Foundations of Engineering Economy

Foundations of Engineering Economy
Basics of Engineering Economy Chapter 1 Foundations of Engineering Economy 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. © 2008 by McGraw-Hill All Rights Reserved

Understand the fundamental concepts of engineering economy
Chapter 1 - Foundations PURPOSE Understand the fundamental concepts of engineering economy TOPICS Definition and study approach Interest rate, ROR, and MARR Equivalence Interest – simple and compound Cash flow diagrams Rules of 72 and 100 Spreadsheet introduction © 2008 by McGraw-Hill All Rights Reserved

Sec 1. 1 – Definition of Engineering Economy Sec 1
Sec 1.1 – Definition of Engineering Economy Sec 1.2 – Elements of a Study DEFINITION: Techniques that simplify comparison of alternatives on an economic basis Most project decisions consider additional factors – safety, environmental, political, public acceptance, etc. Fundamental terminology: Alternative -- stand-alone solution Cash flows -- estimated inflows (revenues) and outflows (costs) for an alternative Evaluation criteria -- Basis used to select ‘best’ alternative; usually money (currency of the country) Time value of money -- Change in amount of money over time (Most important concept in Eng. Econ.) © 2008 by McGraw-Hill All Rights Reserved

Sec 1.3 - Interest Rate, ROR, MARR
Interest is a manifestation of time value of money Calculated as difference between an ending amount and a beginning amount of money Interest = end amount – original amount Interest rate is interest over specified time period based on original amount Interest rate (%) = Interest rate and rate of return (ROR) have same numeric value, but different interpretations © 2008 by McGraw-Hill All Rights Reserved

Sec 1.3 - Interest Rate and ROR Interpretations
Borrower’s perspective Take loan of $5,000 for one year; repay $5,350 Interest paid = $350 Interest rate = 350/5,000 = 7% INTEREST RATE Investor’s perspective Invest (or lend) $5,000 for one year; receive $5,350 Interest earned = $350 Rate of return = 350/5,000 = 7% RATE OF RETURN © 2008 by McGraw-Hill All Rights Reserved

Sec ROR and MARR Cost of capital (COC) – interest rate paid for funds to finance projects MARR – Minimum ROR needed for an alternative to be justified and economically acceptable. MARR ≥ COC. If COC = 5% and 6% must be realized, MARR = 11% Always, for acceptable projects ROR ≥ MARR > COC ROR © 2008 by McGraw-Hill All Rights Reserved

Sec Equivalence Different sums of money at different times may be equal in economic value Interest rate = 6% per year $106 one year from now $94.34 last year $100 now Interpretation: $94.34 last year, $100 now, and $106 one year from now are equivalent only at an interest rate of 6% per year © 2008 by McGraw-Hill All Rights Reserved

Sec 1.5 – Simple and Compound Interest
Simple interest is always based on the original amount, which is also called the principal Interest per period = (principal)(interest rate) Total interest = (principal)(n periods)(interest rate) Example: Invest $250,000 in a bond at 5% per year simple Interest each year = 250,000(0.05) = $12,500 Interest over 3 years = 250,000(3)(0.05) = $37,500 © 2008 by McGraw-Hill All Rights Reserved

Sec 1.5 – Simple and Compound Interest
Compound interest is based on the principal plus all accrued interest Interest per period = (principal + accrued interest)(interest rate) Total interest = (principal)(1+interest rate)n periods - principal Example: Invest $250,000 at 5% per year compounded Interest, year 1 = 250,000(0.05) = $12,500 Interest, year 2 = 262,500(0.05) = $13,125 Interest, year 3 = 275,625(0.05) = $13,781 Interest over 3 years = 250,000(1.05)3 – 250,000 = $39,406 © 2008 by McGraw-Hill All Rights Reserved

Sct 1.6 - Terminology and Symbols
t = time index in periods; years, months, etc. P = present sum of money at time t = 0; $ F = sum of money at a future time t; $ A = series of equal, end-of-period cash flows; currency per period, $ per year n = total number of periods; years, months i = compound interest rate or rate of return; % per year © 2008 by McGraw-Hill All Rights Reserved

Sct 1.6 - Terminology and Symbols
Example: Borrow $5,000 today and repay annually for 10 years starting next year at 5% per year compounded. Identify all symbols. Given: P = $5,000 Find: A = ? per year i = 5% per year n = 10 years t = year 1, 2, …, 10 (F not used here) © 2008 by McGraw-Hill All Rights Reserved

Sec 1.7 – Cash Flow Estimates
Cash inflow – receipt, revenue, income, saving Cash outflows – cost, expense, disbursement, loss Net cash flow (NCF) = inflow – outflow End-of-period convention: all cash flows and NCF occur at the end of an interest period © 2008 by McGraw-Hill All Rights Reserved

Sec 1.7 – Cash Flow Diagrams
Year 1 Year 5 Typical time scale or 5 years Time, t + Cash flow P = ? Find P in year 0, given 3 cash flows - Cash flow © 2008 by McGraw-Hill All Rights Reserved

Sec 1.7 – Cash Flow Diagrams
Example: Find an amount to deposit 2 years from now so that $4,000 per year can be available for 5 years starting 3 years from now. Assume i = 15.5% per year © 2008 by McGraw-Hill All Rights Reserved

Sec 1.8 – Rule of 72 ( and 100) Approximate n = 72 / i
Estimates # of years (n) for an amount to double (2X) at a stated compound interest rate e.g., at i = 10%, $1,000 doubles to $2,000 in ~7.2 years Solution for i estimates compound rate to double in n years Approximate i = 72 / n For simple interest, doubling time is exact, using rule of 100 n = 100/i or i = 100/n $1,000 doubles in 10 years at 10% simple interest © 2008 by McGraw-Hill All Rights Reserved

Sec 1.9 – Introduction to Spreadsheet Functions
To display Excel Function Present value, P = PV(i%,n,A,F) Future value, F = FV(i%,n,A,P) Annual amount, A = PMT(i%,n,P,F) # of periods, n = NPER(i%,A,P,F) Compound rate, i = RATE(n,A,P,F) i for input series = IRR(first_cell:last_cell) P for input series = NPV(i%,second_cell: last_cell)+first_cell © 2008 by McGraw-Hill All Rights Reserved

