Foundations Of Engineering Economy Lecture slides to accompany

Foundations Of Engineering Economy Lecture slides to accompany
Chapter 1 Foundations Of Engineering Economy Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin At this point: 1. Introduce yourself - your students are likely to want to know something about your qualifications and interests - overall, where you are coming from. 2. Have students introduce themselves. Ask why they are taking this class. If you are fortunate enough to have a Polaroid camera, take pictures of each student for later posting on a class “board” so both they and you get to know each other. 3. Discuss both choice of textbook and development of syllabus. 4. If you are expecting students to work in teams, at east introduce the choice of team members. If at all possible, have students participate in a team building or team study exercise. It works wonders. Most student have been told to work in teams in prior classes, but have never examined exactly what a team is and how it works. One hour spent in a team building/examination exercise saves many hours and avoids many problems later on. © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

LEARNING OUTCOMES Role in decision making Economic equivalence
Study approach Ethics and economics Interest rate Terms and symbols Cash flows Economic equivalence Simple and compound interest Minimum attractive rate of return Spreadsheet functions © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Why Engineering Economy is Important to Engineers
Engineers design and create Designing involves economic decisions Engineers must be able to incorporate economic analysis into their creative efforts Often engineers must select and implement from multiple alternatives Understanding and applying time value of money, economic equivalence, and cost estimation are vital for engineers A proper economic analysis for selection and execution is a fundamental task of engineering © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Time Value of Money (TVM)
Description: TVM explains the change in the amount of money over time for funds owed by or owned by a corporation (or individual) Corporate investments are expected to earn a return Investment involves money Money has a ‘time value’ The time value of money is the most important concept in engineering economy © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Engineering Economy Engineering Economy involves
Formulating Estimating, and Evaluating expected economic outcomes of alternatives designed to accomplish a defined purpose Easy-to-use math techniques simplify the evaluation Estimates of economic outcomes can be deterministic or stochastic in nature © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

General Steps for Decision Making Processes
Understand the problem – define objectives Collect relevant information 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 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Steps in an Engineering Economy Study
© 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Ethics – Different Levels
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 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Code of Ethics for Engineers
All 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 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Interest and Interest Rate
Interest – the manifestation of the time value of money Fee that one pays to use someone else’s money Difference between an ending amount of money and a beginning amount of money Interest = amount owed now – principal Interest rate – Interest paid over a time period expressed as a percentage of principal © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Rate of Return Interest earned over a period of time is expressed as a percentage of the original amount (principal) Borrower’s perspective – interest rate paid Lender’s or investor’s perspective – rate of return earned © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Interest paid Interest earned
Interest rate Rate of return © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Commonly used Symbols t = time, usually in periods such as years or months P = value or amount of money at a time t designated as present or time 0 F = value or amount of money at some future time, such as at t = n periods in the future A = series of consecutive, equal, end-of-period amounts of money n = number of interest periods; years, months i = interest rate or rate of return per time period; percent per year or month © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

NCF = cash inflows – cash outflows = R – D
Cash Flows: Terms Cash Inflows – Revenues (R), receipts, incomes, savings generated by projects and activities that flow in. Plus sign used Cash Outflows – Disbursements (D), costs, expenses, taxes caused by projects and activities that flow out. Minus sign used Net Cash Flow (NCF) for each time period: NCF = cash inflows – cash outflows = R – D End-of-period assumption: Funds flow at the end of a given interest period © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Cash Flows: Estimating
Point estimate – A single-value estimate of a cash flow element of an alternative Cash inflow: Income = $150,000 per month Range estimate – Min and max values that estimate the cash flow Cash outflow: Cost is between $2.5 M and $3.2 M Point estimates are commonly used; however, range estimates with probabilities attached provide a better understanding of variability of economic parameters used to make decisions © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Cash Flow Diagrams What a typical cash flow diagram might look like
Draw a time line Always assume end-of-period cash flows … … … n n Time One time period F = $100 Show the cash flows (to approximate scale) … … … n n Cash flows are shown as directed arrows: + (up) for inflow - (down) for outflow P = $-80 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Cash Flow Diagram Example
Plot observed cash flows over last 8 years and estimated sale next year for $150. Show present worth (P) arrow at present time, t = 0 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Economic Equivalence Definition: Combination of interest rate (rate of return) and time value of money to determine different amounts of money at different points in time that are economically equivalent How it works: Use rate i and time t in upcoming relations to move money (values of P, F and A) between time points t = 0, 1, …, n to make them equivalent (not equal) at the rate i © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Example of Equivalence
Different sums of money at different times may be equal in economic value at a given rate $110 Year Rate of return = 10% per year $100 now $100 now is economically equivalent to $110 one year from now, if the $100 is invested at a rate of 10% per year. © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Simple and Compound Interest
Simple Interest Interest is calculated using principal only Interest = (principal)(number of periods)(interest rate) I = Pni Example: $100,000 lent for 3 years at simple i = 10% per year. What is repayment after 3 years? Interest = 100,000(3)(0.10) = $30,000 Total due = 100, ,000 = $130,000 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Simple and Compound Interest
Interest is based on principal plus all accrued interest That is, interest compounds over time Interest = (principal + all accrued interest) (interest rate) Interest for time period t is © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Compound Interest Example
Example: $100,000 lent for 3 years at i = 10% per year compounded. What is repayment after 3 years? Interest, year 1: I1 = 100,000(0.10) = $10,000 Total due, year 1: T1 = 100, ,000 = $110,000 Interest, year 2: I2 = 110,000(0.10) = $11,000 Total due, year 2: T2 = 110, ,000 = $121,000 Interest, year 3: I3 = 121,000(0.10) = $12,100 Total due, year 3: T3 = 121, ,100 = $133,100 Compounded: $133,100 Simple: $130,000 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Minimum Attractive Rate of Return
MARR is a reasonable rate of return (percent) established for evaluating and selecting alternatives An investment is justified economically if it is expected to return at least the MARR Also termed hurdle rate, benchmark rate and cutoff rate © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

MARR Characteristics MARR is established by the financial managers of the firm MARR is fundamentally connected to the cost of capital Both types of capital financing are used to determine the weighted average cost of capital (WACC) and the MARR MARR usually considers the risk inherent to a project © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

For an economically justified project
Types of Financing Equity Financing –Funds either from retained earnings, new stock issues, or owner’s infusion of money. Debt Financing –Borrowed funds from outside sources – loans, bonds, mortgages, venture capital pools, etc. Interest is paid to the lender on these funds For an economically justified project ROR ≥ MARR > WACC © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Opportunity Cost Definition: Largest rate of return of all projects not accepted (forgone) due to a lack of capital funds If no MARR is set, the ROR of the first project not undertaken establishes the opportunity cost Example: Assume MARR = 10%. Project A, not funded due to lack of funds, is projected to have RORA = 13%. Project B has RORB = 15% and is funded because it costs less than A Opportunity cost is 13%, i.e., the opportunity to make an additional 13% is forgone by not funding project A © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Introduction to Spreadsheet Functions
Excel financial functions Present Value, P: = PV(i%,n,A,F) Future Value, F: = FV(i%,n,A,P) Equal, periodic value, A: = PMT(i%,n,P,F) Number of periods, n: = NPER((i%,A,P,F) Compound interest rate, i: = RATE(n,A,P,F) Compound interest rate, i: = IRR(first_cell:last_cell) Present value, any series, P: = NPV(i%,second_cell:last_cell) + first_cell Example: Estimates are P = $5000 n = 5 years i = 5% per year Find A in $ per year Function and display: = PMT(5%, 5, 5000) displays A = $ © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Chapter Summary Engineering Economy fundamentals
Time value of money Economic equivalence Introduction to capital funding and MARR Spreadsheet functions Interest rate and rate of return Simple and compound interest Cash flow estimation Cash flow diagrams End-of-period assumption Net cash flow Perspectives taken for cash flow estimation Ethics Universal morals and personal morals Professional and engineering ethics (Code of Ethics) © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved

Chapter 2 Factors: How Time and Interest Affect Money
Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 2-29 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 29

LEARNING OUTCOMES F/P and P/F Factors P/A and A/P Factors
F/A and A/F Factors Factor Values Arithmetic Gradient Geometric Gradient Find i or n 2-30 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 30

Single Payment Factors (F/P and P/F)
Single payment factors involve only P and F Cash flow diagrams are as follows: Formulas are as follows: F = P(1 + i ) n P = F[1 / (1 + i ) n] Terms in parentheses or brackets are called factors. Values are in tables for i and n values Factors are represented in standard factor notation such as (F/P,i,n), where letter to left of slash is what is sought; letter to right represents what is given 2-31 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 31

F/P and P/F for Spreadsheets
Future value F is calculated using FV function: = FV(i%,n,,P) Present value P is calculated using PV function: = PV(i%,n,,F) Note the use of double commas in each function 2-32 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 32

Example: Finding Future Value
A person deposits $5000 into an account which pays interest at a rate of 8% per year. The amount in the account after 10 years is closest to: (A) $2, (B) $9, (C) $10, (D) $12,165 The cash flow diagram is: Solution: F = P(F/P,i,n ) = 5000(F/P,8%,10 ) = 5000(2.1589) = $10,794.50 Answer is (C) 2-33 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 33

Example: Finding Present Value
A small company wants to make a single deposit now so it will have enough money to purchase a backhoe costing $50,000 five years from now. If the account will earn interest of 10% per year, the amount that must be deposited now is nearest to: (A) $10, (B) $ 31, (C) $ 33, (D) $319,160 Solution: The cash flow diagram is: P = F(P/F,i,n ) = 50,000(P/F,10%,5 ) = 50,000(0.6209) = $31,045 Answer is (B) 2-34 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 34

Uniform Series Involving P/A and A/P
The uniform series factors that involve P and A are derived as follows: (1) Cash flow occurs in consecutive interest periods (2) Cash flow amount is same in each interest period The cash flow diagrams are: 1 2 3 4 5 A = Given P = ? 1 2 3 4 5 A = ? P = Given P = A(P/A,i,n) A = P(A/P,i,n) Standard Factor Notation Note: P is one period Ahead of first A value 2-35 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 35

Example: Uniform Series Involving P/A
A chemical engineer believes that by modifying the structure of a certain water treatment polymer, his company would earn an extra $5000 per year. At an interest rate of 10% per year, how much could the company afford to spend now to just break even over a 5 year project period? (A) $11, (B) 13, (C) $15, (D) $18,950 Solution: The cash flow diagram is as follows: 1 2 3 4 5 A = $5000 P = ? i =10% P = 5000(P/A,10%,5) = 5000(3.7908) = $18,954 Answer is (D) 2-36 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 36

Uniform Series Involving F/A and A/F
The uniform series factors that involve F and A are derived as follows: (1) Cash flow occurs in consecutive interest periods (2) Last cash flow occurs in same period as F Cash flow diagrams are: 1 2 3 4 5 F = ? A = Given 1 2 3 4 5 F = Given A = ? F = A(F/A,i,n) A = F(A/F,i,n) Standard Factor Notation Note: F takes place in the same period as last A 2-37 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 37

Example: Uniform Series Involving F/A
An industrial engineer made a modification to a chip manufacturing process that will save her company $10,000 per year. At an interest rate of 8% per year, how much will the savings amount to in 7 years? (A) $45, (B) $68, (C) $89, (D) $151,500 A = $10,000 F = ? i = 8% Solution: The cash flow diagram is: F = 10,000(F/A,8%,7) = 10,000(8.9228) = $89,228 Answer is (C) 2-38 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 38

Factor Values for Untabulated i or n
3 ways to find factor values for untabulated i or n values Use formula Use spreadsheet function with corresponding P, F, or A value set to 1 Linearly interpolate in interest tables Formula or spreadsheet function is fast and accurate Interpolation is only approximate 2-39 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 39

Example: Untabulated i
Determine the value for (F/P, 8.3%,10) Formula: F = ( )10 = OK Spreadsheet: = FV(8.3%,10,,1) = OK Interpolation: 8% 8.3% x 9% x = [( )/( )][ – ] = (Too high) Absolute Error = – = 2-40 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 40

Arithmetic Gradients Arithmetic gradients change by the same amount each period The cash flow diagram for the PG of an arithmetic gradient is: G starts between periods 1 and 2 (not between 0 and 1) This is because cash flow in year 1 is usually not equal to G and is handled separately as a base amount (shown on next slide) 1 2 3 n G 2G 4 3G (n-1)G PG = ? Note that PG is located Two Periods Ahead of the first change that is equal to G Standard factor notation is PG = G(P/G,i,n) 2-41 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 41

Typical Arithmetic Gradient Cash Flow
PT = ? i = 10% 400 450 Amount in year 1 is base amount 500 550 600 This diagram = this base amount plus this gradient PA = ? i = 10% PG = ? i = 10% + Amount in year 1 is base amount 400 400 400 400 400 50 100 PA = 400(P/A,10%,5) PG = 50(P/G,10%,5) 150 200 PT = PA + PG = 400(P/A,10%,5) + 50(P/G,10%,5) 2-42 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 42

Converting Arithmetic Gradient to A
Arithmetic gradient can be converted into equivalent A value using G(A/G,i,n) i = 10% i = 10% G A = ? 2G 3G 4G General equation when base amount is involved is A = base amount + G(A/G,i,n) For decreasing gradients, change plus sign to minus 4G 3G A = base amount - G(A/G,i,n) 2G G 2-43 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 43

Example: Arithmetic Gradient
The present worth of $400 in year 1 and amounts increasing by $30 per year through year 5 at an interest rate of 12% per year is closest to: (A) $ (B) $1, (C) $1, (D) $1,829 Solution: 1 2 3 Year 430 460 4 490 520 PT = ? 5 400 i = 12% G = $30 PT = 400(P/A,12%,5) + 30(P/G,12%,5) = 400(3.6048) + 30(6.3970) = $1,633.83 Answer is (B) The cash flow could also be converted into an A value as follows: A = (A/G,12%,5) = (1.7746) = $453.24 2-44 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 44

Geometric Gradients Geometric gradients change by the same percentage each period Cash flow diagram for present worth of geometric gradient There are no tables for geometric factors Use following equation for g ≠ i: 1 2 3 n A1 A 1(1+g)1 4 A 1(1+g)2 A 1(1+g)n-1 Pg = ? Pg = A1{1- [(1+g)/(1+i)]n}/(i-g) where: A1 = cash flow in period 1 g = rate of increase If g = i, Pg = A1n/(1+i) Note: g starts between periods 1 and 2 Note: If g is negative, change signs in front of both g values 2-45 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 45

Example: Geometric Gradient
Find the present worth of $1,000 in year 1 and amounts increasing by 7% per year through year 10. Use an interest rate of 12% per year. (a) $5, (b) $7, (c) $12, (d) $13,550 1 2 3 10 1000 1070 4 1145 1838 Pg = ? Solution: i = 12% Pg = 1000[1-(1+0.07/1+0.12)10]/( ) = $7,333 Answer is (b) g = 7% To find A, multiply Pg by (A/P,12%,10) 2-46 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 46

Unknown Interest Rate i
Unknown interest rate problems involve solving for i, given n and 2 other values (P, F, or A) (Usually requires a trial and error solution or interpolation in interest tables) Procedure: Set up equation with all symbols involved and solve for i A contractor purchased equipment for $60,000 which provided income of $16,000 per year for 10 years. The annual rate of return of the investment was closest to: (a) 15% (b) 18% (c) 20% (d) 23% Solution: Can use either the P/A or A/P factor. Using A/P: 60,000(A/P,i%,10) = 16,000 (A/P,i%,10) = Answer is (d) From A/P column at n = 10 in the interest tables, i is between 22% and 24% 2-47 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 47

Unknown Recovery Period n
Unknown recovery period problems involve solving for n, given i and 2 other values (P, F, or A) (Like interest rate problems, they usually require a trial & error solution or interpolation in interest tables) Procedure: Set up equation with all symbols involved and solve for n A contractor purchased equipment for $60,000 that provided income of $8,000 per year. At an interest rate of 10% per year, the length of time required to recover the investment was closest to: (a) 10 years (b) 12 years (c) 15 years (d) 18 years Solution: Can use either the P/A or A/P factor. Using A/P: 60,000(A/P,10%,n) = 8,000 (A/P,10%,n) = From A/P column in i = 10% interest tables, n is between 14 and 15 years Answer is (c) 2-48 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 48

Summary of Important Points
In P/A and A/P factors, P is one period ahead of first A In F/A and A/F factors, F is in same period as last A To find untabulated factor values, best way is to use formula or spreadsheet For arithmetic gradients, gradient G starts between periods 1 and 2 Arithmetic gradients have 2 parts, base amount (year 1) and gradient amount For geometric gradients, gradient g starts been periods 1 and 2 In geometric gradient formula, A1 is amount in period 1 To find unknown i or n, set up equation involving all terms and solve for i or n 2-49 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 49

Chapter 3 Combining Factors and Spreadsheet Functions
Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 3-50 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 50

LEARNING OUTCOMES Shifted uniform series
Shifted series and single cash flows Shifted gradients 3-51 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 51

Shifted Uniform Series
A shifted uniform series starts at a time other than period 1 The cash flow diagram below is an example of a shifted series Series starts in period 2, not period 1 1 2 3 4 5 A = Given PA = ? FA = ? Shifted series usually require the use of multiple factors Remember: When using P/A or A/P factor, PA is always one year ahead of first A When using F/A or A/F factor, FA is in same year as last A 3-52 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 52

Example Using P/A Factor: Shifted Uniform Series
The present worth of the cash flow shown below at i = 10% is: (a) $25, (b) $29, (c) $34, (d) $37,908 P0 = ? A = $10,000 i = 10% P1 = ? Actual year Series year Solution: (1) Use P/A factor with n = 5 (for 5 arrows) to get P1 in year 1 (2) Use P/F factor with n = 1 to move P1 back for P0 in year 0 P0 = P1(P/F,10%,1) = A(P/A,10%,5)(P/F,10%,1) = 10,000(3.7908)(0.9091) = $34,462 Answer is (c) 3-53 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 53

