a large scale BI (business intelligence) project (hardware, software, …) Private sector Special case: C/E analysis Cost‐effectiveness analysis: basics Cost are (of course) in monetary units, Benefits are expressed in non‐monetary, but quantitative measures, e.g. % of risk reduction; % of throughput increase, response time, amount of pollution removed, number of occupational accidents avoided “tangibles” Of course aim is to have measures that are: relevant, quantitative/objective, practical (limited treatment in new book) Private sector Maximize “effect” Effectiveness budget threshold E K Minimize “cost” Private sector Costs Example C/E Three projects were filed for enhancing occupational safety and working conditions beyond what is legally required. Which project represents the best investment ? Relevant Quantitative Objective Measurable Project PW (euro) Number of occupational accidents avoided Days away from work (accident) avoided Project A 100000 10 15 Project B 150000 12 18 Project C 75000 10 12 Private sector How to choose ? Example C/E A hospital service in Brooklyn (NY) is considering some alternatives to reorganize its services: keep the number of ambulances (currently 7) but relocate them to a satellite garage, increase the number of ambulances to 10 and relocate them to a satellite garage, have a number of cruising ambulances (7, 8, or 10) instead of ambulances stationed at a garage. How to choose between these alternatives ? Private sector Costs identified Investment in ambulances + equipment, Investment in garage, Operations and maintenance costs, Personnel (ambulance, garage – supervisor cruising) Benefits (effects) identified Quality of healthcare (!), Intervention time {Average time Tavg ‐ Number of quick interventions (within 20 min)} Calculation of costs Rather straightforward Calculation of benefits With the help of simulation Waiting time Private sector Dispatch delay Travel time to Pick up delay Travel Drop‐off time from delay (Back to base) Example C/E Thresholds: no general rule exists e.g. Cohen, 2008 CE < $50000 (>$1000000) per life year gained: attractive (unattractive) Effectiveness can be measured disease/intervention specific: e.g. LOS, infection rate, mortality or more general: e.g. QALY LOS=length‐of‐stay; QALY=quality adjusted life years C/E – C/B analysis Example An automotive plant has investigated four options for a production line for accessories. Besides money (PW) they also wanted to include aspects like speed (capacity), reliability, safety, flexibility (other products) and standardization (equipment components) into their analysis. The results of their analysis was summarized in the table below. Private sector Criterion Criterion weight Installation option 1 Installation option 2 Installation option 3 Installation option 4 Abs. score Rel. score Abs. score Rel. score Abs. score Rel. score Abs. score Rel. score Speed 0.32 85 27.2 80 25.6 75 24.0 60 19.2 Reliability 0.24 85 20.4 60 14.4 80 19.2 95 22.8 Safety 0.24 85 20.4 50 12.0 70 16.8 90 21.6 Flexibility 0.12 50 6.0 80 9.6 80 9.6 99 11.9 Standardization 0.08 85 6.8 90 7.2 70 5.6 50 4.0 Total effectiveness score 80.8 68.8 75.2 79.5 Present worth (103 euro) 450 250 300 350 Private sector C/B analysis: issues Criteria Multiple criteria Conflicting criteria No dominance Important issues Definition Measurement Exhaustiveness Exclusiveness Private sector Example Setting manufacturer of industrial vehicles, painting department Problem at hand buying new masks (occupational safety/diseases ‐ solvents) Analysis technical risks + legisla on + market study → three alterna ves remaining Criteria cost (life time !) Purchasing cost O&M cost non‐cost criteria Easy to move around with Functionality (protection) What about these criteria ? Exhalation (resistance) Heat build‐up in mask How to move on ? Easy to put on General level of comfort Air supply (extra filter needed ?) Readability manual Robustness Spare parts replacement Other (e.g. hygiene, mounting mechanism) Private sector Weights – Scoring Rating/Budget Ranking Pairwise comparison Delphi like method e.g. Rate according to the importance on a scale from 0‐10 (normalize afterwards) or Distribute a budget of 100 point over the criteria according to the importance Rank (sort) in order of decreasing importance Iterative panel effort Likert‐type scales Very bad Bad Private sector Average Compare the criteria two by two, each time indicate which is the most important one and also to which degree this holds Good Very good (Numerical values !!) With N number of criteria and Ri rank of criterium (ranking 1 best ) # criterua uniformweights 1 wi N 7 weights uniform criterium rank 1 2 3 4 5 6 7 rank sumweights wi N Ri 1 N R N k 1 k 1 rank sum rank reciprocal 28 2.593 0.1429 0.250 0.386 0.1429 0.214 0.193 0.1429 0.179 0.129 0.1429 0.143 0.096 0.1429 0.107 0.077 0.1429 0.071 0.064 0.1429 0.036 0.055 1.0000 1.000 1.000 0 0 0 0.4500 0.4000 0.3500 0.3000 weight rank reciprocal weights 1 Ri wi N k 1 1Rk 0.2500 uniform 0.2000 rank sum 0.1500 rank reciprocal 0.1000 0.0500 0.0000 1 Private sector 2 3 4 rank 5 6 7 Pairwise comparison Typically used in AHP Normalize (add up columns and divide elements by column sum) Weights Private sector Critical reflection Rating/budget Ranking Pairwise comparison Delphi like C/E analysis (CEA) ‐ multiple projects Similar to procedures seen before CER=cost‐effectiveness ratio (note: in chapter on cost estimation CER=cost estimation relationship) Example: independent projects Example: mutual exclusive projects C/B analysis: MCDM; MADM Steps Decomposition step Alternatives (mutual exclusive !!!) Criteria Weights Scores Synthesis step Aggregation of information Optimization Conclusion Private sector Decision support !! Often one of ‐‐ Multi criteria/atribute decision making Also e.g. “goal programming” Decision support models (MCDM) Mathematical optimization model Choose one criterion (=objective function) and set min/max requirements for the other criteria (= constraints) Absolute approach All alternatives are scored and alternative with the best score is “rank and rate” determined E.g. LAM (linear additive method) E.g. Consensus, TOPSIS (distance functions) Relative approach Alternatives are pairwise compared for all criteria E.g. PROMETHEE E E.g. AHP (analytic hierarchy process) Other option: C/E analysis C MCDM example: Consensus method Type: outranking method (absolute) Start Basic data Criterion 1 Criterion 2 Criterion 3 Criterion 4 … Private sector i Alt A Alt B ... Utility function 1.00 Utility 0.50 0.00 -0.50 0 1 2 3 4 5 6 7 8 9 10 -1.00 Performance scores Ideal situation U2j +1 A Alt.1 -1 Alt.2 Alt.4 +1 Alt.3 Worst situation B Private sector -1 U1j Weight for criterion i Distance functions elite U p u1 j , u 2 j ,..., u mj ; 1 ,..., m 1 p i 1 u ij m i 1 1 p Utility score of alternative j for criterion i anti-elite ~ U p u1 j , u 2 j ,..., u mj ; 1 ,..., m p i 1 u ij m i 1 1 p 1 Distance parameter Private sector Example Private sector Gaussian utility function (can be different for each criterion) Linear additive method Distance parameter (the larger p, the larger the difference between elite and anti‐elite) Private sector Consensus method MADM: AHP AHP Analytic hierarchy process (Saaty) – extension: ANP (analytic network process) Relative method (note: consensus = absolute method) Extension: ANP (analytic network process) Tutorial: http://people.revoledu.com/kardi/tutorial/AHP/index.html Private sector Indicator for work order evaluation Criteria network (strategic operational) Weighting factors (operational strategic) Class scores (1: excellent to 3: improvement needed) Extra‐1: Paper Paper: The use of alternative solvent purification techniques Cournoyer & Dare – Chemical Health & Safety – July/August 2003, pp 15‐18 ook: http://www.physsci.edu/news/entries/202022.html Background University research lab Lab activities: organic solvent (e.g. benzene) purification to remove moisture and oxygen Reason for C/B analysis Explosion of solvent distillation still Student injured (second degree burn wounds) Material damage of $3.500.000 (2 labs) Research question “C/B” of alternatives: column method – purchase ultra‐dry solvents Technical analysis Still method: traditional, batch process Column method: continuous process Purchasing pure solvents: possible, but expensive Risk analysis Risks entailed in organic solvent purification Legislation Probability – Severity “Damage”: injuries (death) – material losses Severity Comparable for still and column method Probability Typical problem for working with flammable organic solvents Different for still and column method Still: active metal and flammable solvents – ignition sources: heating mantles, vacuum pumps, high temperature water lines – (safety shutdowns) Column: no initiators – but stored energy hazards – peroxide accumulation – larger quantities of solvent … User validation Column method: accepted method, O2 concentration mostly OK, problem is Cu cannot be used (Ar, N) Technical risk (reduction) Still method when accident occurred: amount of solvent>legally allowed – fire door blocked open – no sprinklers; But, in normal circumstances: “occasional” for likelihood (1996, 1997) Severity: “catastrophic” Residual risk (with controls): “high” Management decision needed (case by case) Column method Likelihood: “remote” Severity: “high” Risk: “low” Management: acceptable Costs/Benefits C/B = business view: “acceptable level may be achieved when the costs of decreasing a given risk further are greater than the costs realized from the occupational exposure to hazardous chemicals” Problem No useful numbers published for (lowering) chemical risks Solution Inspiration from nuclear industry: “$1000‐2000‐10000 per person‐rem” Costs Purchasing: 20‐50% more expensive than regular solvents Conclusion article Still method: acceptable, but… Column method: cost friendly alternative Purchasing: costly, but recommendable for small amounts C/B approach in paper Interesting analysis, dictated by an accident Elements taken up in the analysis User requirements Technical aspects – Risks Cost elements Conclusions of the analysis Some data issues Recommendations No closed formula Decision support, not decision making Quid situation? Instructions for use “user knows what he/she is doing” Blocking safety/fire door No sprinklers Volumes chemicals more than needed (allowed) … Extra‐2: Case discussion Setting Medium‐sized company, LiBa Production line for special effect light bulbs Recent inspection (governmental regulations): potential problems concerning occupational safety → Threats: paying fines, losing license‐to‐operate) “Engineering has a number of alternative solutions” Assignment Use a MCDM approach to choose the “best” alternative Alternatives Sensibilization Posters and signs Protection screens Limited intervention Control system Rather extensive intervention Renovation Safety as part of Relocation (international) Closing down and moving Outsourcing activities Social impact New line Large investments Stakeholders Project engineer Production Maintenance Safety Management Trade unions How to tackle this problem … Assemble project team, make sure all stakeholders are represented Have a first look at the options (alternatives), do you want to take up all of them in your analysis? Determine the criteria definition (consensus?), measure exhaustive (not too many!), exclusive (see definitions)? Determine the criteria weights choose a method to determine the weights look for a way to “combine” the different stakeholder views Determine the scores idem Bring everything together in one MCDM method which one do you choose? why? Note on MCDM Increasing interest Industry & Government & Healthcare Analysis Complex – (time consuming) Data collection hypotheses – scenarios Benefits in decision process Sensitivity analysis easy – danger for manipulation Completeness Tangibles & intangibles Transparancy Scientific tool Acceptance Involvement stakeholders C/B report Sometimes “extra” Sometimes “funneling” Role in decision process Incorporation decision attitude Decision support: rather supporting than final conclusion Risk analysis can be combined perfectly with MCDM Scoring & weights (utility funcion) Tools Models and software available Wrap‐up Chapter 9 Extras Public sector Type of projects – B/C ratios ‐ Single and multiple projects Ethical considerations Private sector Type of projects – C/E analysis – C/B issues – MCDM – Consensus method Engineering Economy Project Financing and Noneconomic Attributes 1 Overview Cost of capital – MARR D-E mix 1-2 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved LEARNING OUTCOMES 1. Explain cost of capital and MARR 2. Calculate weighted average cost of capital 3. Estimate cost of debt capital 4. Estimate cost of equity capital 5. Understand high D-E mix and risk 6. Develop weights for multiple attributes 7. Apply