An alternate solution may be obtained by examining each separable increment of investment. Reject A. Conclusion: Select B. Reject B. Conclusion: Select A. Reject 5 stories. The 10 stories rather than 2 stories is desirable. Conclusion: Choose the story alternative. Conclusion: Choose B. An incremental Uniform Annual Benefit becomes a cost rather than a benefit.
Reject D. Conclusion: Select C. Reject E. Similarly, D, with greater benefits and identical cost, is preferred over B. Hence B may be rejected. On this basis C may be rejected. Therefore, do A. Alternative A may be considered if the investor is very short of cash and the short payback period is of importance to him.
The Present Worth method requires common analysis period, which is virtually impossible for this problem. The problem is easy to solve by Annual Cash Flow Analysis.
In future worth analysis there must be a common future time for all calculations. In this case 12 years hence is a practical future time.
C Alt. Increment B- C Year Alt. B Alt. Solutions for part b : Choose Alternative C. Thus each generates uniform annual benefits in excess of the cost, during the life of the alternative. From this is must follow that the alternative with a 2-year life has a payback period less than 2 years.
The alternative with a 4-year life has a payback period less than 4 years, and the alternative with a 6-year life has a payback period less than 6 years. Thus we see that the shorter-lived asset automatically has an advantage over longer-lived alternatives in a situation like this. While Alternative A takes the shortest amount of time to recover its investment, Alternative C is best for long- term economic efficiency. The problem may be solved by inspection. Alternative z dominates Alternative y.
Alternative z has a positive rate of return actually Choose Alternative z. To maximize NFW, select F. To minimize payback period, select F. Here, however, we will include it. Chapter Uncertainty in Future Events a Some reasons why a pole might be removed from useful service: 1. The pole has deteriorated and can no longer perform its function of safely supporting the telephone lines 2.
The telephone lines are removed from the pole and put underground. The poles, no longer being needed, are removed. Poles are destroyed by damage from fire, automobiles, etc. The street is widened and the pole no longer is in a suitable street location. The pole is where someone wants to construct a driveway.
Typical values for Pacific Telephone Co. The salvage value drops by 5x8,x. The salvage value increases by 5x8,x. So the total probability of higher rates for year 2 is. An inspection of the Regular Season situation reveals that the sum of the probabilities for the outcomes enumerated is 0. Thus one outcome win less than three games , with probability 0.
This is not a faulty problem statement. The student is expected to observe this difficulty. Die 1 Die 2 2 6 3 5 4 4 5 3 6 2 The five ways of throwing an 8 have equal probability of 0. The probability of winning is 0. Here it is ignored. Remember, only the differences between alternatives are relevant. In any year period, for example, there are 4 chances in 10 that a year flood or greater will occur. Using a probabilistic approach, however, Alternative I is most likely to result in the least equivalent uniform annual cost.
The P loss is unchanged at. For example, the first and second rows' PWs are unchanged. The probability of a negative PW is. This illustrates why standard deviation alone is not the best measure of risk. The standard deviation is higher, but the P loss has dropped by half. One can also see this by inspection of the depreciation schedules above. Cost of Proceeds Undep.
This is not a correct analysis of the situation. This may be illustrated by computing the Straight Line depreciation for Year 3, if DDB depreciation had been used in the prior years. One would naturally choose to continue with DDB depreciation. While the depreciation charges in any year may be different for different methods, the sum of the depreciation charges will be the same.
The difference is not the amount of the taxes, but their timing. XYZ, Inc. Further calculations show actual rate of return to be approximately 4.
It does change the timing of these items. Calculator solution is Therefore the project should not be undertaken. The cash investment is greatly reduced. Since the truck rate of return Two items worth noting: 1. The truck and the loan are independent decisions and probably should be examined separately.
There is increased risk when investments are leveraged. Choose Alternative 1. Choose B. By NPW one can see that A is the better of the two undesirable alternatives.
Select Alternative A. Chapter eplacement nalysis For the Replacement Analysis Decision Map, the appropriate analysis method is a function of the cash flows and assumptions made regarding the defender and challenger assets.
Thus, the answer would be the last it depends on the data and the assumptions The replacement decision is a function of both the defender and the challenger. The statement is false. This is such a common situation that the early versions of the MAPI replacement analysis model were based on a one year remaining life for the defender.
The answer is one year. The EUAC of maintenance is constant. Thus total EUAC is declining over time. The book indicates that trade-in value may be purposely inflated as a selling strategy, hence it may or may not represent market value. Retraining in operation and maintenance may be required. High comfort of operation. High purchase price. May not be immediately available. Sales taxes to be paid. Can be depreciated. Supplier warranty and spare parts backup available.
