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Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques.

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1 Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques

2 9-2 Chapter 9 … Develops and uses future value, benefit-cost ratio, payback period, and sensitivity analysis methods to solve engineering economic problems. Links the use of the future value method to the present value and annual value methods. Uses spreadsheets to perform sensitivity and breakeven analyses.

3 9-3 Present and Future Value Methods Net Present Value (NPV) What is the present value of future activity? Project cash flows are discounted to an equivalent value, usually at the start of the project (see Chapter 5). Net Future Value (NFV) What is the future value of current activity? Project cash flows are converted to a future value, usually at the end of the project.

4 9-4 012345 $750 $1500 $1000 $500 $1000 ≡ 0 FV = ? 5 Future Value Find the value of the cash flows at the end of the 5th year if the rate of interest is 10% compounded annually. (Answer= $5854.58.)

5 9-5 Economic Criteria Projects are judged against an economic criterion. SituationCriterion Fixed inputMaximize output Fixed outputMinimize input Neither fixedMaximize difference (output  input)

6 9-6 Economic Criteria Restated for Future Value Techniques SituationCriterion Fixed inputAmount of capital available is fixed Maximize future value of benefits Fixed outputAmount of benefits is fixed Minimize future value of costs Neither fixedNeither capital nor benefits is fixed Maximize net future value (NFV)

7 9-7 Type of Projects Independent The selection of a project is independent of the decision to undertake any other project(s). Mutually exclusive At most one project (including the status quo, or “ do nothing ” option) can be selected amongst competing alternatives. Contingent (dependent) The selection of a project is dependent on the selection of at least one other project.

8 9-8 YEAR CASH FLOW 0$0 1$1000 2$2000 3$3000 4$2500 5$2000 6$1500 7$1000 8$500 Problem: Find NFV (i = 10%) NFV = $1000(1.10) 7 + $2000(1.10) 6 + $3000(1.10) 5 + $2500(1.10) 4 + $2000(1.10) 3 + $1500(1.10) 2 + $1000(1.10) + $500 = $20,060.62 (or find NPV, then  (1.10) 8 )

9 9-9 NFV and Independent Projects Select all projects with non-negative net future value (NFV). if NFV > 0, accept project if NFV < 0, reject project if NFV = 0, accept project If all projects have NFV < $0, select the status quo (do nothing) option; invest available funds at the prevailing risk-free interest rate. The selection of independent projects is usually constrained by a capital budget (limited funds).

10 9-10 NFV and Independent Projects … The NFV is the amount that remains at the end of a project after the annual cash flows are accumulated into the unrecovered capital, plus required returns, throughout the project. Example: find the NFV and the NPV of a project that requires an investment of $500K and has the cash flows shown below if the MARR is fourteen percent. (Answer: NFV = $149,131.52; NPV = $88,297.83). Year1234 Cash Flow$150K$300K$225K$125K

11 9-11 Future Value and Present Value Analysis PARAMETERSPROJECT APROJECT BPROJECT C Investment$2500$3500$5000 Annual cost$900$700$1000 + (G=$100) Salvage value$200$350$600 Life (years)555 Annual revenue$1800$1900$2100 (15% growth) MARR10%

12 9-12 Calculate the NPV and the NFV for Projects A, B, and C. NPV A = $1035.89; NFV A = $1668.32. NPV B = $1266.27; NFV B = $2039.34. NPV C = $1349.16; NFV C = $2172.83. Future Value and Present Value Analysis …

13 9-13 Useful Lives  Analysis Period Project A has a 3-year life. Project B has a 4- year life. Both projects are valid because their NFVs are both > $0 (i = 10%). Which project is better? YearProject AProject B 0  $1000  $2000 1$500$720 2$500$720 3$500$720 4 NFV$324.00$413.32