Example Using F/A Factor: Shifted Uniform Series
How much money would be available in year 10 if $8000 is deposited each year in years 3 through 10 at an interest rate of 10% per year? Cash flow diagram is: FA = ? i = 10% Actual year Series year A = $8000 Solution: Re-number diagram to determine n = 8 (number of arrows) FA = 8000(F/A,10%,8) = 8000( ) = $91,487 3-54 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 54

Shifted Series and Random Single Amounts
For cash flows that include uniform series and randomly placed single amounts: Uniform series procedures are applied to the series amounts Single amount formulas are applied to the one-time cash flows The resulting values are then combined per the problem statement The following slides illustrate the procedure 3-55 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 55

Example: Series and Random Single Amounts
Find the present worth in year 0 for the cash flows shown using an interest rate of 10% per year. PT = ? i = 10% A = $5000 $2000 PT = ? i = 10% Actual year Series year A = $5000 $2000 Solution: First, re-number cash flow diagram to get n for uniform series: n = 8 3-56 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 56

Example: Series and Random Single Amounts
PA PT = ? i = 10% Actual year Series year A = $5000 $2000 Use P/A to get PA in year 2: PA = 5000(P/A,10%,8) = 5000(5.3349) = $26,675 Move PA back to year 0 using P/F: P0 = 26,675(P/F,10%,2) = 26,675(0.8264) = $22,044 Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933 Now, add P0 and P2000 to get PT: PT = 22, = $22,977 1-57 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 57

Example Worked a Different Way
(Using F/A instead of P/A for uniform series) The same re-numbered diagram from the previous slide is used PT = ? FA = ? i = 10% A = $5000 $2000 Solution: Use F/A to get FA in actual year 10: FA = 5000(F/A,10%,8) = 5000( ) = $57,180 Move FA back to year 0 using P/F: P0 = 57,180(P/F,10%,10) = 57,180(0.3855) = $22,043 Move $2000 single amount back to year 0: P2000 = 2000(P/F,10%,8) = 2000(0.4665) = $933 Now, add two P values to get PT: PT = 22, = $22,976 Same as before As shown, there are usually multiple ways to work equivalency problems 3-58 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 58

Example: Series and Random Amounts
Convert the cash flows shown below (black arrows) into an equivalent annual worth A in years 1 through 8 (red arrows) at i = 10% per year. A = ? i = 10% A = $3000 $1000 1. Convert all cash flows into P in year 0 and use A/P with n = 8 2. Find F in year 8 and use A/F with n = 8 Approaches: Solve for F: F = 3000(F/A,10%,5) (F/P,10%,1) = 3000(6.1051) (1.1000) = $19,415 Solution: Find A: A = 19,415(A/F,10%,8) = 19,415( ) = $1698 3-59 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 59

Shifted Arithmetic Gradients
Shifted gradient begins at a time other than between periods 1 and 2 Present worth PG is located 2 periods before gradient starts Must use multiple factors to find PT in actual year 0 To find equivalent A series, find PT at actual time 0 and apply (A/P,i,n) 3-60 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 60

Example: Shifted Arithmetic Gradient
John Deere expects the cost of a tractor part to increase by $5 per year beginning 4 years from now. If the cost in years 1-3 is $60, determine the present worth in year 0 of the cost through year 10 at an interest rate of 12% per year. i = 12% PT = ? Actual years 1 2 3 4 5 10 Gradient years 60 60 60 65 70 G = 5 95 Solution: First find P2 for G = $5 and base amount ($60) in actual year 2 P2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 P0 = P2(P/F,12%,2) = $295.29 Next, move P2 back to year 0 Next, find PA for the $60 amounts of years 1 and 2 PA = 60(P/A,12%,2) = $101.41 Finally, add P0 and PA to get PT in year 0 PT = P0 + PA = $396.70 3-61 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 61

Shifted Geometric Gradients
Shifted gradient begins at a time other than between periods 1 and 2 Equation yields Pg for all cash flows (base amount A1 is included) Equation (i ≠ g): Pg = A 1{1 - [(1+g)/(1+i)]n/(i-g)} For negative gradient, change signs on both g values There are no tables for geometric gradient factors 3-62 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 62

Example: Shifted Geometric Gradient
Weirton Steel signed a 5-year contract to purchase water treatment chemicals from a local distributor for $7000 per year. When the contract ends, the cost of the chemicals is expected to increase by 12% per year for the next 8 years. If an initial investment in storage tanks is $35,000, determine the equivalent present worth in year 0 of all of the cash flows at i = 15% per year. 3-63 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 63

Example: Shifted Geometric Gradient
Gradient starts between actual years 5 and 6; these are gradient years 1 and 2. Pg is located in gradient year 0, which is actual year 4 Pg = 7000{1-[(1+0.12)/(1+0.15)]9/( )} = $49,401 Move Pg and other cash flows to year 0 to calculate PT PT = 35, (P/A,15%,4) + 49,401(P/F,15%,4) = $83,232 1-64 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 64

Negative Shifted Gradients
For negative arithmetic gradients, change sign on G term from + to - General equation for determining P: P = present worth of base amount - PG Changed from + to - For negative geometric gradients, change signs on both g values Changed from + to - Pg = A1{1-[(1-g)/(1+i)]n/(i+g)} Changed from - to + All other procedures are the same as for positive gradients 3-65 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 65

Example: Negative Shifted Arithmetic Gradient
For the cash flows shown, find the future worth in year 7 at i = 10% per year F = ? PG = ? i = 10% Actual years Gradient years 450 500 550 600 650 700 G = $-50 Solution: Gradient G first occurs between actual years 2 and 3; these are gradient years 1 and 2 PG is located in gradient year 0 (actual year 1); base amount of $700 is in gradient years 1-6 PG = 700(P/A,10%,6) – 50(P/G,10%,6) = 700(4.3553) – 50(9.6842) = $2565 F = PG(F/P,10%,6) = 2565(1.7716) = $4544 3-66 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 66

P for shifted uniform series is one period ahead of first A; n is equal to number of A values F for shifted uniform series is in same period as last A; n is equal to number of A values For gradients, first change equal to G or g occurs between gradient years 1 and 2 For negative arithmetic gradients, change sign on G from + to - For negative geometric gradients, change sign on g from + to - 3-67 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 67

Nominal and Effective Interest Rates Lecture slides to accompany
Chapter 4 Nominal and Effective Interest Rates Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 4-68 © 2012 by McGraw-Hill All Rights Reserved 68

© 2012 by McGraw-Hill All Rights Reserved
LEARNING OUTCOMES Understand interest rate statements Use formula for effective interest rates Determine interest rate for any time period Determine payment period (PP) and compounding period (CP) for equivalence calculations Make calculations for single cash flows Make calculations for series and gradient cash flows with PP ≥ CP Perform equivalence calculations when PP < CP Use interest rate formula for continuous compounding Make calculations for varying interest rates 4-69 © 2012 by McGraw-Hill All Rights Reserved 69

Interest Rate Statements
The terms ‘nominal’ and ‘effective’ enter into consideration when the interest period is less than one year. New time-based definitions to understand and remember Interest period (t) – period of time over which interest is expressed. For example, 1% per month. Compounding period (CP) – Shortest time unit over which interest is charged or earned. For example,10% per year compounded monthly. Compounding frequency (m) – Number of times compounding occurs within the interest period t. For example, at i = 10% per year, compounded monthly, interest would be compounded 12 times during the one year interest period. 4-70 © 2012 by McGraw-Hill All Rights Reserved 70

Understanding Interest Rate Terminology
A nominal interest rate (r) is obtained by multiplying an interest rate that is expressed over a short time period by the number of compounding periods in a longer time period: That is: r = interest rate per period x number of compounding periods Example: If i = 1% per month, nominal rate per year is r = (1)(12) = 12% per year) Effective interest rates (i) take compounding into account (effective rates can be obtained from nominal rates via a formula to be discussed later). IMPORTANT: Nominal interest rates are essentially simple interest rates. Therefore, they can never be used in interest formulas. Effective rates must always be used hereafter in all interest formulas. 4-71 © 2012 by McGraw-Hill All Rights Reserved 71

More About Interest Rate Terminology
There are 3 general ways to express interest rates as shown below Sample Interest Rate Statements Comment i = 2% per month i = 12% per year (1) When no compounding period is given, rate is effective i = 10% per year, comp’d semiannually i = 3% per quarter, comp’d monthly (2) When compounding period is given and it is not the same as interest period, it is nominal When compounding period is given and rate is specified as effective, rate is effective over stated period i = effective 9.4%/year, comp’d semiannually i = effective 4% per quarter, comp’d monthly (3) 4-72 © 2012 by McGraw-Hill All Rights Reserved

Effective Annual Interest Rates
Nominal rates are converted into effective annual rates via the equation: ia = (1 + i)m – 1 where ia = effective annual interest rate i = effective rate for one compounding period m = number times interest is compounded per year Example: For a nominal interest rate of 12% per year, determine the nominal and effective rates per year for (a) quarterly, and (b) monthly compounding Solution: (a) Nominal r / year = 12% per year Nominal r / quarter = 12/4 = 3.0% per quarter Effective i / year = ( )4 – 1 = 12.55% per year (b) Nominal r /month = 12/12 = 1.0% per year Effective i / year = ( )12 – 1 = 12.68% per year 4-73 © 2012 by McGraw-Hill All Rights Reserved

Effective Interest Rates
Nominal rates can be converted into effective rates for any time period via the following equation: i = (1 + r / m)m – 1 where i = effective interest rate for any time period r = nominal rate for same time period as i m = no. times interest is comp’d in period specified for i Spreadsheet function is = EFFECT(r%,m) where r = nominal rate per period specified for i Example: For an interest rate of 1.2% per month, determine the nominal and effective rates (a) per quarter, and (b) per year Solution: (a) Nominal r / quarter = (1.2)(3) = 3.6% per quarter Effective i / quarter = ( /3)3 – 1 = 3.64% per quarter (b) Nominal i /year = (1.2)(12) = 14.4% per year Effective i / year = ( / 12)12 – 1 = 15.39% per year 4-74 © 2012 by McGraw-Hill All Rights Reserved 74

Equivalence Relations: PP and CP
New definition: Payment Period (PP) – Length of time between cash flows In the diagram below, the compounding period (CP) is semiannual and the payment period (PP) is monthly $1000 Similarly, for the diagram below, the CP is quarterly and the payment period (PP) is semiannual F = ? i = 10% per year, compounded quarterly Years Semi-annual periods A = $8000 Semi-annual PP 4-75 © 2012 by McGraw-Hill All Rights Reserved 75

Single Amounts with PP > CP
For problems involving single amounts, the payment period (PP) is usually longer than the compounding period (CP). For these problems, there are an infinite number of i and n combinations that can be used, with only two restrictions: (1) The i must be an effective interest rate, and (2) The time units on n must be the same as those of i (i.e., if i is a rate per quarter, then n is the number of quarters between P and F) There are two equally correct ways to determine i and n Method 1: Determine effective interest rate over the compounding period CP, and set n equal to the number of compounding periods between P and F Method 2: Determine the effective interest rate for any time period t, and set n equal to the total number of those same time periods. 4-76 © 2012 by McGraw-Hill All Rights Reserved 76

Example: Single Amounts with PP ≥ CP
How much money will be in an account in 5 years if $10,000 is deposited now at an interest rate of 1% per month? Use three different interest rates: (a) monthly, (b) quarterly , and (c) yearly. For monthly rate, 1% is effective [n = (5 years)×(12 CP per year = 60] F = 10,000(F/P,1%,60) = $18,167 months i and n must always have same time units effective i per month (b) For a quarterly rate, effective i/quarter = ( /3)3 –1 = 3.03% F = 10,000(F/P,3.03%,20) = $18,167 quarters i and n must always have same time units effective i per quarter (c) For an annual rate, effective i/year = ( /12)12 –1 = % F = 10,000(F/P,12.683%,5) = $18,167 years i and n must always have same time units effective i per year 4-77 © 2012 by McGraw-Hill All Rights Reserved 77

Series with PP ≥ CP For series cash flows, first step is to determine relationship between PP and CP Determine if PP ≥ CP, or if PP < CP When PP ≥ CP, the only procedure (2 steps) that can be used is as follows: First, find effective i per PP Example: if PP is in quarters, must find effective i/quarter (2) Second, determine n, the number of A values involved Example: quarterly payments for 6 years yields n = 4×6 = 24 Note: Procedure when PP < CP is discussed later 4-78 © 2012 by McGraw-Hill All Rights Reserved 78

Example: Series with PP ≥ CP
How much money will be accumulated in 10 years from a deposit of $500 every 6 months if the interest rate is 1% per month? Solution: First, find relationship between PP and CP PP = six months, CP = one month; Therefore, PP > CP Since PP > CP, find effective i per PP of six months Step i /6 months = ( /6)6 – 1 = 6.15% Next, determine n (number of 6-month periods) Step 2: n = 10(2) = 20 six month periods Finally, set up equation and solve for F F = 500(F/A,6.15%,20) = $18,692 (by factor or spreadsheet) 4-79 © 2012 by McGraw-Hill All Rights Reserved 79

Series with PP < CP Two policies: (1) interperiod cash flows earn no interest (most common) (2) interperiod cash flows earn compound interest For policy (1), positive cash flows are moved to beginning of the interest period in which they occur and negative cash flows are moved to the end of the interest period Note: The condition of PP < CP with no interperiod interest is the only situation in which the actual cash flow diagram is changed For policy (2), cash flows are not moved and equivalent P, F, and A values are determined using the effective interest rate per payment period 4-80 © 2012 by McGraw-Hill All Rights Reserved 80

Example: Series with PP < CP
A person deposits $100 per month into a savings account for 2 years. If $75 is withdrawn in months 5, 7 and 8 (in addition to the deposits), construct the cash flow diagram to determine how much will be in the account after 2 years at i = 6% per year, compounded quarterly. Assume there is no interperiod interest. Solution: Since PP < CP with no interperiod interest, the cash flow diagram must be changed using quarters as the time periods F = ? F = ? 75 150 75 75 75 from to Months 23 24 Quarters this this 100 300 300 300 300 300 4-81 © 2012 by McGraw-Hill All Rights Reserved 81

Continuous Compounding
When the interest period is infinitely small, interest is compounded continuously. Therefore, PP > CP and m increases. Take limit as m → ∞ to find the effective interest rate equation i = er – 1 Example: If a person deposits $500 into an account every 3 months at an interest rate of 6% per year, compounded continuously, how much will be in the account at the end of 5 years? Solution: Payment Period: PP = 3 months Nominal rate per three months: r = 6%/4 = 1.50% Effective rate per 3 months: i = e0.015 – 1 = 1.51% F = 500(F/A,1.51%,20) = $11,573 4-82 © 2012 by McGraw-Hill All Rights Reserved 82

Varying Rates When interest rates vary over time, use the interest rates associated with their respective time periods to find P Example: Find the present worth of $2500 deposits in years 1 through 8 if the interest rate is 7% per year for the first five years and 10% per year thereafter. For the cash flow shown below, find the future worth (in year 7) at i = 10% per year. Solution: P = 2,500(P/A,7%,5) + 2,500(P/A,10%,3)(P/F,7%,5) = $14,683 An equivalent annual worth value can be obtained by replacing each cash flow amount with ‘A’ and setting the equation equal to the calculated P value 14,683 = A(P/A,7%,5) + A(P/A,10%,3)(P/F,7%,5) A = $2500 per year 4-83 © 2012 by McGraw-Hill All Rights Reserved

Must understand: interest period, compounding period, compounding frequency, and payment period i = (1 + r / m)m – 1 Always use effective rates in interest formulas Interest rates are stated different ways; must know how to get effective rates For single amounts, make sure units on i and n are the same 4-84 © 2012 by McGraw-Hill All Rights Reserved 84

Important Points (cont’d)
For uniform series with PP ≥ CP, find effective i over PP For uniform series with PP < CP and no interperiod interest, move cash flows to match compounding period For continuous compounding, use i = er – 1 to get effective rate For the cash flow shown below, find the future worth (in year 7) at i = 10% per year. For varying rates, use stated i values for respective time periods 4-85 © 2012 by McGraw-Hill All Rights Reserved

Present Worth Analysis Lecture slides to accompany
Chapter 5 Present Worth Analysis Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 5-86 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 86

LEARNING OUTCOMES Formulate Alternatives PW of equal-life alternatives PW of different-life alternatives Future Worth analysis Capitalized Cost analysis 5-87 © 2012 by McGraw-Hill All Rights Reserved 87

Formulating Alternatives
Two types of economic proposals Mutually Exclusive (ME) Alternatives: Only one can be selected; Compete against each other Independent Projects: More than one can be selected; Compete only against DN Do Nothing (DN) – An ME alternative or independent project to maintain the current approach; no new costs, revenues or savings 5-88 © 2012 by McGraw-Hill All Rights Reserved 88

Formulating Alternatives
Two types of cash flow estimates Revenue: Alternatives include estimates of costs (cash outflows) and revenues (cash inflows) Cost: Alternatives include only costs; revenues and savings assumed equal for all alternatives; also called service alternatives 5-89 © 2012 by McGraw-Hill All Rights Reserved 89

PW Analysis of Alternatives
Convert all cash flows to PW using MARR Precede costs by minus sign; receipts by plus sign EVALUATION For one project, if PW > 0, it is justified For mutually exclusive alternatives, select one with numerically largest PW For independent projects, select all with PW > 0 5-90 © 2012 by McGraw-Hill All Rights Reserved

Selection of Alternatives by PW
For the alternatives shown below, which should be selected selected selected if they are (a) mutually exclusive; (b) independent? Project ID Present Worth A $30,000 B $12,500 C $-4,000 D $ 2,000 Solution: (a) Select numerically largest PW; alternative A (b) Select all with PW > 0; projects A, B & D 5-91 © 2012 by McGraw-Hill All Rights Reserved

