Lecture No.15 Chapter 5 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010 Chapter Opening Story – GE’s Healthymagination Project GE Unveils $6 Billion Health-Unit Plan: • Goal: Increase the market share in the healthcare sector. • Strategies: Develop products that will lower costs, increase access and improve health-care quality. • Investment required: $6 billion over six years • Desired project outcome: Would help GE’s health-care unit grow at least twice as fast as the broader economy. Contemporary Engineering Economics, 5th edition, © 2010 Ultimate Questions GE’ s Point of View: Would there be enough demand for their products to justify the investment required in new facilities and marketing? What would be the potential financial risk if the actual demand is far less than its forecast or adoption of technology is too slow? If everything goes as planned, how long does it take to recover the initial investment? Contemporary Engineering Economics, 5th edition, © 2010 Bank Loan vs. Project Cash Flows Contemporary Engineering Economics, 5th edition, © 2010 Example 5.1 Describing Project Cash Flows – A Computer-Process Control Project Year (n) Cash Inflows (Benefits) 0 0 1 Cash Outflows (Costs) Net Cash Flows $650,000 -$650,000 215,500 53,000 162,500 2 215,500 53,000 162,500 … … … … 8 215,500 53,000 162,500 Contemporary Engineering Economics, 5th edition, © 2010 Cash Flow Diagram for the Computer Process Control Project Contemporary Engineering Economics, 5th edition, © 2010 Payback Period Principle: How fast can I recover my initial investment? Method: Based on the cumulative cash flow (or accounting profit) Screening Guideline: If the payback period is less than or equal to some specified bench-mark period, the project would be considered for further analysis. Weakness: Does not consider the time value of money Contemporary Engineering Economics, 5th edition, © 2010 Example 5.3 Payback Period N Cash Flow Cum. Flow 0 1 2 3 4 5 6 -$105,000+$20,000 $15,000 $25,000 $35,000 $45,000 $45,000 $35,000 -$85,000 -$70,000 -$45,000 -$10,000 $35,000 $80,000 $115,000 Payback period should occurs somewhere between N = 3 and N = 4. Contemporary Engineering Economics, 5th edition, © 2010 Annual cash flow $45,000 $45,000 $35,000 $35,000 $25,000 $15,000 0 1 2 Years 3 4 5 6 Cumulative cash flow ($) $85,000 150,000 3.2 years Payback period 100,000 50,000 0 -50,000 -100,000 0 1 2 3 4 Contemporary Engineering Economics, 5th edition, © 2010 5 Years (n) 6 Practice Problem How long does it take to recover the initial investment for the computer process control system project in Example 5.1? Initial Cost Payback Period = Uniform annual benefit $650,000 $162,500 4 years Contemporary Engineering Economics, 5th edition, © 2010 Discounted Payback Period Principle: How fast can I recover my initial investment plus interest? Method: Based on the cumulative discounted cash flow Screening Guideline: If the discounted payback period (DPP) is less than or equal to some specified bench-mark period, the project could be considered for further analysis. Weakness: Cash flows occurring after DPP are ignored Contemporary Engineering Economics, 5th edition, © 2010 Discounted Payback Period Calculation Period (n) Cash Flow (An) Cost of Funds (15%)* Ending Cash Balance 0 -$85,000 0 1 15,000 -$85,000(0.15) = -$12,750 -82,750 2 25,000 -$82,750(0.15) = -12,413 -70,163 3 35,000 -$70,163(0.15) = -10,524 -45,687 4 45,000 -$45,687(0.15) =-6,853 -7,540 5 45,000 -$7,540(0.15) = -1,131 36,329 6 35,000 $36,329(0.15) = 5,449 76,778 * Cost of funds = (Unrecovered beginning balance) X (interest rate) Contemporary Engineering Economics, 5th edition, © 2010 -$85,000 Illustration of Discounted Payback Period Contemporary Engineering Economics, 5th edition, © 2010 Payback periods can be used as a screening tool for liquidity, but we need a measure of investment worth for profitability. Contemporary Engineering Economics, 5th edition, © 2010 Lecture No.16 Chapter 5 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010 Net Present Worth Measure Principle: Compute the equivalent net surplus at n = 0 for a given interest rate of i. Decision Rule for Single Project Evaluation: Accept the project if the net surplus is positive. Decision Rule for Comparing Multiple Alternatives: Select the alternative with the largest net present worth. Inflow 0 1 2 3 4 5 Net surplus Outflow PW(i) > 0 PW(i)inflow 0 PW(i)outflow Contemporary Engineering Economics, 5th edition, © 2010 0 Example 5.5 - Tiger Machine Tool Company $35,560 $37,360 $31,850 $34,400 inflow 0 1 2 3 outflow $76,000 PW (12%)inflow $35,560(P / F ,12%,1) $37,360(P / F ,12%,2) $31,850(P / F ,12%,3) $34,400(P / F ,12%,4) PW (12%)outflow $ $76,000 PW (12%) $106,065 $76,000 $30,065 0, Accept Contemporary Engineering Economics, 5th edition, © 2010 Present Worth Amounts at Varying Interest Rates Contemporary Engineering Economics, 5th edition, © 2010 Can you explain what $30,065 really means? 1. 2. Project Balance Concept Investment Pool Concept Contemporary Engineering Economics, 5th edition, © 2010 Project Balance Concept Suppose that the firm has no internal funds to finance the project, so will borrow the entire investment from a bank at an interest rate of 12%. Then, any proceeds from the project will be used to pay off the bank loan. Then, our interest is to see if how much money would be left over at the end of the project period. Contemporary Engineering Economics, 5th edition, © 2010 Calculating Project Balances End of Year (n ) 0 1 2 3 4 Beginning Project Balance $ (76,000) $ (49,560) $ (18,147) $ 11,525 Interest Charged (12%) $ (9,120) $ (5,947) $ (2,178) $ 1,383 Payment $ (76,000) $ 35,560 $ Ending Project Balance $ (76,000) $ (49,560) $ (18,147) $ 11,525 $ 47,308 Contemporary Engineering Economics, 5th edition, © 2010 37,360 $ 31,850 $ 34,400 Project Balance Diagram – Four Pieces of Information The exposure to financial risk The discounted payback period The profit potential The net future worth Contemporary Engineering Economics, 5th edition, © 2010 Investment Pool Concept Suppose the company has $76,000. It has two options. (1)Take the money out and invest it in the project or (2) leave the money in the pool and continue to earn a 12% interest. If you take Option 1, any proceeds from the project will be returned to the investment pool and earn 12% interest yearly until the end of the project period. Let’s see what the consequences are for each option. Contemporary Engineering Economics, 5th edition, © 2010 Option A If $76,000 were left in the investment pool for 4 years $76,000(F/P,12%,4) $119,587 Option B If $76,000 withdrawal from the investment pool were invested in the project 3) 1 • $49,959 $37,360(F/P,12% ,2) 2 • $46,864 3 • $35,672 $35,650(F/P,12%, $31,850(F/P,12%, 1) Year • Amount 1 • $35,650 2 • $37,360 3 • $31,850 4 • $34,400 4 • $34,400 $166,896 $34,400(F/P,12%, 0) $166,896 Investment Pool $47,309 $119,587 PW(12%) = $47,309(P/F,12%,4) = $30.065 Contemporary Engineering Economics, 5th edition, © 2010 The net benefit of investing in the project What Factors Should the Company Consider in Selecting a MARR in Project Evaluation? Cost of capital Contemporary Engineering Economics, 5th edition, © 2010 Cost of capital MARR make an investment project worthwhile. Viewed as the rate of return that a firm would receive if it invested its money someplace else with a similar risk Risk premium The additional risk associated with the project if you are dealing with a project with higher risk than normal project Risk premium The required return necessary to Practice Problem An electrical motor rated at 15HP needs to be purchased for $1,000. The service life of the motor is known to be 10 years with negligible salvage value. Its full load efficiency is 85%. The cost of energy is $0.08 per Kwh. The intended use of the motor is 4,000 hours per year. Find the total present worth cost of owning and operating the motor at 10% interest. Contemporary Engineering Economics, 5th edition, © 2010 Solution 1HP=0.7457kW 15HP = 15 0.7457 = 11.1855kW Required input power at 85% efficiency rating: 11.1855kW 13.1594kW 0.85 Required total kWh per year 13.1594kW 4,000 hours/year =52,638 kWh/yr Total