Lecture slides to accompany Leland Blank and Anthony Tarquin
Basics of Engineering Economy by Leland Blank and Anthony Tarquin Chapter 3 Nominal and Effective Interest Rates 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. Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Chapter 3 – Nominal & Effective Interest
TOPICS Recognize nominal and effective rates Effective interest rates Payment period (PP) and compounding period (CP) Single amounts with PP ≥ CP Series with PP ≥ CP Single and series with PP < CP Spreadsheet use PURPOSE Perform calculations for interest rates and cash flows that occur on a time basis other than yearly Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.1 – Nominal and Effective Rate Statements
Nominal rates Interest rate per time period without regard to compounding frequency Some nominal statements: 8% per year compounded monthly 2% per month compounded weekly 8% per year compounded quarterly 5% per quarter compounded monthly Effective rates Interest rate is compounded more frequently than once per year Some statements indicating an effective rate: 15% per year effective 8.3% per year compounded monthly 2% per month compounded monthly effective 1% per week compounded continuously Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.2 – Effective Interest Rate Formula
i = effective rate per some stated period, e.g., quarterly, annually r = nominal rate for same time period m = frequency of compounding per same time period Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.2 – Effective Interest Rate
Compounding frequency Period for effective i Time period for r m must equal Annual annual year 1 Semi-annual 2 Quarterly 4 Monthly 12 Daily 365 semi-annual 6 months 6 Weekly quarterly quarter Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Example: Find i per year, if m = 4 for quarterly compounding, and r = 12% per year Stated period for i is YEAR i = ( /4) = 12.55% Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.2 – Nominal and Effective Rates
r = rate/period × periods Example: Rate is 1.5% per month. Determine nominal rate per quarter, year, and over 2 years Qtr: r = 1.5 × 3 mth = 4.5% Year: r = 1.5 ×12 mth = 18% = 4.5 × 4 qtr = 18% 2 yrs: r =1.5 × 24 mth = 36% = 18 × 2 yrs = 36% Effective Example: Credit card rate is 1.5% per month compounded monthly. Determine effective rate per quarter and per year Period is quarter: r = 1.5 × 3 mth = 4.5% m = 3 i = ( /3)3 – 1 = 4.57% per quarter Period is year: r = 18% m = 12 i = ( /12)12 - 1) = 19.6% per year Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

As m → ∞, continuous compounding is approached effective i = (℮r – 1) Example: r = 14% per year compounded continuously i = (℮ ) = 15.03% per year Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.2 – Nominal and Effective Rates
Using Excel functions to find rates Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.3 – Payment Periods (PP) and Compounding Periods (CP)
PP – how often cash flows occur CP – how often interest in compounded If PP = CP, no problem concerning effective i rate Examples where effective i is involved: Monthly deposit, quarterly compounding (PP < CP) Semi-annual payment, monthly compounding (PP > CP) Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.3 – Payment Periods (PP) and Compounding Periods (CP)
Initial things to observe about cash flows Compare length of PP with CP PP = CP PP > CP PP < CP Determine types of cash flows present Only single amounts (P and F) Series (A, G, g) Determine correct effective i and n (same time unit on both) Remember: An effective i rate must be used in all factors Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.4 – Equivalence with Single Amounts
If only P and F cash flows are present, equivalence relations are P = F(P/F, effective i per period, # of periods) [1] F = P(F/P, effective i per period, # of periods) [2] Example: Find equivalent F in 10 years if P is $1000 now. Assume r = 12% per year compounded semi-annually. - PP = year and CP = 6 months; period is 6 months - Only single amount cash flows - Use relation [2] above to find F F = 1000(F/P, 6% semi-annually, 20 periods) = 1000(3.2071) = $3207 Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.5 – Equivalence with Series and PP ≥ CP
Count number of payments. This is n Determine effective i over same time period as n Use these i and n values in factors Example: $75 per month for 3 years at 12% per year compounded monthly PP = CP = month n = 36 months effective i = 1% per month Relation: F = A(F/A,1%,36) Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Count number of payments. This is n Determine effective i over same time period as n Use these i and n values in factors Example: $5000 per quarter for 6 years at 12% per year compounded monthly PP = quarter and CP = month → PP > CP n = 24 quarters i = 1% per month or 3% per quarter m = 3 CP per quarter effective i per quarter = ( /3)3 – 1 = 3.03% Relation: F = A(F/A,3.03%,24) Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

P = $3M First step: Find P for n = 10 annual payments Period is year CP = 6 months; PP = year; PP > CP Effective i per year = ( /2)2 – 1 = 8.16% Relation: P = 3M + 200,000(P/A,8.16%,10) = $4,332,400 (continued →) Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

P = $3M Second step: Find A for n = 20 semi-annual amounts Period is six months CP = 6 months; PP = 6 months; PP = CP Effective i per 6 months = 8%/2 = 4% Relation: A = 4,332,400(A/P,4%,20) = $318,778 Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.6 – Equivalence with Series and PP < CP
Example: deposits monthly (PP) with interest compounded semi-annually (CP) Result: PP < CP Usually, interest is not paid on interperiod deposits For equivalence computations: Cash flows are ‘moved’ to match CP time period Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.6 – Equivalence with Series and PP < CP
APPROACH NORMALLY TAKEN Move cash flows not at end of a compounding period: Deposits ( minus cash flows) - to end of period Withdrawals (plus cash flows) - to beginning of same period (which is the end of last period) Example (next slide): move monthly deposits to match quarterly compounding. Now, PP = CP = quarter Find P, F or A using effective i per quarter Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.6 – Equivalence with Series and PP < CP Moving cash flows turns top cash flow diagram into bottom Qtr Qtr Qtr 3 Qtr4 Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 3.7 – Spreadsheet Usage
Spreadsheet function format and structure: Fine effective rate: = EFFECT(nom r%, m) Nominal r is over same time period as effective i Find nominal rate: = NOMINAL(eff i%, m) Result of nominal is always per year Example: Deposits are planned as follows: $1000 now, $3000 after 4 years, $1500 after 6 years. Find F after 10 years. Interest is 12% per year compounded semiannually Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Basics of Engineering Economy
Lecture slides to accompany Basics of Engineering Economy by Leland Blank and Anthony Tarquin Chapter 4 Present Worth Analysis © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Chapter 4 – Present Worth Analysis
PURPOSE Identify types of alternatives; and compare alternatives using a present worth basis TOPICS Formulating alternatives Single and equal-life alternatives Different-life alternatives Capitalized cost alternative evaluation Independent alternatives Spreadsheet usage © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.1 – Formulating Alternatives
Types of alternatives Mutually exclusive (ME) - only one viable project can be accepted. Do-nothing (DN) alternative is selected if none are justified economically Independent - more than one project can be selected. DN is one of the projects Do-nothing – maintain status quo/current approach Types of cash flow estimates for an alternative Revenue – estimates include costs, revenues and (possibly) savings Cost – only cost estimates included; revenues assumed equal for all alternatives © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.1 – Formulating Alternatives
Much of the emphasis in professional engineering practice is on ME, cost alternatives. However, all tools in Eng Econ can be used to evaluate ME and independent alternatives that are revenue- or cost-based. Examples of both are included later. Notes: P value of cash flows is now called PW, or present worth P now represents first cost of an alternative © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.2 – PW of a Single Alternative
Example: MARR = 10% First cost, P = $-2500 Annual revenue, R = $2000 Annual cost, AOC = $-900 Salvage value, S = $200 Life, n = 5 years PW = P +S(P/F,10%,5) + (R-AOC)(P/A,10%,5) = (P/F,10%,5) + ( )(P/A,10%,5) = $1794 PW > 0; project is economically justified Single project analysis Calculate PW at stated MARR Criterion: If PW ≥ 0, project is economically justified © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.2 – Single Alternative Example – Bond Investment
Bond – like an IOU; issued by corporations and all levels of government to raise capital Face value, V – Value of bond; this amount is returned at end of bond’s life. Purchase price may be discounted Life, n – years to bond maturity, e. g., 5, 10, 20+ Dividend, I – periodic interest payments to purchaser based on coupon rate, b © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.2 – Single Alternative Example – Bond Investment
Example: A 10-year $10,000 6% coupon rate bond is purchased at 5% discount. Bond dividend is paid semi-annually. Will the investor make 7% per year compounded semiannually? Calculate PW at i = 3.5% per 6-month period for n = 20 periods I = (10,000)(0.06)/2 = $300 PW = - 10,000(0.95) ,000(P/F,3.5%,20) + 300(P/A,3.5%,20) = $ Bond will not make the required return $300 per 6-months period © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.2 – Equal-life ME Alternatives
Calculate PW of each alternative at MARR Equal-service of alternatives is assumed Selection criterion: Select alternative with most favorable PW value, that is, numerically largest PW value PW1 PW2 Select $-1,500 $-500 2 -2,500 500 2,500 1,500 1 Note : Not the absolute value © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Example: Two ME cost alternatives for traffic analysis. Revenues are equal. MARR is % per year. Select one (cont→) Estimate Electric-powered Solar-powered P, $/unit -2,500 -6,000 AOC, $/year -900 -50 S, $ 200 100 n, years 5 © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Determine PWE and PWS; select larger PW PWE = (P/A,10%,5)+200(P/F,10%,5) = $-5788 PWS = (P/A,10%,5)+100(P/F,10%,5) = $-6127 Conclusion: PWE > PWS; select electric-powered © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.3 – Different-life Alternatives
PW evaluation always requires equal- service between all alternatives Two methods available: Study period (same period for all alternatives) Least common multiple (LCM) of lives for alternatives Study period method is recommended Evaluation approach: Determine each PW at stated MARR; select alternative with numerically largest PW © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Study Period of length n years (periods) n is same for each alternative If life > n, use market value estimate in year n for salvage value If life < n, estimate costs for remaining years Estimates outside time frame of the study period are ignored © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

