# ENGG 401 X2 Fundamentals of Engineering Management Spring 2008 Chapter 7: Other Analysis Techniques Dave Ludwick Dept. of Mechanical Engineering University.

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ENGG 401 X2 Fundamentals of Engineering Management Spring 2008 Chapter 7: Other Analysis Techniques Dave Ludwick Dept. of Mechanical Engineering University of Alberta http://members.shaw.ca/dave_ludwick/

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 2 ENGG 401 X2 – Fundamentals of Engineering Management Benefit-Cost Ratio Analysis Benefit-cost ratio analysisBenefit-cost ratio analysis compares the value gained from a project to the cost of the project. –We expect the benefits to outweigh the costs (i.e., the total value of the benefits should be greater than the total value of the costs). benefit-cost ratioThe idea is that a project with a large benefit-cost ratio (BCR) will be a good project (but bigger isn’t always better). Benefit-cost ratio analysis is frequently used for analyzing pubic sector projects. –and can include costs or benefits that may not necessarily be considered if it was a private investment

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 3 ENGG 401 X2 – Fundamentals of Engineering Management Benefit-Cost Ratio Example You can select between the Baseline model or Gold Standard model for a new piece of equipment your company needs. The equivalent uniform annual costs and benefits of each are as follows: –Baseline model: EUAC = \$50k, EUAB = \$75k –Gold Standard model: EUAC = \$125k, EUAB = \$140k Which model should you choose? –Since the BCR of each is greater than one, then each is individually acceptable, so calculate the incremental BCR: Choose the Baseline model. –Since ∆BCR < 1, the extra benefits we get from the Gold Standard model isn’t worth the extra costs. Choose the Baseline model.

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 4 ENGG 401 X2 – Fundamentals of Engineering Management Benefit-Cost Ratio In-Class Problem #1 You can consider two types of equipment for your company, each with different costs, net annual benefits, salvage values, and useful lives: #1 #2 Purchase cost\$ 200k\$ 700k Annual benefit\$ 95k\$ 120k Salvage value \$ 50k\$ 150k Useful life 6 years12 years Assuming a 10% interest rate, what are their benefit-cost ratios? Which would you choose? Why?

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 5 ENGG 401 X2 – Fundamentals of Engineering Management Benefit-Cost Ratio Analysis with Multiple Options When comparing more than one alternative project, we don’t necessarily want to choose the one with the largest BCR. –For reasons similar to those for not automatically selecting the project with the largest IRR. The method is to calculate incremental costs (∆C) and incremental benefits (∆B) for each progressively more expensive pair, which will allow us to calculate incremental benefit-cost ratios (∆BCR) for those pairs. –If ∆BCR ≥ 1, then the more expensive project is worthwhile, since the incremental benefit is worth more than the incremental cost.

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 6 ENGG 401 X2 – Fundamentals of Engineering Management Benefit-Cost Ratio In-Class Problem #2 Consider that you can choose one of six alternatives, each with the same useful life and with no salvage value: A B C D E F PW cost \$4000\$2000\$6000\$1000\$9000\$10000 PW benefit \$7330\$4700\$8730\$1340\$9000 \$9500 BCR 1.83 2.35 1.46 1.34 1.00 0.95 Question:Question: Which alternative should you select? –Note: We don’t need to know i in this case because we’re given the PW of all of our cash flows. But if all we had were the actual cash flows, we’d need to calculate the PW cost and PW benefit ourselves, so we’d need to know i.

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 7 ENGG 401 X2 – Fundamentals of Engineering Management Payback Period Payback period is the amount of time that elapses before the net benefits of an investment equal its cost. –For projects with high uncertainty, payback period becomes very important. Two main types of payback period analysis: –Discounted payback Uses the specified discount rate, MARR, interest rate, etc. –Simple payback Doesn’t use any discount rate at all (i.e., i = 0, or FV = PV) There are other versions payback: –can consider depreciation, inflation, taxes, etc.

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 8 ENGG 401 X2 – Fundamentals of Engineering Management Payback Period Example Payback of 4 years (actually 3.71 years) Payback of 6 years (actually 5.24 years)

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 9 ENGG 401 X2 – Fundamentals of Engineering Management Payback Period – In-Class Problem What is the payback period for the following two investment options?

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 10 ENGG 401 X2 – Fundamentals of Engineering Management Sensitivity and Break-Even Analysis Sensitivity and break-even analysisSensitivity and break-even analysis is used to determine which value of a particular parameter will result in a break- even scenario (and which parameters a decision is sensitive to). –At break-even, costs will equal revenues, NPV equals zero, two options are equivalent, etc. Example uses: –What cost do we set for a particular project so that it is equivalent to another project? –What timing should be used to build a multi-phase projects? –How will the useful life of a piece of equipment impact a decision?

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 11 ENGG 401 X2 – Fundamentals of Engineering Management Sensitivity and Break-Even Analysis Example Your company needs to build a new plant. –Option A is to build all at once: The plant has the capacity you will need years from now, at a cost of \$140k. –Option B is to build in two phases: Phase 1 provides the capacity you need for the first few years at a cost of \$100k. Phase 2 provides the remaining additional capacity at a cost of \$120k. –Both options have the same total useful lifetime, the same operation and maintenance costs, and no salvage value. With a WACC of 8%, at what time will the cost of both options be equivalent? –What does this mean?

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 12 ENGG 401 X2 – Fundamentals of Engineering Management Sensitivity and Break-Even Analysis Example (2) both options equivalent here (between 14 & 15 years out) Option B Option A The decision of which option to use is only sensitive to the timing if the range of estimates is in the area of 15 years.

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 13 ENGG 401 X2 – Fundamentals of Engineering Management Sensitivity and Break-Even Analysis Example (3) with i = 10%, options are equivalent here (13 years) Option B Option A The decision will also depend on our WACC, or the interest rate we use to calculate our discounted cash flows.

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 14 ENGG 401 X2 – Fundamentals of Engineering Management Sensitivity and Break-Even Analysis Example (4) with i = 6%, options are equivalent here (19 years) Option B Option A

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 15 ENGG 401 X2 – Fundamentals of Engineering Management Break-Even – In-Class Problem #1 Suppose you can choose between two options. –Option A has a cost that is known with some precision to be \$5000, and will provide an net annual benefit of \$700. –Option B has an unknown cost, but we know it will provide an net annual benefit of \$639. –Both options have a 20-year useful life with no salvage value. With a WACC of 6%, which option should we choose?

Dave Ludwick, Dept. of Mech. Eng. Ch 7 – Other Analysis Techniques Summer 2008 16 ENGG 401 X2 – Fundamentals of Engineering Management Break-Even – In-Class Problem #2 You need to replace a component in a piece of equipment used in an environment highly conducive to corrosion. An ordinary part has a cost of \$350, a useful life of only 6 years, and no salvage value. How long a useful life must a more expensive (\$500) corrosion resistant part have if it is preferred over the ordinary part? Assume a WACC of 10%.

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