14 9-14 Because Projects A and B have different lives, the NFV criterion requires that they be evaluated over a common period of analysis (12 years). Hence, Project A will be repeated four times (after the initial invest- ment at t=0) and Project B three times. YearProject AProject B 0  $1000  $2000 1$500$720 2$500$720 3  $500 $720 4$500  $1280 5$500$720 6  $500 $720 7$500$720 8$500  $1280 9  $500 $720 10$500$720 11$500$720 12$500$720 NFV =$2093.20$1904.45 Useful Lives ≠ Analysis Period …

15 9-15 Benefit-Cost Ratio By restating the NPV, NFV, and EACF criteria, we know that an investment is acceptable at a given MARR provided: PV(positive CFs) – PV(negative CFs) ≥ 0 FV(positive CFs) – FV(negative CFs) ≥ 0 EACF(positive CFs) – EACF(negative CFs) ≥ 0 These provisions can be put in terms of the benefit-cost ratio (BCR): BCR = PV(positive CFs)/PV(negative CFs) = EACF(positive CFs)/EACF(negative CFs) If its BCR  1, an investment is acceptable.

16 9-16 Benefit-Cost Ratio … Example: a firm is considering a project that requires an investment of $600K and has the cash flows shown below. Decide whether the firm should accept this project if its MARR is 15%. Use NPV, EACF, NFV, IRR, and BCR criteria. (Answer: the firm should not accept the project. NPV =  $23,311.81 < 0; EACF =  $8165.32 < 0; NFV =  $40,772.50 < 0; IRR = 13.024% < 15%; BCR = 0.96115 < 1). Year1234 Cash Flow$150K$300K$225K$125K

17 9-17 Economic Criteria Restated for Benefit-Cost Ratio Method SituationCriterion Fixed input Amount of capital avail- able is fixed Maximize BCR Fixed output Amount of benefit is fixed Maximize BCR Neither fixed Neither capital nor benefits are fixed 1 candidate: BCR ≥ 1. 2 or more alternatives: use incremental BCR.

18 9-18 BCR and Incremental BCR (∆BCR) The BCR and other criteria such as the NPV, NFV, EACF, and IRR lead to the same conclusion regarding whether to accept individual projects. Similar to the IRR case, we use incremental BCR (∆BCR) to determine the best alternative among two or more competing alternatives. The BCRs of competing alternatives can not be compared directly even if the lifetimes or analysis periods are equal, as they can for the NPV and NFV cases.

19 9-19 BCR and ∆BCR … Example: a company needs a new network and it is considering two systems. System M costs $120K to acquire/install; they estimate its annual net benefit will be $45K for four years. System N ’ s estimated net benefit will be $40K per year for three years at a cost of $85K to acquire/install. Determine which system the company should acquire if they use a 10½ percent rate of return. Use the BCR. (Answer: acquire System M. BCR M = 1.2194; BCR N = 1.2274;  BCR M  N = 1.0493)

20 9-20 BCR and Competing Alternatives Example: Moose County has three alterna- tives for a new road. Benefits and costs are shown below. Expected road life is 50 years and MARR = 10%. Apply BCR and ∆BCR. Road First Cost Annual Benefits Annual Operating Costs A$25,000$3200$200 B$35,000$3800$250 C$50,000$6050$350

21 9-21 Payback Period Defined as the time taken for the cumulative cash flows of an investment to reach zero. Accept an investment if its payback period ≤ the maximum allowable payback period. Notes about payback period: Payback is an approximate method that ignores the time value of money. After the maximum allowable payback period, all cash flows are ignored. The payback method will not necessarily produce a recommended alternative that is consistent with valuation and rate of return methods.

22 9-22 Payback Period … Exclusive focus on liquidity (no concern for profitability). Liquidity: how quickly can the investment be recovered from the project’s cash flows? All other methods of project evaluation focus on profitability (not liquidity). No time-value discounting, i.e. implicitly assumes that MARR = 0%. The opportunity cost of the project balance, or the unrecovered portion of the capital investment, is ignored. Project Balance = – P + ∑(OR i – OC i ), i = 0,1, … N.