LCM Method Assumptions (may be unrealistic at times) Same service needed for LCM years (e.g., LCM of 5 and 9 is 45 years!) Alternatives available for multiple life cycles Estimates are correct over all life cycles (true only if cash flow estimate changes match inflation/deflation rate) Evaluation approach: obtain LCM, repeat purchase and life cycle for LCM years; calculate PW over LCM; select alternative with most favorable PW © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.3 - Different-life Analysis - Example
Use PW to select lower-cost alternative: For 5-year study period Using LCM of alternatives’ lives Assume MARR = 15% per year (cont →) © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Study period of 5 years Assume deposit returns are good estimates after 5 years F = 1,000 PWA = ? Location A P = -15, A = -3,500 F = 2,000 P = -18, A = -3,100 PWB = ? Location B For 5 years at i = 15%: PWA = $-26,236 and PWB = $-27,397 Select Location A with lower PW of costs © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

LCM evaluation LCM is 18 years Repurchase A twice (years 6 and 12) Repurchase B once (year 9) Assume all cash flow estimates (including first cost end-of-lease ‘deposit return’) are correct for repeated life cycles to total 18 years (cont →) © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

For 18 years at MARR = 15%: PWA = $-45,036 For 18 years at MARR = 15%: PWB = $-41,384 Select location B Note: Selection changed from 5-year study period © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.3 – Future Worth Evaluation
FW evaluation of alternatives is especially applicable for LARGE capital investment situations when maximizing the future worth of a corporation is important e.g., buildings, power generation, acquisitions Evaluation approach: Determine FW value from cash flows or PW with an n value in F/P factor equal to study period, or equal to LCM of alternatives’ lives © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.3 – Life Cycle Costing (LCC)
Another application of PW analysis Useful when entire life cycle of a system is under evaluation e.g., new car model or aircraft model; introducing new technology PW evaluation must include cost estimates for all stages of the product or service: Design (initial and detail) Development Production cost Marketing cost Operating costs Warranty commitments Phase-out costs etc. © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.4 – Capitalized Cost (CC)
PW of alternative that will last ‘forever’ Especially applicable to public project evaluation (dams, bridges, irrigation, hospitals, police, etc.) CC relation is derived using the limit as n → ∞ for the P/A factor PW = A(P/A,i%,n) = PW = A[1/i ] © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.4 – Capitalized Cost Refer to PW as CC when n is large (can be considered infinite). Then and AW = CC × i Example: If $10,000 earns 10% per year, $1,000 is interest earned annually for eternity. Principal remains in tact Cash flows for CC computations are of two types -- recurring and nonrecurring © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.4 – Capitalized Cost Procedure to find CC
Draw diagram for 2 cycles of recurring cash flows and any nonrecurring amounts Calculate PW (CC) for all nonrecurring amounts Find AW for 1 cycle of recurring amounts; then add these to all A series applicable for all years 1 to ∞ (or long life) Find CC for amount above using CC = AW/i Add all CC values (steps 2 and 4) © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.4 – CC Computation - Example
Find CC and A values at i = 5% of long-term public project with cash flows below. Cycle time is 13 years. Nonrecurring costs: first $150,000; one-time of $50,000 in year 10 Recurring costs: annual maintenance of $5000 (years 1-4) and $8000 thereafter; upgrade costs $15,000 each 13 years Step 1 (cont →) © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.4 – CC Computation - Example
CC of nonrecurring costs: CC1 = -150,000 – 50,000(P/F,5%,10) = $-180,695 AW of recurring $15,000 upgrade: AW = -15,000(A/F,5%,13) = $-847 per year AW of recurring maintenance costs years 1 to ∞: AW = $-5000 per year forever CC of extra $3000 maintenance for years 5 to ∞: CC2 = -3000(P/F,5%,4)/0.05 = $-49,362 CC for recurring upgrade and maintenance costs: CC3 = ( )/0.05 = $-116,940 Total CC obtained by adding all three CC components CCT = -180,695 – 49,362 – 116,940 = $-346,997 The AW value is the annual cost forever: AW = CC × i = -346,997(0.05) = $-17,350 © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.4 – CC Evaluation of Alternatives
For two long-life or infinite-life alternatives: SELECT ALTERNATIVE WITH LOWER CC OF COSTS For one infinite life and one finite life: Determine CC for finite life alternative using AW of 1 life cycle and relation CC = AW/i © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.4 – CC Evaluation of Alternatives - Example
1 long-term (assumed infinite); 1 finite life Long-term alternative (LT): $8 million now; $25,000 renewal annual contract Short-term alternative (ST): $2.75 million now; $120,000 AOC; life is n = 5 years Select better at MARR = 15% per year CCLT = -8,000,000 – 25,000/0.15 = $-8.17 million CCST = AW/0.15 = [-2,750,000(A/P,15%,5) – 120,000]/0.15 = $-6.27 million Conclusion: Select ST with lower CC of costs © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.5 – Independent Projects
Situation: Select from several (m) projects. Revenue and costs are estimated for each Solution approach: Basically different from that for ME alternatives One-time projects; no equal-service evaluation necessary; LCM not necessary Two types of budget situations are possible -- no limit or stated limit No limit: select from none (DN alternative) to all m projects using criterion SELECT ALL PROJECTS WITH PW ≥ 0 AT MARR © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.5 – Independent Projects
Procedure for stated budget limited evaluation No more than specified amount (b) can be invested and each project must demonstrate PW ≥ 0 at MARR Form ME bundles of projects which do not exceed limit. Include DN alternative. There are 2m ME bundles Procedure: Determine all bundles with total investment ≤ b Calculate PW of all projects included in bundles. (Note: any bundle with a PW < 0 project can be eliminated now) Add project PW values to get total PW for each viable bundle Select bundle with largest PW value. These are the projects to accept © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.5 – Independent Projects - Example
(cont →) Select from 4 independent projects at MARR of 15% per year; b = $15,000 PROCEDURE: 1. Total of 24 = 16 bundles. Only 6 require $15,000 or less: F, G, H, J, FH, DN 2. PW = investment + NCF(P/A,15%,n) Project Life, n PW at 15% F 6 $6646 G 9 -1019 (out) H 5 973 J 3 1553 © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.5 – Independent Projects - Example
3. PW of viable bundles (after G is removed) PWF = $6,646 PWH = $973 PWJ = $1,553 PWFH = 6, = $7,619 PWDN = $0 4. Bundle with largest PW is FH. Select these two projects © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 4.6 – Spreadsheet Evaluation of ME Alternatives
For one project evaluation, equal-life and study period comparisons with same annual amounts A, use the single-cell PV function = P – PV(i%,n,A,F) (Note minus sign on PV function) For different-life alternatives or when annual amounts vary, enter cash flow (CF) series and use the NPV function = P + NPV(i%,year_1_CF_cell, last_year_CF_cell) (Note that initial cost P is not included in NPV function) © 2008 McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Purchase generator Lease generator P, $ -120,000 Extra -40,000 now and extra -20,000 in year 6 S, $ 40,000 None n, years 3 6 AOC, $/year -8,000 -20,000 Determine which alternative is cheaper at MARR = 12% per year (cont →) Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008 © 2008 McGraw-Hill All rights reserved