weighted attribute method to alternative evaluations 10-3 © 2012 by McGraw-Hill All Rights Reserved Cost of Capital and MARR Cost of capital is the weighted average interest rate paid based on debt and equity sources Debt capital represents borrowing outside company Equity capital is from owners’ funds and retained earnings MARR is set relative to cost of capital 10-4 © 2012 by McGraw-Hill All Rights Reserved Factors Affecting MARR Project risk: higher risk leads to higher MARR Investment opportunity: in order to capture perceived opportunity, MARR may be temporarily lowered Government intervention: gov’t actions such as tariffs, subsidies, etc. can cause companies to raise or lower MARR Tax structure: rising corporate tax rates lead to higher MARR Limited capital: as capital becomes limited, MARR increases Rates at other corporations: competition can cause companies to raise or lower MARR 10-5 © 2012 by McGraw-Hill All Rights Reserved D-E Mix and Weighted Average COC Debt-to equity (D-E) mix identifies percentages of debt and equity financing for a corporation WACC = (% equity)(cost of equity) + (% debt)(cost of debt) This figure Illustrates WACC If the percentage of equity capital from each source is known, each component of WACC is separately calculated 10-6 © 2012 by McGraw-Hill All Rights Reserved Example: WACC Calculation A company that specializes in producing cold-weather clothing and accessories is expanding its ski-jacket and boot-sock manufacturing facilities. The company plans to borrow $2.5 million at 7% interest, issue stock worth $4 million worth at 5.9%, and use $1.5 million of retained earnings at 5.1% to finance the project. Determine the company’s WACC. Solution: Equity sources are stock and retained earnings Total project cost = 4 + 1.5 + 2.5 = $8 million WACC = 4/8(5.9% + 1.5/8(5.1%) + 2.5/8(7%) = 6.09% 10-7 © 2012 by McGraw-Hill All Rights Reserved Parameters Cost of debt capital Debt capital -- Funds received by borrowing; loans or issuance of bonds Interest to pay to “bank” is actually lower than stated due tax savings Cost of equity capital Equity capital obtained from 4 possible sources: (1) Sale of preferred stock , (2) Use of retained earnings, (3) Owner’s private capital, (4) Sale of common stock No tax advantage or tax savings for equity capital Determined using dividends, stock price, … 1-8 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved Debt-Equity Mix and Risk As proportion of debt capital increases, overall cost of capital decreases Because interest on debt capital offers a tax advantage Corporations that become highly leveraged (large D-E mixes) have increased risk and more difficulty in obtaining project funding Investors take more risk and lenders are leery to provide funds Best: balance between debt and equity funding 10-14 © 2012 by McGraw-Hill All Rights Reserved Multiple Attribute Analysis Attributes other than the economic one are considered in most alternative selections, e.g., public and service sector projects Steps necessary to identify and use multiple attributes are: Identification of attributes Determination of importance weight of each attribute Assignment of a value rating to each attribute Alternative evaluation using a technique that accommodates several attributes (covered in previous chapter in course) 10-15 © 2012 by McGraw-Hill All Rights Reserved Summary of Important Points Cost of capital is weighted average of debt and equity funding There is a tax savings with debt capital because interest is deductible; nothing is tax deductible for equity capital High D-E mixes mean higher risk for lenders and investors; project funding becomes more problematic Multiple attribute analysis brings other factors (besides cost) into the decision-making process Four different techniques for assigning weights to attributes Likert scale is good for assigning value ratings to alternatives Multiple attribute evaluation measure is R j 10-25 n WV j 1 © 2012 by McGraw-Hill i ij All Rights Reserved Wrap-up Cost of capital – MARR D-E mix 1-27 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved Engineering Economy Replacement and Retention Decisions Overview Basics of replacement study Economic service life (ESL) Performing a replacement study No study period specified Study period specified Literature !? Basics of Replacement Study Why replacing a machine, an installation, a plant, … ? Reduced Performance: Wear and tear Decreasing reliability and productivity Increasing operating and maintenance costs Altered Requirements: New production needs, accuracy, speed, etc. Obsolescence: Current assets may be less productive Not state of the art – need to meet competition Basics 11‐3 Terminology Defender Asset: Currently installed asset; Challenger Asset: The potential replacement or “challenging” asset; Under consideration to replace the defender asset. Together, Defender (D) and Challenger (C) Constitute mutually exclusive alternatives; Select one and reject the other. Basics 11‐4 AW values are typically calculated EUAC –Equivalent Uniform Annual Cost Includes mostly costs estimates Economic Service Life (ESL) Not depreciated book value Not “exotic” trade‐in value For a given alternative is: Number of years at which the lowest AW‐of‐cost occurs Defender First Cost Initial investment for the defender – P Current Market Value (MV) now – correct estimate for P to assign to keeping a defender Basics Fair market value obtained from professional appraisers, resellers, ... 11‐5 Actual cost: Opportunity cost (MV) Out‐of‐pocket cost (P‐TIV) Challenger, first cost Amount of capital that must be recovered when replacing a defender with a challenger P (normal situation) or P‐(TIV‐MV) (unrealistic trade‐in value) Sunk cost Cost that has occurred in the past and cannot be altered Sunk costs are generally of no use in a before‐tax replacement analysis Consultant’s viewpoint The analyst assumes that he/she is an outsider (a consultant) Thus, the analyst owns neither asset and must assign reasonable value to the defender and challenger, and is thus unbiased It assumes the services provided by the defender can be purchased now by making an “initial investment” equal to the market value of the defender While it may seem strange to charge an investment cost for keeping one’s own asset (the defender) this is what must occur. Keeping the defender is not free! Why? Because the firm is giving up the opportunity to receive a possible cash flow from selling the current defender. 11‐6 Example 3 years ago Defender (old system) Challenger (new system) P=MV=$70000 P=$100000 AOC=$30000 AOC=$20000 S=$10000 S=$20000 n=3 years n=10 years now Ground leveling system Purchase price: $120000 ‐ Expected service life: 10 years ‐ Expected salvage value: $ 25000 ‐ Annual operating cost: $ 30000 ‐ ‐ Not relevant Current book value: $ 80000 Expected service life left: 3 years Expected salvage value: $ 10000 Annual operating cost: $ 30000 Market value appraisal: $ 70000 New system (laser guided) Purchase price: $ 100000 ($110000 next week) Trade‐in value current system: $ 70000 Expected service life: 10 years Expected salvage value: $ 20000 Annual operating cost: $ 20000 Investment Concerns One must assign an investment cost for KEEPING the defender asset The appropriate investment cost to assign to the defender asset is: The current fair market value of the defender at the time the replacement decision is being examined. Challenger Investment This is the total investment (Pchallenger) required for a new (challenger) asset that will possibly replace the current defender. Basics 11‐11 Economic Service Life (ESL) Definition of ESL is: Number of years n at which the equivalent annual worth (AW) of costs is minimized considering the most current cost estimates over all possible years that the asset may provide a needed service ESL is the n where total AW is a minimum for tabulation at i% over a selected range of years. To find the ESL Plot n versus total AW, where Capital recovery CR=‐P(A/P,i,n)+S(A/F,i;n) ≈ annual worth of investment Usually decreasing with time Total AW = – CR – AW of annual operating costs Seek the time period that minimizes the Total AW for the range of lives selected Usually increasing with time Economic service life (ESL) 11‐12 Complete equation k total Ak P( A / P, i, k ) Sk ( A / F , i, k ) AOC j ( P / F , i, j 1 where j )( A / P, i , k ) P initial investment or current market value ( MV ) Sk salvage value or market value after k years AOC j annual operating cos t for year j with j 1 to k Procedure: Year‐by‐year analysis for “k” years – where “k” is given or assumed. Typical Annual Worth Plot Seek the n value resulting in minimum Total AW Represents the n for Economic Service Life Economic service life (ESL) 11‐15 Example Economic service life (ESL) 11‐16 Solution Defender Asset; 3 years old now; Market value now: $13,000; 5‐year study period assumed; Estimated Future Market Values and AOC’s: increasing decreasing 17 Period – by – period analysis For “k” = 1 year: S1 = $9000 0 1 P=$13,000 AOC1 = ‐2500 AW(10%)1 = (‐$13,000)(A/P,10%,1) + $9000(A/F,10%,1) ‐2500 = ‐$7800 ( for one year!) 18 For “k” = 2 years: S2 = $8000 0 1 2 AOC1 = ‐2500 P=$13,000 AOC2 = ‐$2700 AW(10%)2 = (‐13,000)(A/P,10%,2) + 8000(A/F,10%,2) ‐[2500(P/F,10%,1) + 2700(P/F,10%,2)](A/P,10%,2) = ‐$6276/yr for 2 years. 19 For “k” = 3 years: 0 1 S3 = $6000 2 3 AOC1 = ‐2500 P=$13,000 AOC2 = ‐$2700 AOC3 = ‐$3000 AW(10%)3 = (‐13,000)(A/P,10%,3) +6000(A/F,10%,3) ‐[2500(P/F,10%,1) + 2700(P/F,10%,2) + 3000(P/F,10%,3](A/P,10%,3) = ‐$6132/yr for 3 years. 20 A similar analysis for k = 4 and 5 is conducted; The AW(10)k, K = {1,2,3,4,5} are tabulated as: Total AWk k=1: $‐7800 k=2: $‐6276 k=3: $‐6132 Min. Cost Year = 3 years = Economic Service Life k=4: $‐6556 k=5: $‐6579 21 Example Important Points Marginal costs (MC) are year‐by‐year estimates of the costs to own and operate an asset for one year Three Components of Marginal Costs: Cost of ownership (loss in Market Value/yr); Foregone interest of Market Value at beginning of the year; AOC for each year AW of marginal costs = total AW of costs The analysis is best performed via a spreadsheet model See example 11.3 Economic service life (ESL) 11‐30 DATA Performing a Replacement Study Two approaches D= defender C=challenger ESL=economic service life AW=annual worth Performing study 11‐34 I. No study period specified Example Solution Defender data Current Market Value: $15,000; Future Market Values will decrease by 20%/yr; Keep for no more than 3 years; AOC’s: {$4,000,$8,000,$12,000} Retrofit next year = $16,000; AOC’sD := {$20,000, $8,000,$12,000} (costs). Old m/c Challenger data First Cost: $50,000 Future Market Values decreasing by 20%/year; Retain for no more than 5 years; AOC’sC := {$5,000,$7,000,$9,000,$11,000,$13,000} New m/c Assume the interest rate is set at 10%/year. 39 updated Remark What minimum market value of the defender will make the current challenger economically attractive? If a high enough market value (trade‐in) is possible for the defender asset, one should take it and go with the challenger immediately! Break‐even or replacement value (RV) If the actual market value (trade-in) exceed the breakeven replacement value, the challenger is the better alternative. If this is the case, replace now with the challenger! Performing study 11‐52 Additional Considerations Several additional aspects of a replacement study 1. Future‐year replacement decisions; 2. Opportunity cost vs. Cash flow approaches; 3. Anticipation of improved future challengers. The analyst must understand trends, advances, changes in technology, and competitive pressures that often impact the remaining life of a defender and the challenger Additional considerations 11‐53 II. Specified Study Period Example “P” (MV) P S AOC Option 1 v Option 2 Truck has a life time of 12 years, and is 3 years old at the moment of the study Remark If the defender’s remaining life is shorter than the study period … Focus on upgrading the defender and obtain cost estimates in order to extend the defender out to the study period. These costs become part of retaining the defender out to the prescribed study period and are used in the replacement study Specified study period 11‐66 Remember … Compare the best challenger to the current defender; “Best” is defined as the option with the lowest AW of costs over the specified period of time; Apply the ESL method to find the economic life of the defender and challenger; For ESL, you need estimates of future market values and annual operating costs for both C and D. AW is the best method, especially if alternative lives are unequal. For a fixed study period, apply the traditional AW method for both D and C. Always focus on the estimates of future costs and market values using sound principles of estimation. Avoid the trap of incorporating sunk costs into a replacement analysis 11‐67 “sound