Example: PW Evaluation of Equal-Life ME Alts.
Alternative X has a first cost of $20,000, an operating cost of $9,000 per year, and a $5,000 salvage value after 5 years. Alternative Y will cost $35,000 with an operating cost of $4,000 per year and a salvage value of $7,000 after 5 years. At an MARR of 12% per year, which should be selected? Solution: Find PW at MARR and select numerically larger PW value PWX = -20, (P/A,12%,5) (P/F,12%,5) = -$49,606 PWY = -35, (P/A,12%,5) (P/F,12%,5) = -$45,447 Select alternative Y 5-92 © 2012 by McGraw-Hill All Rights Reserved

PW of Different-Life Alternatives
Must compare alternatives for equal service (i.e., alternatives must end at the same time) Two ways to compare equal service: Least common multiple (LCM) of lives Specified study period (The LCM procedure is used unless otherwise specified) 5-93 © 2012 by McGraw-Hill All Rights Reserved 93

Assumptions of LCM approach
Service provided is needed over the LCM or more years Selected alternative can be repeated over each life cycle of LCM in exactly the same manner Cash flow estimates are the same for each life cycle (i.e., change in exact accord with the inflation or deflation rate) 1-94 © 2012 by McGraw-Hill All Rights Reserved

Example: Different-Life Alternatives
Compare the machines below using present worth analysis at i = 10% per year Machine A Machine B First cost, $ 20,000 30,000 Annual cost, $/year 9000 7000 Salvage value, $ 4000 6000 Life, years 3 6 Solution: LCM = 6 years; repurchase A after 3 years PWA = -20,000 – 9000(P/A,10%,6) – 16,000(P/F,10%,3) (P/F,10%,6) = $-68,961 20,000 – 4,000 in year 3 PWB = -30,000 – 7000(P/A,10%,6) (P/F,10%,6) = $-57,100 Select alternative B 5-95 © 2012 by McGraw-Hill All Rights Reserved 95

PW Evaluation Using a Study Period
Once a study period is specified, all cash flows after this time are ignored Salvage value is the estimated market value at the end of study period Short study periods are often defined by management when business goals are short-term Study periods are commonly used in equipment replacement analysis 1-96 © 2012 by McGraw-Hill All Rights Reserved

Example: Study Period PW Evaluation
Compare the alternatives below using present worth analysis at i = 10% per year and a 3-year study period Machine A Machine B First cost, $ -20,000 -30,000 Annual cost, $/year -9,000 -7,000 Salvage/market value, $ 4,000 6,000 (after 6 years) 10,000 (after 3 years) Life, years 3 6 Solution: Study period = 3 years; disregard all estimates after 3 years PWA = -20,000 – 9000(P/A,10%,3) (P/F,10%,3) = $-39,376 PWB = -30,000 – 7000(P/A,10%,3) + 10,000(P/F,10%,3) = $-39,895 Marginally, select A; different selection than for LCM = 6 years 5-97 © 2012 by McGraw-Hill All Rights Reserved 97

FW exactly like PW analysis, except calculate FW
Future Worth Analysis FW exactly like PW analysis, except calculate FW Must compare alternatives for equal service (i.e. alternatives must end at the same time) Two ways to compare equal service: Least common multiple (LCM) of lives Specified study period (The LCM procedure is used unless otherwise specified) 5-98 © 2012 by McGraw-Hill All Rights Reserved 98

FW of Different-Life Alternatives
Compare the machines below using future worth analysis at i = 10% per year Machine A Machine B First cost, $ -20,000 -30,000 Annual cost, $/year -9000 -7000 Salvage value, $ 4000 6000 Life, years 3 6 Solution: LCM = 6 years; repurchase A after 3 years FWA = -20,000(F/P,10%,6) – 9000(F/A,10%,6) – 16,000(F/P,10%,3) = $-122,168 FWB = -30,000(F/P,10%.6) – 7000(F/A,10%,6) = $-101,157 Select B (Note: PW and FW methods will always result in same selection) 5-99 © 2012 by McGraw-Hill All Rights Reserved 99

Capitalized Cost (CC) Analysis
CC refers to the present worth of a project with a very long life, that is, PW as n becomes infinite A Basic equation is: CC = P = i “A” essentially represents the interest on a perpetual investment For example, in order to be able to withdraw $50,000 per year forever at i = 10% per year, the amount of capital required is 50,000/0.10 = $500,000 For finite life alternatives, convert all cash flows into an A value over one life cycle and then divide by i 5-100 © 2012 by McGraw-Hill All Rights Reserved 100

Example: Capitalized Cost
Compare the machines shown below on the basis of their capitalized cost. Use i = 10% per year Machine 1 Machine 2 First cost,$ -20,000 -100,000 Annual cost,$/year -9000 -7000 Salvage value, $ 4000 ----- Life, years 3 ∞ Solution: Convert machine 1 cash flows into A and then divide by i A1 = -20,000(A/P,10%,3) – (A/F,10%,3) = $-15,834 CC1 = -15,834 / 0.10 = $-158,340 CC2 = -100,000 – 7000/ 0.10 = $-170,000 Select machine 1 5-101 © 2012 by McGraw-Hill All Rights Reserved 101

PW method converts all cash flows to present value at MARR Alternatives can be mutually exclusive or independent Cash flow estimates can be for revenue or cost alternatives PW comparison must always be made for equal service Equal service is achieved by using LCM or study period Capitalized cost is PW of project with infinite life; CC = P = A/i 5-102 © 2012 by McGraw-Hill All Rights Reserved 102

Lecture slides to accompany
Chapter 6 Annual Worth Analysis Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 6-103 © 2012 by McGraw-Hill, New York, N.Y All Rights Reserved 103

Advantages of AW Analysis
AW calculated for only one life cycle Assumptions: Services needed for at least the LCM of lives of alternatives Selected alternative will be repeated in succeeding life cycles in same manner as for the first life cycle All cash flows will be same in every life cycle (i.e., will change by only inflation or deflation rate) 6-105 © 2012 by McGraw-Hill All Rights Reserved 105

Alternatives usually have the following cash flow estimates
Initial investment, P – First cost of an asset Salvage value, S – Estimated value of asset at end of useful life Annual amount, A – Cash flows associated with asset, such as annual operating cost (AOC), etc. Relationship between AW, PW and FW AW = PW(A/P,i%,n) = FW(A/F,i%,n) n is years for equal-service comparison (value of LCM or specified study period) 6-106 © 2012 by McGraw-Hill All Rights Reserved 106

Calculation of Annual Worth
AW for one life cycle is the same for all life cycles!! An asset has a first cost of $20,000, an annual operating cost of $8000 and a salvage value of $5000 after 3 years. Calculate the AW for one and two life cycles at i = 10% AWone = - 20,000(A/P,10%,3) – (A/F,10%,3) = $-14,532 AWtwo = - 20,000(A/P,10%,6) – 8000 – 15,000(P/F,10%,3)(A/P,10%,6) + 5000(A/F,10%,6) = $-14,532 6-107 © 2012 by McGraw-Hill All Rights Reserved

Capital Recovery and AW
Capital recovery (CR) is the equivalent annual amount that an asset, process, or system must earn each year to just recover the first cost and a stated rate of return over its expected life. Salvage value is considered when calculating CR. CR = -P(A/P,i%,n) + S(A/F,i%,n) Use previous example: (note: AOC not included in CR ) CR = -20,000(A/P,10%,3) (A/F,10%,3) = $ – 6532 per year Now AW = CR + A AW = – 6532 – 8000 = $ – 14,532 6-108 © 2012 by McGraw-Hill All Rights Reserved

Selection Guidelines for AW Analysis
6-109 © 2012 by McGraw-Hill All Rights Reserved

ME Alternative Evaluation by AW
Not necessary to use LCM for different life alternatives A company is considering two machines. Machine X has a first cost of $30,000, AOC of $18,000, and S of $7000 after 4 years. Machine Y will cost $50,000 with an AOC of $16,000 and S of $9000 after 6 years. Which machine should the company select at an interest rate of 12% per year? Solution: AWX = -30,000(A/P,12%,4) –18,000 +7,000(A/F,12%,4) = $-26,412 AWY = -50,000(A/P,12%,6) –16, ,000(A/F,12%,6) = $-27,052 Select Machine X; it has the numerically larger AW value 6-110 © 2012 by McGraw-Hill All Rights Reserved

AW of Permanent Investment
Use A = Pi for AW of infinite life alternatives Find AW over one life cycle for finite life alternatives Compare the alternatives below using AW and i = 10% per year C D First Cost, $ , ,000 Annual operating cost, $/year , ,000 Salvage value, $ , ,000 Life, years ∞ Solution: Find AW of C over 5 years and AW of D using relation A = Pi AWC = -50,000(A/P,10%,5) – 20, ,000(A/F,10%,5) = $-32,371 AWD = Pi + AOC = -250,000(0.10) – 9,000 = $-34,000 Select alternative C 6-111 © 2012 by McGraw-Hill All Rights Reserved

Typical Life-Cycle Cost Distribution by Phase

Life-Cycle Cost Analysis
LCC analysis includes all costs for entire life span, from concept to disposal Best when large percentage of costs are M&O Includes phases of acquisition, operation, & phaseout Apply the AW method for LCC analysis of 1 or more cost alternatives Use PW analysis if there are revenues and other benefits considered 6-113 © 2012 by McGraw-Hill All Rights Reserved

AW method converts all cash flows to annual value at MARR Alternatives can be mutually exclusive, independent, revenue, or cost AW comparison is only one life cycle of each alternative For infinite life alternatives, annualize initial cost as A = P(i) Life-cycle cost analysis includes all costs over a project’s life span 6-114 © 2012 by McGraw-Hill All Rights Reserved 114

Rate of Return One Project Lecture slides to accompany
Chapter 7 Rate of Return One Project Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 7-115 © 2012 by McGraw-Hill All Rights Reserved 115

LEARNING OUTCOMES Understand meaning of ROR Calculate ROR for cash flow series Understand difficulties of ROR Determine multiple ROR values Calculate External ROR (EROR) Calculate r and i for bonds 7-116 © 2012 by McGraw-Hill All Rights Reserved 116

with interest considered
Interpretation of ROR Rate paid on unrecovered balance of borrowed money such that final payment brings balance to exactly zero with interest considered ROR equation can be written in terms of PW, AW, or FW Use trial and error solution by factor or spreadsheet Numerical value can range from -100% to infinity 7-117 © 2012 by McGraw-Hill All Rights Reserved 117

ROR Calculation and Project Evaluation
To determine ROR, find the i* value in the relation PW = 0 or AW = 0 or FW = 0 Alternatively, a relation like the following finds i* PWoutflow = PWinflow For evaluation, a project is economically viable if i* ≥ MARR 7-118 © 2012 by McGraw-Hill All Rights Reserved

Finding ROR by Spreadsheet Function
Using the RATE function = RATE(n,A,P,F) P = $-200,000 A = $-15,000 n = F = $435,000 Function is = RATE(12,-15000, ,450000) Display is i* = 1.9% Using the IRR function = IRR(first_cell, last_cell) = IRR(B2:B14) 7-119 © 2012 by McGraw-Hill All Rights Reserved

ROR Calculation Using PW, FW or AW Relation
ROR is the unique i* rate at which a PW, FW, or AW relation equals exactly 0 Example: An investment of $20,000 in new equipment will generate income of $7000 per year for 3 years, at which time the machine can be sold for an estimated $8000. If the company’s MARR is 15% per year, should it buy the machine? Solution:: The ROR equation, based on a PW relation, is: 0 = -20, (P/A,i*,3) (P/F,i*,3) Solve for i* by trial and error or spreadsheet: i* = 18.2% per year Since i* > MARR = 15%, the company should buy the machine 7-120 © 2012 by McGraw-Hill All Rights Reserved 120

Special Considerations for ROR
May get multiple i* values (discussed later) i* assumes reinvestment of positive cash flows earn at i* rate (may be unrealistic) Incremental analysis necessary for multiple alternative evaluations (discussed later) 7-121 © 2012 by McGraw-Hill All Rights Reserved

Multiple ROR Values Two tests for multiple i* values:
Multiple i* values may exist when there is more than one sign change in net cash flow (CF) series. Such CF series are called non-conventional Two tests for multiple i* values: Descarte’s rule of signs: total number of real i* values is ≤ the number of sign changes in net cash flow series. Norstrom’s criterion: if the cumulative cash flow starts off negatively and has only one sign change, there is only one positive root . 7-122 © 2012 by McGraw-Hill All Rights Reserved

Example: Multiple i* Values
Determine the maximum number of i* values for the cash flow shown below Year Expense Income , , ,000 , ,000 , ,000 , ,000 , ,000 Net cash flow -12,000 -2,000 +3,000 +8,000 -1,000 Cumulative CF -12,000 -14,000 -11,000 -3,000 +5,000 +4,000 The cumulative cash flow begins negatively with one sign change Solution: The sign on the net cash flow changes twice, indicating two possible i* values Therefore, there is only one i* value ( i* = 8.7%) 7-124 © 2012 by McGraw-Hill All Rights Reserved

Removing Multiple i* Values
Two new interest rates to consider: Investment rate ii – rate at which extra funds are invested external to the project Borrowing rate ib – rate at which funds are borrowed from an external source to provide funds to the project Two approaches to determine External ROR (EROR) (1) Modified ROR (MIRR) (2) Return on Invested Capital (ROIC) 7-125 © 2012 by McGraw-Hill All Rights Reserved 125

Modified ROR Approach (MIRR)
Four step Procedure: 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 7-126 © 2012 by McGraw-Hill All Rights Reserved 126

Example: EROR Using MIRR Method
For the NCF shown below, find the EROR by the MIRR method if MARR = 9%, ib = 8.5%, and ii = 12% Year NCF Solution: PW0 = -500(P/F,8.5%,1) (P/F,8.5%,2) = $-7342 FW3 = 2000(F/P,12%,3) = $9610 PW0(F/P,i’,3) + FW3 = 0 -7342(1 + i’) = 0 i’ = (9.39%) Since i’ > MARR of 9%, project is justified 7-127 © 2012 by McGraw-Hill All Rights Reserved 127

Return on Invested Capital Approach
Measure of how effectively project uses funds that remain internal to project ROIC rate, i’’, is determined using net-investment procedure Three step Procedure (1) Develop series of FW relations for each year t using: 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’’ (3) If i’’ ≥ MARR, project is justified; otherwise, reject 7-128 © 2012 by McGraw-Hill All Rights Reserved 128

ROIC Example For the NCF shown below, find the EROR by the ROIC method if MARR = 9% and ii = 12% Year NCF Solution: Year 0: F0 = $ F0 > 0; invest in year 1 at ii = 12% Year 1: F1 = 2000(1.12) = $ F1 > 0; invest in year 2 at ii = 12% Year 2: F2 = 1740(1.12) = $ F2 < 0; use i’’ for year 3 Year 3: F3 = -6151(1 + i’’) Set F3 = 0 and solve for i’’ -6151(1 + i’’) = 0 i’’= 10.55% Since i’’ > MARR of 9%, project is justified 7-129 © 2012 by McGraw-Hill All Rights Reserved 129

Important Points to Remember
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 7-130 © 2012 by McGraw-Hill All Rights Reserved

ROR of Bond Investment Bond is IOU with face value (V), coupon rate (b), no. of payment periods/year (c), dividend (I), and maturity date (n). Amount paid for the bond is P. I = Vb/c General equation for i*: 0 = - P + I(P/A,i*,nxc) + V(P/F,i*,nxc) A $10,000 bond with 6% interest payable quarterly is purchased for $8000. If the bond matures in 5 years, what is the ROR (a) per quarter, (b) per year? Solution: (a) I = 10,000(0.06)/4 = $150 per quarter ROR equation is: = (P/A,i*,20) + 10,000(P/F,i*,20) By trial and error or spreadsheet: i* = 2.8% per quarter (b) Nominal i* per year = 2.8(4) = 11.2% per year Effective i* per year = ( )4 – 1 = 11.7% per year 7-131 © 2012 by McGraw-Hill All Rights Reserved Effective i per year = ( )4 – 1 = 11.7% per year 131

ROR equations can be written in terms of PW, FW, or AW and usually require trial and error solution i* assumes reinvestment of positive cash flows at i* rate More than 1 sign change in NCF may cause multiple i* values Descarte’s rule of signs and Norstrom’s criterion useful when multiple i* values are suspected EROR can be calculated using MIRR or ROIC approach. Assumptions about investment and borrowing rates is required. General ROR equation for bonds is 0 = - P + I(P/A,i*,nxc) + V(P/F,i*,nxc) 7-132 © 2012 by McGraw-Hill All Rights Reserved 132

Rate of Return Multiple Alternatives Lecture slides to accompany
Chapter 8 Rate of Return Multiple Alternatives Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 8-133 © 2012 by McGraw-Hill All Rights Reserved 133

LEARNING OUTCOMES Why incremental analysis is required in ROR Incremental cash flow (CF) calculation Interpretation of ROR on incremental CF Select alternative by ROR based on PW relation Select alternative by ROR based on AW relation Select best from several alternatives using ROR method 8-134 © 2012 by McGraw-Hill All Rights Reserved 134

Why Incremental Analysis is Necessary
Selecting the alternative with highest ROR may not yield highest return on available capital Must consider weighted average of total capital available Capital not invested in a project is assumed to earn at MARR Example: Assume $90,000 is available for investment and MARR = 16% per year. If alternative A would earn 35% per year on investment of $50,000, and B would earn 29% per year on investment of $85,000, the weighted averages are: Overall RORA = [50,000(0.35) + 40,000(0.16)]/90,000 = 26.6% Overall RORB = [85,000(0.29) + 5,000(0.16)]/90,000 = 28.3% Which investment is better, economically? 8-135 © 2012 by McGraw-Hill All Rights Reserved 135