Lecture slides to accompany Basics of Engineering Economy by Leland Blank and Anthony Tarquin Chapter 5 Annual Worth Analysis © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Chapter 5 – Annual Worth Analysis
PURPOSE Compare alternatives using an annual worth basis TOPICS AW calculations Alternative evaluation using AW AW of permanent investments Spreadsheet usage © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.1 – AW Advantage AW is also called
AE – annual equivalent EAC – equivalent annual cost EUAC (or EUAW) – equivalent uniform annual cost (or worth) Compare alternatives over only one life cycle – no LCM to meet equal service assumption Same AW amount assumed for future cycles, and estimates change with inflation rate If this assumption not correct; use a study period and specific estimates for cash flows © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.1 – AW and Multiple Life Cycles
AW of cycle 1 with i = 22% AW = -20,000(A/P,22%,3) -8000 = $-17,793 per year AW over 2 cycles AW = -20,000(A/P,22%,3) - 20,000(A/P,22%,6) -8000 Estimated costs over two life cycles Demonstrates that AW will be the same for any number of cycles © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.1 – Calculating Project AW
Use project PW or FW to determine AW with n = LCM for equal service or length of study period AW = PW(A/P,i%,n) = FW(A/F,i%,n) AW is the sum of 2 separate components: Capital recovery (CR) Equivalent annual A of operating costs (A of AOC) AW = CR + A of AOC © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.1 – Capital Recovery (CR)
What does CR mean? CR is the annual equivalent cost A incurred by initially spending an amount P on an asset (project) and using it for n years plus the return on the investment P at i% per year Example: Project costs P = $-13 million Estimated salvage S = $0.5 million Estimated life = 8 years Expected return i = 12% per year © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.1 – Capital Recovery Capital recovery: Find A per year
S = $0.5M Return expected: i = 12% Capital recovery: Find A per year P = $13M Capital recovery is the equivalent annual amount A to recover $13M at 12% per year if the salvage is $0.5M after 8 years CR = -13M(A/P,12%,8) + 0.5M(A/F,12%,8) = $-2,576M per year Conclusion: Project must develop revenue of at least $2.576M per year to recover P and make 12% on the investment © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.1 – Capital Recovery Formula
General formula for CR CR = -P(A/P,i,n) + S(A/F,i,n) Alternative formula to calculate CR CR = -(P-S)(A/P,i,n) + Si For previous estimates, using alternate CR = -(13 – 0.5)(A/P,12%,8) + 0.5(0.12) = $-2,576M per year © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.1 – Calculating AW AW = CR + A of AOC
AW: sum of CR plus A value of annual costs AW = CR + A of AOC If AOC estimate is same each year, add A to CR If AOC varies, find A value first, then add to CR Example: For previous project, estimate AOC at $0.9M each year. AW = M – 0.9M = $-3.476M per year Conclusion: Project must develop revenue of at least $3.476M per year to recover P, A and return 12% per year © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.1 – Example Cash Flows for AW
S = $0.5M AOC = $0.9M AW = $-3.476M per year P = $13M This is the AW for all future life cycles of 8 years each, provided costs estimates change at the inflation or deflation rate © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.2 – AW-based Evaluation
Single project analysis Calculate AW at stated MARR Acceptance criterion: If AW ≥ 0, project is economically justified Multiple alternatives Calculate AW of each alternative at MARR over respective life or study period Selection criterion: Select alternative with most favorable AW value, that is, numerically largest AW value © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.2 – AW Evaluation – Example 1
To select the more economic alternative at i = 12%, compare AWX over 4 years with AW Y over 6 years AWX = -40,000(A/P,12%,4) + 10,000(A/F,12%,4) – 25,000 = $-36,077 AWY = -75,000(A/P,12%,6) + 7,000(A/F,12%,6) – 15,000 = $-32,380 Select Y © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