estimates” “reason for replacement: reduced performance” “Replacement” models in literature I have an asset for which the maintenance costs keep increasing every year and/or the probability of failure keeps increasing with time. What is the best time to replace this asset with a new one ? If I have a set of identical devices, should I replace them as a group or as individuals ? Do I best use a replacement policy based on asset age or based on a fixed period ? Example A small welding robot costs € 7000. The maximum useful life is 5 years. The maintenance costs ck are €1000k, with k being the quipment age. From historical data and supplier information, probabilities of unrepairable failures are known to be: p1=0.1, p2=0.1, p3=0.2, p4=0.3 and p5=0.3. Yearly it can be decided to replace the equipment by a new one. What is the optimum replacement timing ? Solution Typically this type of problem (cfr “repair limit”) is solved by renewal theory (a.k.a. replacement theory), provided that the hypotheses for applying this theory hold (AGAN, N=) The optimum replacement timing is found by minimizing C(n), with E[cos t ] C (n) E [life] 0 time 72 73 Formulas Thus 74 Spot the differences Find the similarities Exercise Wrap‐up Basics of replacement study Economic service life (ESL) Performing a replacement study No study period specified Study period specified Literature?! Critical reflection ! Insight ( Chapter 11 Engineering economy Independent projects under budget limitation Overview Capital rationing among projects Projects with equal lifes Projects with unequal lifes Linear programming model This chapter uses basic Linear Programming Overview of Capital Rationing Capital is a scarce resource; never enough to fund all projects Each project is independent of others; select one, two, or more projects; don’t exceed budget limit b ‘Bundles’ are collections of independent projects that are mutually exclusive (ME) For 3 projects, there are 23 = 8 ME bundles, e.g., A, B, C, AB, AC, BC, ABC, Do nothing (DN) 12-3 © 2012 by McGraw-Hill Reserved All Rights Capital Budgeting Problem Each project selected entirely or not selected at all Budget limit restricts total investment allowed Projects usually are quite different from each other and have different lives Reinvestment assumption: Positive annual cash flows reinvested at MARR until end of life of longest‐lived project 12-4 © 2012 by McGraw-Hill Reserved All Rights Capital Budgeting for Equal‐Life Projects Procedure Develop ≤ 2m ME bundles that do not exceed budget b Determine NCF for projects in each viable bundle Calculate PW of each bundle j at MARR (i) Note: Discard any bundle with PW < 0; it does not return at least MARR Select bundle with maximum PW (numerically largest) 12-5 © 2012 by McGraw-Hill Reserved All Rights Example: Capital Budgeting for Equal Lives Select projects to maximize PW at i = 15% and b = $70,000 Project Initial investment, $ Annual NCF, $ Life, years Salvage value, $ A ‐25,000 +6,000 4 +4,000 B ‐20,000 +9,000 4 0 C ‐50,000 +15,000 4 +20,000 Solution: Five bundles meet budget restriction. Calculate NCF and PW values Conclusion: Select projects B and C with max PW value Bundle, j Projects 1 A ‐25,000 +6,000 2 B ‐20,000 +9,000 0 +5,695 3 C ‐50,000 +15,000 +20,000 +4,261 4 A, B ‐45,000 +15,000 +4,000 +112 5 B, C ‐70,000 +24,000 +20,000 +9,956 6 DN 0 0 0 0 12-6 NCFj0 , $ NCFjt , $ SV, $ PWj , $ +4,000 © 2012 by McGraw-Hill Reserved ‐5,583 All Rights Capital Budgeting for Unequal‐Life Projects LCM is not necessary in capital budgeting; use PW over respective lives to select independent projects Same procedure as that for equal lives Example: If MARR is 15% and b = $20,000 select projects 12-7 © 2012 by McGraw-Hill Reserved All Rights Example: Capital Budgeting for Unequal Lives Solution: Of 24 = 16 bundles, 8 are feasible. By spreadsheet: Reject with PW < 0 Conclusion: Select projects A and C 12-8 © 2012 by McGraw-Hill Reserved All Rights Capital Budgeting Using LP Formulation Why use linear programming (LP) approach? ‐‐ Manual approach not good for large number of projects as 2m ME bundles grows too rapidly Apply 0‐1 integer LP (ILP) model to: Objective: Maximize Sum of PW of NCF at MARR for projects Constraints: Sum of investments ≤ investment capital limit Each project selected (xk = 1) or not selected (xk = 0) LP formulation strives to maximize Z 12-9 © 2012 by McGraw-Hill Reserved All Rights Example: LP Solution of Capital Budgeting Problem MARR is 15%; limit is $20,000; select projects using LP PW @ 15%, $ 6646 -1019 984 -748 LP formulation for projects A, B, C, D labeled k = 1, 2, 3, 4 and b = $20,000 is: Maximize: 6646x1 ‐ 1019x2 + 984x3 ‐ 748x4 Constraints: 8000x1 + 15,000x2 +8000x3 + 8000x4 ≤ 20,000 x1, x2, x3, and x4 = 0 or 1 12-10 © 2012 by McGraw-Hill Reserved All Rights Different Project Ranking Measures Possible measures to rank and select projects: PW (present worth) – previously used to solve capital budgeting problem; maximizes PW value) IROR (internal ROR) – maximizes overall ROR; reinvestment assumed at IROR value PWI (present worth index) – same as PI (profitability index); provides most money for the investment amount over life of the project, i.e., maximizes ‘bang for the buck’. PWI measure is: Projects selected by each measure can be different since each measure maximizes a different parameter 12-12 © 2012 by McGraw-Hill Reserved All Rights Wrap‐up Capital rationing among projects Projects with equal lifes Projects with unequal lifes Linear programming model Chapter 12 Engineering Economy Breakeven and payback analysis 1 Overview For a single project With two or three alternatives (Spreadsheet tool: Excel’s Solver) 2 Breakeven analysis Performed to determine the value of a variable or parameter of a project or alternative that make to elements “equal” Illustration That specific replacement value of the defender in a replacement study that makes the challenger an equally good choice That sales figure that will allow to cover all costs made; if we sell less we have a loss; if we sell more we make a profit … Tools Direct solution (one factor). Trial and error (multiple factors). Spreadsheets (using PV, FV, RATE, IRR, NPV, PMT, and NPER and Goal Seek or Solver in Excel). Estimates are considered to be certain →Part of sensitivity analysis →Can be complemented with e.g. simula ons Introduction 3 Breakeven Analysis for a Single Project Given P, F, A, i, n If all of the parameters (variables) shown above are known except one, then the unknown parameter can be calculated or approximated A breakeven value can be determined by setting PW, FW, or AW = 0 and solve for or approximate the unknown parameter Single project 1‐4 Cost – Revenue Model Approach A popular application of Breakeven (BE) is where cost‐ revenue‐volume relationships are studied Define cost and revenue functions and assume some linear or non‐linear cost or revenue relationships One objective: Find a parameter value ‐‐ termed QBE ‐‐ that will minimize costs or maximize profits Concepts: fixed costs, variable costs, total costs Single project 1‐5 • see See also first class 6 Total Costs Total Cost (TC) = Fixed Costs (FC) + Variable Costs (VC) TC = FC + VC Profit (P) = Revenue (R) – Total Cost (TC) P = R – TC = R – (FC + VC) Single project 1‐9 Cost – Revenue Relationships Linear Models and Non‐linear models are used as approximations to reality Single project 1‐10 11 Breakeven point The breakeven (BE) point QBE is the point where the revenue and total cost relationships intersect For non‐linear relations, it is possible to have more than one QBE point Revenue and total cost relationships tend to be static in nature May not truly reflect reality of the dynamic firm However, the breakeven point(s) can be useful for planning purposes Single project 1‐12 Example: One Project Breakeven Point A plant produces 15,000 units/month. Find breakeven level if FC = $75,000 /month, revenue is $8/unit and variable cost is $2.50/unit. Determine expected monthly profit or loss. Solution: Find QBE and compare to 15,000; calculate Profit QBE = 75,000 / (8.00‐2.50) = 13,636 units/month Production level is above breakeven Profit Profit = R – (FC + VC) = rQ – (FC + vQ) = (r‐v)Q – FC = (8.00 – 2.50)(15,000) – 75,000 = $ 7500/month 13-6 Breakeven analysis with more than one alternative Given two alternatives (assume mutually exclusive) Need to determine a common variable or economic parameter common to both alternatives Parameter could be Interest rate, First cost (investment), Annual operating cost, etc. More than 1 alternative 1‐17 Example: Two Alternative Breakeven Analysis Perform a make/buy analysis where the common variable is X, the number of units produced each year. AW relations are: AWmake = ‐18,000(A/P,15%,6) +2,000(A/F,15%,6) – 0.4X AW, 1000 $/year Breakeven value of X AWbuy 8 7 AWbuy = ‐1.5X 6 Solution: Equate AW relations, solve for X 5 AWmake 4 ‐1.5X = ‐4528 ‐ 0.4X X = 4116 per year 3 2 1 If anticipated production > 4116, select make alternative (lower variable cost) 0 1 2 3 4 5 X, 1000 units per year 13-8 Three Alternative Analysis If three alternatives are present, compare the alternatives pair‐wise, or Use a spreadsheet model to plot the present worth or annual worth over a specified range of values. A typical three‐alternative BE plot might look like …. More than 1 alternative 1‐22 Using Excel’s SOLVER for Breakeven Analysis SOLVER is one of many built‐in Excel analysis tools; SOLVER has been designed to aid in more complex forms of “goal seeking” and performing “what‐if” evaluations of properly constructed models. For a properly constructed model SOLVER will require that the analyst: Specify a target cell (the objective); Identify one or more changing cell(s) that will have to change to achieve the desired target cell value Self‐study SOLVER 1‐24 Remember BE analysis can be a form of sensitivity analysis Breakeven point for a variable X is normally expressed as: Units per time period; Hours per month; etc. At exactly breakeven (QBE) one is indifferent regarding a project Typical breakeven models are: Linear Non‐linear Two or more alternatives can be compared using breakeven analysis Complex models can be evaluated using Excel’s SOLVER tool 1‐28 Exercise 29 Real‐life example 31 Wrap‐up Chapter 13 For a single project With two or three alternatives [Spreasheet tool: Excel’s Solver] 33 Engineering Economy Effects of Inflation 1 Overview Impact of inflation Rates: i, if, f PW (FW, AW) with inflation considered 2 Impact of Inflation Inflation is an increase in the amount of money necessary to obtain the same amount of product or service before the inflated price was present. Impact because the value of the currency changes downward in value Opposite: deflation Firms should set their MARR rate to: Cover the cost of capital; Cover or buffer the inflationary aspects perceived to exist; Account for risk. 1‐3 Equating “Value” Money in time period t1 can be related to money in time period t2 by the following: Dollars t1 = Dollars t 2 inflation rate between t1 and t 2 1‐5 The Inflation rate ‐‐ f The inflation rate, f, is a percent per time period; stated in a manner similar to interest rates. Example: If f = 5.0% per year inflation, $100 today requires $105 to buy the same amount next year. Inflation is on top of the interest rate 1‐6 The Basic Inflation Relationship Let n represent the amount of time between t1 and t2 then . . . . Future Dollars = Today’s dollars(1+f)n Dollars in period t1 are termed: Constant‐value dollars or today’s dollars Dollars in time period t2 are termed: Future dollars or then‐current dollars. 1‐7 Example Assume a firm desires to purchase an asset that costs $209,000 in today’s dollars. Assume a reasonable inflation rate of, say, 4% per year In 10 years, that same piece of equipment would cost: $209,000(1.04)10 = $309,371 Inflation is not an interest rate or rate of return consideration !! 1‐8 Impact The impact of inflation can be significant. From the previous example we see that even at a modest 4% rate of inflation, the future impact on cost can and is significant! The previous example does not consider the time value of money. A proper engineering economy analysis should consider both inflation and time value of money. 1‐9 Three Important Rates Real or inflation‐free interest rate Denoted as “i”. Inflation‐adjusted interest rate – market rate Denoted as “if ” Inflation rate Denoted as “f” 1‐10 Real or Inflation‐free Interest Rate ‐ i Rate at which interest is earned; Effects of any inflation have been removed; Represents the actual or real gain received/charged on investments or borrowed funds. 