Why Incremental Analysis is Necessary
If selection basis is higher ROR: Select alternative A (wrong answer) If selection basis is higher overall ROR: Select alternative B Conclusion: Must use an incremental ROR analysis to make a consistently correct selection Unlike PW, AW, and FW values, if not analyzed correctly, ROR values can lead to an incorrect alternative selection. This is called the ranking inconsistency problem (discussed later) 8-136 © 2012 by McGraw-Hill All Rights Reserved 136

Calculation of Incremental CF
Incremental cash flow = cash flowB – cash flowA where larger initial investment is Alternative B Example: Either of the cost alternatives shown below can be used in a grinding process. Tabulate the incremental cash flows. A B -20,000 +6000 +2000 B - A First cost, $ , ,000 Annual cost, $/year , ,000 Salvage value, $ , ,000 The incremental CF is shown in the (B-A) column The ROR on the extra $20,000 investment in B determines which alternative to select (as discussed later) 8-137 © 2012 by McGraw-Hill All Rights Reserved 137

Interpretation of ROR on Extra Investment
Based on concept that any avoidable investment that does not yield at least the MARR should not be made. Once a lower-cost alternative has been economically justified, the ROR on the extra investment (i.e., additional amount of money associated with a higher first-cost alternative) must also yield a ROR ≥ MARR (because the extra investment is avoidable by selecting the economically-justified lower-cost alternative). This incremental ROR is identified as ∆i* For independent projects, select all that have ROR ≥ MARR (no incremental analysis is necessary) 8-138 © 2012 by McGraw-Hill All Rights Reserved

ROR Evaluation for Two ME Alternatives
Order alternatives by increasing initial investment cost (2) Develop incremental CF series using LCM of years (3) Draw incremental cash flow diagram, if needed (4) Count sign changes to see if multiple ∆i* values exist (5) Set up PW, AW, or FW = 0 relation and find ∆i*B-A Note: Incremental ROR analysis requires equal-service comparison. The LCM of lives must be used in the relation (6) If ∆i*B-A < MARR, select A; otherwise, select B If multiple ∆i* values exist, find EROR using either MIRR or ROIC approach. 8-139 © 2012 by McGraw-Hill All Rights Reserved

Example: Incremental ROR Evaluation
Either of the cost alternatives shown below can be used in a chemical refining process. If the company’s MARR is 15% per year, determine which should be selected on the basis of ROR analysis? A B First cost ,$ , ,000 Annual cost, $/year , ,000 Salvage value, $ , ,000 Life, years Initial observations: ME, cost alternatives with equal life estimates and no multiple ROR values indicated 8-140 © 2012 by McGraw-Hill All Rights Reserved

Example: ROR Evaluation of Two Alternatives
Solution, using procedure: A B B - A First cost , $ , ,000 -20,000 Annual cost, $/year , ,000 +6000 Salvage value, $ , ,000 +2000 Life, years Order by first cost and find incremental cash flow B - A Write ROR equation (in terms of PW, AW, or FW) on incremental CF 0 = -20, (P/A,∆i*,5) (P/F,∆i*,5) Solve for ∆i* and compare to MARR ∆i*B-A = 17.2% > MARR of 15% ROR on $20,000 extra investment is acceptable: Select B 8-141 © 2012 by McGraw-Hill All Rights Reserved 141

Breakeven ROR Value An ROR at which the PW, AW or FW values:
Of cash flows for two alternatives are exactly equal. This is the i* value Of incremental cash flows between two alternatives are exactly equal. This is the ∆i* value If MARR > breakeven ROR, select lower-investment alternative © 2012 by McGraw-Hill All Rights Reserved 8-142

ROR Analysis – Multiple Alternatives
Six-Step Procedure for Mutually Exclusive Alternatives (1) Order alternatives from smallest to largest initial investment (2) For revenue alts, calculate i* (vs. DN) and eliminate all with i* < MARR; remaining alternative with lowest cost is defender. For cost alternatives, go to step (3) (3) Determine incremental CF between defender and next lowest-cost alternative (known as the challenger). Set up ROR relation (4) Calculate ∆i* on incremental CF between two alternatives from step (3) (5) If ∆i* ≥ MARR, eliminate defender and challenger becomes new defender against next alternative on list (6) Repeat steps (3) through (5) until only one alternative remains. Select it. For Independent Projects Compare each alternative vs. DN and select all with ROR ≥ MARR 8-143 © 2012 by McGraw-Hill All Rights Reserved 143

Example: ROR for Multiple Alternatives
The five mutually exclusive alternatives shown below are under consideration for improving visitor safety and access to additional areas of a national park. If all alternatives are considered to last indefinitely, determine which should be selected on the basis of a rate of return analysis using an interest rate of 10%. A B C D E_ First cost, $ millions -40 -35 -90 -70 Annual M&O cost, $ millions -1.5 -1.9 -1.1 -1.3 Solution: Rank on the basis of initial cost: A,C,B,E,D; calculate CC values C vs. A: 0 = / ∆i* = 6.7% (eliminate C) B vs. A: 0 = / ∆i* = 25% (eliminate A) E vs. B: 0 = / ∆i* = 6.7% (eliminate E) D vs. B: 0 = / ∆i* = 8% (eliminate D) Select alternative B 8-144 © 2012 by McGraw-Hill All Rights Reserved 144

Must consider incremental cash flows for mutually exclusive alternatives Incremental cash flow = cash flowB – cash flowA where alternative with larger initial investment is Alternative B Eliminate B if incremental ROR ∆i* < MARR; otherwise, eliminate A Breakeven ROR is i* between project cash flows of two alternatives, or ∆i* between incremental cash flows of two alternatives For multiple mutually exclusive alternatives, compare two at a time and eliminate alternatives until only one remains For independent alternatives, compare each against DN and select all that have ROR ≥ MARR 8-145 © 2012 by McGraw-Hill All Rights Reserved 145

Benefit/Cost Analysis Lecture slides to accompany
Chapter 9 Benefit/Cost Analysis Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 9-146 © 2012 by McGraw-Hill All Rights Reserved 146

LEARNING OUTCOMES Explain difference in public vs. private sector projects Calculate B/C ratio for single project Select better of two alternatives using B/C method Select best of multiple alternatives using B/C method Use cost-effectiveness analysis (CEA) to evaluate service sector projects 6. Describe how ethical compromises may enter public sector projects 9-147 © 2012 by McGraw-Hill All Rights Reserved 147

Differences: Public vs. Private Projects
Characteristic Public Private Size of Investment Large Small, medium, large Life Longer (30 – 50+ years) Shorter (2 – 25 years) Annual CF No profit Profit-driven Funding Taxes, fees, bonds, etc. Stocks, bonds, loans, etc. Interest rate Lower Higher Selection criteria Multiple criteria Primarily ROR Environment of evaluation Politically inclined Economic 9-148 © 2012 by McGraw-Hill All Rights Reserved 148

Types of Contracts 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 9-149 © 2012 by McGraw-Hill All Rights Reserved

Cash Flow Classifications and B/C Relations
Must identify each cash flow as either benefit, disbenefit, or cost Benefit (B) -- Advantages to the public Disbenefit (D) -- Disadvantages to the public Cost (C) -- Expenditures by the government Note: Savings to government are subtracted from costs Conventional B/C ratio = (B–D) / C Modified B/C ratio = [(B–D) – C] / Initial Investment Profitability Index = NCF / Initial Investment Note 1: All terms must be expressed in same units, i.e., PW, AW, or FW Note 2: Do not use minus sign ahead of costs 9-150 © 2012 by McGraw-Hill All Rights Reserved

Decision Guidelines for B/C and PI
Benefit/cost analysis If B/C ≥ 1.0, project is economically justified at discount rate applied If B/C < 1.0, project is not economically acceptable Profitability index analysis of revenue projects If PI ≥ 1.0, project is economically justified at discount rate applied If PI < 1.0, project is not economically acceptable 9-151 © 2012 by McGraw-Hill All Rights Reserved

B/C Analysis – Single Project
Conventional B/C ratio = B - D C If B/C ≥ 1.0, accept project; otherwise, reject Modified B/C ratio = B – D – M&O C Denominator is initial investment PW of NCFt PI = PW of initial investment If PI ≥ 1.0, accept project; otherwise, reject 9-152 © 2012 by McGraw-Hill All Rights Reserved 152

Example: B/C Analysis – Single Project
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? Solution: Express all values in AW terms and find B/C ratio B = $175,000 D = $25,000 C = 1,400,000(A/P,6%,10) + $40,000 = $230,218 B/C = (175,000 – 25,000)/230,218 = < 1.0 Do not build project 9-153 © 2012 by McGraw-Hill All Rights Reserved

Defender, Challenger and Do Nothing Alternatives
When selecting from two or more ME alternatives, there is a: Defender – in-place system or currently selected alternative Challenger – Alternative challenging the defender Do-nothing option – Status quo system General approach for incremental B/C analysis of two ME alternatives: Lower total cost alternative is first compared to Do-nothing (DN) If B/C for the lower cost alternative is < 1.0, the DN option is compared to ∆B/C of the higher-cost alternative If both alternatives lose out to DN option, DN prevails, unless overriding needs requires selection of one of the alternatives 9-154 © 2012 by McGraw-Hill All Rights Reserved

Alternative Selection Using Incremental B/C Analysis – Two or More ME Alternatives
Procedure similar to ROR analysis for multiple alternatives Determine equivalent total cost for each alternative Order alternatives by increasing total cost Identify B and D for each alternative, if given, or go to step 5 Calculate B/C for each alternative and eliminate all with B/C < 1.0 Determine incremental costs and benefits for first two alternatives Calculate ∆B/C; if >1.0, higher cost alternative becomes defender Repeat steps 5 and 6 until only one alternative remains 9-155 © 2012 by McGraw-Hill All Rights Reserved

Example: Incremental B/C Analysis
Compare two alternatives using i = 10% and B/C ratio Alternative X Y First cost, $ 320, ,000 M&O costs, $/year , ,000 Benefits, $/year 110, ,000 Disbenefits, $/year , ,000 Life, years Solution: First, calculate equivalent total cost AW of costsX = 320,000(A/P,10%,10) + 45,000 = $97,080 AW of costsY = 540,000(A/P,10%,20) + 35,000 = $98,428 Order of analysis is X, then Y X vs. DN: (B-D)/C = (110,000 – 20,000) / 97,080 = Eliminate X Y vs. DN: (150,000 – 45,000) / 98,428 = Eliminate DN Select Y 9-156 © 2012 by McGraw-Hill All Rights Reserved

Example: ∆B/C Analysis; Selection Required
Must select one of two alternatives using i = 10% and ∆B/C ratio Alternative X Y First cost, $ 320, ,000 M&O costs, $/year , ,000 Benefits, $/year 110, ,000 Disbenefits, $/year , ,000 Life, years Solution: Must select X or Y; DN not an option, compare Y to X AW of costsX = $97,080 AW of costsY = $98,428 Incremental values: ∆B = 150,000 – 110,000 = $40,000 ∆D = 45,000 – 20,000 = $25,000 ∆C = 98,428 – 97,080 = $1,348 Y vs. X: (∆B - ∆D) / ∆C = (40,000 – 25,000) / 1,348 = Eliminate X Select Y 9-157 © 2012 by McGraw-Hill All Rights Reserved

B/C Analysis of Independent Projects
Independent projects comparison does not require incremental analysis Compare each alternative’s overall B/C with DN option No budget limit: Accept all alternatives with B/C ≥ 1.0 Budget limit specified: capital budgeting problem; selection follows different procedure (discussed in chapter 12) 9-158 © 2012 by McGraw-Hill All Rights Reserved

Cost Effectiveness Analysis Cost-effectiveness ratio (CER)
Service sector projects primarily involve intangibles, not physical facilities; examples include health care, security programs, credit card services, etc. Cost-effectiveness analysis (CEA) combines monetary cost estimates with non-monetary benefit estimates to calculate the Cost-effectiveness ratio (CER) Equivalent total costs Total effectiveness measure = C/E CER = 9-159 © 2012 by McGraw-Hill All Rights Reserved

CER Analysis for Independent Projects
Procedure is as follows: (1) Determine equivalent total cost C, total effectiveness measure E and CER (2) Order projects by smallest to largest CER (3) Determine cumulative cost of projects and compare to budget limit b (4) Fund all projects such that b is not exceeded Example: The effectiveness measure E is the number of graduates from adult training programs. For the CERs shown, determine which independent programs should be selected; b = $500,000. Program CER, $/graduate Program Cost, $ A ,000 B ,000 C ,000 D ,000 9-160 © 2012 by McGraw-Hill All Rights Reserved 160

Example: CER for Independent Projects
First, rank programs according to increasing CER: Cumulative Program CER, $/graduate Program Cost, $ Cost, $ B , ,000 A , ,000 D , ,000 C , ,000 Next, select programs until budget is not exceeded Select programs B and A at total cost of $403,000 Note: To expend the entire $500,000, accept as many additional individuals as possible from D at the per-student rate 9-161 © 2012 by McGraw-Hill All Rights Reserved 161

CER Analysis for Mutually Exclusive Projects Procedure is as follows
(1) Order alternatives smallest to largest by effectiveness measure E Calculate CER for first alternative (defender) and compare to DN option Calculate incremental cost (∆C), effectiveness (∆E), and incremental measure ∆C/E for challenger (next higher E measure) If ∆C/Echallenger < C/Edefender challenger becomes defender (dominance); otherwise, no dominance is present and both alternatives are retained (5) Dominance present: Eliminate defender and compare next alternative to new defender per steps (3) and (4). Dominance not present: Current challenger becomes new defender against next challenger, but old defender remains viable (6) Continue steps (3) through (5) until only 1 alternative remains or only non-dominated alternatives remain (7) Apply budget limit or other criteria to determine which of remaining non-dominated alternatives can be funded 9-162 © 2012 by McGraw-Hill All Rights Reserved 162

Example: CER for ME Service Projects
The effectiveness measure E is wins per person. From the cost and effectiveness values shown, determine which alternative to select. Cost (C) Effectiveness (E) CER Program $/person wins/person $/win A B C 9-163 © 2012 by McGraw-Hill All Rights Reserved

Example: CER for ME Service Projects
Solution: Order programs according to increasing effectiveness measure E Cost (C) Effectiveness (E) CER Program $/person wins/person $/win B , A , C , B vs. DN: C/EB = 1400/2 = 700 A vs. B: ∆C/E = (2200 – 1400)/(4 – 2) = Dominance; eliminate B C vs. A: ∆C/E = (6860 – 2200)/(7 – 4) = No dominance; retain C Must use other criteria to select either A or C 9-164 © 2012 by McGraw-Hill All Rights Reserved

Ethical Considerations
Engineers are routinely involved in two areas where ethics may be compromised: Public policy making – Development of strategy, e.g., water system management (supply/demand strategy; ground vs. surface sources) Public planning - Development of projects, e.g., water operations (distribution, rates, sales to outlying areas) Engineers must maintain integrity and impartiality and always adhere to Code of Ethics 9-165 © 2012 by McGraw-Hill All Rights Reserved

B/C method used in public sector project evaluation Can use PW, AW, or FW for incremental B/C analysis, but must be consistent with units for B,C, and D estimates For multiple mutually exclusive alternatives, compare two at a time and eliminate alternatives until only one remains For independent alternatives with no budget limit, compare each against DN and select all alternatives that have B/C ≥ 1.0 CEA analysis for service sector projects combines cost and nonmonetary measures Ethical dilemmas are especially prevalent in public sector projects 9-166 © 2012 by McGraw-Hill All Rights Reserved 166

Making Choices: The Method, MARR, and Multiple Attributes
Developed By: Dr. Don Smith, P.E. Department of Industrial Engineering Texas A&M University College Station, Texas Executive Summary Version Chapter 10 Making Choices: The Method, MARR, and Multiple Attributes At this point: 1. Introduce yourself - your students are likely to want to know something about your qualifications and interests - overall, where you are coming from. 2. Have students introduce themselves. Ask why they are taking this class. If you are fortunate enough to have a Polaroid camera, take pictures of each student for later posting on a class “board” so both they and you get to know each other. 3. Discuss both choice of textbook and development of syllabus. 4. If you are expecting students to work in teams, at east introduce the choice of team members. If at all possible, have students participate in a team building or team study exercise. It works wonders. Most student have been told to work in teams in prior classes, but have never examined exactly what a team is and how it works. One hour spent in a team building/examination exercise saves many hours and avoids many problems later on.