In the previous example, assume the selected equipment will be retained for a total of 8 years. Additionally, assume after 8 years S = 0 for X and Y; and AOCX continues at $25,000, but AOCY doubles starting in year 7 AOC = $25,000 AOC = $15,000 AOC = $30,000 P = $40,000 P = $75,000 Alternative X Alternative Y © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Study period is n = 8 years AWX = -40,000(A/P,12%,8) – 25,000 = $-33,052 AWY = -75,000(A/P,12%,8) – 15,000 - 15,000(F/A,12%,2)(A/F,12%,8) = $-32,683 Still select Y as the cheaper choice; however the advantage of Y over X is now only 1/10 of the previous AW values © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.3 – AW of Permanent Investment
AW of alternative that will last ‘forever’ This is the annual worth equivalent of capitalized cost (CC) Solve for AW in relation PW = AW(1/i) from chapter 4 AW = PW(i) = CC(i) Procedure: Regular interval cash flows – find AW over one cycle Non-regular intervals – find P, then calculate AW = P(i) for long-term AW value © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Example: How long must $10,000 remain invested at 5% per year so that $2000 per year can be withdrawn forever? P = A/i determines total in year n to generate $2000 forever P = 2000/0.05 = $40,000 F/P factor determines n if money grows at 5%, with no withdrawals 40,000 = 10,000(F/P,5%,n) n = years © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Example 5.6 demonstrates comparison of short-lived and long-lived (forever) alternatives at i = 5% For each proposal, determine CR and AW values Prop P and S AOC Life CR and AW A P = $650,000 S = $17,000 A = $170,000 10 CR over 10; add AOC B P = $4 million A = $5,000 $30,000 every 5 years ‘forever’ CR over ∞; add AOC; add periodic repair over 5 years C P = $6 million A = $3,000 50 CR over 50; add AOC © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.3 – Example 5.6 (cont) AWA = - 650,000(A/P,5%,10) + 17,000(A/F,5%,10) - 170,000 = $-252,824 AWB = - 4,000,000(0.05) - 5, ,000(A/F,5%,5) = $-210,429 AWC = - 6,000,000(A/P,5%,50) - 3,000 = $-331,680 Select proposal B © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.4 – Spreadsheet Evaluation Using AW Analysis
Use PMT function with n = life of alternative or study period = PMT(i%,n,P,-S) – A This provides sum of CR and A of AOC, if AOC is uniform Note signs on P, S and A to obtain correct sign on result n values are one life cycle for each alternative, since LCM is not necessary for AW-based evaluation If AOC is not uniform, enter annual estimates, find their P value, then use PMT. Or, use embedded NPV function = PMT(i%,n, P+NPV(i%,year_1_cell,year_n_cell)) © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.4 – Spreadsheet Evaluation Using AW
Spreadsheet-based AW analysis in Example 5.6 – 3 alternatives with different lives AWA; n = 10: PMT(5%,10,650000,-17000) - 170,000 = $-252,826 AWB; n = ∞: -4,000,000* ,000 + PMT(5%,5,,30000) = $-210,429 AWC; n = 50: PMT(5%,50, ) - 3,000 = $-331,660 © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 5.4 – AW Spreadsheet Evaluation – Sign Usage
On the spreadsheet, note the careful use of minus signs to ensure a correct PMT function response. First costs and expenses have positive signs in PMT statement Alternatives A and C: = PMT(i%,n,P,-S) Alternative B periodic expense: = PMT(i%,n,,30000) PMT function is preceded by + sign © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Lecture slides to accompany Basics of Engineering Economy by Leland Blank and Anthony Tarquin Chapter 7 Benefit/Cost Analysis and Public Sector Projects © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Chapter 7 – Benefit/Cost Analysis
PURPOSE Learn about public sector projects compared to private sector projects, and perform a benefit/cost analysis TOPICS Characteristics of public and private sector projects B/C analysis of single project Incremental B/C method for 2 or more projects Spreadsheet usage for B/C analysis © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.1 – Public Sector Projects
Owned, used and financed by citizens of government units. Some examples are: Highways Universities Hospitals Sports arenas Prisons Public housing Emergency relief Utilities Public projects provide service to citizenry at no profit Partnerships of public entities and private enterprise are more prevalent now as funding for large public projects becomes more difficult © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.1 – Public Sector Project Characteristics
Size: Usually large compared to private projects with initial investment distributed over several years Life: Long-lived (often years); capitalized cost method is useful with A = Pi estimating annual costs Cash flows: No profits allowed; estimates are in form of costs paid by government unit, benefits to the citizenry (can include revenues or ‘savings’), and disbenefits (descriptions on later slide) © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.1 – Public Sector Project Characteristics
Funding: Public projects use taxes, fees, bonds (and gifts) for funding; taxes and fees are collected from ‘users’ of project services; funding examples are federal/state taxes of various sorts, tolls, surcharge fees Interest rate: Called discount rate, it is considerably lower than for private projects since no profit is considered and governments are exempt from taxes; typical rates in the 3 to 6% per year range Alternative selection: Politics and special interest groups make selection more complex for public projects; B/C method developed to put more objectivity into the analysis process © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.1 – Public Sector Project Estimates
Description Costs: expenditures to the government to build, maintain, & operate project; salvage/sales value possible Benefits: advantages to public; income and savings Analysis requires estimates as accurate as possible for costs, benefits, and disbenefits Example Bridge construction cost Annual cost of drug abusers’ treatment program New jobs and salary money Reduced property taxes Lower transportation costs due to less gas used © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.1 – Public Sector Project Estimates
Disbenefits: expected undesirable, negative consequences of project to owners – the public; usually these are economic disadvantages estimable in monetary units Disbenefits are not always included in the analysis; subject to political and special interest argumentation Examples $55M school bond issue -- Increased property taxes Tourist amusement park -- Higher local car insurance premiums based on increased traffic accidents New state prison – Reduced property values for houses in adjacent subdivisions © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.1 – Viewpoint for Public Sector Project Analysis
Determine viewpoint (perspective) before costs, benefits, and disbenefits are estimated Choose one and maintain it throughout estimation and analysis. Sample viewpoints Citizen Tax base Creation/retention of jobs Economic development Specific industry Read Example 7.1 for some specific samples © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.1 – Types of Public Project Contracts
Traditional Construction Contract Government funding via taxes, user fees and bonds Constructed through fixed price or cost plus contract with a profit margin specified for contractor Owned and operated by government unit CONTRACTOR SHARES NO RISK ON FINANCING OR OPERATION Examples: Design and construct a toll road Install a networked IT system between 4 county offices Design and build public housing for 400 families © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Public-Private Partnership Often called BOT (Build-Operate-Transfer) contract Contractor partially or completely responsible for financial arrangements Contractor operates and maintains system for specified time period. Contract includes these funds Ownership transferred to government in future. This stage is often negotiated in different ways Profit margin is specified for contractor during time of involvement © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Public-Private Partnership CONTRACTOR SHARES RISK ON FINANCING AND OPERATION Examples: Design, finance construct operate nuclear power plant for 15 years Recondition and operate state hospital for mental health patients Organize and operate a municipal security (police) force for a 20-year period; contract renewable each 5 years © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.2 – B/C Analysis – Single Project
Convert all estimates to PW, AW or FW value at discount rate i%. If PW used PW of benefits PW of costs Same formula for AW or FW All + signs, costs included Salvage has – sign; subtracted from costs If disbenefits are estimated, subtract from benefits Use PW, AW or FW for B/C If D is added to costs in denominator, the B/C value changes, but economic decision is the same © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Guideline for economic justification If B/C ≥ 1.0 accept project If B/C < project not acceptable Example: P = $15 M A = $500 K per year B = $1,500 K per year D = $200 K per year i = 6% n = 10 years AW equivalent of P = $15M(A/P,6%,10) = $2,038 K B/C = = 0.51 © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

