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.
Makes a wonderful companion to both the text and the Casebook. On-line quiz questions for self study are also included. Following the In this effective Healthiest Way of E Engineering Engineerin PartTwo: The Basics Econom Design and Economics.
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