1‐11 PW adjusted for inflation In prior chapters, present worth was calculated assuming that all cash flows were in constant value dollars; Example: $5,000 (now) inflated at 4% per year with a interest of 10% per year. verification 1‐12 Derivation of a Combined Interest Rate Start with: 1 PF (1 i ) n Real interest rate Assume F is a future dollar amount with inflation built in. Then F 1 P (1 f ) n (1 i ) n 1 PF (1 i f if ) n Inflation rate Or 1 PF F ( P / F , i f , n) n (1 i f ) if Market rate i 1‐14 FW adjusted for inflation Solving F=f(i,if,n) • There are four different interpretations for future worth Current value calculations. 1. Actual future $$; 2. Purchasing power of future $$ stated in constant‐value $$; 3. Future $$ required at t = n to maintain t = 0 purchasing power; 4. $$ at t = n to maintain purchasing power and earn a stated interest rate of i% per time period. 1‐15 17 Example P = $1,000 now and the market rate of interest (if) is 10% per year. If n = 7, what is the future value of the $1,000 now? F = $1,000(F/P,10%,7) = $1,948. Remember, the market rate of interest includes both the inflation rate and the discount rate. 1‐19 Example Find the real interest rate for a market rate of 10% and inflation at 4%. 0.10 0.04 Solution 0.0577 5.77% i 1 0.04 Purchasing power of $1000 (cv), with a market rate of 10% and inflation at 4%? F = $1,000(F/P,5.77%,7) = $1481 Inflation of 4%/year has reduced to a real rate that is less than 6% per year! 1‐24 Example P = $1,000 now Inflation of 4% per year F = $1,000(1.04)7 = 1,000(F/P,4%,7) = $1316 26 Example Given P = $1000 inflation rate f = 4% Real interest rate i of 5.77%; Calculate if as; if = 0.0577 + 0.04 + 0.0577(0.04) = 0.10 Then: F = $1000(F/P,10%,7) = $1948 $1948 seven years out is equivalent to $1,000 now with a real return of 5.77% per year and inflation at 4% per year. 1‐28 Importance of Inflation Impacts Most countries: inflation is from 2% to 8% per year; Some countries with weak currencies, political instability, poor balance of payments can have hyperinflation (as high as 100% per year). 1‐30 When Hyperinflation is Present Spend money almost immediately; Money looses value quickly; Very difficult to perform engineering economy calculations in a hyper‐inflated economy; Future values are unreliable and, Future availability of investment capital is very uncertain. 1‐31 Capital Recovery Calculations Adjusted for Inflation With inflation present: Current dollars invested in a productive asset must be recovered over time with future inflated dollars. With loss of future purchasing power, future dollars will have less buying power than current dollars; More dollars will be required to recover a present investment is a productive asset. 1‐32 33 AW 40 42 Chapter 14 Wrap‐up Impact of inflation Rates: i, if, f PW (FW, AW) with inflation considered 45 Engineering Economy Cost estimation Overview Cost estimation: what ? Need. Cost types. Approaches (↑↓). Accuracy. Extra: Learning curve. Cost estimation: some techniques “Per unit” method. Cost indexes. Cost‐capacity equations (CER). Factor method (CER). Or professional organizations … Indirect costs: Traditional rates and allocation vs Activity based costing. Cost estimation: what ? For the most part Engineering economy analysis tends to be “cost driven”. Future costs are very critical to the analysis of a project. Engineers generally have the responsibility of cost estimation. Revenue generation generally comes from marketing/sales areas. Questions to be resolved: What cost components must be estimated? What approaches will be utilized? How accurate should the estimates be? What estimation techniques will be utilized? Introduction Accuracy issue Estimation software Two major cost types initial or first cost P Delivered‐equipment cost Installed‐equipment cost Such as equipment cost, delivery charges, installation costs, insurance coverage, initial training annual operating costs AOC (M&O costs) Such as direct labor, direct materials, energy, maintenance, rework & rebuild Annual Operating Costs Maintenance & Operating costs Introduction Approaches Bottom‐up Approach Treats the final cost as an independent (output) variable and the associated costs as input or dependent variables. Design‐to‐Cost Approach Treats the competitive cost as an input variable and the associated cost estimates as the output variables. Introduction Management accounting Introduction Bottom‐up Approach Treats the final cost as an output variable and the associated costs as input variables. Cost components are first identified; Cost elements are estimated; Summed to = total direct cost; (DC) Factor in estimated indirect costs. (TIDC) Factor in required margin or profit. Price = DC + TIDC + Profit. Traditional costing approach; Applied in industries where pricing the product or the service is not the dominant factor in competition . Introduction ! Value engineering ! Design‐to‐Cost Approach Treats the competitive cost as an input variable and the associated cost estimates as the output variables. Applied in the early stages of new or enhanced product design. Price estimates are conducted to set “target” values. Detailed design and equipment needs are not yet defined. Becoming more popular especially in high competition markets; Places greater emphasis on the accuracy of the estimation process; Target cost estimates must be realistic; Design to the target price; Common practice in Eastern cultures (Japan and other Asian countries). Introduction Illustration: remote control Illustration: facility design Accuracy needed No estimate is intended to be “exact” But must be reasonable and accurate enough to provide a robust economic analysis In preliminary design phase: Estimates are viewed as “first cut” estimates (go – no go) Serve as inputs to the project initial budget The “unit method” is often applied here “Per unit” approaches should fall into the 5% to 15% range for total costs. Estimates outside of the 15% range are not very reliable! Introduction General guidelines for accuracy Conceptual/Feasibility stage – order‐of‐magnitude estimates are in range of ±20% of actual costs Detailed design stage ‐ Detailed estimates are in range of ±5% of actual costs Characteristic curve of accuracy vs. time to make estimates Work breakdown structure Physical work elements Functional work elements (Sullivan) Learning curves or unit production cost A learning curve is a line displaying the relationship between unit production time and the cumulative number of units produced. Learning (or experience) curve theory has a wide range of application in the business world. e.g. In airplane industry (origin), in manufacturing1,2 and distribution, healthcare, etc. Types Individual learning: “practice makes perfect” Organizational learning: “practice makes perfect” plus changes in administration, equipment and product design Learning curve theory is based on three assumptions: 1. 2. 3. 1 The amount of time required to complete a given task or unit of a product will be less each time the task is undertaken. The unit time will decrease at a decreasing rate. The reduction in time will follow a predictable pattern. for new tasks – trade‐off in JIT systems with short runs 2 also in machine learning Typical formula Also: learning rate, improvement rate Assuming a constant percentage reduction in input resources each time the output quantity is doubled Example Illustration “(Per) Unit method” Examples Number of units Introduction Unit cost factor Example (p 390) Cost indexes A cost index is a ratio of the cost of something today to its cost sometime in the past. As such, a cost index is always a dimensionless number. Examples Well‐known daily life example: CPI (consumer price index) Common industrial examples: Plant cost index (chemical engineering), Marshall & Swift index (equipment) Measuring the cost of living Example: Belgium http://economie.fgov.be/nl/statistieken/cijfers/ economie/consumptieprijzen/consumptieprijsi ndexen/#dwtable FOD Economie, K.M.O., Middenstand & Energie Statistiek en Economische Informatie Eenheid Consumptieprijzen Simon Bolivarlaan 30 - 1000 Brussel Tel. : 02/277.51.11 - Fax : 02/277.96.15 Autom. antwoordapparaat : 02/277.56.40 http://economie.fgov.be http://statbel.fgov.be e-mail : [email protected] Formula It Ct C0 I0 where Ct estimated cost at present time C0 estimated cost at time t0 I t index value at time t I 0 index value at time t 0 Bit outdated example, but are still updated Chemical Engineering Engineering News Record Marshall & Swift (Blank & Tarquin, 2018) Example Engineering News Record Example 100.17=56.70/56.60; 100; 56.40/56.60=99.65 Note that In an index different (sub)components (with different weights) are taken into account. The base period is a selected time when the index is defined with a basis value of 100 (or 1). The index each period (year) is determined as the cost divided by the base year cost and multiplied by 100. Future index values may be forecast using simple extrapolation or more refined techniques such as time‐series analysis. Refer to the articles on the next pages to obtain more insights in cost indexes. Example: Read this article to gain more insight in how indices are composed Chemical Engineering Plant Cost Index (CE PCI) Structure Evolution: °1963, only some cosmetic revisions (names) changed till 1982: major revisions like reduction of number of components (110→66) and change in productivity factor and in 2002: major revision, not in structure, but in underlying details Values Not a monotonically increasing curves, but ups & downs ! Weight revisions Normalization Seamless transition between old and new indices Example: ENR Construction Cost Index (CCI) – ENR Building Cost Index (BCI) Both for construction cost, but difference in labor component: CCI more suited if labor costs are high proportion of total costs, BCI more applicable for structures Data 20‐US cities unweighted average Not including e.g. Arizona and Florida – Canadian version for Montreal and Toronto Base year: 1913 For materials “spot prices” from a single source for all materials tracked No seasonal adjustment (! Careful if period of analysis is less than 1 year) Verifiability: published Evolution Makeup of index components, e.g. 1996: price structural steel →beams Revisions, if necessary CCI and BCI do not capture all factors influencing construction costs Included are cement, lumber, structural steel and labor Not all factors influencing project costs: snapshot of cost trends ! Example: Read this article to gain insight in the “local” character of an index Focus on methodology and concept, not on numerical values Belgian chemical engineering plant cost index Idea: need for index because of lack of historical data and of large variety of equipment – estimation, quality not so good as cost engineering calculations or cost estimation software, but faster ! A lot of indices are available Chemical industry Average process industries (cement, chemical, clay, glass, paint, paper, petroleum, rubber) General industry Petroleum refineries Rapid growth in activities in CPI in mid and late 1970s (fast increase) – 1980s over‐ supply of constructors (price drops) Large difference Inflation dropped, thus lower increase rate after 1981 Research question: need for “Belgian” index (BCEPC) Methodology Start with a simple index (few components) and add components until it is a good fit with the official American CEPC • • • • General 2 parameters: labor + material 3 parameters: + productivity 4 parameters: + inflation 5 parameters: + crude oil price Finetune the new cost index for the Belgian situation on hand (company !) taking into account the fact that construction projects in Belgium also include non‐Belgian inputs Company‐ specific country weight Belgium 67 % Netherlands 16 % Germany 8 % France 6 % UK 3 % USA 0% Implementation Compare new, improved index with current practice in CPI Company Index 1 Langfrist 2 In‐house (German) 3 Nelson‐Farrar 4 2‐parameter – UK model 5 Individual indices only Analyse what the impact of the new index will be on established management ratios * Maintenance ratio =(total maintenance cost per year)/capital replacement value * Stores ratio=(MRO inventory value)/ capital replacement value Capital replacement value = Equipment replacement value Belgian chemical engineering plant cost index (c’d) maintenance ratio (%) based on old cost index new cost index 3.8 4.9 3.3 4.1 3.0 3.6 year X X+1 X+2 stores ratio (%) based on old cost index new cost index 1.0 1.3 0.9 1.1 0.8 0.9 Trend is similar for old and new index, numerical values however are quite different ! 