TWO ALTERNATE BENEFIT-COST MEASURES Determine PW, AW or FW equivalent; place any salvage in denominator with only initial investment cost Same selection guideline: accept if B/C ≥ 1.0 If difference relation is desired, subtract net C from net B, once equivalents are determined Difference B-C = B – C Selection guideline: accept if B - C ≥ 0 © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.3 – B/C Analysis for ≥ 2 Alternatives
Technique similar to incremental ROR evaluation using ∆i* for ME alternatives ME are the mutually exclusive alternatives Find equivalent PW, AW or FW; calculate ∆B/C Selection guideline If ∆B/C ≥ 1.0 → select larger-cost alternative Otherwise → select lower-cost alternative Decision is based on incrementally justified total project cost, not incremental initial investment © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

To perform the ∆B/C analysis correctly Order alternatives by increasing equivalent total costs. Note: If this is not done, a larger-cost alternatives that is actually justifiable may be rejected because ∆B/C < 1 results Equal service requirement is necessary. If lives are short, use LCM method Note: Public projects usually have long lives (> 25 or 30 years), so lives are equal, or long enough to use capitalized cost © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Two types of benefits that require different treatment during ∆B/C analysis USAGE COST ESTIMATES DEFINITION Implied benefits based on difference in cost estimates of alternatives TREATMENT Comparison of alternatives is against each other only DIRECT BENEFIT ESTIMATES DEFINITION Benefits estimated for each alternative TREATMENT Comparison of alternatives is against DN first, then each other (Like revenue alternatives in ROR analysis) © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.3-B/C Analysis for ≥ 2 ME Alternatives
PROCEDURE FOR ∆B/C OF MUTUALLY EXCLUSIVE ALTERNATIVES Determine equivalent values for costs, benefits (and disbenefits, if estimated) Order alternatives by increasing total equivalent cost (for direct benefit alternatives, add DN first) For each pair 2 and 1, determine incremental C and B over LCM. For usage cost alternative, use ∆B = usage cost2 – usage cost1 4. Determine ∆B/C or ∆(B-D)/C 5. If ∆B/C ≥ 1.0, eliminate A; B is survivor Otherwise, A is survivor 6. Compare survivor with next alternative; continue steps (3) – (5) until only 1 alternative survives © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.3 – Book Correction to ∆B/C Procedure
Please note these corrections on pages 168 and 169 of textbook Page 168, bottom: should read “If ∆B/C ≥ 1.0, …” Page 169, step 5: should read “If ∆B/C ≥ 1.0, …” © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.3 – B/C Analysis – Example 1
Equal 30-year life; i = 5%; direct benefit alternatives Step 1. No disbenefits; use equivalent AW of costs AW1 = 10 M(A/P,5%,30) + 35,000 = $685,500 AW2 = 15 M(A/P,5%,30) + 55,000 = $1,030,750 Step 2. Add DN option with C = $0 and B = $0; comparison order is DN, 1, 2 Step 3. Compare 1-to-DN over 30 years cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Step 4. ∆B/C = 800,000/685,500 = 1.17 Step > 1, eliminate DN; 1 is survivor Step 6. Compare 2-to-1 (back to step 3) Step 3. ∆B = 1,050,000 – 800,000 = $250,000 ∆C = 1,030,750 – 685,500 = $345,250 Step 4. ∆B/C = 250,000/345,250 = 0.72 Step < 1, eliminate 2; 1 is survivor Step 6. Select design 1 © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

8-year study period; i = 7%; usage cost alternatives Step 1. Total cost is sum of two incentives. Determine AW over 8 years. For proposal 1 AW1 = 250,000(A/P,7%,8) + 25,000 = $66,867 cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Step 2. Order alternatives by increasing AW of total costs Step 3. Compare 2-to-1 over 8 years; use ∆usage costs for ∆B ∆B = entrance fee decrease + sales tax receipt increase = 50, ,000 = $60,000 ∆C = 93,614 – 66,867 = $26,747 Step 4. ∆B/C = 60,000/26,747 = 2.24 Step > 1.0; eliminate 1; 2 survives cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Results of comparisons Compare 3-to-2: ∆B/C = 25,000/40,120 = 0.62 Proposal 2 survives Compare 4-to-2: ∆B/C = 220,000/120,360 = 1.83 Proposal 2 eliminated; 4 survives Conclusion: Select Proposal 4 © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.3 - B/C Analysis-Additional Comments
Long lives (consider infinite for analysis purposes) Use capitalized cost to determine PW or AW. Incremental analysis is performed after using the equivalency relations A = P(i) or P = A/i Independent Projects (with no budget limitation) No incremental analysis needed Compare each project to DN option Select all projects with B/C ratio ≥ 1.0 Example 7.5 illustrates both of these situations © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.4 – Spreadsheet Usage for B/C Analysis for ME Alternatives
Spreadsheet format is same as that for incremental ROR evaluation Common approach is to use PV and NPV functions to find PW equivalents, then order alternatives by increasing total equivalent cost Use 6-step procedure to calculate pairwise ∆B/C; select one best alternative with ∆B/C ≥ 1.0 Remember minus sign convention on PV function to retain correct sign sense for responses © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 7.4 – Spreadsheet Example
15-year equal lives; usage-cost alternatives; i = 5% Step 1. PV function determines PW over 15 years: Total costs (initial and maintenance costs) Benefits (utility bill differences) Disbenefits (back-up system cost) cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Note minus sign on PV functions Step 2. Evaluation order is G, H, C. Note that G has lower PW costs, though H has a lower initial cost cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Steps H-to-G comparison Requires ∆B calculation as difference in PW of usage costs (utility bills) ∆B = H bill – G bill = 10,379,658 – 9,964,472 Value of ∆B = $- 415,186 for ∆(B-D)/C equation, since H has higher utility bill ∆(B-D)/C will be < 0. It is actually -0.51 Step 5. Eliminate H; G survives; Step 6. Compare C-to-G (back to step 3) Complete analysis on spreadsheet Conclusion: Select C (Crumbley) with ∆(B-D)/C = 1.22 cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Lecture slides to accompany Basics of Engineering Economy by Leland Blank and Anthony Tarquin Chapter 10 Effects of Inflation © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Chapter 10 – Effects of Inflation
PURPOSE Consider the effects of inflation in PW, AW and FW equivalence calculations TOPICS Definition and impact PW adjusted for inflation FW with inflation; real interest rate; inflation adjusted MARR AW with inflation Spreadsheet usage © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.1 – Understanding Inflation
Inflation definition – A decrease in the value of a currency Inflation impact – An increase in the amount of money required to purchase the same amount of goods or services over time Directly associated with inflation – an increase in the money supply, i. e., more (government) money is printed to counteract inflationary impacts To perform economic comparisons - conversion to a common basis of money at different times must be included What to do in engineering economy? © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Constant-value (CV) – Money represents same purchasing power regardless of when in time Another term used for CV: Today’s dollars Future dollars – amount of money in the future Other terms used for future dollars: Then-current dollars Inflated dollars © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