LEARNING OBJECTIVES Choose a Method Cost of capital and MARR
WACC – Weighted Average Cost of Capital Cost of debt capital Cost of equity capital High D-E mixes Multiple attributes Weighted attribute methods

Different information is available from different evaluation methods
Sct 10.1 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 See Tables 10-1 & 10-2 for a concise summary

Evaluation Times Equal lives of the alternatives LCM of lives
PW, AW, FW LCM of lives PW approach Specified study period Normally exercised in industry Infinity (capitalized cost)

Decision Guidelines Select the alternative with: Numerically largest
PW, FW, or AW value For ROR and B/C Apply the incremental analysis approach

Sct 10.2 MARR Relative to the Cost of Capital
Establishing the MARR within the enterprise Requires: Cost of equity capital (cost of corporate funds) Cost of retained earnings included here Cost of debt capital (cost of borrowed funds) Debt Capital $$ acquired from borrowing outside of the firm Equity Capital $$ acquired from the owners and retained earnings

Cost of Capital and the MARR
Established MARR is the sum of: (expressed as a % cost) Cost of capital + Expected return + Risk factor MARR will vary from firm to firm and from project to project Established MARR Risk factor added (R%) ER + R% Expected return (%) ER CC + x% = ER Min. MARR Cost of capital (%) CC

Factors Impacting the MARR
Perceived project risk Higher the risk – higher the MARR for that project Investment opportunity Expansion opportunity – may set a lower MARR Maintain flexibility Tax structure Higher tax rate – higher MARR Federal reserve monetary policy – interest rates Limited capital Tighter constraints on capital – higher MARR Market rates of other firms Competitors alter their MARR - the firm could follow suit

Sct 10.3 Debt-Equity Mix and WACC
D/E ratio (Debt to Equity mix) Ex.: DE = {40% from debt, 60% from equity} Weighted Average Cost of Capital (WACC) WACC = (equity fraction)(cost of equity capital) + (debt fraction)(cost of debt capital) Both ‘costs’ are expressed as a percentage cost Example: WACC = 0.6(4%) + 0.6(9%) = 7.8% A variety of “models” exist that will approximate the WACC for a given firm

WACC: Example 10.3 Source of Capital Amount ($) Cost (%) Common Stock
$5 million 13.7% Retained Earnings $2 million 8.9% Debt from bonds $3 million 7.5% Sum: $10 million CS = 50%; RE = 20%; Bonds = 30% WACC = (0.50)(13.7) + (0.20)(8.9) + (0.30)(7.5) = 10.88% This firm’s MARR must be > 10.88%

Tax Implications (detailed in Chapter 17)
WACC values are computed: Before-tax basis After-tax basis After-tax WACC = (Before Tax WACC)(1- Te) Where Te represents the effective tax rate composed of: Federal rate State rate Local rate(s)

Sct 10.4 Determining Cost of Debt Capital
Debt financing Loans (borrowing) $ borrowed from banks $ borrowed from Insurance companies, etc Issuance of bonds (borrowing) Interest on loans and bonds are tax deductible in the US Bonds are sold (floated) within a bond market by investment bankers on behalf of the firm Subject to extensive state and federal regulations Interest payments from the firm to the lenders is tax deductible – important cost consideration

Tax Savings from Debt Financing
The cost of financing by debt is lower than the actual interest rate charged because of the tax deductibility of the interest payments Assume Te = the effective tax rate (%) Tax Savings = ($ expenses)(Te) Net Cash Flow = {$ expenses - $ tax savings} NCF = expenses (1 – Te) See Example 10.4

Example of Tax Deductibility Impact on Cost of Debt Capital
Assume a loan has a 10% interest rate charged to the borrower The effective tax rate is 30% The after-tax cost of borrowing at 10% is (0.10)(1 – 0.30) = (0.10)(0.70) = 0.07 or 7% Observations Due to tax deductibility the effective cost is 7% after tax Higher tax rates result in lower after-tax borrowing rates

Sct 10.5 Determination of the Cost of Equity Capital and the MARR
Sources of equity capital Sale of preferred stock (PS) Sale of common stock (CS) Use of retained earnings (RE) RE = past profits retained within the firm This money belongs to the owners of the firm Sale of new stock is handled by investment bankers and brokerage firms – highly regulated – charge the firm for these sales

Types of Stock Preferred Stock Common Stock A form of ownership
Pays a stated dividend per share periodically Generally a conservative type of stock Common Stock Carries more risk than preferred No guarantee of dividends to be paid

Cost of Equity Capital Cost of equity capital generally applies some form of a dividend growth model or valuation model Basic model “g” is the estimated annual increase in returns to the shareholders

Capital Asset Pricing Model -- CAPM
Re for equity capital is specified by Re = risk-free return + premium above risk-free return Re = Rf + (Rm – Rf)  = volatility of firm’s common stock relative to other stocks  = 1 is the norm Rm = return on stocks is a defined market portfolio as measured by a prescribed index Rf = quoted US Treasury Bill rate (considered a safe investment) (Rm – Rf) = premium paid above the safe or “risk-free” rate See Example 10.6

Sct 10.6 Effect of Debt-Equity Mix on Investment Risk
D-E mix (Review Section 10.3) As the proportion of debt increasesDue to t the calculated cost of capital tends to decrease Tax advantage of deducting interest But…..leverage offered by larger percentage of debt capital increases the risks of funding future projects within the company Too much debt is a “bad thing” Objective – strive for a balance between debt and equity funding

Too Much Debt….. Use of larger percentages of debt capital increases the risk that is assumed by Investors (owners) and Lenders Over time, investor confidence in the firm may diminish and the value of the stock could well decline Difficult to attract new investment funds Lenders will charge higher and higher interest rates to hedge the risk

Refer back to Chapter 1 and
Sct 10.7 Multiple Attribute Analysis: Identification and Importance of Each Attribute Refer back to Chapter 1 and 7-steps in Figure 10-5 Up to now we have focused on one attribute of a decision making problem Economic attribute! Complex problems possess more than one attribute Multiple attribute analysis is often required Quantitative attributes Subjective attributes

Identification of Key Attributes
Must ID the key attributes Comparison Input from experts Surveys Group discussion Delphi methods Tabulate and then agree on the critical mix of subjective and objective attributes

Importance of Each Attribute
Determine the extent of importance of each attribute Implies some form of weighting – wi Given m attributes we want: Weights for each attribute Tabular format of attributes vs. alternatives Value ratings Vi j

Weighting Methodologies
Equal Weighting All defined attributes are assigned equal weights Default model May or may not be appropriate Rank Order m attributes are ranked in order of increasing importance (1 = least important; 2, 3, ….) Weighted Rank Order m attributes ranked in order of importance and apply:

Value Rating of Attributes
Each alternative is assigned a value rating – Vij for each attribute i Can apply a scale of 0-100 Can apply a Likert Scale 4-5 graduations (prefer an even number of choices)e.g. Very Poor Poor Good Very good See Table 10.4

Sct 10.8 Evaluation Measure for Multiple Attributes
Weighted Attribute Method Selection guideline Choose the alternative with the largest Rj value Assumes increasing weights mean more important attributes Increasing Vij mean better performance for a given alternative See Example 10.10

Chapter Summary Best methods for economic evaluation Public projects
PW and AW at the stated MARR Public projects Use the B/C ratio The interest rate used is based upon the cost of capital, mix between equity and debt, and risk levels Multiple attributes incorporate more than objective measures and permit the incorporation of criteria that is not totally economic based

Chapter 10 End of Set

Replacement & Retention Lecture slides to accompany
Chapter 11 Replacement & Retention Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 11-195 © 2012 by McGraw-Hill All Rights Reserved 195

LEARNING OUTCOMES Explain replacement terminology and basics
Determine economic service life (ESL) Perform replacement/retention study Understand special situations in replacement Perform replacement study over specified time Calculate trade-in value of defender 11-196 © 2012 by McGraw-Hill All Rights Reserved 196

Replacement Study Basics
Reasons for replacement study Reduced performance Altered requirements Obsolescence Terminology Defender – Currently installed asset Challenger – Potential replacement for defender Market value (MV) – Value of defender if sold in open market Economic service life – No. of years at which lowest AW of cost occurs Defender first cost – MV of defender; used as its first cost (P) in analysis Challenger first cost – Capital to recover for challenger (usually its P value) Sunk cost – Prior expenditure not recoverable from challenger cost Nonowner’s viewpoint – Outsider’s (consultant’s) viewpoint for objectivity 11-197 © 2012 by McGraw-Hill All Rights Reserved 197

Example: Replacement Basics
An asset purchased 2 years ago for $40,000 is harder to maintain than expected. It can be sold now for $12,000 or kept for a maximum of 2 more years, in which case its operating cost will be $20,000 each year, with a salvage value of $9,000 two years from now. A suitable challenger will have a first cost of $60,000 with an annual operating cost of $4,100 per year and a salvage value of $15,000 after 5 years. Determine the values of P, A, n, and S for the defender and challenger for an AW analysis. Solution: Defender: P = $-12,000; A = $-20,000; n = 2; S = $9,000 Challenger: P = $-60,000; A = $-4,100; n = 5; S = $15,000 11-198 © 2012 by McGraw-Hill All Rights Reserved

Overview of a Replacement Study
Replacement studies are applications of the AW method Study periods (planning horizons) are either specified or unlimited Assumptions for unlimited study period: Services provided for indefinite future Challenger is best available now and for future, and will be repeated in future life cycles Cost estimates for each life cycle for defender and challenger remain the same If study period is specified, assumptions do not hold Replacement study procedures differ for the two cases 11-199 © 2012 by McGraw-Hill All Rights Reserved 199

Economic Service Life Economic service life (ESL) refers to the asset retention time (n) that yields its lowest equivalent AW Determined by calculating AW for 1, 2, 3,…n years General equation is: Total AW = capital recovery – AW of annual operating costs = CR – AW of AOC 11-200 © 2012 by McGraw-Hill All Rights Reserved 200

Example: Economic Service Life
Determine the ESL of an asset which has the costs shown below. Let i = 10% Year Cost,$/year Salvage value,$ , , ,000 , ,000 , ,000 , ,000 , ,000 Solution: AW1 = - 20,000(A/P,10%,1) – 5000(P/F,10%,1)(A/P,10%,1) + 10,000(A/F,10%,1) = $ -17,000 AW2 = - 20,000(A/P,10%,2) – [5000(P/F,10%,1) (P/F,10%,2)](A/P,10%,2) + 8000(A/F,10%,2) = $ -13,429 Similarly, AW3 = $ -13, AW4 = $ -12, AW5 = $ -13,623 Economic service life is 4 years 11-201 © 2012 by McGraw-Hill All Rights Reserved

Performing a Replacement Study

Performing a Replacement Study -- Unlimited Study Period
1. Calculate AWD and AWC based on their ESL; select lower AW 2. If AWC was selected in step (1), keep for nC years (i.e., economic service life of challenger); if AWD was selected, keep defender one more year and then repeat analysis (i.e., one-year-later analysis) 3. As long as all estimates remain current in succeeding years, keep defender until nD is reached, and then replace defender with best challenger If any estimates change before nD is reached, repeat steps (1) through (4) Note: If study period is specified, perform steps (1) through (4) only through end of study period (discussed later) 11-203 © 2012 by McGraw-Hill All Rights Reserved

Example: Replacement Analysis
An asset purchased 2 years ago for $40,000 is harder to maintain than expected. It can be sold now for $12,000 or kept for a maximum of 2 more years, in which case its operating cost will be $20,000 each year, with a salvage value of $10,000 after 1 year or $9000 after two years. A suitable challenger will have an annual worth of $-24,000 per year. At an interest rate of 10% per year, should the defender be replaced now, one year from now, or two years from now? Solution: First, determine ESL for defender AWD1 = -12,000(A/P,10%,1) – 20, ,000(A/F,10%,1) = $-23,200 AWD2 = -12,000(A/P,10%,2) – 20, ,000(A/F,10%,2) = $-22,629 ESL is n = 2 years; AWD = $-22,629 AWC = $-24,000 Lower AW = $-22, Replace defender in 2 years Note: conduct one-year-later analysis next year 11-204 © 2012 by McGraw-Hill All Rights Reserved

Additional Considerations
Opportunity cost approach is the procedure that was previously presented for obtaining P for the defender. The opportunity cost is the money foregone by keeping the defender (i.e., not selling it). This approach is always correct Cash flow approach subtracts income received from sale of defender from first cost of challenger. Potential problems with cash flow approach: Provides falsely low value for capital recovery of challenger Can’t be used if remaining life of defender is not same as that of challenger 11-205 © 2012 by McGraw-Hill All Rights Reserved 205

Replacement Analysis Over Specified Study Period
Same procedure as before, except calculate AW values over study period instead of over ESL years of nD and nC It is necessary to develop all viable defender-challenger combinations and calculate AW or PW for each one over study period Select option with lowest cost or highest income 11-206 © 2012 by McGraw-Hill All Rights Reserved 206

Example: Replacement Analysis; Specified Period
An asset purchased 2 years ago for $40,000 is harder to maintain than expected. It can be sold now for $12,000 or kept for a maximum of 2 more years, in which case its operating cost will be $20,000 each year, with a salvage value of $10,000 after 1 year or $9000 after two. A suitable challenger will have an annual worth of $-24,000 per year. At an interest rate of 10% per year and over a study period of exactly 2 years, determine when the defender should be replaced. Solution: From previous analysis, AWD for 1 and 2 years, and AWC are: AWD1 = $-23,200 AWD2 = $-22,629 AWC = $-24,000 Option Year 1, $ Year 2, $ Year 3, $ AW, $ 1 (C, C, C) -24,000 2 (D, C, C) -23,200 -23,708 3 (D, D, C) -22,629 -23,042 Decision: Option 3; Keep D for 2 years, then replace 11-207 © 2012 by McGraw-Hill All Rights Reserved

Replacement Value Replacement value (RV) is market/trade-in value of
defender that renders AWD and AWC equal to each other Set up equation AWD = AWC except use RV in place of P for the defender; then solve for RV If defender can be sold for amount > RV, challenger is the better option, because it will have a lower AW value 11-208 © 2012 by McGraw-Hill All Rights Reserved 208

Example: Replacement Value
An asset purchased 2 years ago for $40,000 is harder to maintain than expected. It can be sold now for $12,000 or kept for a maximum of 2 more years, in which case its operating cost will be $20,000 each year, with a salvage value of $10,000 at the end of year two. A suitable challenger will have an initial cost of $65,000, an annual cost of $15,000, and a salvage value of $18,000 after its 5 year life. Determine the RV of the defender that will render its AW equal to that of the challenger, using an interest rate of 10% per year. Recommend a course of action. Solution: Set AWD = AWC - RV(A/P,10%,2) - 20, ,000(A/F,10%,2) = - 65,000(A/P,10%,5) -15,000 +18,000(A/F,10%,5) RV = $24,228 Thus, if market value of defender > $24,228, select challenger 11-209 © 2012 by McGraw-Hill All Rights Reserved

In replacement study, P for presently-owned asset is its market value Economic service life is the n value that yields lowest AW In replacement study, if no study period is specified, calculate AW over the respective life of each alternative Opportunity cost approach is correct, it recognizes money foregone by keeping the defender, not by reducing challenger’s first cost When study period is specified, must consider all viable defender-challenger combinations in analysis Replacement value (RV) is P value for defender that renders its AW equal to that of challenger 11-210 © 2012 by McGraw-Hill All Rights Reserved 210

Chapter 12 Independent Projects with Budget Limitation
Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin 12-1 © 2012 by McGraw-Hill All Rights Reserved 211

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 ‘Bundle’ is a collection 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 All Rights Reserved

Capital Budgeting Problem
Each project selected entirely or not selected at all Budget limit restricts total investment allowed Projects usually 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 Objective of Capital Budgeting Maximize return on project investments using a specific measure of worth, usually PW 12-4 © 2012 by McGraw-Hill All Rights Reserved

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 All Rights Reserved

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 C -50,000 +15,000 +20,000 Solution: Five bundles meet budget restriction. Calculate NCF and PW values Bundle, j Projects NCFj0 , $ NCFjt , $ SV, $ PWj , $ 1 A -25,000 +6,000 +4,000 -5,583 2 B -20,000 +9,000 +5,695 3 C -50,000 +15,000 +20,000 +4,261 4 A, B 45,000 +112 5 B, C 70,000 +24,000 +9,956 6 DN Conclusion: Select projects B and C with max PW value 12-6 © 2012 by McGraw-Hill All Rights Reserved

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 All Rights Reserved

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 All Rights Reserved

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 All Rights Reserved

Example: LP Solution of Capital Budgeting Problem
MARR is 15%; limit is $20,000; select projects using LP LP formulation for projects A, B, C, D labeled k = 1, 2, 3, 4 and b = $20,000 is: Maximize: x x x x4 Constraints: 8000x1 + 15,000x x x4 ≤ 20,000 x1, x2, x3, and x4 = 0 or 1 15%, $ 6646 -1019 984 -748 12-10 © 2012 by McGraw-Hill All Rights Reserved

Example: LP Solution of Capital Budgeting Problem
Use spreadsheet and Solver tool to solve LP problem Select projects A (x1 = 1) and C (x3 = 1) for max PW = $7630 12-11 © 2012 by McGraw-Hill All Rights Reserved

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 All Rights Reserved

Capital budgeting requires selection from independent projects. The PW measure is commonly maximized with a limited investment budget Equal service assumption is not necessary for a capital budgeting solution Manual solution requires development of ME bundles of projects Solution using linear programming formulation and the Solver tool on a spreadsheet is used to maximize PW at a stated MARR Projects ranked using different measures, e.g., PW, IROR and PWI, may select different projects since different measures are maximized 12-13 © 2012 by McGraw-Hill All Rights Reserved

Breakeven and Payback Analysis Lecture slides to accompany
Chapter 13 Breakeven and Payback Analysis Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin © 2012 by McGraw-Hill All Rights Reserved 13-1 224

Value of a parameter that makes two elements equal
Breakeven Point Value of a parameter that makes two elements equal The parameter (or variable) can be an amount of revenue, cost, supply, demand, etc. for one project or between two alternatives One project - Breakeven point is identified as QBE. Determined using linear or non-linear math relations for revenue and cost Between two alternatives - Determine one of the parameters P, A, F, i, or n with others constant Solution is by one of three methods: Direct solution of relations Trial and error Spreadsheet functions or tools (Goal Seek or Solver) 13-3 © 2012 by McGraw-Hill All Rights Reserved 226

Cost-Revenue Model ― One Project
Quantity, Q — An amount of the variable in question, e.g., units/year, hours/month Breakeven value is QBE Fixed cost, FC — Costs not directly dependent on the variable, e.g., buildings, fixed overhead, insurance, minimum workforce cost Variable cost, VC — Costs that change with parameters such as production level and workforce size. These are labor, material and marketing costs. Variable cost per unit is v Total cost, TC — Sum of fixed and variable costs, TC = FC + VC Revenue, R — Amount is dependent on quantity sold Revenue per unit is r Profit, P — Amount of revenue remaining after costs P = R – TC = R – (FC+VC) © 2012 by McGraw-Hill All Rights Reserved 13-4 227

Breakeven for linear R and TC
Set R = TC and solve for Q = QBE R = TC rQ = FC + vQ FC r – v When variable cost, v, is lowered, QBE decreases (moves to left) QBE = 13-5 © 2012 by McGraw-Hill All Rights Reserved 228