annual energy cost to operate the motor 52,638kWh $0.08/kWh =$4,211/yr The total present worth cost of owning and operating the motor PW (10%) $1,000 $4,211(P / A,10%,10) = $26,875 Contemporary Engineering Economics, 5th edition, © 2010 Cash Flow Series Associated with Owning and Operating the Motor 0 1 2 3 4 5 6 7 8 9 10 $1,000 $4,211 PW (10%) $1,000 $4,211(P / A,10%,10) $26,875 Contemporary Engineering Economics, 5th edition, © 2010 Lecture No.17 Chapter 5 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010 Future Worth Criterion Given: Cash flows and MARR (i) Find: The net equivalent worth at a specified period other than “present”, commonly the end of project life Decision Rule: Accept the project if the equivalent worth is positive. $47,309 $35,560 $37,360 $31,850 $34,400 0 1 $76,000 Contemporary Engineering Economics, 5th edition, © 2010 2 3 Project life Example 5.6 Future Equivalent at an Intermediate Time Contemporary Engineering Economics, 5th edition, © 2010 Example 5.8 Project’s Service Life is Extremely Long • Q1: Was Bracewell's $800,000 investment a wise one? •Q2: How long does he have to wait to recover his initial investment, and will he ever make a profit? Contemporary Engineering Economics, 5th edition, © 2010 Ex. 5.8: Mr. Bracewell’s Hydroelectric Project V1 V2 $1,101K $1, 468 K $367 K 0 V2 120 K ( P / A,8%,50) $1, 468K V1 $50 K ( F / P,8%,9) $50 K ( F / P,8%,8) $100 K ( F / P,8%,1) 60 K $1,101K Contemporary Engineering Economics, 5th edition, © 2010 How Would You Find P for a Perpetual Cash Flow Series, A? Contemporary Engineering Economics, 5th edition, © 2010 Capitalized Equivalent Worth Principle: PW for a project with an annual receipt of A over infinite service life A 0 Equation: CE(i) = A(P/A, i, ) = A/i P =CE(i) Contemporary Engineering Economics, 5th edition, © 2010 Practice Problem Given: i = 10%, N = ∞ Find: P or CE (10%) $2,000 $1,000 0 10 P = CE (10%) = ? Contemporary Engineering Economics, 5th edition, © 2010 ∞ Solution: $2,000 $1,000 0 10 P = CE (10%) = ? ∞ $1,000 $1,000 CE(10%) (P / F ,10%,10) 0.10 0.10 $10,000(1 0.3855) $13,855 Contemporary Engineering Economics, 4th edition, © 2007 A Bridge Construction Project Construction cost = $2,000,000 Annual Maintenance cost = $50,000 Renovation cost = $500,000 every 15 years Planning horizon = infinite period Interest rate = 5% Contemporary Engineering Economics, 5th edition, © 2010 Cash Flow Diagram for the Bridge Construction Project 15 Years 30 45 60 $500,000 $500,000 $500,000 $500,000 0 $50,000 $2,000,000 Contemporary Engineering Economics, 5th edition, © 2010 Solution: Construction Cost P1 = $2,000,000 Maintenance Costs P2 = $50,000/0.05 = $1,000,000 Renovation Costs P3 = $500,000(P/F, 5%, 15) + $500,000(P/F, 5%, 30) + $500,000(P/F, 5%, 45) + $500,000(P/F, 5%, 60) . = {$500,000(A/F, 5%,15)}/0.05 = $463,423 Total Present Worth P = P1 + P2 + P3 = $3,463,423 Contemporary Engineering Economics, 5th edition, © 2010 Alternate way to calculate P3 Concept: Find the effective interest rate per payment period Effective interest rate for a 15-year period Effective interest 0 15 30 45 60 rate for a 15-year cycle i = (1 + 0.05)15 - 1 = 107.893% $500,000 $500,000 $500,000 $500,000 Capitalized equivalent worth P3 = $500,000/1.0789 = $463,423 Contemporary Engineering Economics, 5th edition, © 2010 Lecture No.18 Chapter 5 Contemporary Engineering Economics Copyright © 2010 Contemporary Engineering Economics, 5th edition, © 2010 Comparing Mutually Exclusive Projects – Basic Terminologies Mutually Exclusive Projects Alternative vs. Project Do-Nothing Alternative Contemporary Engineering Economics, 5th edition, © 2010 Revenue versus Service Projects Revenue Projects Service Projects Projects whose revenues Projects whose revenues do depend on the choice of alternatives Revenue and cost streams vary with the choice of alternatives. not depend on the choice of alternatives Fixed (or constant) revenues for all alternatives Contemporary Engineering Economics, 5th edition, © 2010 Analysis Period versus Required Service Period Analysis Period Required Service Period The time