f - inflation rate per period (year), e.g., 5% per year n – number of periods (years) considered Example: A $100 bill five years from now at 4% inflation will purchase only $82.20 worth of goods © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Viewed from a future perspective Future amount = CV amount (1+f)n Example: If it costs $82.20 today, in five years it will cost $100 if inflation is 4% per year 82.20(1+0.04)5 = 82.20(1.2167)= $100 Two (equivalent) approaches to adjust for inflation: Convert all amounts (estimates) to have same value (purchasing power) by equivalency relations (this uses CV relation shown above prior to time-value-of-money calculations) Change interest rates to account for inflation, as well as time value of money (this approach follows) © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Consider the erosive impact of even moderate inflation Save $25,000/year for 40 years of employment (ages 22 to 62) Assume 7% per year return on your investments Future total amount of retirement portfolio No return considered: 25,000(40) = $1 million 7% return considered FW = 25,000(F/A,7%,40) = 25,000( ) = $4.99 million  A return of 7% per year means a lot over time © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Now, assume 4% per year inflation What is the effective purchasing power of your retirement portfolio 40 years from now? FW = $1.835 million This means that purchasing power is $1.835 million, not $4.99 million  Inflation is a real killer of investment returns Conclusions: The real return on savings is significantly lower when inflation is accounted for If inflation rate exceeds return rate ( f > i), money loses purchasing power over time Computations discussed later © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.1 – Inflation-related Terminology
i Real (inflation-free) rate – Rate at which interest is earned or paid when effects of inflation are removed Usually considered the ‘safe investment’ rate. Approximately 3.5% per year, historically, but varies depending upon health of economy if Inflation-adjusted rate – Also called market rate. Interest rate with effect of inflation accounted for This rate is the one quoted daily, .e.g., 7% per year or effective 7.08%. Combination of real rate i and inflation rate f f Inflation rate – Rate of change in value of currency Formulas for and use of these terms follow © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.1 - Deflation Deflation is opposite of inflation
Deflation rate is -f in % per year Purchasing power of deflated currency is greater in future than at present Sounds good after inflationary period, however … There are fewer jobs, less credit and fewer loans available; overall ‘tighter’ money situation Can disrupt national economies faster and more severely than an equivalent inflation rate © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Equivalency computations use -f, instead of f International Dumping
Sec Deflation Equivalency computations use -f, instead of f Example: Asset costs $10,000 today. If 2% per year deflation is assumed, future estimated cost of repurchase in 5 years is FW = 10,000(1-0.02)5 = 10,000(0.9039) = $9,039 International Dumping Import materials (e.g., cars, cement, steel) into a country from international sources at very low prices Causes temporary deflation in the targeted sector Domestic producers must reduce costs; some go broke, if financially weak Once competition is beaten, prices return to normal or above previous levels © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.2 – PW Adjusted for Inflation
As described earlier, there are two (equivalent) approaches to adjust for inflation. These are correct for all PW computations 1. Convert everything to current value amounts by PW equivalency relations before applying the real interest rate i in PW computations 2. Determine the inflation-adjusted interest rate if and use it in all PW calculations Example: Assume a first cost of $5,000 now (CV amount). Will purchase item during next 4 years. Let Inflation = 4% per year f = 4% Real rate of return = 10% per year i = 10% © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Approach 1--Use CV dollars and real rate i CV = PW = $5,000 now Future costs at f = 4% are in column (3) FW = 5,000(1+ f)n = 5,000(1.04)n © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Approach 1 – Now find equivalent PW of $5,000 CV over 4 years at i = 10% Year, t Future cost in CV $ PW at real i = 10% 5,000(P/F,10%,t) 5,000 1 4,545 2 4,132 3 3,757 4 3,415 © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

This approach is more commonly used Approach 2 – Use inflation-adjusted interest rate if in all PW calculations To derive a formula for if, calculate PW from FW amounts using real i and determine CV amounts using f Let if = i + f + if be the inflation-adjusted (market) interest rate © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Resulting PW formula Interest rate formula Requires no conversion to CV amounts to obtain PW values Example: Real i = 8% Inflation = 6% Inflation-adjusted (market) rate = (0.08)(0.06) if = or % © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

From previous example P = $5,000 is cost now; cost increases to $5,849 in 4 years  Inflation rate = 4%/year  Real interest rate = 10%/year Use if to determine equivalent PW now of the cost 4 years in future: if = (0.10)(0.04) = 14.4% P = 5,849(P/F,14.4%,4) = 5,849(0.5838) = $3,415 Conclusion: At market rate (inflation considered) of 14.4%, paying $5,849 in four years for something that costs $5,000 now is equivalent to paying $3,415 now in CV terms © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.2 – PW Inflation-Adjusted Example #1
Lottery winnings can be taken in 1 of 3 ways. Which is better financially if f = 4% and i = 6%? 1: $100,000 immediately 2: 8 payments of $15,000 each starting next year 3: 3 payments of $45,000 each in years 0, 4 and 8 Use approach 2 with if: if = (0.06)(0.04) = 10.24% Plan 1 PW Plan 2 PW Plan 3 PW $100,000 15,000(P/A,10.24%,8) = $79,329 45,000[( (P/F,10.24%,4) + (P/F,10.24%,8)] = $96,099 Winner © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Use (P/A,g,i,n) = (P/A,12%,interest rate,9) factor for geometric series to determine PW, where ‘interest rate’ is 1. i = 15% (without inflation) 2. if = (0.15)(0.11) = 27.65% (with inflation) Without inflation considered PW = -35, ,000(P/A,15%,4) ,000(P/A,12%,15%,9)(P/F,15%,4) = $-83,232 cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