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 / ( ) = 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 © 2012 by McGraw-Hill All Rights Reserved 229

Breakeven Between Two Alternatives
To determine value of common variable between 2 alternatives, do the following: Define the common variable Develop equivalence PW, AW or FW relations as function of common variable for each alternative Equate the relations; solve for variable. This is breakeven value Selection of alternative is based on anticipated value of common variable: Value BELOW breakeven; select higher variable cost Value ABOVE breakeven; select lower variable cost 13-7 © 2012 by McGraw-Hill All Rights Reserved 230

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 AWbuy = -1.5X Solution: Equate AW relations, solve for X -1.5X = X X = 4116 per year Breakeven value of X AW, 1000 $/year AWbuy 8 7 6 5 4 3 2 1 AWmake If anticipated production > 4116, select make alternative (lower variable cost) X, 1000 units per year 13-8 © 2012 by McGraw-Hill All Rights Reserved 231

Breakeven Analysis Using Goal Seek Tool
Spreadsheet tool Goal Seek finds breakeven value for the common variable between two alternatives Problem: Two machines (1 and 2) have following estimates. a) Use spreadsheet and AW analysis to select one at MARR = 10%. b) Use Goal Seek to find the breakeven first cost. Machine P, $ , ,000 NCF, $/year , ,000 S, $ , ,000 n, years Solution: a) Select machine A with AWA = $193 13-9 © 2012 by McGraw-Hill All Rights Reserved 232

Breakeven Analysis Using Goal Seek Tool
Solution: b) Goal Seek finds a first-cost breakeven of $96,669 to make machine B economically equivalent to A Changing cell Spreadsheet after Goal Seek is applied Target cell 13-10 © 2012 by McGraw-Hill All Rights Reserved 233

Payback Period Analysis
Payback period: Estimated amount of time (np) for cash inflows to recover an initial investment (P) plus a stated return of return (i%) Types of payback analysis: No-return and discounted payback No-return payback means rate of return is ZERO (i = 0%) Discounted payback considers time value of money (i > 0%) Caution: Payback period analysis is a good initial screening tool, rather than the primary method to justify a project or select an alternative (Discussed later) 13-11 © 2012 by McGraw-Hill All Rights Reserved 234

Payback Period Computation
Formula to determine payback period (np) varies with type of analysis. NCF = Net Cash Flow per period t Eqn. 1 Eqn. 2 Eqn. 3 Eqn. 4 13-12 © 2012 by McGraw-Hill All Rights Reserved 235

Points to Remember About Payback Analysis
No-return payback neglects time value of money, so no return is expected for the investment made No cash flows after the payback period are considered in the analysis. Return may be higher if these cash flows are expected to be positive. Approach of payback analysis is different from PW, AW, ROR and B/C analysis. A different alternative may be selected using payback. Rely on payback as a supplemental tool; use PW or AW at the MARR for a reliable decision Discounted payback (i > 0%) gives a good sense of the risk involved 13-13 © 2012 by McGraw-Hill All Rights Reserved 236

Example: Payback Analysis
System 1 System 2 First cost, $ 12,000 8,000 NCF, $ per year 3,000 1,000 (year 1-5) 3,000 (year 6-14) Maximum life, years 7 14 Problem: Use (a) no-return payback, (b) discounted payback at 15%, and (c) PW analysis at 15% to select a system. Comment on the results. Solution: (a) Use Eqns. 1 and 2 np1 = 12,000 / 3,000 = 4 years np2 = -8, (1,000) + 1(3,000) = 6 years Select system 1 13-14 © 2012 by McGraw-Hill All Rights Reserved 237

Example: Payback Analysis (continued)
System System 2 First cost, $ 12,000 8,000 NCF, $ per year 3,000 1,000 (year 1-5) 3,000 (year 6-14) Maximum life, years Solution: (b) Use Eqns. 3 and 4 System 1: = -12, ,000(P/A,15%,np1) np1 = 6.6 years System 2: = -8, ,000(P/A,15%,5) + 3,000(P/A,15%,np2 - 5)(P/F,15%,5) np1 = 9.5 years Select system 1 (c) Find PW over LCM of 14 years PW1 = $663 PW2 = $2470 Select system 2 Comment: PW method considers cash flows after payback period Selection changes from system 1 to 2 13-15 © 2012 by McGraw-Hill All Rights Reserved 238

Breakeven amount is a point of indifference to accept or reject a project One project breakeven: accept if quantity is > QBE Two alternative breakeven: if level > breakeven, select lower variable cost alternative (smaller slope) Payback estimates time to recover investment. Return can be i = 0% or i > 0% Use payback as supplemental to PW or other analyses, because np neglects cash flows after payback, and if i = 0%, it neglects time value of money Payback is useful to sense the economic risk in a project 13-16 © 2012 by McGraw-Hill All Rights Reserved 239

LEARNING OUTCOMES Understand inflation/deflation
Calculate PW of cash flows with inflation Calculate FW with inflation considered Calculate AW with inflation considered 14-241 © 2012 by McGraw-Hill All Rights Reserved 241

Understanding Inflation
Inflation: Increase in amount of money needed to purchase same amount of goods or services. Inflation results in a decrease in purchasing power, i.e., one unit of money buys less goods or services Two ways to work problems when considering inflation: Convert to constant value (CV) dollars, then use real rate i. If f = inflation rate (% per year), the equation is: Constant-value dollars = future dollars = then-current dollars (1+ f)n (1+ f)n Leave money amounts as is and use interest rate adjusted for inflation, if if = i + f + (i)(f) 14-242 © 2012 by McGraw-Hill All Rights Reserved 242

Example: Constant Value Dollars
How much would be required today to purchase an item that increased in cost by exactly the inflation rate? The cost 30 years ago was $1000 and inflation has consistently averaged 4% per year. Solution: Solve for future dollars Future dollars = constant value dollars(1 + f)n = 1000( )30 = $3243 Note: This calculation only accounts for the decreased purchasing power of the currency. It does not take into account the time value of money (to be discussed) Deflation: Opposite of inflation; purchasing power of money is greater in future than at present; however, money, credit, jobs are ‘tighter’ 14-243 © 2012 by McGraw-Hill All Rights Reserved 243

Relation between three rates is derived using the relation
Three Different Rates Real or inflation rate i – Rate at which interest is earned when effects of inflation are removed; i represents the real increase in purchasing power Market or inflation-adjusted rate if – Rate that takes inflation into account. Commonly stated rate everyday Inflation rate f – Rate of change in value of currency Relation between three rates is derived using the relation Market rate is: if = i + f + (i)(f) 14-244 © 2012 by McGraw-Hill All Rights Reserved

Example: Market vs. Real Rate
Money in a medium-risk investment makes a guaranteed 8% per year. Inflation rate has averaged 5.5% per year. What is the real rate of return on the investment? Solution: Solve for the real rate i in relation for if if = i + f + (i)(f) if – f 1 + f 0.08 – 0.055 0.024 Investment pays only 2.4% per year in real terms vs. the stated 8% i = = = 14-245 © 2012 by McGraw-Hill All Rights Reserved

PW Calculations with Inflation
Two ways to account for inflation in PW calculations Convert cash flow into constant-value (CV) dollars and use regular i where: CV = future dollars/(1 + f)n = then-current dollars/(1 + f)n f = inflation rate (Note: Calculations up to now have assumed constant-value dollars) Express cash flow in future (then-current ) dollars and use inflated interest rate where if = i + f + (i)(f) ( Note: Inflated interest rate is the market interest rate) 14-246 © 2012 by McGraw-Hill All Rights Reserved

Example: PW with Inflation
A honing machine will have a cost of $25,000 (future cost) six years from now. Find the PW of the machine, if the real interest rate is 10% per year and the inflation rate is 5% per year using (a) constant-value dollars, and (b) future dollars. Solution: (a Determine constant-value dollars and use i in PW equation CV = 25,000 / ( )6 = $18,655 PW = 18,655(P/F,10%,6) = $10,530 (b) Leave as future dollars and use if in PW equation if = (0.10)(0.05) = 15.5% PW = 25,000(P/F,15.5%,6) = $10,530 14-247 © 2012 by McGraw-Hill All Rights Reserved

FW Calculations with Inflation
FW values can have four different interpretations (1) The actual amount accumulated Use if in FW equation FW = PW(F/P, if, n) (2) The purchasing power in terms of CV dollars of the future amount Use if in FW equation and divide by (1+f)n or use real i where real i = (if – f)/(1 + f) FW = PW(F/P,i,n) (3) The number of future dollars required to have the same purchasing power as a dollar today with no time value of money considered Use f instead of i in F/P factor FW = PW(F/P,f,n) (4) The amount required to maintain the purchasing power of the present sum and earn a stated real rate of return Use if in FW equation FW = PW(F/P, if, n) 14-248 © 2012 by McGraw-Hill All Rights Reserved

Example: FW with Inflation
An engineer invests $15,000 in a savings account that pays interest at a real 8% per year. If the inflation rate is 5% per year, determine (a) the amount of money that will be accumulated in 10 years, (b) the purchasing power of the accumulated amount (in terms of today’s dollars), (c) the number of future dollars that will have the same purchasing power as the $15,000 today, and (d) the amount to maintain purchasing power and earn a real 8% per year return. Solution: The amount accumulated is a function of the market interest rate, if if = (0.08)(0.05) = 13.4% Amount Accumulated = 15,000(F/P,13.4%,10) = $52,750 14-249 © 2012 by McGraw-Hill All Rights Reserved 249

Example: FW with Inflation (cont’d)
(b) To find the purchasing power of the accumulated amount deflate the inflated dollars Purchasing power = 15,000(F/P,13.4%,10) / ( )10 = $32,384 (c) The number of future dollars required to purchase goods that cost $15,000 now is the inflated cost of the goods Number of future dollars = 15,000(F/P,5%,10) = $24,434 (d) In order to maintain purchasing power and earn a real return, money must grow by the inflation rate and the interest rate, or if = 13.4%, as in part (a) FW = 15,000(F/P,13.4%,10) = $52,750 14-250 © 2012 by McGraw-Hill All Rights Reserved 250

Capital Recovery with Inflation
The A/P and A/F factors require the use of if when inflation is considered If a small company invests $150,000 in a new production line machine, how much must it receive each year to recover the investment in 5 years? The real interest rate is 10% and the inflation rate is 4% per year. Solution: Capital recovery (CR) is the AW value if = (0.10)(0.04) = 14.4% CR = AW = 150,000(A/P,14.4%,5) = $44,115 per year 14-251 © 2012 by McGraw-Hill All Rights Reserved 251

Inflation occurs because value of currency has changed Inflation reduces purchasing power; one unit buys less goods ort services Two ways to account for inflation in economic analyses: Convert all cash flows into constant-value dollars and use i Leave cash flows as inflated dollars and use if During deflation, purchasing power of money is greater in future than at present Future worth values can have four different interpretations, requiring different interest rates to find FW Use if in calculations involving A/P or A/F when inflation is considered 14-252 © 2012 by McGraw-Hill All Rights Reserved 252

Cost Estimation and Indirect Costs Lecture slides to accompany
Chapter 15 Cost Estimation and Indirect Costs Lecture slides to accompany Engineering Economy 7th edition Leland Blank Anthony Tarquin © 2012 by McGraw-Hill All Rights Reserved 15-1 253

LEARNING OUTCOMES Approaches to estimation Unit method Cost indexes Cost-capacity equations Factor method Indirect cost rates and allocation ABC allocation Ethical considerations 15-2 © 2012 by McGraw-Hill All Rights Reserved 254

Direct and Indirect Cost Estimates
Direct cost examples Physical assets Maintenance and operating costs (M&O) Materials Direct human labor (costs and benefits) Scrapped and reworked product Direct supervision of personnel Indirect cost examples Utilities IT systems and networks Purchasing Management Taxes Legal functions Warranty and guarantees Quality assurance Accounting functions Marketing and publicity 15-3 © 2012 by McGraw-Hill All Rights Reserved 255

What Direct Cost Estimation Includes
Direct costs are more commonly estimated than revenue in an engineering environment. Preliminary decisions required are: What cost components should be estimated? What approach to estimation is best to apply? How accurate should the estimates be? What technique(s) will be applied to estimate costs? Sample direct cost components: first costs and its elements (P); annual costs (AOC or M&O); salvage/market value (S) Approaches: bottom-up; design-to-cost (top down) Accuracy: feasibility stage through detailed design estimates require more exacting estimates Some techniques: unit; factor; cost estimating relations (CER) 15-4 © 2012 by McGraw-Hill All Rights Reserved 256

Different Approaches to Cost Estimation
15-5 © 2012 by McGraw-Hill All Rights Reserved 257

Accuracy of Cost Estimates
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 15-6 © 2012 by McGraw-Hill All Rights Reserved 258

Unit Method Commonly used technique for preliminary design stage estimates Total cost estimate CT is per unit cost (u) times number of units (N) CT = u × N Example uses: Cost to operate a car at 60¢/mile for 500 miles: CT = 0.60 × 500 = $300 Cost to build a 250 m2 house at $2250/m2: CT = 2250 × 250 = $562,500 Cost factors must be updated periodically to remain timely When several components are involved, estimate cost of each component and add to determine total cost estimate CT 15-7 © 2012 by McGraw-Hill All Rights Reserved 259

Cost Indexes Definition: Cost Index is ratio of cost today to cost in the past Indicates change in cost over time; therefore, they account for the impact of inflation Index is dimensionless CPI (Consumer Price Index) is a good example Formula for total cost is Formula for total cost is 15-8 © 2012 by McGraw-Hill All Rights Reserved 260

Example: Cost Index Method
Problem: Estimate the total cost of labor today in US dollars for a maritime construction project using data from a similar project in Europe completed in 1998. Labor index, 1998: Cost in 1998: €3.9 million Labor index, current: Currently, 1 € = 1.5 US$ Solution: Let t = today and 0 = 1998 base Ct = 3.9 million × (1165.8/789.6) = €5.76 million = €5.76 × 1.5 = $8.64 million 15-9 © 2012 by McGraw-Hill All Rights Reserved 261

Finding Cost Indexes Cost indexes are maintained in areas such as construction, chemical and mechanical industries Updated monthly and annually; many include regionalized and international project indexes Indexes in these areas are often subdivided into smaller components and can be used in preliminary, as well as detailed design stages Examples are: Chemical Engineering Plant Cost Index (CEPCI) McGraw-Hill Construction Index US Department of Labor, Bureau of Labor Statistics 15-10 © 2012 by McGraw-Hill All Rights Reserved 262

Cost-Estimating Relationships (CER)
CER equations are used in early design stages to estimate plant, equipment and construction costs CERs are generically different from index relations, because they estimate based on design variables (weight, thrust, force, pressure, speed, etc.) Two commonly used CERs Cost-capacity equation (relates cost to capacity) Factor method (total plant cost estimator, including indirect costs) 15-11 © 2012 by McGraw-Hill All Rights Reserved 263

Cost-Capacity Equation
Also called power law and sizing model Exponent defines relation between capacities x = 1, relationship is linear x < 1, economies of scale (larger capacity is less costly than linear) x > 1, diseconomies of scale 15-12 © 2012 by McGraw-Hill All Rights Reserved 264

Cost-Capacity Combined with Cost Index
Multiply the cost-capacity equation by a cost index (It/I0) to adjust for time differences and obtain estimates of current cost (in constant-value dollars) Example: A 100 hp air compressor costs $3000 five years ago when the cost index was 130. Estimate the cost of a 300 hp compressor today when the cost index is 255. The exponent for a 300 hp air compressor is 0.9. Solution: Let C300 represent the cost estimate today C300 = 3000(300/100)0.9(255/130) = $15,817 15-13 © 2012 by McGraw-Hill All Rights Reserved 265

Overall cost factor h is determined using one of two bases:
Factor Method Factor method is especially useful in estimating total plant cost in processing industries Both direct and indirect costs can be included Total plant cost estimate CT is overall cost factor (h) times total cost of major equipment items (CE) CT = h × CE Overall cost factor h is determined using one of two bases: Delivered-equipment cost (purchase cost of major equipment) Installed-equipment cost (equipment cost plus all make-ready costs) 15-14 © 2012 by McGraw-Hill All Rights Reserved 266

Cost Factor h The cost factor is commonly the sum of a direct cost component and an indirect cost component, that is, h = 1 + Σfi for i = 1, 2, …, n components, including indirect costs Example: Equipment is expected to cost $20 million delivered to a new facility. A cost factor for direct costs of 1.61 will make the plant ready to operate. An indirect cost factor of 0.25 is used. What will the plant cost? Solution: h = = 2.86 CT = 20 million (2.86) = $57.2 million 15-15 © 2012 by McGraw-Hill All Rights Reserved 267

Cost Factor h If indirect costs are charged separately against all direct costs, the indirect cost component is added separately, that is, h = 1 + Σfi (direct costs components) and CT = hCE(1 + findirect) Example: Conveyor delivered-equipment cost is $1.2 million. Factors for installation costs (0.4) and training (0.2) are determined. An indirect cost factor of 0.3 is applied to all direct costs. Estimate total cost. Solution: h = = 1.6 CT = hCE(1 + findirect) = 1.6(1.2 million)( ) = $2.5 million 15-16 © 2012 by McGraw-Hill All Rights Reserved 268

Indirect Costs Indirect costs (IDC) are incurred in production, processes and service delivery that are not easily tracked and assignable to a specific function. Indirect costs (IDC) are shared by many functions because they are necessary to perform the overall objective of the company Indirect costs make up a significant percentage of the overall costs in many organizations – 25 to 50% Sample indirect costs IT services Quality assurance Human resources Management Safety and security Purchasing; contracting Accounting; finance; legal 15-17 © 2012 by McGraw-Hill All Rights Reserved 269