23 9-23 Payback Period … Project balance at any point during its life is: Simple payback: a project ’ s unrecovered investment without discounting. Discounted payback: a project ’ s unrecovered investment taking into account the opportunity cost (i.e., interest charges) of the unrecovered investment. Generally, annual net cash flows (revenues less costs) will cause a project ’ s balance to become more positive over time. Project balance is negative until the initial investment has been fully recovered.

24 9-24 PROJECT BALANCE Project Balance = $0 + 0 – t1t1 t2t2 t3t3 t* First Cost Year 1: Benefits – Costs Payback Period …

25 9-25 Payback Period … YearCash Flow: Project A Cash Flow: Project B 0–$15,000 1$1000$3000 2$2000$3000 3 4$4000$3000 5$5000$3000 6$6000$3000 7$7000$3000 8$8000$3000

26 9-26 Payback Period … Projects A and B both have payback period = 5 years (project balance = 0 after five years). They have very different profitability profiles, however, with Project A being more profitable than Project B by the end of its lifetime. Since we focus only on liquidity (how fast the initial investment is recovered), the payback method makes us indifferent between projects A and B.

27 9-27 Payback Period … Payback period = 3 + 300/1000 = 3.3 years. Compare 3.3 years to an industry threshold to determine if the project is acceptable. If similar investments require, on average, three years to recover, this project would be rejected. YearCash FlowProject Balance 0 – $1000 1 – $500 – $1500 2+$500 – $1000 3+$700 – $300 4+$1000+$700

28 9-28 Payback Period … Discounted Payback Period Year Cash Flow Opportunity Cost (10%) Project Balance 0 – $15000 1$3500 – $1500 – $13000 2$3500 – $1300 – $10800 3$3500 – $1080 – $8380 4$3500 – $838 – $5718 5$3500 – $571.8 – $2789.8 6$3500 – $278.98$431.22

29 9-29 Payback Period … Discounted payback period = 5 + $2789.8(1.1)/$3500 = 5.877 years. If the industry average discounted payback period for this type of project is > 5.877 years, this project is acceptable. Otherwise, it is not. We can see that the NFV is $431.22 (> 0), therefore the project should be considered acceptable.

30 9-30 Payback Period … We can also use a present value approach for the discounted payback period. Find the point at which the cumulative PVs of the cash flows becomes 0. For the same example. Discounted payback period = 5.877 years (same as above). Note, the NPV = $243.41. If the discounted payback period ≤ the life- time, a project should be considered accep- table since this means that NPV ≥ 0. Regret- tably, such a criterion is not always specified.

31 9-31 Sensitivity & Break-even Analysis Economic data represent projections of expenditures and returns. These projections ultimately affect our decisions. To consider our choice of a decision more fully, we should play a “ what if ” game to determine the amount of change in a project parameter that might alter the decision. This allows us to make an assessment of the inherent risk in a project, i.e. the probability of different outcomes due to uncertainty in the project variables.

32 9-32 Sensitivity & break-even analysis … Projected and actual cash flows may differ due to: Technological change: changes to production costs Changes in the size and number of competing firms Introduction of new products: substitutes or complements Changes to key macroeconomic variables, e.g. inflation, unemployment, economic growth, exchange rate International events

33 9-33 Sensitivity & break-even analysis … Sensitivity example: System cost = $200,000; annual profit = $36,000; salvage value = $23,500; lifetime = 10 years; cost of capital = 12%. Use sensitivity analysis to determine which project components affect NPV most.

34 9-34 Sensitivity & break-even analysis … Break-even example: using the information from the sensitivity example, find the break- even values for the initial investment and for the annual profit (the parameters to which the NPV is most sensitive).

35 9-35 Sensitivity & break-even analysis … Scenario example: using the information from the sensitivity example, perform a scenario analysis. Worst case: $210,000 investment, $37,500 annual profit, $17,500 salvage value, lifetime seven years, cost of capital 13%. Best case: $190,000 investment, $40,500 annual profit, $30,000 salvage value, lifetime ten years, cost of capital 11%.

36 9-36 Suggested Problems 9-22, 23, 29, 32, 41, 43, 45, 56, 59 (use 0.74570 kW/hp).


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