3 .5 Used for benchmarking Beware of value of benchmarks ! stores ra tio (%) 3 2 .5 2 1 .5 world class 1 0 .5 0 U1 U2 U3 U4 U5 U6 U7 U8 units Very low ! Too high ? Briefly describe how cost indices work … What do you see as benefits of cost indices for management ? What challenges do you see ? (Parametric methods) Cost‐estimating relationships (CER) Design variables (speed, weight, thrust, physical size, etc.) for plants, equipment, and construction are determined in the early design stages. Cost estimating relationships (CER) use these design variables to predict costs. A CER is thus generically different from the cost index method, because the index is based on the cost history of a defined quantity and quality of a variable. Examples Cost‐capacity equation Factor method Statistical and other mathematical models are used! Cost‐capacity equations One of the most widely used CER models Cost‐capacity equations are also called power law and sizing models. Sources like e.g. Chemical Engineers’ Handbook, US Formula x Straight line on log‐log paper Q C2 C1 2 Q1 where C1 cost at capacity Q1 Environment Protection Agency, handbooks, professional journals, consultants, equipment manufacturers, … C2 cost at capacity Q2 Average x = 0.6 If 0<x≤1: economies of scale If x=1: linear relationship If x>1: diseconomies of scale x correlating exponent Assume C1=1000, Q1=1000 and Q2=2000, then C2? x 1 x 1/ 2 Q2 Q2 C2 C1 C1 2C1 2000 linear relationship Q1 Q1 Q Q C2 C1 2 C1 2 Q1 Q1 x 21 / 2 C1 1.73C1 1730 economies of scale 2 Q Q C2 C1 2 C1 2 22 C1 4C1 4000 di sec onomies of scale Q1 Q1 Cost‐capacity equations can be time adjusted using cost indexes. Formula Q x I C2 ,t C1,0 2 t Q1 I 0 Example Factor method Originally developed for estimating total plant costs; Identify the major equipment items then multiply by certain factors; Premise: Major equipment costs are readily available; Using the proper factors, then a rapid estimate of total plant cost can be achieved. Factors: called “Lang factors” (1947). Formula CT hCE where CT total plant cost CE total cost of major equipment h overall cost factor or sum of individual cost factors Example Recall: Direct & indirect costs manufacturing Direct and indirect costs Recall Direct costs – identifiable with product, function, process Indirect costs – general administration, computer services, quality, safety, taxes, security, … Lang showed that direct cost factors and indirect cost factors can be combined into one overall factor for some types of plants: 3.1 for solid process plants; 3.63 for solid‐fluid process plants and 4.74 for fluid process plants. Subsequent refinements of the factor method have led to the development of separate factors for direct and indirect cost components. formulas n h 1 fi i 1 where f i factor for each cost component direct & indirect fi indirect cost factor applied to total direct cost i 1 to n components, including indirect cost n CT CE 1 f i (1 f I ) i 1 where f I indirect cost factor f i factors for direct cost components Examples Example 1 Example 2 Direct cost factors Indirect cost factors “Issue: Indirect costs” Traditional indirect cost rates and allocation indirect cos t rate estimated indirect cos ts estimated basis level Activity‐based costing Define cost pools. Usually these are support functions. Identify cost drivers. These help to trace costs to the cost pools. Management Accounting Why ABC makes sense … Setting: 3rd party warehouse Assume warehousing costs (building maintenance, equipment, packaging materials, personnel, …) are € 300000 per year There are three customers, each of the customers has a contract for 1500 pallet position (including m2 needed and all incoming/outgoing movements. It seems logical to split the € 300000 evenly over the three customers (€ per pallet position) or not? Only logical when the three customers have similar activities. If e.g. for customers A and B receptions and shipments are always in full pallets, but for customer C only the receipts are in full pallets and the shipping occurs in individual boxes, it seems fair that C pays more for the warehousing services. A more detailed analysis on warehousing activities is needed then. Exercises Have a look at it Wrap‐up Chapter 15 Extras Cost estimation: what ? Need. Cost types. Approaches (↑↓), learning curve. Accuracy. Cost estimation: some techniques “Per unit” method. Cost indexes. Cost‐capacity equations (CER). Factor method (CER). Or professional organizations … Indirect costs: Traditional rates and allocation vs Activity based costing. Engineering Economy Depreciation methods 1 Overview Terminology Depreciation methods Depletion methods 2 Terminology Depreciation is the reduction in value over time of an asset (due to wear & tear, use, deterioration, obsolescence, …) . The method used to depreciate an asset is a way to account for the decreasing value of the asset to the owner and to represent the diminishing value (amount) of capital funds invested in it. The annual depreciation amount Di does not represent an actual cash flow, nor does it necessarily reflect the actual use pattern of the asset during ownership. Terminology 3 Book depreciation and tax depreciation Book depreciation: used by a corporation or business for internal financial accounting Indicates the reduced investment in an asset based upon the usage pattern and expected useful life of an asset Classical, internationally accepted methods: straight line, declining balance, sum‐of‐year digits Tax depreciation: used in tax calculations per government regulations. In the USA and many industrialized countries, the annual tax depreciation is tax deductible, i.e. it is subtracted from income when calculating the amount of taxes due each year. However the tax depreciation amount must be calculated using a government‐ approved method. Terminology 4 First Cost or Unadjusted Basis ‐ B Initial purchase price + all costs incurred in placing the asset in service Book Value ‐ BV Remaining undepreciated capital investment on the accounting books (B ‐ sum of all depreciation claimed in the past) Recovery Period – n Depreciable life of the asset in question – often set by law Market Value ‐ MV Amount realized by sale on the open market Salvage Value ‐ S Estimated trade‐in value or market value at the end the asset’s useful life Depreciation Rate ‐ dt The fraction of the first cost removed by depreciation each year Personal Property All property except real estate use in the pursuit of profit or gain Real Property Real estate and improvements, buildings and certain structures Note 1: Half‐year convention: assets are placed in service or disposed of in midyear Note2: Land is Real Property, but by law is NOT depreciable for tax purposes Terminology 5 Depreciation methods Single line (SL) Declining balance (DB) and double declining balance (DDB) Sum‐of‐years digits (SYD) Modified accelerated cost recovery system (MACRS) Depreciation 7 Straight line (SL) deprecation Book value decreases linearly with time. Depreciation rate d=1/n is the same each year of recovery period n Formulas SLN(B,S,n) Dt ( B S )d BS d BV t B tD t where t year ( t 1,2 ,..., n ) Dt annual depreciation charge Book value after t years of service B first cost or unadjusted basis S estimated salvage value n recovery period Depreciation d d t depreciation rate 1 / n 8 Example (16.1) Depreciation 9 Declining balance (DB) and double declining balance (DDB) depreciation Accelerated depreciation method; also known as fixed percentage or uniform percentage method Annual depreciation is determined by multiplying the book value at the beginning of the year by a fixed percentage d, thus the depreciation amount decreases every year. DB does not directly use the estimated salvage value; DB has its own implied salvage value. The pure DB method will never depreciate an asset down to a “0” Depreciation 10 Double declining Max. depr. rate by law: balance d max 2 n Depreciation for year t: Dt ( d ) BVt 1 Actual depr. rate for year t: d t d (1 d )t 1 If BVt‐1 not known, apply: Book Value amounts (two ways): BVt B (1 d )t BVt BVt 1 Dt Implied Salvage Value: impS BVn B (1 d ) n Implied d for S >0: 1/ n S implied d 1 B Dt dB (1 d )t 1 DDB(B,S,n,t,d) Depreciation 16‐11 Example Depreciation 12 Depreciation 13 Depreciation 14 Sum‐of‐years digits (SYD) depreciation Classical accelerated depreciation technique that removes much of the basis in the first one‐third of this recovery period; however write‐off in not as rapid as for DDB n t 1 Formulas D (BS ) t SUM j n SUM j j 1 n( n 1 ) 2 t( n t / 2 0.5 ) (BS ) SUM n t 1 dt SUM BVt B Depreciation SYD(B,S,n,t) 16 Example B‐S=25000‐4000=21000 n‐t+1=8‐2+1 SUM=8*9/2=36 Depreciation 17 18 Modified accelerated cost recovery system (MACRS) By US Federal Tax Law, all assets placed in service and eligible for depreciation MUST use the current MACRS methods of calculation of depreciation amounts. Tax Law permits states to have their own respective depreciation methods for state income tax purposes (complicating factor) Formula Dt d t B From table ! Depreciation 19 Depreciation 20 Example Depreciation 22 Depreciation 23 Depreciation 24 MACRS recovery periods Alternative depreciation system General depreciation system Depreciation 25 Compared … 26 Depletion methods Depreciation is applied to assets that can be replaced. Depletion applies to (natural) resources that are not easily replaced, like: timber, mineral deposits, oil and gas, etc. Cost and percentage depletion Percentage depletion percentage depletion amount percentage* gross income from property Cost depletion or factor depletion first cos t pt resource capacity Examples 16.5 – 16.6 Depletion 27 28 Remember Depreciation may be determined for internal company records (book depreciation) or for income tax purposes (tax depreciation). Depreciation is a book method by which the capital investment in tangible property is recovered. Depreciation does not result in actual cash flows directly – rather, tax savings are the result of depreciation. The annual depreciation amount is tax deductible, which can result in actual cash flow changes. In the US, the MACRS method is the only one allowed for tax depreciation. Depletion methods are used to recover investment in the extraction or harvesting of natural resources. 16‐29 Wrap‐up Chapter 16 Terminology Depreciation methods Depletion methods 31 Engineering Economy After‐tax economic analysis 1 Overview Book focuses on USA !! Terminology Before‐Tax and After‐Tax Cash Flow Taxes and Depreciation method After‐Tax analysis 2 Income Tax Terms and Relations (Corporations) Income taxes are real cash flow payments to governments levied against income and profits. The (noncash) allowance of asset depreciation is used in income tax computations. Two fundamental relations: NOI and TI Net operating income = gross revenue – operating expenses NOI = GI – OE (only actual cash involved) NOI is also call EBIT (earnings before interest and taxes) Taxable income = gross revenue – operating expenses – depreciation TI = GI – OE – D (involves noncash item) Note: All terms and relations are calculated for each year t, but the subscript is often omitted for simplicity 17‐3 © 2012 by McGraw-Hill Reserved All Rights Tax Terms and Relations ‐ Corporations Gross Income GI or operating revenue R ‐‐ Total income for the tax year realized from all revenue producing sources Operating expenses OE ‐‐ All annual operating costs (AOC) and maintenance & operating (M&O) costs incurred in transacting business; these are tax deductible; depreciation not included here Income Taxes and tax rate T ‐‐ Taxes due annually are based on taxable income TI and tax rates, which are commonly graduated (or progressive) by TI level. Taxes = tax rate × taxable income = T × (GI – OE – D) Net operating profit after taxes NOPAT – Money remaining as a result of capital invested during the year; amount left after taxes are paid. NOPAT = taxable income – taxes = TI – T × (TI) = TI × (1 – T) 17‐4 © 2012 by McGraw-Hill Reserved All Rights Average and Effective Tax Rates Marginal tax rates change as TI increases. Calculate an average tax rate using: total taxes paid taxes Average tax rate = = taxable income TI To approximate a single‐figure tax rate that combines local (e.g., state) and federal rates calculate the effective tax rate Te Te = local rates + (1‐ local rates) × federal rate Then, Taxes = Te × TI 17‐6 © 2012 by McGraw-Hill Reserved All Rights Cash Flow After Taxes (CFAT) NCF is cash inflows – cash outflows. Now, consider taxes and deductions, such as depreciation Cash Flow Before Taxes (CFBT) CFBT = gross income – expenses – initial investment + salvage value = GI – OE – P + S A negative TI value is Cash Flow After Taxes (CFAT) CFAT = CFBT – taxes = GI – OE – P + S – (GI – OE – D)(Te) considered a tax savings for the project Once CFAT series is determined, economic evaluation using any method is performed the same as before taxes, now using estimated CFAT values 17‐9 © 2012 by McGraw-Hill Reserved All Rights Effects on Taxes of Depreciation Method and Recovery Period Goal is to minimize PW of taxes, which is equivalent to maximizing PW of depreciation Compared … DEPRECIATION METHOD All methods have the same amount of total taxes due RECOVERY PERIOD All lengths have the same amount of total taxes due Accelerated depreciation methods result in lower PWtaxes Shorter recovery periods result in lower PWtaxes General observation for SL, DDB and MACRS