Indirect Cost Allocation - Traditional Method
Cost center -- Department, function, or process used by the cost accounting system to collect both direct and indirect costs Indirect-cost rate – Traditionally, a predetermined rate is used to allocate indirect costs to a cost center using a specified basis. General relation is: Estimated total indirect costs Indirect-cost rate = Estimated basis level Example: Allocation rates for $50,000 to each machine Machine 1: Rate = $50,000/100,000 = $0.50 per DL $ Machine 2: Rate = $50,000/2,000 = $25 per DL hour Machine 3: Rate = $50,000/250,000 = $0.20 per DM $ 15-18 © 2012 by McGraw-Hill All Rights Reserved 270

Example: AW Analysis - Traditional IDC Allocation
MAKE/BUY DECISION Buy: AW = $-2.2 million per year Make: P = $-2 million S = $50,000 n = 10 years MARR = 15% Direct costs of $800,000 per year are detailed below Indirect cost rates are established by department © 2012 by McGraw-Hill All Rights Reserved 15-19 271

Example: Indirect Cost Analysis - Traditional Method
INDIRECT COST ALLOCATION FOR MAKE ALTERNATIVE Dept A: Basis is -- Direct labor hours 25,000(10) = $250,000 Dept B: Basis is -- Machine hours ,000(5) = $125,000 Dept C: Basis is -- Direct labor hours 10,000(15) = $150,000 ECONOMIC COMPARISON AT MARR = 15% AOCmake = direct labor + direct materials + indirect allocation = 500, , ,000 = $1.325 M AWmake = - 2 M(A/P,15%,10) + 50,000(A/F,15%,10) M = $-1.72 M AWbuy = $-2.2 M $525,000 Conclusion: Cheaper to make 15-20 © 2012 by McGraw-Hill All Rights Reserved 272

ABC rate = total activity cost/volume of cost driver
ABC Allocation Activity-Based Costing ─ Provides excellent allocation strategy and analysis of costs for more advanced, high overhead, technologically-based systems Cost Centers (cost pools) ─ Final products/services that receive allocations Activities ─ Support departments that generate indirect costs for distribution to cost centers (maintenance, engineering, management) Cost drivers ─ These are the volumes that drive consumption of shared resources (# of POs, # of machine setups, # of safety violations, # of scrapped items) Steps to implement ABC: Identify each activity and its total cost (e.g., maintenance at $5 million/year) Identify cost drivers and expected volume (e.g., 3,500 requested repairs and 500 scheduled maintenances per year) 3. Calculate cost rate for each activity using the relation: ABC rate = total activity cost/volume of cost driver 4. Use ABC rate to allocate IDC to cost centers for each activity 15-21 © 2012 by McGraw-Hill All Rights Reserved 273

Example: ABC Allocation
Use ABC to allocate safety program costs to plants in US and Europe Cost centers: US and European plants Activity and cost: Safety program costs $200,200 per year Cost driver: # of accidents Volume: 560 accidents; 425 in US plants and 135 in European plants Solution: ABC rate for accident basis = 200,200/560 = $357.50/accident US allocation: (425) = $151,938 Europe allocation: (135) = $48,262 15-22 © 2012 by McGraw-Hill All Rights Reserved 274

Example: Traditional Allocation Comparison
Use traditional rates to allocate safety costs to US and EU plants Cost centers: US and European plants Activity and cost: Safety program costs $200,200 per year Basis: # of employees Volume: employees; 900 in US plants and 500 in European plants Solution: Rate for employee basis = 200,200/1400 = $143/employee US allocation: 143(900) = $128,700 Europe allocation: 143(500) = $71,500 Comparison: US allocation went down; European allocation increased 15-23 © 2012 by McGraw-Hill All Rights Reserved 275

Traditional vs. ABC Allocation
Traditional method is easier to set up and use Traditional method is usually better when making cost estimates ABC is more accurate when process is in operation ABC is more costly, but provides more information for cost analysis and decision making Traditional and ABC methods complement each other: Traditional is good for cost estimation and allocation ABC is better for cost tracking and cost control 15-24 © 2012 by McGraw-Hill All Rights Reserved 276

Ethics and Cost Estimating
Unethical practices in estimation may be the result of: Personal gain motivation Bias Deception Favoritism toward an individual or organization Intentional poor accuracy Pre-arranged financial favors (bribes, kickbacks) When making any type of estimates, always comply with the Code of Ethics for Engineers Avoid deceptive acts 15-25 © 2012 by McGraw-Hill All Rights Reserved 277

Required accuracy of cost estimates depends on the stage of a system design; accuracy varies from ±20% to ±5% of actual cost Costs can be updated using the unit method and cost indexes, where time differences are considered (inflation over time) The factor method estimates total plant costs, including indirect costs Indirect costs comprise a large percentage of product and service costs Traditional indirect cost allocation use bases such as direct labor hours, costs, and direct materials The ABC method of indirect cost allocation uses cost drivers to allocate to cost centers; it is better for understanding and analyzing cost accumulation Unethical practices in cost estimation result from personal financial motives, deception, financial pre-arrangements. Avoid deceptive acts 15-26 © 2012 by McGraw-Hill All Rights Reserved 278

LEARNING OUTCOMES Understand basic terms of asset depreciation
Apply straight line method of depreciation Apply DB and DDB methods of depreciation; switch between DDB and SL methods Apply MACRS method of depreciation Select asset recovery period for MACRS Explain depletion and apply cost depletion & percentage depletion methods 16-280 © 2012 by McGraw-Hill All Rights Reserved 280

Depreciation Terminology
Definition: Book (noncash) method to represent decrease in value of a tangible asset over time Two types: book depreciation and tax depreciation Book depreciation: used for internal accounting to track value of assets Tax depreciation: used to determine taxes due based on tax laws In USA only, tax depreciation must be calculated using MACRS; book depreciation can be calculated using any method 16-281 © 2012 by McGraw-Hill All Rights Reserved 281

Common Depreciation Terms
First cost P or unadjusted basis B: Total installed cost of asset Book value BVt: Remaining undepreciated capital investment in year t Recovery period n: Depreciable life of asset in years Market value MV: Amount realizable if asset were sold on open market Salvage value S: Estimated trade-in or MV at end of asset’s useful life Depreciation rate dt: Fraction of first cost or basis removed each year t Personal property: Possessions of company used to conduct business Real property: Real estate and all improvements (land is not depreciable) Half-year convention: Assumes assets are placed in service in midyear 16-282 © 2012 by McGraw-Hill All Rights Reserved 282

Straight Line Depreciation
Book value decreases linearly with time B - S Dt = Where: Dt = annual depreciation charge t = year B = first cost or unadjusted basis S = salvage value n = recovery period n BVt = B - tDt Where: BVt = book value after t years SL depreciation rate is constant for each year: d = dt = 1/n 16-283 © 2012 by McGraw-Hill All Rights Reserved

Example: SL Depreciation
An argon gas processor has a first cost of $20,000 with a $5,000 salvage value after 5 years. Find (a) D3 and (b) BV3 for year three. (c) Plot book value vs. time. Solution: (a ) D3 = (B – S)/n = (20,000 – 5,000)/5 = $3,000 (c) Plot BV vs. time BVt 20,000 (b) BV3 = B – tDt = 20,000 – 3(3,000) = $11,000 11,000 5,000 3 5 Year, t 16-284 © 2012 by McGraw-Hill All Rights Reserved

Declining Balance (DB) and Double Declining Balance (DDB) Depreciation
Determined by multiplying BV at beginning of year by fixed percentage d Max rate for d is twice straight line rate, i.e., d ≤ 2/n Cannot depreciate below salvage value Depreciation for year t is obtained by either relation: Dt = dB(1 – d)t-1 = dBVt-1 Where: Dt = depreciation for year t d = uniform depreciation rate (2/n for DDB) B = first cost or unadjusted basis BVt -1 = book value at end of previous year Book value for year t is given by: BVt = B(1 – d)t 16-285 © 2012 by McGraw-Hill All Rights Reserved

Example: Double Declining Balance
A depreciable construction truck has a first cost of $20,000 with a $4,000 salvage value after 5 years. Find the (a) depreciation, and (b) book value after 3 years using DDB depreciation. Solution: (a) d = 2/n = 2/5 = 0.4 D3 = dB(1 – d)t-1 = 0.4(20,000)(1 – 0.40)3-1 = $2880 (b) BV3 = B(1 – d)t = 20,000(1 – 0.4)3 = $4320 16-286 © 2012 by McGraw-Hill All Rights Reserved 286

Spreadsheet Functions for Depreciation
Straight line function: SLN(B,S,n) Declining balance function: DB(B,S,n,t) Double declining balance function: DDB(B,S,n,t,d) Note: It is better to use the DDB function for DB and DDB depreciation. DDB function checks for BV < S and is more accurate than the DB function. 16-287 © 2012 by McGraw-Hill All Rights Reserved 287

Switching Between Depreciation Methods
Switch between methods to maximize PW of depreciation PWD = ∑ Dt (P/F,i%,t) A switch from DDB to SL in latter part of life is usually better Can switch only one time during recovery period t = n t = 1 Procedure to switch from DDB to SL: Each year t compute DDB and SL depreciation using the relations DDDB = d(BVt-1) and DSL = BVt-1 / (n-t+1) 2) Select larger depreciation amount, i.e., Dt = max[DDDB, DSL] 3) If required, calculate PWD Alternatively, use spreadsheet function VDB(B,S,n,start_t,end_t) to determine Dt 16-288 © 2012 by McGraw-Hill All Rights Reserved

MACRS Depreciation Dt = dtB BVt = B - ∑Dj
Required method to use for tax depreciation in USA only Originally developed to offer accelerated depreciation for economic growth Where: Dt = depreciation charge for year t B = first cost or unadjusted basis dt = depreciation rate for year t (decimal) Dt = dtB Get value for dt from IRS table for MACRS rates j = t BVt = B - ∑Dj Where: Dj = depreciation in year j ∑ Dj = all depreciation through year t j = 1 16-289 © 2012 by McGraw-Hill All Rights Reserved 289

MACRS Depreciation Always depreciates to zero; no salvage value considered Incorporates switching from DDB to SL depreciation Standardized recovery periods (n) are tabulated MACRS recovery time is always n+1 years; half-year convention assumes purchase in midyear No special spreadsheet function; can arrange VDB function to display MACRS depreciation each year 16-290 © 2012 by McGraw-Hill All Rights Reserved 290

Example: MACRS Depreciation
A finishing machine has a first cost of $20,000 with a $5,000 salvage value after 5 years. Using MACRS, find (a) D and (b) BV for year 3. Solution: (a) From table, d3 = 19.20 D3 = 20,000(0.1920) = $3,840 (b) BV3 = 20, ,000( ) = $5,760 Note: Salvage value S = $5,000 is not used by MACRS and BV6 = 0 16-291 © 2012 by McGraw-Hill All Rights Reserved 291

MACRS Recovery Period Recovery period (n) is function of property class Two systems for determining recovery period: general depreciation system (GDS) – fastest write-off allowed alternative depreciation system (ADS) – longer recovery; uses SL IRS publication 946 gives n values for an asset. For example: MACRS n value Asset description GDS ADS range Special manufacturing devices, racehorses, tractors Computers, oil drilling equipment, autos, trucks, buses Office furniture, railroad car, property not in another class – 15 Nonresidential real property (not land itself) 16-292 © 2012 by McGraw-Hill All Rights Reserved

Unit-of-Production (UOP) Depreciation
Depreciation based on usage of equipment, not time Depreciation for year t obtained by relation Dt = (B – S) Example: A new mixer is expected to process 4 million yd3 of concrete over 10-year life time. Determine depreciation for year 1 when 400,000 yd3 is processed. Cost of mixer was $175,000 with no salvage expected. Solution: D1 = (175,000 – 0) = $17,500 actual usage for year t expected total lifetime usage 400,000 4,000,000 16-293 © 2012 by McGraw-Hill All Rights Reserved

Depletion Methods Depletion: book (noncash) method to represent decreasing value of natural resources Two methods: cost depletion (CD) and percentage depletion (PD) Cost depletion: Based on level of activity to remove a natural resource Calculation: Multiply factor CDt by amount of resource removed Where: CDt = first cost / resource capacity Total depletion can not exceed first cost of the resource Percentage depletion: Based on gross income (GI) from resource Calculation: Multiply GI by standardized rate (%) from table Annual depletion can not exceed 50% of company’s taxable income (TI) 16-294 © 2012 by McGraw-Hill All Rights Reserved

Example: Cost and Percentage Depletion
A mine purchased for $3.5 million has a total expected yield of one million ounces of silver. Determine the depletion charge in year 4 when 300,000 ounces are mined and sold for $30 per ounce using (a) cost depletion, and (b) percentage depletion. (c) Which is larger for year 4? Solution: Let depletion amounts equal CDA4 and PDA4 (a) Factor, CD4 = 3,500,000/ 1,000,000 = $3.50 per ounce CDA4 = 3.50(300,000) = $1,050,000 (b) Percentage depletion rate for silver mines is 0.15 PDA4 = (0.15)(300,000)(30) = $1,350,000 (c) Claim percentage depletion amount, provided it is ≤ 50% of TI 16-295 © 2012 by McGraw-Hill All Rights Reserved

Two types for depreciation: tax and book Classical methods are straight line and declining balance In USA only, MACRS method is required for tax depreciation Determine MACRS recovery period using either GDS or ADS Switching between methods is allowed; MACRS switches automatically from DDB to SL to maximize write-off Depletion (instead of depreciation) used for natural resources Two methods of depletion: cost (amount resource removed × CDt factor) and percentage (gross income × tabulated %) 16-296 © 2012 by McGraw-Hill All Rights Reserved 296

Executive Summary Version After-Tax Economic Analysis
Developed By: Dr. Don Smith, P.E. Department of Industrial Engineering Texas A&M University College Station, Texas Executive Summary Version Chapter 17 After-Tax Economic Analysis At this point: 1. Introduce yourself - your students are likely to want to know something about your qualifications and interests - overall, where you are coming from. 2. Have students introduce themselves. Ask why they are taking this class. If you are fortunate enough to have a Polaroid camera, take pictures of each student for later posting on a class “board” so both they and you get to know each other. 3. Discuss both choice of textbook and development of syllabus. 4. If you are expecting students to work in teams, at east introduce the choice of team members. If at all possible, have students participate in a team building or team study exercise. It works wonders. Most student have been told to work in teams in prior classes, but have never examined exactly what a team is and how it works. One hour spent in a team building/examination exercise saves many hours and avoids many problems later on.

LEARNING OBJECTIVES Terminology and rates CFBT and CFAT
Taxes and depreciation Depreciation recapture and capital gains After-tax analysis Spreadsheets After-tax replacement Value-added analysis Taxes outside the United States

Sct 17.1 Income Tax Terminology and Relations for Corporations (and Individuals)
Gross Income Total income for the tax year from all revenue producing function of the enterprise. Sales revenues, Fees, Rent, Royalties, Sale of assets Income Tax The total amount of money transferred from the enterprise to the various taxing agencies for a given tax year. Federal corporate taxes are normally paid at the end of every quarter and a final adjusting payment is submitted with the tax return at the end of the fiscal year. This tax is based upon the income producing power of the firm.

Terms - continued TI = GI – E – D Operating Expenses Taxable Income
All legally recognized costs associated with doing business for the tax year. Real cash flows, Tax deductible for corporations: Wages and salaries Utilities Other taxes Material expenses etc. Taxable Income Calculated amount of money for a specified time period from which the tax liability is determined. Calculated as: TI = Gross Income – expenses – depreciation TI = GI – E – D

Terms - continued Net Profit After Tax (NPAT) Tax rate T
Amount of money remaining each year when income taxes are subtracted from taxable income. NPAT = TI – {(TI)(T)} = (TI)(1-T) Equivalent tax rate Te combines federal and local rates: Te = state rate + (1 state rate)(federal rate) Tax rate T A percentage or decimal equivalent of TI. For Federal corporate income tax T is represented by a series of tax rates. The applicable tax rate depends upon the total amount of TI. Taxes owed equals: Taxes = (taxable income) x (applicable rate) = (TI)(T).

U.S. Individual Federal Tax Rates (2003)
Taxable Income, $ Tax Rate (1) Filing Single (2) Filing Married and Jointly (3) 0.10 0-7,000 0-14,000 0.15 7,001-28,400 14,001-56,800 0.25 28,401-68,800 56, ,650 0.28 68, ,500 114, ,700 0.33 143,501 – 311,950 174, ,950 0.35 Over 311,950

Basic Tax Equations - Individual
Gross Income GI = salaries + wages + interest and dividends + other income Taxable Income TI = GI – personal exemptions – standard or itemized deductions Tax T = (taxable income)(applicable tax rate)

Sct 17.2 Before-Tax and After-Tax Cash Flow
NCF = cash inflows – cash outflows Cash Flow before Tax (CFBT) CFBT = gross income – expenses – initial investment + salvage value = GI – E – P + S Cash Flow After Tax (CFAT) CFAT = CFBT – taxes Add Depreciation CFAT = GI – E – P + S – (GI – E – D)(Te) An evaluation format See Table 17 – 3 and Example 17.3 for a computational format

Sct 17.3 Effect on Taxes of Different Depreciation Methods and Recovery Periods
Criteria used to compare different depreciation methods – compute --- Objective – Minimize the PW of future taxes paid owing to a given depreciation method The total taxes paid are equal for all depreciation models The PW of taxes paid is less for accelerated depreciation methods Shorter depreciation periods result in lower PW of future taxes paid over longer time periods See Examples 17.4 and 17.5

Sct 17.4 Depreciation Recapture and Capital Gains (Losses) for Corporations
Capital gain (CG) CG = selling price – first cost CG = SP – P Depreciation Recapture (DR) DR = selling priceyear t – book valuetime of sale DR – SP – BVt Capital Loss (CL) CL = book value – selling price CL = BVt - SP

DR Summary - Outcomes CG SP1 plus SP2 DR SP3 Zero, $0 First Cost P DR
For and AT study the tax effect is: If SP at time of sale is.. The CG, DR or CL is: CG SP1 CG: Taxed at Te after any CL offset First Cost P plus SP2 DR DR: taxed at Te DR Book Value BVt SP3 CL CL: Can only offset CG Zero, $0 DR occurs when a productive asset is sold for more than its current BV

General TI Equation – for Corporations
The basic TI equation is: TI = GI – E – D + DR + CG – CL The basic spreadsheet format is Year GI E P DEPR BV TI Taxes 1 2 … n See Figure 17-4 and associated Example 17.6

Sct 17.5 After-Tax PW, AW, and ROR Evaluation
One project Apply PW or AW = 0 Accept the project if after-tax MARR is met or exceeded Two or More Projects Select the alternative with the largest PW or AW value Assume discounting occurs at the firm’s after-tax MARR rate See Example 17.7

ROR Analysis The Before-tax ROR For ROR analysis -- review Chapter 8
Selection rules Apply incremental ROR Select the one alternative that requires the largest initial investment provided the incremental investment is justified relative to another justified alternative

Sct 17.6 Spreadsheet Applications – After-Tax Incremental ROR Analysis
Two spreadsheet examples for after-tax ROR are presented Examples and 17.11

Example 17.10 – Comparison of S and B
The interest rate at which the two alternatives are economically equal (6.36%)

Sct 17.7 After-Tax Replacement Study
After-tax treatment of a replacement problem will generate a different data set than a before-tax replacement analysis Year of replacement Could have DR, CG, CL situations After-tax replacement considers Depreciation Operating expenses See Examples and Table 17-6 for the formats After-tax replacement analysis is more involved An after-tax analysis could reverse a before-tax analysis on some problems

Format for After-Tax Replacement
Analysis with a 5-year straight line depreciation method applied

Warnings Always beware of using the ROR method for selecting from among alternatives. DO NOT use computed ROR! This means the ROR computed on each separate investment alternative. Rather, form the incremental cash flow and make a determination on the i* value. Need to design a spreadsheet model to effectively evaluate.