2. Inflation accounted for using if = 27.65% PW = -35, ,000(P/A,27.65%,4) ,000(P/A,12%,27.65%,9)(P/F,27.65%,4) = $-62,436 Conclusions: As f and if increase, the P/A and P/F factors decrease, thus making the PW values smaller Moving debt to the future and paying with future (inflated) money seems financially ‘smart’, BUT … If cash is not available in the future, debt-laden people, companies and countries will suffer and may go bankrupt ‘Buy now, pay later’ philosophy can be a dangerous strategy © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.3 – FW Adjusted for Inflation
Like PW computations, possible to consider or neglect inflation rate f on top of real interest rate i Let P = $1,000; n = 7; find F under different conditions Market return if = 10% Inflation rate f = 4% Cases 1 and 4: Actual future amount accumulated at market rate if (both purchasing power and return included) F = P(F/P,if,n) = 1,000(F/P,10%,7) = $1,948 Interpretation: This case correctly projects amount of capital needed in future to overcome ongoing inflation plus make a stated return on the investment © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Case 2: Purchasing power maintained, but no inflation considered requires use of real interest rate i Usually the market (inflation-adjusted) rate if and inflation rate f are estimated Determine real interest rate i by dividing market rate by inflation factor (1+f) © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Case 2 (cont): Let P = $1,000; n = 7; find F Market return if = 10% Inflation rate f = 4% Therefore, real return i = (0.10 – 0.04)/(1.04) = 5.77% F = 1,000(F/P,5.77%,7) = $1,481 Interpretation: This case correctly estimates future purchasing power with inflation effects removed. An inflation rate of 4% reduces the future 10% return amount that can be purchased from $1,948 to $1, a 24% reduction © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Case 3: Future amount with no return uses only the inflation rate f Let P = $1,000; n = 7; find F Inflation rate f = 4% F = P(F/P,f%,n) = 1,000(F/P,4%,7) = $1,316 Interpretation: This case correctly estimates the future amount needed to keep up with inflation only. A price of $1,000 grows by 31+% in 7 years at 4% per year inflation © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.3 – Inflation-Adjusted MARR
Most corporations determine MARR using both inflation and a return to cover capital increases and expected return This is if used in cases 1 and 4 Inflation-adjusted MARR is: MARRf = i + f + if The real rate of return i is the corporate return requirement at or above the following: ‘safe’ investment, which is usually ~3.5%, or cost of capital © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.3 – Inflation-Adjusted MARR
Example: Cost of capital = 10% Required return = 3% Inflation rate projected = 4% Perform PW, FW and AW computations at MARRf = (0.13)(0.04) = 17.52% Real return i = 13% © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.3 – FW Inflation-Adjusted Example
Alternatives: purchase now for $200,000, or in 3 years at an estimated $340,000 MARR is 12% and inflation averages 6.75% per year INFLATION CONSIDERED MARRf = 12% % + 12%(6.75%) = 19.56% Purchase now FW = -200,000(F/P,19.56%,3) = $-341,812 Purchase later FW = $-340,000 Buy later is slightly more economic WITHOUT INFLATION MARR = 12% Purchase now FW = -200,000(F/P,12%,3) = $-280,986 Purchase later FW = $-340,000 Buy now is more economic © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.3 - Hyperinflation Inflation usually averages 2% to 8% per year
Political and/or financial instability, overspending, serious trade imbalance can increase inflation dramatically Inflation rates above 50%, 75% and 100% per year are considered hyperinflation FW estimates skyrocket as inflation increases become larger But, future estimates are so uncertain that a dependable economic analysis can not be performed © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

FW = -200,000(F/P,146.4%,3) = $-2.99 million
Sec Hyperinflation Previous example: P = $200, MARR = 12% per year If f = 10%/month = 120%/ year (without considering compounding of inflation) MARRf = 12% + 120% + 12%(120%) = 146.4% Future equivalent cost estimate with inflation is huge FW = -200,000(F/P,146.4%,3) = $-2.99 million Yet, estimated cost in 3 years is unpredictable What is to be done from the economic perspective???? © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.4 - AW Adjusted for Inflation
Include inflation in AW computations because: Capital must be recovered with future, inflated dollars Less buying power in future means more money needed to recover present investments plus a return To find future inflated amount needed per year – Use market rate if to determine A, given P If future amount is known, fewer annual dollars are needed, since their current buying power is greater – Again, use if to determine A, given F © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Example: Spend $1,000,000 now for 3-year IT service contract Expect 4% return over cost of capital of 8.5% Inflation averages 6% per year What annual amount is needed to recover cost? Expected real return = 4% + 8.5% = 12.5% MARRf = 12.5% + 6% %(6%) = 19.25% Required annual recovery amount: AW = 1 million(A/P,19.25%,3) = $469,159 / year © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

For comparison purposes only, assume contract provider agreed to one payment of $1 million after 3 years. What is AW now? Again, MARRf = 19.25% Use A/F factor to find AW AW = 1 million(A/F,19.25%,3) = $276,659 There is a large reduction of ~$192,000 per year to recover the (fixed) $1 million expenditure after 3 years © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.4 - AW Inflation-Adjusted Example
For retirement, desire is to invest equal amounts annually for 5 years to maintain purchasing power of $10,000 today Expected (market) return is 10% Assume inflation is relatively high at 8% Step 1: Required total after 5 years when f = 8% (case 3): F = 10,000(F/P,8%,5) = $14,693.28 Step 2: Annual amount for 5 years at if = 10% (cases 1, 4): A = 14,693.28(A/F,10%,5) = $ Note: real return is only i = ( )/(1.08) = 1.85%/year © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Sec 10.4 – Inflationary Period Reactions
THINGS THAT OCCUR WHEN INFLATION INCREASES More needed annually to recover capital investments + required return Lenders tend to increase interest rates to individuals (credit cards, mortgages) and corporations (loans) People make lower payments on cards and loans (money is used to buy other, necessary items to live) Lenders need more money to cover their higher costs of making loans Bankruptcies increase Spiraling effects of inflation can lead to national recession © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Lottery winnings can be taken in 1 of 3 ways 1: $100,000 immediately 2: 8 payments of $15,000 each starting next year 3: 3 payments of $45,000 each in years 0, 4 and 8 Assume f = 4% and i = 6% per year Questions about the 3 plans: Best plan based on PW values? FW in 8 years with inflation considered? FW in 8 years in terms of today’s purchasing power? cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

Calculate if = 10.24%, enter cash flows and use NPV function (row 13); select Plan 1 cont → © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

B. Determine FW with inflation considered using FV function at 10.24% (row 16 – note minus sign) C. Determine FW without inflation considered using FV function at 6% (row 18- note minus sign) As expected, Plan 1 has best PW and FW values Note: The impact of inflation is very clear. For example, Plan 3 generates equivalent value in 8 years of ~$209,600. However, this will purchase only about ~ $153,100 of goods in terms of today’s dollars © 2008, McGraw-Hill All rights reserved Slide to accompany Blank and Tarquin Basics of Engineering Economy, 2008

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