Chapter 6: Annual Cash Flow Analysis Engineering Economic Analysis Canadian Edition.

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Chapter 6: Annual Cash Flow Analysis Engineering Economic Analysis Canadian Edition

6-2 Chapter 6 … nDefine equivalent annual cash flow (EACF) as annual (benefits  costs). nConduct an economic analysis of an investment based on EACF. nDetermine when an analysis based on EACF, rather than NPV, is required. nUse EACF to compare alternatives with equal or unequal lives over a fixed analysis period.

6-3 Annual Cash Flow Calculations nResolving a present value to an annual value: the simplest approach is to convert the NPV to a series of equivalent annual cash flows or EACFs (previously we used the symbol A). nThe text calls these EUACs (equivalent uniform annual costs) or EUABs (equivalent uniform annual benefits). We will use the EACF terminology since it is obvious whether a cash flow is a cost or a benefit.

6-4 Annual Cash Flow Calculations … nExample: Find the EACF of an investment that requires $1.35m today, has net profits of $0.45m for six years and a salvage value of $0.25m if the required rate of return is 11½%. nExample: Calculate the EACF of a project that costs $280K today and generates the following cash flows for the next four years, consecutively: $50K, $100K, $125K, $75K. Use 9¾% as the required rate of return.

6-5 Annual Cash Flow Calculations … nSome points regarding cash flow calculations: 1.The EACF is the annuity payment that gives the present value (NPV) when discounted by the required rate of return, and vice versa. 2.The EACF increases (becomes more positive) when benefits increase, and decreases when costs increase. 3.The EACF is a simple, convenient representation for a series of irregular cash flows or for a series of cash flows with regular changes, i.e. an arithmetic gradient series or geometric series.

6-6 Annual Cash Flow Analysis Input/outputSituationCriterion Fixed inputAmount of capital available is fixed Maximize EACF Fixed outputAmount of benefit is fixed Maximize EACF Neither fixedNeither amount of capital nor amount of benefit is fixed Maximize EACF

6-7 Annual Cash Flow Analysis … nAnalysis Period Cases: 1.Analysis period is equal to alternative lives 2.Analysis period is a common multiple of alternative lives 3.Analysis period for a continuing requirement 4.Infinite analysis period 5.Some other analysis period such as project life

6-8 Annual Cash Flow Analysis … nCase 1: Analysis period = alternative lives Base the comparison on the common lifetime. We can compare NPVs directly or compare EACFs of alternatives. This is rarely the case in real life. nCase 2: Analysis period = a common multiple of alternative lives When the lives of the equipment in the alternatives vary, use a common multiple of the lives. We can compare NPVs or EACFs of alternatives. This is often unrealistic because we must assume that benefits and costs apply to multiple lives.

6-9 Annual Cash Flow Analysis … nCase 3: Analysis period for a continuing requirement It can be assumed that the project will last for a long or indefinite time period. Use a common multiple of the lives of the alternatives since the length of the analysis period is not important. nCase 4: Infinite analysis period This is the most straightforward case since the EACF for an infinite analysis period = EACF for one lifetime (or for any finite period of time as long as all relevant costs and benefits are considered).

6-10 Annual Cash Flow Analysis … nCase 5: Some other period An appropriate analysis period is chosen, e.g. the life of one of the alternatives, the project life, or some other period. For each alternative, we repeat lifetimes, including a possible partial lifetime at the end, to match the analysis period. We can compare the NPVs or the EACFs of the alternatives with repeated lifetimes. This is the most common case in real life.

6-11 Annual Cash Flow Analysis … nExample: a city needs a water purification system for a 75-year period. System X costs $35m today and will require annual mainten- ance costing $1.75m. System X must be re- placed after 25 years. System Y costs $28m today, will last 15 years, and will cost $2.15m annually to maintain. Recommend one of the two systems if the city uses a 10¼% rate of return for systems like this. Calculate the difference in the values today for making the better choice over the planning period.

6-12 Suggested Problems n6-16, 18, 19, 22, 29, 31, 32, 33, 38, 43, 45, 47.