methods: MACRS PWtaxes < SL PWtaxes < DDB PWtaxes General goal: use shortest (MACRS) recovery period allowed (Note: with same single tax rate, recovery period and salvage value) 17‐10 (Note: with same single tax rate, depreciation method and salvage value) © 2012 by McGraw-Hill Reserved All Rights 12 13 Depreciation Recapture (DR) and Capital Gain (CG) DR, also called ordinary gain, in year t occurs when an asset is sold for more than its BVt DR = selling price – book value = SP – BVt CG occurs when an asset is sold for more than its unadjusted basis B (or first cost P) CG = selling price – basis = SP – B CL occurs when an asset is sold for less than its current BVt CL = book value – selling price = BVt – SP 17‐11 © 2012 by McGraw-Hill Reserved All Rights Effects of DR, CG and CL on TI and Taxes CG = SP1 ‐ B DR = SP2 ‐ BV CL = BV – SP3 Update of TI relation: TI = GI – OE – D + DR + net CG – net CL 17‐12 © 2012 by McGraw-Hill Reserved All Rights Example: Depreciation Recapture A laser‐based system installed for B = $150,000 three years ago can be sold for SP = $180,000 now. Based on 5‐year MACRS recovery, BV3 = $43,200. GI for year is $800,000 and annual operating expenses average $50,000. Determine TI and taxes if Te = 34% and the system is sold now. Solution: Depreciation recapture (DR) and capital gain (CG) are present DR = SP‐BVt = 180,000 – 43,200 = $136,800 CG = SP‐B = 180,000 – 150,000 = $30,000 MACRS D3 = 0.192(150,000) = $28,800 TI = GI – OE – D + DR + CG = 800,000 – 50,000 – 28,800 + 136,800 + 30,000 = $888,000 Taxes = TI × Te = 888,000 × 0.34 = $301,920 Note: If not sold now, taxes = (800,000 – 50,000 – 28,800) × (0.34) = $245,208 17‐13 © 2012 by McGraw-Hill Reserved All Rights After‐Tax Evaluation Use CFAT values to calculate PW, AW, FW, ROR, B/C or other measure of worth using after‐tax MARR Same guidelines as before‐tax; e.g., using PW at after‐tax MARR: One project: PW ≥ 0, project is viable Two or more alternatives: select one ME alternative with best (numerically largest) PW value For costs‐only CFAT values, use + sign for OE, D, and other savings and use same guidelines Remember: equal‐service requirement for PW‐based analysis ROR analysis is same as before taxes, except use CFAT values: One project: if i* ≥ after‐tax MARR, project is viable Two alternatives: select alternative with ∆i* ≥ after‐tax MARR for incremental CFAT series 17‐14 © 2012 by McGraw-Hill Reserved All Rights Approximating After‐Tax ROR Value To adjust a before‐tax ROR without details of after‐tax analysis, an approximating relation is: After‐tax ROR ≈ before‐tax ROR × (1 – Te) Example: P = $‐50,000 GI – OE = $20,000/year n = 5 years D = $10,000/year Te = 0.40 Estimate after‐tax ROR from before‐tax ROR analysis Solution: Set up before‐tax PW relation and solve for i* 0 = ‐ 50,000 + 20,000(P/A,i*%,5) i* = 28.65% After‐tax ROR ≈ 28.65% × (1 – 0.40) = 17.19% (Note: Actual after‐tax analysis results in i* = 18.03%) 17‐15 © 2012 by McGraw-Hill Reserved All Rights Example: After‐Tax Analysis Asset: B = $90,000 Per year: R = $65,000 S = 0 n = 5 years OE = $18,500 D = $18,000 Effective tax rate: Te = 0.184 Find ROR (a) before‐taxes, (b) after‐taxes actual and (c) approximation Solution: (a) Using IRR function, i* = 43% (b) Using IRR function, i* = 36% (c) By approximation: after‐tax ROR = 43% × (1 – 0.1840) = 35% 17‐16 © 2012 by McGraw-Hill Reserved All Rights After‐Tax Replacement Analysis Consider depreciation recapture (DR) or capital gain (CG), if challenger is selected over defender Can include capital loss, if trade occurs at very low (‘sacrifice’) trade‐in for defender An after‐tax analysis can reverse the selection compared to before‐tax analysis, but more likely it will provide information about differences in PW, AW or ROR value when taxes are included Apply same procedure as before‐tax replacement evaluation once CFAT series is estimated 17‐17 © 2012 by McGraw-Hill Reserved All Rights Economic Value Added (EVA)TM Analysis Definition: The economic worth added by a product or service from the perspective of the consumer, owner or investor In other words, it is the contribution of a capital investment to the net worth of a corporation after taxes Example: The average consumer is willing to pay significantly more for potatoes processed and served at a fast-food restaurant as fries (chips) than as raw potatoes in the skin from a supermarket. Value‐added analysis is performed in a different way than CFAT analysis, however… Selection of the better economic alternative is the same for EVA and CFAT analysis, because it is always correct that … AW of EVA estimates = AW and CFAT estimates TM At this time, the term EVA is a registered trademark of Stern Stewart & Co. 17‐19 © 2012 by McGraw-Hill Reserved All Rights International Tax Structures Tax related questions for internationally located projects concentrate on items such as: Depreciation methods approved by host country Capital investment allowances Business expense deductibility Corporate tax rates Indirect tax rates – Value‐Added Tax and Goods and Service Tax (VAT/GST) Rules and laws vary considerably from country to country 17‐21 © 2012 by McGraw-Hill Reserved All Rights Summary of International Corporate Tax Rates Tax Rate Levied on TI, % ≥ 40 For These Countries United States, Japan 35 to < 40 Pakistan, Sri Lanka 32 to < 35 France, India, South Africa 28 to < 32 Australia, United Kingdom, Canada, New Zealand, Spain, Germany, Mexico 24 to < 28 China, Indonesia, South Korea, Israel 20 to < 24 Russia, Turkey, Saudi Arabia < 20 Singapore, Hong Kong, Taiwan, Chile, Ireland, Iceland, Hungary Source: KPMG Corporate and Indirect Tax Rate Survey, 2010 17‐22 © 2012 by McGraw-Hill Reserved All Rights Summary of Important Points For a corporation’s taxable income (TI), operating expenses and asset depreciation are deductible items Income tax rates for corporations and individuals are graduated by increasing TI levels CFAT indirectly includes (noncash) depreciation through the TI computation Depreciation recapture (DR) occurs when an asset is sold for more than the book value; DR is taxed as regular income in all after‐tax evaluations After‐tax analysis uses CFAT values and the same guidelines for alternative selection as before‐tax analysis EVA estimates extra worth that an alternative adds to net worth after taxes; it mingles actual cash flows and noncash flows A VAT system collects taxes progressively on unfinished goods and services; different than a sales tax system where only end users pay 17‐25 © 2012 by McGraw-Hill Reserved All Rights Exercises 30 32 17.36 34 Wrap‐up Chapter 17 Terminology Before‐Tax and After‐Tax Cash Flow Taxes and Depreciation method After‐Tax analysis 37 Engineering Economy Sensitivity analysis Decision making under risk and under uncertainty Staged decisions 1 Overview Structure of chapter is not following the handbook (extras !) Basics Risk – (Un)Certainty Decision making under uncertainty Decision trees Basics – VOPI, VOII, VOC Case study Simulation 2 Basics Chapter 18 (8th edition) Sensitivity analysis Parameter: a variable or factor for which an estimated or stated value is necessary Sensitivity analysis: an analysis to determine how a measure of worth (e.g. PW) changes when one or more parameters vary over a selected range of values Alternatives Pessimistic (P), most likely (ML), optimistic (O) estimate Expected value calculation 3 Expected value – standard deviation Probability distribution (discrete, continuous) Random sample Selection in a random fashion of n values for a variable from a population with assumed or known probability distribtion 4 Simulation analysis uses random samples from the probability distribution of selected variables for alternative evaluation using a measure of worth Monte Carlo sampling is a commonly used simulation approach 5 7 Illustration Data (in €) Cash flows (present worth) CF in probability CF i n 50000 30 % 60000 40 % 70000 30 % CF out 50000 70000 probability CF out 45 % 55 % Model PW=CF in - CF out Parameters and their cumulative distribution CF in (€) cum. prob. CF out (€) 50000 30% 50000 60000 70% 70000 70000 100% cum. prob. 45% 100% Random generator (use RN – random numbers between 0 en 99) and corresponding parameters CF in RN1 CF out RN2 50000 50000 029 044 60000 70000 3069 4599 70000 7099 8 Results (simulation runs) Run RN1 1 46 2 30 3 14 4 35 5 9 6 19 7 72 8 20 9 75 10 16 … CF in 60000 60000 50000 60000 50000 50000 70000 50000 70000 50000 RN2 81 8 88 21 73 77 1 46 97 43 CF uit 70000 50000 70000 50000 70000 70000 50000 70000 70000 50000 PW -10000 +10000 -20000 +10000 -20000 -20000 +20000 -20000 0 0 Average Std dev Reliable result needed! (confidence intervals) 9 Risk – (Un)certainty Deterministic (no variation !) estimates for all parameters Certainty For some parameters there is a probability distribution available for the estimates. Expected value or simulation approach Decision making “Probabilities” are unknown Uncertainty Risk 10 Decision making under uncertainty Chances are not known for the identified states of nature (values) of some parameters. It is possible to agree that each state of nature is equally likely to occur (cfr risk) Other methods exist; e.g. Laplace criterion, Minimax (Wald) approach, Hurwicz criterion, Savage criterion, … Certainty ?! Uncertainty Risk 12 Θj with j=1, …n Assume What can happen … we don’t know anything about the chances, we can’t influence anything The choices we have, i.e. investment alternatives ai with i=1, …m The outcome vij: what willl happen if we decided for 13 a given action i and a specific state of nature j occurs Criterion Decision rule Wald – maximin return Hurwicz – maximax return optimism‐pessimism index Savage – minimax regret Laplace – insufficient reason j i sk=security level ok=optimism level rij=regret ρk=index 15 16 17 18 19 These criteria may very well all suggest a different cause of action. Example vij θ1 θ2 θ3 θ4 2 2 0 1 1 1 1 1 0 4 0 0 1 3 0 0 20 i=1, ... m j=1, ... n vij si θ1 θ2 θ3 θ4 2 2 0 1 0 1 1 1 1 1 0 4 0 0 0 1 3 0 0 0 Wald ‐ maximin: For each action: worst possible outcome = security level si Choose action with largest security level: a2 22 i=1, ... m j=1, ... n vij θ1 θ2 θ3 2 2 0 1 1 0 1 θ4 si 1 0 oi 1 1 1 1 4 0 0 0 4 3 0 0 0 3 2 Hurwicz ‐ maximax Optimism level oi = best consequence that can result if ai is chosen Choose action with best oi : a3 or User preference (optimistic‐pessimistic) with the use of weighted sum introducing α α*0+(1‐ α)*2; α*1+(1‐ α)*1; α*0+(1‐ α)*4; α*0+(1‐ α)*3 2*(1‐ α); 1; 4*(1‐ α); 3*(1‐ α) Als α <3/4: a3 23 i=1, ... m j=1, ... n vij θ1 θ2 θ3 θ4 2 2 0 1 1 1 1 1 0 4 0 0 1 3 0 0 Savage – minimax regret rij Compute regret θ1 θ2 θ3 θ4 2 2‐2=0 2 4‐2=2 0 1‐0=0 1 1‐1=0 1 2‐1=1 1 4‐1=3 1 1‐1=0 1 1‐1=0 0 2‐0=2 4 4‐4=0 0 1‐0=0 0 1‐0=1 1 2‐1=1 3 4‐3=1 0 1‐0=1 0 1‐0=1 Compute Max rij Find i with Min Max rij 24 i=1, ... m j=1, ... n vij θ1 θ2 θ3 θ4 2 2 0 1 1 1 1 1 0 4 0 0 1 3 0 0 ¼*(2+2+0+1)=5/4 ¼*(1+1+1+1)=1 ¼*(0+4+0+0)=1 ¼*(1+3+0+0)=1 Laplace * Equal probabilities !! 25 27 Certainty Decision trees ?! Uncertainty Risk When to use decision trees More than one stage of alternative selection Selection of an alternative at one stage that leads to another stage Probability estimates for each outcome Estimates of economic value (cost or revenue) for each outcome Measure of worth (expected value of ‐) as the selection criterion, e.g. E(PW) 28 You decide Basic concept “Nature” decides Build tree Solve tree 29 Steps to take Build up the tree structure from the left (final decision to make) to the right, including each possible decision and outcome. A square represents a decision node with the possible alternatives indicated on the branches. A circle represents a probability node with the possible outcomes and estimated probabilities on the branches. Since outcomes always follow decisions, a treelike structure results. Fill out all quantitative information on the tree. Economic value (payoff) for each branch of the decision tree. Cost, revenue or benefit. Expressed in PW, AW or FW. Probabilities for each of the mutually exclusive outcomes emanating from each decision node. These probabilities must sum to 1 for each set of outcomes (branches) that are possible from a given decision Solve the decision tree (foldback), going from the right to the left Start at the top right node. Calculate the expected value for each decision alternative E[decision]=∑ (outcome es mate) p(outcome) At each decision node, select the best E( decision ) value Continue moving to the left of the tree back to the root in order to select the best alternative Trace the best decision path back through the tree 30 Example Probability – not sure what will happen Sequence of two decisions (staged decisions) 32 Probability nodes Decision nodes 33 Example 34 43 CFBT=cash flow before tax Exercise A company needs to take a decision concerning product M9 which has been developed in one of its R&D labs. The decision is about either putting M9 on a test market or to stop all activities for M9. Launching M9 on a test market would cost around € 100K. Based on historical data the company knows that only about 30% of products tested on the market prove to be successful. If M9 would be a success on the test market, another decision needs to be made, i.e. how large should the production plant for M9 be designed, if any at all. A small plant would cost € 150K to build and produce 2000 units per year. A large plant would cost € 250K and produce 4000 units per year. Marketing estimates a 40% probability that competitors put a similar product on the market, this would of course influence the selling price. Assuming that everything produced will be sold the following estimates are made Unit price (€) With competition No competition Large plant Small plant 20 35 50 65 The commercial life time of M9 is estimated to be 7 years. Yearly operating costs for both plants are equal, being € 50K. Would you recommend to continue or to abandon the M9 project? 