span over which The time span over which the economic effects of an investment will be evaluated (study period or planning horizon). the service of an equipment (or investment) will be needed. Contemporary Engineering Economics, 5th edition, © 2010 Road Map – A Process of Making a Choice among Mutually Exclusive Alternatives Contemporary Engineering Economics, 5th edition, © 2010 Comparing Mutually Exclusive Projects Principle: Projects must be compared over an equal time span. Rule of Thumb: If the required service period is given, the analysis period should be the same as the required service period. Contemporary Engineering Economics, 5th edition, © 2010 Case 1 Analysis Period Equals Project Lives What to do: Compute the NPW for each project over its life and select the project with the largest NPW. Contemporary Engineering Economics, 5th edition, © 2010 Comparing projects requiring different levels of investment Assume that the unused funds will be invested at MARR. This portion of investment will earn a 10% return on investment. PW(10%)A $4,000 $450( P / F ,10%,1) $600( P / F ,10%, 2) $4, 493( P / F ,10%,3) $283 Contemporary Engineering Economics, 5th edition, © 2010 Analysis Period Implied in Comparing Mutually Exclusive Alternatives Contemporary Engineering Economics, 5th edition, © 2010 Case 2 – Ex. 5.10 Project Lives Longer than the Analysis Period What to do: Estimate the salvage value at the end of the required service period. Compute the NPW for each project over the required service period. PW(15%)A = -$362 PW(15%)B = -$364 Contemporary Engineering Economics, 5th edition, © 2010 Case 3A – Service Projects – Ex. 5.11 Project Lives Shorter than the Analysis Period What to do: Come up with replacement projects that match or exceed the required service period. Compute the NPW for each project over the required service period. PW(15%)A = -$34,359 PW(15%)B = -$31,031 Contemporary Engineering Economics, 5th edition, © 2010 Case 3B – Revenue Projects – Ex. 5.12 Analysis Period Coincides with the Project with the Longest Life in the Mutually Exclusive Group What to do: Compute the NPW of each project over its analysis period, assuming that no cash flows after the service life for the shorter-lived project. PW(15%)Drill = $2,208,470 PW(15%)Lease = $2,180,210 Contemporary Engineering Economics, 5th edition, © 2010 Case 4 – Ex. 5.13 Analysis Period Is Not Specified What to do: Come up with replacement projects that serve out the least common multiple period (LCM). Compute the NPW for each project over the LCM. PW(15%)A = -$53,657 PW(15%)B = -$48,534 Contemporary Engineering Economics, 5th edition, © 2010 Summary Present worth is an equivalence method of analysis in which a project’s cash flows are discounted to a lump sum amount at present time. The MARR or minimum attractive rate of return is the interest rate at which a firm can always earn or borrow money. MARR is generally dictated by management and is the rate at which NPW analysis should be conducted. Two measures of investment, the net future worth and the capitalized equivalent worth, are variations to the NPW criterion. Contemporary Engineering Economics, 5th edition, © 2010 The term mutually exclusive means that, when one of several alternatives that meet the same need is selected, the others will be rejected. Revenue projects are those for which the income generated depends on the choice of project. Service projects are those for which income remains the same, regardless of which project is selected. The analysis period (study period) is the time span over which the economic effects of an investment will be evaluated. The required service period is the time span over which the service of an equipment (or investment) will be needed. The analysis period should be chosen to cover the required service period. When not specified by management or company policy, the analysis period to use in a comparison of mutually exclusive projects may be chosen by the individual analyst. Contemporary Engineering Economics, 5th edition, © 2010
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