Sct 17.8 After-Tax Value Added Analysis
Value added is a term to indicate that a product or a service: Has added value to the consumer or buyer. Popular concept in Europe; Value-added taxes are imposed in Europe on certain products and paid to the government. Rule: The decision concerning an economic alternative will be the same for a value added analysis and a CFAT analysis. Because, the AW of economic value added estimates is the same as the AW and CFAT estimates!

Value Added To start, apply Eq. 17.3:
NPAT = Taxable Income – taxes NPAT = (TI)(1-T) Value added or Economic Value Added ( EVA) is: The amount of NPAT remaining after removing the cost of invested capital during the time period in question. EVA indicates the project’s contribution to the net profit of the corporation after taxes have been paid. The cost of invested capital is normally the firm’s after-tax required MARR value. One multiplies the after-tax MARR by the current level of capital (investment). Charge interest on the unrecovered capital investment at the after-tax MARR rate.

Value Added Recall, firms often have two sets of books relating to depreciation: One for tax purposes and, One for internal management use. (book depreciation). For EVA, book depreciation is more often used. More closely represent the true rate of usage of the assets in question. The annual EVA is the NPAT remaining on the books after removing the cost of invested capital during the year. EVA indicates the project’s contribution to the net profit after taxes EVA = NPAT – cost of invested capital = NPAT – (after-tax interest book rate)(book value in year t-1) EVA = TI(1-Te) – (i)(BVt-1)

Sct 17.9 After-Tax Analysis for International Projects - Canada
Depreciation – DB or SL with ½ yr convention Capital Cost Allowance (CCA) Standard recovery rates as in US Expenses – deductible in calculating TI Expenses related to capital investment are not deductible and are handles under CCA

Mexico SL method with inflation indexing
Assets generally classified with annual recovery rates that vary 5% for machinery to 100% for environmental assets Profit tax with most expenses deductible Tax of Net Assets (TNA) of 1.8% of the average value of assets locating in Mexico

Japan Depreciation – SL or DB with 95% of the unadjusted basis used
Class and life – 4 to 24 years by law; up to 50 years for certain structures Expenses are deductible

Chapter Summary After-tax (AT) analysis is a more thorough approach in the evaluation of industrial projects In some cases, AT analysis will show a reversal in before-tax decision, but not always Tax rates in the US are graduated – higher taxable incomes pay higher taxes Operating expenses are tax deductible Depreciation amounts represent non-cash flows -- but do generate tax savings

Summary - continued In the US, the MACRS method is required on federal corporate tax returns and recovery lives are mandated by law and by class In replacement analysis, the impact of depreciation recapture, capital gain or loss is incorporated into the analysis For AT replacement, the decision to replace will generally follow the before-tax analysis AT replacement will show substantially different CFAT than the before-tax analysis

Formalized Sensitivity Analysis and Expected Value Decisions
Developed By: Dr. Don Smith, P.E. Department of Industrial Engineering Texas A&M University College Station, Texas Executive Summary Version Chapter 18 Formalized Sensitivity Analysis and Expected Value Decisions At this point: 1. Introduce yourself - your students are likely to want to know something about your qualifications and interests - overall, where you are coming from. 2. Have students introduce themselves. Ask why they are taking this class. If you are fortunate enough to have a Polaroid camera, take pictures of each student for later posting on a class “board” so both they and you get to know each other. 3. Discuss both choice of textbook and development of syllabus. 4. If you are expecting students to work in teams, at east introduce the choice of team members. If at all possible, have students participate in a team building or team study exercise. It works wonders. Most student have been told to work in teams in prior classes, but have never examined exactly what a team is and how it works. One hour spent in a team building/examination exercise saves many hours and avoids many problems later on.

LEARNING OBJECTIVES Sensitivity to variation Three estimates
Expected value Expected value of cash flows Decision trees

Sct 18.1 Determining Sensitivity to Parameter Variation
A parameter is a variable or factor for which an estimate or stated value is required to conduct the analysis at hand. Examples: P, F, A; i, n; Future costs, salvages, etc. Sensitivity analysis Seeks to determine what parameters matter most in an economic analysis

Sensitivity Sensitivity is concerned with variability
Variance associated with input parameters impact the output variable the most The MARR as a parameter Interest rates and other interest factors tend to be more stable from project to project The analyst can limit the range over which these type of parameters vary

Visualizing the Impact of Parameters
Plot the PW, AW, or ROR vs. input parameters Steps Pre-select the desired input parameters Select the probable range and increment of variation for each parameter Select the measure of worth Compute the results for each parameter Graphically display the results by plotting the parameter vs. the measure of worth

Example $PW Low High Parameter
Assume a parameter of interest (show on the X-axis). Vary that parameter from some low value to an assumed high value. Plot the resultant values on the Y-axis.

Sensitivity of PW: Example 18.1
One can better visualize the relationship on PW vs. a selected range of discount rates. As the discount rate increases from 10% to 25% the resultant PW is substantially lowered at a rather accelerated rate.

Sensitivity of Several Parameters
For several parameters with one alternative Graph the percentage change for each parameter vs. the measure of worth One will plot the percent deviation from the most likely estimate on the x-axis This type of plot results in what is termed a spider plot

Spider Plot: Figure 18-3 Those plots with positive slope have a positive correlation on the output variable: Those plots with negative slope have a inverse relationship with the output variable

Sct 18.2 Formalized Sensitivity Analysis Using Three Estimates
Given an input parameter of interest Provide three estimates for that parameter A pessimistic estimate, P A most likely estimate, ML An optimistic estimate, O Note: This approach comes from PERT/CPM analysis and is based upon the beta distribution

Three Estimates: Example 18.3
Three alternatives (A, B, C) with 4 Parameters First cost, salvage value, AOC, and life For each parameter we formulate Parameter P pessimistic estimate ML most likely estimate O optimistic estimate See Table 18.2 and observe the dominance by alternative B (cost problem so lower cost is preferred)

Example 18.3 -- Setup Alt A. Alt. C Strategy First Cost SV AOC Life P
-20,000 -11,000 3 ML -9,000 5 O -5,000 8 Alt. B -15,000 500 -4,000 2 1,000 -3,500 4 2,000 -2,000 7 Alt. C -30,000 3,000 -8,000 -7,000 9

Plot for Example 18.3 Observe the dominance by alternative B over A and C. A plot like this clearly shows the relationships.

Sct 18.3 Economic Variability and The Expected Value
Long-run average based upon occurrence and probability of occurrence Definition of Expected Value Xi = value of the variable X for i from 1 to m different values P(Xi) = probability that a specific value of X will occur Subject to: See Example 18.4

Sct 18.4 Expected Value Computations for Alternatives
Two applications for use of Expected Value (EV) Prepare information for a more complete analysis of an economic analysis To evaluate expected utility of a fully formulated alternative Examples 18.5 and 18.6 illustrate this concept

Sect 18.5 Staged Evaluation of Alternatives Using Decision Trees
Some problems involve staged decisions that occur in sequence Define the staged decisions and assign the respective probabilities to the various defined outcomes Useful tool for modeling such a process involves Decision Trees The objective: Make Risk More Explicit

Decision Tree Attributes
More than one stage of alternative selection Selection of an alternative at one stage that leads to another stage Expected results from a decision at each stage Probability estimates for each outcome Estimates of economic value (cost or revenue) for each outcome Measure of worth as the selection criterion, e.g. E(PW)

Examples Decision Node Probability Node Marginal Probabilities
Outcomes

Solving Decision Trees
Once designed, Decision Tree is solved by folding back the tree First, define all of the decision and the decision points Define the various outcomes given the decision Assign the probabilities to the mutually exclusive outcomes emanating from each decision node See Example 18-8 for a comprehensive analysis illustrating the decision tree approach

Important Points for Decision Trees
Estimate the probabilities associated with each outcome These probabilities must sum to 1 for each set of outcomes (branches) that are possible from a given decision Required economic information for each decision alternative are investments and estimated cash flows

Starting Out Assuming the tree logic has been defined…
Start at the top right of the tree Determine the PW for each outcome branch applying the time value of money Calculate the expected value for each decision alternative as: 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

Example 18.8

Chapter Summary The emphasis is on sensitivity to variation in one or more parameters using a specific measure of worth. When two alternatives are compared compute and graph the measure of worth for different values of the parameter to determine when each alternative is better. When several parameters are expected to vary over a predictable range, the measure of worth is plotted and calculated using three estimates for a parameter: Most likely Pessimistic Optimistic

Summary - continued The combination of parameter and probability estimates results in the expected value relations E(X) = (X)P(X) This expression is also used to calculate E(revenue), E(cost), E(cash flow) E(PW), and E(i) for the entire cash flow sequence of an alternative. E9X0 is a measure of central tendency

Summary - continued Decision trees are used to make a series of alternative selections. This is a way to explicitly take risk into account. It is necessary to make several types of estimates for a decision tree: Outcomes for each possible decision, cash flows, and probabilities. Expected value computations are coupled with those for the measure of worth to “solve” the tree structure. Assist is identifying the best alternative stage-by-stage.

Chapter 19 More on Variation and Decision Making Under Risk
Developed By: Dr. Don Smith, P.E. Department of Industrial Engineering Texas A&M University College Station, Texas Executive Summary Version Chapter 19 More on Variation and Decision Making Under Risk At this point: 1. Introduce yourself - your students are likely to want to know something about your qualifications and interests - overall, where you are coming from. 2. Have students introduce themselves. Ask why they are taking this class. If you are fortunate enough to have a Polaroid camera, take pictures of each student for later posting on a class “board” so both they and you get to know each other. 3. Discuss both choice of textbook and development of syllabus. 4. If you are expecting students to work in teams, at east introduce the choice of team members. If at all possible, have students participate in a team building or team study exercise. It works wonders. Most student have been told to work in teams in prior classes, but have never examined exactly what a team is and how it works. One hour spent in a team building/examination exercise saves many hours and avoids many problems later on.

LEARNING OBJECTIVES Certainty and risk Variables and distributions
Random samples Average and dispersion Monte Carlo simulation

Sct 19.1 Interpretation of Certainty, Risk, and Uncertainty
Certainty – Everything know for sure; not present in the real world of estimation, but can be ‘assumed’ Risk – a decision making situation where all of the outcomes are know and the associated probabilities are defined Uncertainty – One has two or more observable values but the probabilities associated with the values are unknown Observable values – states of nature See Example 19.1 – about risk

Types of Decision Making
Decision Making under Certainty Process of making a decision where all of the input parameters are known or assumed to be known Outcomes – known Termed a deterministic analysis Parameters are estimated with certainty Decision Making under Risk Inputs are viewed as uncertain, and element of chance is considered Variation is present and must be accounted for Probabilities are assigned or estimated Involves the notion of random variables

Two Ways to Consider Risk in Decision Making
Expected Value (EV) analysis Applies the notion of expected value (Chapter 18) Calculation of EV of a given outcome Selection of the outcome with the most advantageous outcome Simulation Analysis Form of generating artificial data from assumed probability distributions Relies on the use of random variables and the laws associated with the algebra of random variables

Sct 19.2 Elements Important to Decision Making Under Risk
The concept of a random variable A decision rule that assigns an outcome to a sample space Discrete variable or Continuous variable Discrete variable – finite number of outcomes possible Continuous variable – infinite number of outcomes Probability Number between 0 and 1 Expresses the “chance” in decimal form that a random variable will take on any specific value

Types of Random Variables
Continuous Discrete

Distributions - Continuous Variables
Probability Distribution (pdf) A function that describes how probability is distributed over the different values of a variable P(Xi) = probability that X = Xi Cumulative Distribution (cdf) Accumulation of probability over all values of a variable up to and including a specified value F(Xi) = sum of all probabilities through the value Xi = P(X  Xi)

Three Common Random Variables
Uniform – equally likely outcomes Triangular Normal Study Example 19.3

Discrete Density and Cumulative Example
pdf cdf

Sct 19.3 Random Samples Random Sample Can sample from:
A random sample of size n is the selection in a random fashion with an assumed or known probability distribution such that the values of the variable have the same chance of occurring in the sample as they appear in the population Basis for Monte Carlo Simulation Can sample from: Discrete distributions … or Continuous distributions

Sampling from a Continuous Distribution
Form the cumulative distribution in closed form from the pdf Generate a uniform random number on the interval {0 – 1}, called U(0,1) Locate U(0,1) point on y-axis Map across to intersect the cdf function Map down to read the outcome (variable value) on x-axis

Sct 19.4 Expected Value and Standard Deviation
Two important parameters of a given random variable: Mean -  Measure of central tendency Standard Deviation -  Measure of variability or spread Two Concepts to work within Population Sample from a population

Population vs Sample Sample Population  - population mean
2 - population variance  - population standard deviation Often sample from a population in order to make estimates Sample mean Sample variance Sample standard deviation These values, properly sampled, attempt to estimate their population counterparts

Important Relationships
Population Mean  Distribution E(x) = Sample Measure of the central tendency of the population If one samples from a population the hope is that sample mean is an unbiased estimator of the true, but unknown, population mean

Variance and Standard Deviation
Notes relating to variance and standard deviation properties Illustration of variances for discrete and continuous distributions

Population vs. Sample Variance of a population
Standard deviation of a population: Variance of a sample Standard deviation of a sample S is termed an unbiased estimator of the population standard deviation

Combining the Average and Standard Deviation
Determine the percentage or fraction of the sample that is within ±1, ±2, ±3 standard deviations of the sample mean In terms of probability… Virtually all of the sample values will fall within the ±3s range of the sample mean See example 19.6

Continuous Random Variables
Expected value: Variance: For uniform pdf in Example 19.3: R represents the defined range of the variable in question.

Sct 19.5 Monte Carlo Sampling and Simulation Analysis
Simulation involves the generation of artificial data from a modeled system Monte Carlo Sampling The generation of samples of size n for selected parameters of formulated alternatives The sampled parameters are expected to vary according to a stated probability distribution (assumed)

Key Assumption - Independence
For a given problem: All parameters are assumed to be independent One variable’s distribution in no way impacts any other variable’s distribution Termed: Property of independent random variables The modeling approach follows a 7-step process (page 678)

Simulation Monte Carlo Sampling is a traditional approach (method) for generating pseudo-random numbers (RN) to sample from a prescribed probability distribution. Pseudo-random refers to the fact that a digital computer can generate approximately random numbers due to fixed word size and round off problems.

Sampling Process Requires: Why require the cdf?
The cdf of the assumed pdf; A uniform random number generator; Application of the inverse transform approach. Why require the cdf? The y-axis of a cdf is scaled from 0 to 1. That is the same as the range of U(0,1). Facilitates mapping a RN to achieve the outcome value on the x-axis. The U(0,1) selects a X-value from the cdf.

x-axis: Outcome values
The Need for the cdf x-axis: Outcome values Cumulative probability on the y-axis Generate a U(0,1) random number: Locate that value on the y-axis: Map across to the cdf then map down to the x-axis to obtain the outcome

Summary of the Modeling Steps
Formulate the economic analysis: The alternatives – if more than one; Define which parameters are “constants” and which are to be viewed as random variables. For the random variables, assign the appropriate pdf: Discrete and or continuous. Apply Monte Carlo sampling – a sample size of “n” where it is suggested that n = 30. Compute the measure of worth (PW, AW, . . ) Evaluate and draw conclusions.

Examples to Study Examples 19.7 and 19.8 offer detailed analyses of a simulated economic analysis Also, refer to the Additional Examples at the end of the chapter Example – applying the normal distribution

Chapter Summary To perform decision making under risk implies that some parameters of an engineering alternative are treated as random variables. Assumptions about the shape of the variable's probability distribution are used to explain how the estimates of parameter values may vary. Additionally, measures such as the expected value and standard deviation describe the characteristic shape of the distribution.

Summary - continued In this chapter, we learned several of the simple, but useful, discrete and continuous population distributions used in engineering economy -uniform and triangular - as well as specifying our own distribution or assuming the normal distribution. It is important to note that a sound background in applied statistics is vital to the complete understanding of the simulation process

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