05 Lecture 5

# 05 Lecture 5 - Efficiency Measures So now that you can...

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Efficiency Measures So, now that you can develop business scenarios, estimate cashflows, discount cashflows, etc. How do you know if this is a "good" project? This is where efficiency measures come in. We will not be focusing on typical engineering efficiency measures, but instead financial efficiency measures. Such as, does the project recover its costs and provide a suitable return on capital. There is not one specific measure of efficiency that covers all cases, but there are commonly used measures that taken together can go a long way toward providing an answer to the question "is this a good project?" To determine the answer to this question, we will cover Payback Period, Present Worth, Rate of Return, and Net Present Value Index. Efficiency Measures: Payback Period The simplest measure of project financials is how long it takes to recover the initial investment dollar for dollar on an undiscounted basis. This period of time is called the payback period or simply, payback . This is measured by creating a cumulative cashflow account where all negative cashflows are accumulated to a total negative cashflow and then as positive cashflows are earned they are accumulated until the initial negative cashflow is recovered, the total accumulation account nets to zero. The payback period is best illustrated by the following example. Question: Determine the payback time for each of the following simple investment projects: Project Identifier Required Investment Returns Salvage Value A \$1000 \$300/yr eoy for 6 years \$400 B \$1000 \$150/yr eoy for 6 years \$400

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C \$1000 \$150/yr eoy for 6 years \$50 D \$1000 \$300/yr continuous for 6 years \$400 E \$1000 \$150/yr eoy forever - F \$1000 \$150/yr continuous forever - Answer: Project A: nA = 6(300) = \$1800 > \$1000 Y = é (1000/300) = é 3.33 = 4 years Project B: nA = 6(150) = \$900. nA + S = 900 +400 = \$1300 Y = n = 6 years Project C: nA + S = 900 + 50 = \$950 < I. Y does not exist Project D: nA = 6(300) = \$1800 > \$1000. Y = 1000/300 = 3.33 years Project E: Y = é (1000/150) = é 6.66 = 7 years Project F: Y = (1000.150) = 6.66 years