No sales tax applies. Retraining in operation and maintenance is not required. Production will be lost during the rebuilding period.
Cost may be substantially lower than in previous options. The rebuild costs can be expensed. Retraining in operation and maintenance may be required if the new unit is different from the previous one. Immediate delivery is a possibility. The sales tax applies. Equipment can be depreciated. This would lead to the use of Replacement Analysis Technique 2. We chose the options with the smallest EUAC.
If the challenger is superior, then the defender tank probably will be replaced. It will cost a substantial amount of money to remove the existing tank from the plant, sell it to someone else, and then buy and install another one. As a practical matter, it seems unlikely that this will be economical. The best useful life will be the one in which EUAC is a minimum. Year Reconditioned New New vs. The Thus we use Replacement Analysis Technique 1 and compare the marginal cost data of the defender against the min.
EUAC of the challenger. EUAC at its 5-year life. This is because after three years the marginal costs of the Defender become greater than the min. EUAC of the Challenger. Thus, we use Replacement Analysis Technique 1 and compare the marginal cost data of the defender against the min. We would keep the defender asset for two more years and then replace it with the new automated shearing equipment.
From Figure the marginal cost data is available, and it is not strictly increasing see Total MC column in the table below. Thus, we recommend keeping the defender for at least one more year and reviewing the data for changes. Here we are comparing the min. EUAC def vs.
This is because the min. The problem says the challenger economic life is 10 years. Using the data provided this fact could be verified, but that is not part of the problem.
Keep the old forklift another year. A first step is to compute an after-tax cash flow for each alternative. Choose Alternative C. Conclusion: Choose Alternative C. Keep Machine A. In this way the currency itself is less valuable on a per unit basis. These are the dollars that we carry around in our wallets and purses, and have in our savings accounts.
Real dollars are expressed as of purchasing power base, such as Yearbased-dollars. The inflation rate captures the loss in purchasing power of money in a percentage rate form. The market interest rate, also called the combined rate, combines the inflation and real rates into a single rate. Dollars, and interest rates, are used in engineering economic analyses to evaluate projects. As such, the purchasing power of dollars, and the effects of inflation on interest rates, are important.
The important principle in considering effects of inflation is not to mix-and-match dollars and interest rates that include, or do not include, the effect of inflation. A constant dollar analysis uses real dollars and a real interest rate, a then-current or actual dollar analysis uses actual dollars and a market interest rate.
In much of this book actual dollars cash flows are used along with a market interest rate to evaluate projects this is an example of the later type of analysis. The goods included in this index are those commonly purchased by consumers in the US economy e. Composite indexes measure a collection of items that are related. The PPI measures the cost to produce goods and services by companies in our economy items in the PPI include materials, wages, overhead, etc.
Both commodity specific and composite indexes can be used in engineering economic analyses. Their use depends on how the index is being used to measure or predict cash flows. If, in the analysis, we are interested in estimating the labor costs of a new production process, we would use a specific labor cost commodity index to develop the estimate.
Much along the same lines, if we wanted to know the cost of treated lumber 5 years from today, we might use a commodity index that tracks costs of treated lumber. Allowable depreciation charges are based on the original equipment cost and do not increase. Thus the stable price assumption may be suitable in some before-tax computations, but is not satisfactory where depreciation affects the income tax computations.
So purchase pads of paper- one for immediate use plus 4 extra pads. Whether one will profit from owning the house depends somewhat on an examination of the alternate use of the money. Only the differences between alternatives are relevant. Find i'. This problem illustrates the fact that the prospect of future inflation encourages current expenditures to be able to avoid higher future expenditures. This is the year of which the index has a value of One simple rule might involve using the average of the last 5 years inflation rates.
This rate would be 1. If the money available from depreciation charges is inadequate to purchase needed replacement equipment, then the firm may need also to use after-tax profit for this purpose. Depreciation charges produce a tax-free source of money; profit has been subjected to income taxes. Thus substantial inflation forces a firm to increasingly finance replacement equipment out of costly after-tax profit.
Stated in Year 0 dollars, the total receipts are less than the cost, hence there is no positive rate of return. Choose the higher cost alternative: choose Alternative A. Chapter Selection of a Minimum ttractive ate of eturn The interest rates on these securities vary greatly over time, making it impossible to predict rates. Three factors that distinguish the securities: Bond Duration Bond Safety Municipal Bond 20 years Safe Corporate Bond 20 years Less Safe The importance of the non-taxable income feature usually makes the municipal bond the one with the lowest interest rate.