45 Value of control Decision trees Value of perfect information Informational analysis: How much are we willing to pay for …? Value of imperfect information 50 Value of perfect information (VOPI) VOPI on a variable is calculated as Expected value with perfect information – Expected value without perfect information VOPI is an upper bound on the value of additional research to improve the probability assessment on a uncertain variable In a more complicated problem, the variables can be ranked according to VOPI, providing guidance for additional research. 51 Value of imperfect information (VOII) The value of imperfect information (VOII) can also be determined with decision analysis. An example of imperfect information is e.g. a forecast report with a known accuracy of 80%. This is a more complex calculation and requires the use of Bayesian updating. 52 Value of control (VOC) The value of control determines the upper bound on the value of controlling an uncertainty. Value of Control is calculated as Expected Value With Control ‐ Expected Value Without Control This value can be used to gauge the cost effectiveness of new alternatives. 53 Example: Informational analysis Orange Grower’s Decision Problem: Frost could occur overnight Frost protection costs money Total crop loss if frost occurs without protection measures in place 54 Helps analyzing and structuring the decision problem Uncertainty Impact Decision Influence on Net benefit of frost protection decision = Crop value – Frost protection cost 55 Data Frost protection cost = 25 Value of undamaged crop = 100 Value of crop if frost occurs, but with frost protection = 75 Value of crop if frost occurs, no frost protection = 0 Probability(frost)=0.4; Probability(no frost)=0.6 Tree building 56 Tree solving 57 Value of perfect information Tree structure is different ! Probabilities are still there (nature !) Initial problem 58 80% accuracy Value of imperfect information P ( Frost | " Frost ") P(" Frost " | Frost ) * P ( Frost ) 0.32 0.73 P(" Frost " | Frost ) * P( Frost ) P(" Frost " | No frost ) * P ( No frost ) 0.32 0.12 59 80% accuracy Bayes P ( Frost | " Frost ") P(" Frost " | Frost ) * P ( Frost ) 0.32 0.73 P(" Frost " | Frost ) * P( Frost ) P(" Frost " | No frost ) * P ( No frost ) 0.32 0.12 60 80% accuracy Bayes P ( Frost | " Frost ") P(" Frost " | Frost ) * P ( Frost ) 0.32 0.73 P(" Frost " | Frost ) * P( Frost ) P(" Frost " | No frost ) * P ( No frost ) 0.32 0.12 61 New tree 62 Value of control 63 Case study Interesting case study – full text available on Toledo, please read 66 Case study: Underground pipeline Case source AIChE Problem at hand Potential leakage of flammable and toxic material in old pipelines Action alternatives Do nothing Replace old pipeline Install automated shut‐off valves Truck feed stock to the plant 67 Brainstorm First feasibility check of alternatives elimination of unrealistic alternatives do nothing replace current (aging) pipeline insert automated shutoff valves truck feedstock to plant List of uncertainties (to be taken into account in decision) risk of pipeline failure whether a failure would result in a slow leak or a rupture the amount of chemical released due to failure the risk of ignition if there is a rupture the cost to clean up a chemical release from the pipeline 68 Preliminary list with concerns and their priorities cost of modifying current pipeline or installing new pipeline cost of cleaning up a spill worker safety public safety public concern Purely economic decision Model value measure for decision = NPV (net present value), i.e. PW (present worth) Additional assumptions and simplifications horizon = 10 years E[NPV]=E[PW] Salvage value old pipeline after 10 years = 0 p(ignition) and cost (fire) f(amount chemical released) 69 Constructie van influence diagram Legend Decision to take Value measure (criterium) Dependency, influence Uncertainties (independent of alternative) 70 Constructie van influence diagram Legend Decision to take Value measure (criterium) Dependency, influence Uncertainties (independent of alternative) 71 Deterministic phase Calculate the value function total cost = CA +CC +C F -V, where CC =CG *Q with C A cost of pipeline action CC cost of cleanup CF cost of fire V value of remaining life of new pipeline after 10 years CG cleanup cost per gallon Q quantity released due to failure Collect numerical values for all variables (certain and uncertain) 72 Calculate the nominal value (expected value) for all alternatives best nominal alternative NPV for each alternative under nominal scenario Do nothing Replace pipeline Insert valves $ 1.96 M $ 2.73 M $ 2.04 M Carry out a sensitivity analysis and visualise in a tornado‐diagram variables for which further study is recommended at nominal setting in further analysis 73 Probabilistic phase Determine all probability functions for the appropriate variables Build and solve decision tree(*) best alternative $ 1.95 M $ 2.42 M $ 1.52 M 75 Compute the value of information en the value of control upperbound on cost for monitoring system to control size of leak (limited to 250 gallons) = $ 1.2 M upperbound on cost study about remaining life current pipeline = $ 111000 77 Conclusions From this rough study If immediate action is required and there is no possibility for further study install automated shut‐off valves If no immediate action is necessary additional research to obtain more information on the failure probability of the current pipeline (years to failure) ‐ max. cost = $ 111000 A monitoring system that would limit a slow leak to 250 gallon needs to be investigated ‐ max. cost = $ 1.2 M real life 78 Conclusions (c’d) real life No immediate action required: ultrasonic study of current pipeline (cost $ 20000) info to model Investigate “monitoring system” add to list of feasible alternatives Refine estimation of probabilities on "type of failure" info to model Uncertainty on system effectiveness and amount of release (prevented by early warning) resp. add to model Update analysis and formulate final recommendation on “best alternative” 79 Wrap‐up Chapters 18 & 19 Extras Risk – (Un)Certainty Decision making under uncertainty Decision trees Basics – VOPI, VOII, VOC Case study Simulation 88 Course material Blank & Tarquin, 2018 Work shops [email protected] Course content 7 Will be provided Evaluation – Exam Closed book Written part (+/‐ 25%) Small questions (insight?!) – multiple choice (guess correction)/open questions Oral part with (written) preparation time (+/‐ 75%) Exercise(s) Knowing which method to use and drawing the right conclusions from the results Formularium available Case study – Journal article Proving “maturity” in the subject Being able to evaluate a chosen approach As discussed in first class (see “Introduction” slides on Toledo) https://eng.kuleuven.be/studenten/lijstRM.html FEATURE The use of alternative solvent purification techniques A recent accident at the University of California, Irvine involved the purification of an organic solvent using a solvent still resulting in a fire in the lab. A graduate student was seriously burned and $3.5 million in property damage was incurred. In this report, lessons learned from this accident are used to improve the safety of solvent purification operations. Column methods of aprotic solvent purification processes, as well, as the purchasing of ultra-dry organic synthesis solvents are evaluated. Residual risks associated with the solvent still operation are compared with the alternative purification methods. While severity remains unchanged using the column method, employing the column method reduces the likelihood of an accident. For most applications, the column method and the purchase of ultra-dry solvents remove moisture and oxygen at least to the same level as the solvent still method. This is not without additional costs. Column systems have higher upfront capital costs and ultra-dry solvents cost 20–50% more than certified solvents. Cost–benefit analysis argues that the additional costs make alternative purification methods acceptable. In conclusion, column methods of aprotic solvent purification processes and the purchasing of ultra-dry organic synthesis solvents are, in many applications, a cost-effective alternative to solvent stills for producing a moisture- and oxygen-free product. By Michael E. Cournoyer and Jeffrey H. Dare INTRODUCTION In order to obtain satisfactory results in many syntheses involving air moisture sensitive reactions, it may be necessary to purify solvents to remove reactive impurities such as water, other protic/ acidic materials, or atmospheric contaminants such as oxygen. A commonly employed purification method is the solvent still, which involves the reflux/distillation of an organic solvent in the presence of a dehydrating/deoxygenating reagent. A recent accident at the University of California, Irvine (UCI) involved the purification of an organic solvent using a solvent still.1 The consequence of the accident was a fire in the lab that seriously burned a graduate student, and caused $3.5 million in property damage. An accident of this magnitude warrants an Michael E. Cournoyer, Ph.D., and Jeffrey H. Dare are affiliated with Nuclear Materials Technology Division, Los Alamos National Laboratory, Los Alamos, NM 87545 (Tel.: 505-665-7616; fax: 505-6657170; e-mail: [email protected]). 1074-9098/03/$30.00 doi:10.1016/S1074-9098(03)00055-8 evaluation of this traditional method of purification to assess the adequacy of the control measures currently in place and to consider alternative methods of producing a quality solvent. In 1996, the column method of solvent purification was introduced which eliminates hazards associated with distillation.2 This report consists of a critical evaluation of solvent still and column methods of aprotic solvent purification processes, as well, justification for the purchasing of ultra-dry organic synthesis solvents. Since the sodium benzophenone ketyl solution is among the most common solvent still methods to prepare pure, anhydrous, oxygen-free solvents, it will be used as an example. Typically, an organic solvent is refluxed in the presence of sodium and benzophenone in an inert atmosphere, as shown in Figure 1. The reactive metal removes moisture from the solvent, and the ketyl intermediate that forms upon reaction of the ketone and the metal acts as an oxygen scavenger. The blue color of benzophenone ketyl is used as an indicator that the solvent is ready for use. Usually the reflux process is continued for several hours to ensure solvent purity. Alternatives to the traditional solvent still include the column method in which dry nitrogen or argon is used to force commercially purified solvents through a stainless steel column. The column is packed with activated alumina and copper catalysts to remove trace amounts of water (Figure 2). Alternatively, the solvent must be run through a bubble degassing process when not suited for copper catalyst oxygen removal. Last, if only small quantities of pure anhydrous solvent are needed, it is sometimes considered more cost effective to buy these solvents directly from a supplier. TECHNICAL VALIDITY The technical validity of the solvent purification process is the most important criteria of the review process. The column method and the purchase of ultra-dry solvents must remove moisture and oxygen at least to the same The technical validity of the solvent purification process is the most important criteria of the review process. ß Division of Chemical Health and Safety of the American Chemical Society Elsevier Science Inc. All rights reserved. 