Engineering economic analysis 13th edition pdf free download reddit

Therefore, maximize PW of benefits. By inspection, one can see that C, with its greater benefits, is preferred over A and B. Similarly, E is preferred over D. The problem is reduced to choosing between C and E. But the student should recognize that this is a faulty criterion.

It would buy 83 shares of Spartan Products, but only 42 shares of Western House. The criterion, therefore, is to maximize NPW for the amount invested. Buy Spartan Products. Thus we use 12 years and assume repeatability of the cash flows.

Problem has the same effective interest rate as , but the rate on is lower. Solving this series for A gives us the A for the infinite series. In this situation the annual capital recovery cost equals interest on the investment. This problem is much harder than it looks! The equipment purchase did not turn out to be desirable.

The problem must be segmented to use the 1. The bailer probably should be installed. The fact that the tax payments are for the fiscal year, July 1 Through June 30, does not affect the computations. Ruarterly interest payments to the savings account could have an impact on the solution, but they do not in this problem. The solution may be verified by computing the amount in the savings account on Dec. Now solve for the unknown n. Note: The analysis period is seven years, hence one cannot compare three years of A vs.

Seventy or seventy- five years might be the range of reasonable estimates. Here we will use 71 years. Two possible solutions are provided below.

Whether working or at school there are living expenses. Available interest tables obviously are useless. Convenience, improved quality of life, increased value of the dwellings, etc. Thus, the pipeline appears justified. Therefore, the increment is desirable. Select X. Therefore it is not a desirable increment of investment.

Choose A. Select A. Select B. Buy Kicko. There is external investment until the end of the tenth year. To search for positive rates of return compute the NPW for the cash flow at several interest rates. This is done on the next page by using single payment present worth factors to compute the PW for each item in the cash flow.

Then, their algebraic sum represents NPW at the stated interest rate. Even though there is only one rate of return, there still exists the required external investment in Ruarter 1 for Ruarter 2. On this basis the Part b solution appears to have more realistic assumptions than Part a. Before proceeding, we will check for multiple rates of return. This, of course, is not necessary here. For further computations, see the solution to Problem This is only slightly different from the Tables could be produced, of course, for negative values.

Reject D and retain A. Reject A and accept B. Conclusion: Select Plan B. The rate of return for each Plan is computed. Two incremental analyses are performed.

Reject Plan A. Retain Plan B. Reject Plan C. Since at the same cost B produces a greater annual benefit, it will always be preferred over C. C may, therefore, be immediately discarded. Retain A. Conclusion: Select Alternative A. C- A increment satisfactory Choose C. C- A increment satisfactory. Choose C. C- A increment unsatisfactory. Reject D and retain C. The B- C increment is undesirable. Reject B and retain C.

The A- C increment is undesirable. Reject A and retain C. Conclusion: Select alternative C. Alternative B is preferred over Alternative A. The C- B incremental rate of return of 6. Reject C. Select Alternative B.

Select Y. Do nothing. Assuming this is not recognized, one would first compute the rate of return on the increment B- A and then C- B.

The problem has been worked out to make the computations relatively easy. Decision Reject A. Keep B. Select C. Effective Reject 2. Reject 1 and select 3 continue as is. Select Pump 1. Proceed with incremental analysis. Examine increments of investment. So Alternative 1 can be rejected. This leaves alternatives 2 and 4. Examine increment. Choose Alternative D. Reject Company A. The correct choice is the Regular model.

Since the B-A increment is not acceptable, Alternative B should not be adopted. Compute the amount of annual income for each alternative situation. To maximize annual income, choose C. 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'. The print version of this textbook is ISBN: ,. Anyone have this in pdf version?? I would definitely share something else if I have something; lets help each other out! Get an ad-free experience with special benefits, and directly support Reddit.

Get reddit premium. Anyone have Engineering Economic Analysis 11th edition by D. Preview Download. His co-authors, Donald Newnan and Ted Eschenbach, have both taken home the prize in the past. This eleventh edition of the market-leading Engineering Economic Analysis offers comprehensive coverage of financial and economic decision-making for engineers, with an emphasis on problem solving, life-cycle costs, and the time value of money.

And with the most extensive support package , this is the easiest book to teach from. Each is designed to promote independent, self-paced instruction in this vital tool. Makes a wonderful companion to both the text and the Casebook.