The corporate bond generally will have the highest interest rate. C is preferred for 4. Leasing is therefore preferred at all interest rates above One can make more than One does not have to be concerned about the resale value of the car at the end of two years. Deposit of the money in a Bank.
Purchase of common stock, US Treasury bonds, or corporate bonds. Investment in a new business, or an existing business. The process of identifying and selecting investments is a time-consuming and hence costly process. The opportunity cost of capital is Then the table is sorted with IRR as the key. Projects 3, 1, 4, and 6 should be done. The opportunity cost of capital based on the first rejected project is Often there are those who are advocating for particular projects, those who oppose projects, those who will be immediately affected by such project, and those who may be affected in the future.
Private decision making, on the other hand, is generally focused on increasing stakeholder wealth or investment. This is not to say that private decision-making is entirely focuses on financials, clearly private decision-making focuses on non-monetary issues. However, the goal and objective of the enterprise is economic survival and growth and thus the primary objective is financial in nature for without success financially all other objectives are moot is the firm dissolves.
This approach balances local decisions, which may sub-optimize decision making if not taken. The conflict arises when some regions, states, municipalities perceive that they are consistently passed over for projects that would benefit their region, state, municipality.
Politics have an effect in this regard. However, many projects, including the US parks system, the interstate highway, and others reach many beyond even regional levels.
The Recommended Concept is to select the largest of the cost of capital, the government opportunity cost, or the taxpayer opportunity cost. This quantity is all in the numerator. These leaves only the projects initial costs in the denominator. That is, for any problem, both ratios will either be greater than or less than 1. Advocates of a project may use the method with the larger ratio to bolster their advocacy. Longer durations spread the large initial costs over a greater number of years.
Although this is true for all engineering economy estimates it is particularly true for public projects. It is much easier to estimate labor savings in a production environment than it is to estimate the impact on local hotels of new signage along a major route through town.
Because of this it is not uncommon for there to be turnover in public policy makers. Here the problem specifies a present worth analysis.
Yes No Recommend investing in the Low road, it is the last justified increment. Choose Plan A. Plan C Autos 19, x x 0. This indicates that construction should not be done prior to 19x5 as NPW is not positive. The problem thus reduces to deciding whether to proceed in or The appropriate criterion is to maximize NPW at some point. But, of course, the NPWs must be computed at a common point in time, like Year 0. So the decision to repair the road four years hence is correct. This is a situation of providing the necessary capacity when it is needed- in other words Fixed Output.
Computing the cost is easy, but what is the benefit? Build the half-capacity tunnel now. Our rank order is B, A, D. Yes Yes No Choose Alternative A because it is associated with the last justified increment of investment.
Our rank order is 4, 2, 1, 5, and 6. Rank order is A, then C. Yes No We recommend Alternative A. Our rank order is A then C. Chapter ationing Capital mong Competing Projects a With no budget constraint, do all projects except Project 4. There may be a better solution than simply taking the first six projects, with total NPW equal to There is in this problem.
By trial and error we see that if we forego Projects 1 and 7, we have ample money to fund Project 6. Incremental analysis is required. Project 1B, with a Rate of Return of Therefore the cutoff rate of return is actually Here, each person must be provided a gift. The best solution is to simplify the problem as much as possible and then to proceed with incremental analysis. The other three and four oh alternatives cost more, and the two alternative costs the same as the three oh alternatives.
This can be found by reading across the top of the table on the previous page. Further adjustments are required, first on Mother, then Sister, then Father and finally a further adjustment of Sister. In this situation the replacements do not enter into the computation of NPW. See the data and computations of NPW for this problem. For each project select the alternative which maximizes NPW. The NPW is computed for each alternatives together with any identical replacements. With 22 different alternatives, the problem could be lengthy.
By careful examination, most of the alternatives may be eliminated by inspection. By inspection we can see there must be an external investment prior to year 8 actually in years 6 and 7. Select Alternative 1E. Select Alternative 2D. Reject 4D. Select Alternative 4A.
Select Alternative 5B. Note that this is also the answer to This problem may be solved by the method outlined in Figure The first step is to compute the rate of return for each increment of investment. This appears to be far less than the actual useful life of the tank to Raleigh. Install the Liquid Storage Tank. Project II: nother sulfonation unit There is no alternative available, so the project must be undertaken to provide the necessary plant capacity.
Install Solfonation Unit. Reject the packaging department expansion and plan on two-shift operation. Build the new warehouse. Accounting principles guide the reporting of cash flows for the firm. Engineers and managers can access this information through formal and informal education means, both within and outside the firm.
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