15 level as the solvent still method. Qualitatively, both the column method and the purchase of ultra-dry solvents meet the tests for assessing oxygen and protic contaminants by using the sodium benzophenone-ketyl test or titanocene dichloride/zinc dust test. Quantitative values are shown in Table 1.3 RISK The risk associated with the purification of an organic solvent is a function of the likelihood and potential severity of injury, harm, incurred liability, damage or loss. As the accident at UCI clearly illustrated, the primary consequences (severity) associated with purifying flammable solvents are injuries and property losses due to fire and explosions. It should be noted that the severity remains unchanged using the column method. The main difference in risk between the solvent still and column method is in the likelihood of an accident. Procedures reacting active metals and flammable liquids, such as sodium and most organic solvents increase the likelihood that a fire or an explosion will occur. This is not limited to the purification procedure, but extends to the quenching of spent sodium for disposal, as well. The likelihood of an explosion or a fire increases further with the nearness of the following ignition sources: heating mantles, vacuum pumps, and water lines from condensers and the high temperatures needed for distillation. Stills with automatic controls that shut down the system conditions, such as loss of cooling or overheating of the still pot, do enhance the safety of the distillation operation greatly. While the column method has none of these fire or explosion initiators, it introduces some hazards of its own. Since the columns are pressurized (5–50 psi), a stored energy hazard is present. Peroxides may accumulate on the columns, as well, introducing an explosive hazard. In addition, the column method does involve larger quantities of solvent. Figure 1. A typical solvent still. Figure 2. A commercially available column system. Table 1. Quality of Purified Solvent Impurities Method Solvent still Column Purchasing 16 Water (ppm) Oxygen (ppb) Peroxides (ppm) 10 <14 5 100 10000 <100 <1 <2 <1 DISCUSSION With several prominent universities and companies using the column Chemical Health & Safety, July/August 2003 method to purify organic solvents, the ppm level of oxygen is not an issue with most air and moisture sensitive reactions. On the other hand, the copper catalyst is incompatible with some solvents including tetrahydrofuran and methylene chloride. Oxygen must be removed from these solvents by purging them with dry nitrogen or argon. Thus, column method is a valid substitute for the solvent still method within limits. The main goal of implementing a process improvement is to decrease the risk to an acceptable level associated with abnormal occurrences from the purification of an organic solvent. The main goal of implementing a process improvement is to decrease the risk to an acceptable level associated with abnormal occurrences from the purification of an organic solvent. The first issue that this raises is whether solvent stills with their current set of controls are processes that can be run with an acceptable level of risk. An objective method of determining risk with hazardous material operation may be obtained using the risk matrix from the paper titled ‘‘A Risk Determining Model for Hazardous Material Operations,’’ as shown in Table 2.5 Clearly, the severity of the accident that occurred at the UCI is rated ‘‘Catastrophic’’ due to the magnitude of the property damage. It should be noted that several factors contributed to the immediate consequences of the accident, including the following: The quantities of flammable and combustible liquids present in the building at the time of the fire were in excess of the quantities allowed for B-2 occupancy by the 1979 Building Code. Table 2. Risk Determination Matrix Likelihood Severity Catastrophic Critical Moderate Negligible Frequent Probable Occasional Improbable Remote High High High Low High High Medium Minimal High Medium Low Minimal Medium Low Minimal Minimal Low Minimal Minimal Minimal The door (with a 1-hour-rated fire door with self-closing and positivelatching hardware) between the laboratory with the solvent still and a room filled with flammable and combustible liquids was blocked open. Automatic sprinkler system protection was not installed. The fact that this should not have happened during the life of the facility, but did, gives it likelihood rating of ‘‘Occasional.’’ Similar accidents have occurred at the University of California at Berkeley in 1996 and the University of Texas at Austin in 1997. These observations lend further validity to the ‘‘Occasional’’ likelihood rating.6 Using the matrix, the residual risk (the risk remaining after controls are in place, with consideration of reliability and certainty of the controls and risk of control failure) associated with a solvent still is high. Whether operation with high residual risk should be authorized (acceptable) is the responsibility of line management. While this judgment is determined in a case-bycase study, UCI hopes to replace 30 chemical distillation devices with columns.7 The column method lowers the likelihood to a remote level during the purification. Hence, the residual risk associated with the column method is low (acceptable). The second issue that implementing a process improvement raises is whether the cost is acceptable. Many emerging technologies may be technically sound and lower the risk associated with the process they seek to improve and yet may be economically unacceptable. From a business viewpoint, the acceptable level may be achieved when the costs of decreasing a given risk further are greater than the costs realized from the occupational exposure to hazardous chemicals. Tangible costs for implementing technologies that lower risk in a chemical operation have not been published. On the other hand, criteria for establishing cost-effective dose-reduction measure are well established for the nuclear industry.8 Cost–benefit analyses typically apply monetary equivalents of $1,000 to $10,000 per person-rem with the recommended nominal value being $2,000 per person-rem. Optimization analysis are performed whenever the cost of these measures exceeds $50,000 or the collective dose to be avoided is greater that 5 person-rem. In addition, the cost incurred from radiological exposures have been compared to other non-radiological accidents including property losses, as shown in Table 3.9 Using the UCI accident as an example, the following approach is taken to obtain a reasonable cost–benefit analysis between the two methods of solvent purification: Table 3. Criteria that Trigger an Unusual Occurrence Groups of Unusual Occurrences Criteria Facility condition: loss of control of radioactive >100,000 (dpm)/100 cm2 material or spread of radioactive contamination* Personnel radiological protection: radiation exposure >5 rem** Value Base Reporting: cost-based occurrences ¼$1,000,000 * dpm ¼ disintegrations per minute. rem ¼ Roentgen equivalent man. ** Chemical Health & Safety, July/August 2003 17 1. An accident resulting from a solvent still operation is classified as a costbased occurrence. 2. The UCI accident is rated as an ‘‘Unusual Occurrence’’ in the Value Base Reporting category because over 1 million dollars in property damage occurred. 3. This type of occurrence is categorized at the same level as a radiation exposure of greater than 5 Roentgen equivalent man (rem), as shown in Table 3. 4. Recommended nominal value of $2,000 per person-rem typically is applied. Using this recommended nominal value, over $10,000 could be spent on process improvements to reduce the likelihood of fire and explosion hazards and still be cost-effective. Furthermore, a study conducted by the University Wisconsin, Madison (UWM), concluded that over the long term the column method cost less.10 Results of cost comparison of the solvent still and column methods are compiled in Table 4. While the column method has a higher set up cost, the annual operating are lower. The quality of the distilled product needs to be considered. For most purposes (e.g., Grignard reactions) the use of common inorganic drying reagents such as anhydrous magnesium sulfate, MgSO4 (stirred overnight and filtered) are sufficient. The UWM study also concluded that the column method was an ineffective choice for purifying tetrahydrofuran. Another alternative is the purchase of pure anhydrous solvent. In the past, if only small quantities of pure anhydrous solvent were needed, it was sometimes considered more cost effective to buy these solvents directly from a supplier. Cost increase varies per solvent, but in general the cost Table 4. Solvent Purification Cost Comparison Method Solvent still Column 18 Set Up Costs ($) Annual Operating Costs ($) 4800 6300 4000 100 of an anhydrous solvent is two to three times that of a certified American Chemical Society (ACS) solvent and costs of standard high purity solvents are 20–50% more expensive than that of certified ACS solvents.11 Depending on volumes, required costs vary. Once operating costs of the solvent stills, the capital costs of the column, and the liability of both the purification methods are weighed; the additional costs seem more reasonable. Last, the larger quantities of solvent and peroxides on the used adsorbent associated with the column method must be addressed. The larger quantities of solvent are contained in metal cans, which are safer than glass bottles. Since the peroxides remain unchanged on the alumina, the used adsorbent should not be heated. Stirring the adsorbent into an iron sulfate solution can mitigate the peroxide hazard or spent columns can be sent back to the manufacturer for regeneration. CONCLUSIONS The results of this study demonstrate that distillation and purification of flammable solvents, using the solvent still method, will continue to be an integral part of chemical experimentation, if only on a limited basis as alternative methods become available. Active metals increase the likelihood of an explosion and fire with flammable liquids. In the past the solvent still method of purifying organic solvents had an acceptable level of risk because there was no alternative way of obtaining a moisture- and oxygenfree product. Column methods of aprotic solvent purification processes, as well, as the purchasing of ultra-dry organic synthesis solvents are now cost effective alternative way of obtaining a moisture- and oxygen-free product. Acknowledgements The authors would like to acknowledge the Department of Energy and Los Alamos National Laboratory Weapons Engineering & Manufacturing Directorate for support of this work. References 1. University of California, Irvine Independent Accident Investigation, ‘‘Injury and Fire Resulting from Benzene Vapor Explosion in a Chemistry Laboratory Frederick Reines Hall, University of California, Irvine, July 23, 2001,’’ Final Report, January 24, 2002. 2. Robert H. Grubbs, et al., ‘‘Safe and Convenient Procedure for Solvent Purification,’’ Organometallics 1996, 15, 1518. Robert K. Rosen, et al., ‘‘Safe and Convenient Procedure for Solvent Purification,’’ Journal of Chemical Education January, 2001, 78 (1), 64. 3. James Farina and Paul Bouis, ‘‘Use of Ether Solvents in Organometallics Synthesis: The Good, Bad and the Ugly,’’ 210th American Chemical Society Meeting, Las Vegas, NV, 1997. 4. System specification, SolventPurification.com & Innovative Technology, Inc., 2 New Pasture Road, Newburyport, MA 01950. 5. Jeff Dare and Michael E. Cournoyer, Ph.D., ‘‘A Risk Determining Model for Hazardous Material Operations,’’ LA-UR 02-0378, Proceedings from Probabilistic Safety Assessment and Management 6 (PSAM 6), San Juan, Puerto Rico, USA, June 22–29, 2002. 6. Kurt W. Dreger and Steven F. Pedersen, ‘‘Solvent Purification in Academic Research Laboratories: A Comparison of Old and New Technologies,’’ 2002 AIHCE Laboratory Health & Safety Committee Presentations, http:// www2 . umdnj . edu / eohssweb / aiha / administrative / presentations . htm # Decommissioning. 7. Phil Willon, ‘‘Blame in UCI Fire Severity: No Sprinklers,’’ http://www.physsci.uci.edu/news/entries/2002022.html. 8. NUREG-1530 ‘‘Assessment of NRC’s Dollar Per Person-Rem Conversion Factor Policy,’’ 1995. 9. Steven Lee, Robert F. Grundemann, Ph.D., and Michael E. Cournoyer, ‘‘An Independent Analysis of a Glovebox Glove Failure Incident,’’ LA-UR 012233, Journal of the American Society of Mechanical Engineers, Proceeding from ICEM’01, the 8th International Conference on Radioactive Waste Management and Environmental Remediation, September 30–October 4, 2001. 10. ‘‘Research Laboratory Waste Reduction Project: Solvent Purification, Interim Report to DNR,’’ University of Wisconsin, Madison, Submitted May 1, 2001. 11. Rich Motyka, Personal Communication, Mallinckrodt Baker, Phillipsburg, NJ, July 17, 2002. Chemical Health & Safety, July/August 2003 Tools for Making Acute Risk Decisions with Chemical Process Safety Applications CCPS – AIChE; 1995