Presentation on theme: "Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques."— Presentation transcript:
Engineering Economic Analysis Canadian Edition Chapter 9: Other Analysis Techniques
9-2EECE 450 — Engineering Economics Chapter 9 … Develops additional alternate methods to solve engineering economic problems: future value benefit-cost ratio payback period sensitivity analysis Links the future value method to the present value and equivalent annual value methods. Uses spreadsheets to perform sensitivity and breakeven analyses.
9-3EECE 450 — Engineering Economics 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.
9-4EECE 450 — Engineering Economics $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= $ )
9-5EECE 450 — Engineering Economics Economic Criteria Projects are judged against an economic criterion. SituationCriterion Fixed inputMaximize output Fixed outputMinimize input Neither fixedMaximize difference (output input)
9-6EECE 450 — Engineering Economics 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)
9-7EECE 450 — Engineering Economics 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.
9-9EECE 450 — Engineering Economics 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, marginally 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).
9-10EECE 450 — Engineering Economics 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
9-11EECE 450 — Engineering Economics Future Value and Present Value Analysis PARAMETERSPROJECT APROJECT BPROJECT C Investment$2500$3500$5000 Annual cost$900$700$ (G=$100) Salvage value$200$350$600 Life (years)555 Annual revenue$1800$1900$2100 (15% growth) MARR10%
9-12EECE 450 — Engineering Economics Calculate the NPV and the NFV for Projects A, B, and C. NPV A = $ ; NFV A = $ NPV B = $ ; NFV B = $ NPV C = $ ; NFV C = $ Future Value and Present Value Analysis …
9-13EECE 450 — Engineering Economics Useful Lives Analysis Period Project A has a three-year life. Project B has a four-year life. Both projects are valid since each has NFV > $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
9-14EECE 450 — Engineering Economics 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 $ $500 $720 10$500$720 11$500$720 12$500$720 NFV =$ $ Useful Lives ≠ Analysis Period …
9-15EECE 450 — Engineering Economics 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.
9-16EECE 450 — Engineering Economics 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, < 0; EACF = $ < 0; NFV = $40, < 0; IRR = % < 15%; BCR = < 1). Year1234 Cash Flow$150K$300K$225K$125K
9-17EECE 450 — Engineering Economics 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.
9-18EECE 450 — Engineering Economics 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.
9-19EECE 450 — Engineering Economics 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 = ; BCR N = ; BCR M N = )
9-20EECE 450 — Engineering Economics 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
9-21EECE 450 — Engineering Economics 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.
9-22EECE 450 — Engineering Economics Payback Period … Focuses exclusively on liquidity (no concern for profitability). Liquidity: how quickly the investment can be recovered from the project’s cash flows. All other methods of project evaluation focus on profitability (not liquidity). Ignores the time-value of money, 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.
9-23EECE 450 — Engineering Economics 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 or required return) of the unrecovered investment. Generally, annual net cash flows (revenues less costs) will cause a project ’ s balance to become more positive over time. The project balance is negative until the initial investment has been fully recovered.
9-24EECE 450 — Engineering Economics PROJECT BALANCE Project Balance = $0 + 0 – t1t1 t2t2 t3t3 t* First Cost Year 1: Benefits – Costs Payback Period …
9-25EECE 450 — Engineering Economics Payback Period … YearCash Flow: Project A Cash Flow: Project B 0–$15,000 1$1000$3000 2$2000$ $4000$3000 5$5000$3000 6$6000$3000 7$7000$3000 8$8000$3000
9-26EECE 450 — Engineering Economics Payback Period … Projects A and B both have payback period = 5 years (project balance = 0 after five years). They have very different profitability profiles: Project A is 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.
9-27EECE 450 — Engineering Economics Payback Period … Payback period = /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 – $ – $500 – $ $500 – $ $700 – $300 4+$1000+$700
9-29EECE 450 — Engineering Economics Payback Period … Discounted payback period = 5 + $2789.8(1.1)/$3500 = years. If the industry average discounted payback period for this type of project is > years, this project is acceptable. Otherwise, it is not. We can see that the NFV is $ (> 0), therefore the project should be considered acceptable.
9-30EECE 450 — Engineering Economics 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 = years (same as above). Note, the NPV = $ 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.
9-31EECE 450 — Engineering Economics 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.
9-32EECE 450 — Engineering Economics 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
9-33EECE 450 — Engineering Economics Sensitivity & break-even analysis … Sensitivity example: System cost = $200,000; annual profit = $36,000; salvage value = $23,500; lifetime = 10 years; MARR = 12%. Use sensitivity analysis to determine which project components affect the NPV most.
9-34EECE 450 — Engineering Economics 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).
9-35EECE 450 — Engineering Economics Sensitivity & break-even analysis … Scenario example: using the information from the sensitivity example, perform a scenario analysis. Worst case: $210,000 investment, $33,000 annual profit, $17,500 salvage value, lifetime eight years, MARR 13%. Best case: $195,000 investment, $37,500 annual profit, $26,500 salvage value, lifetime eleven years, MARR 11 ½ %.