Replace with treated ties. 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.
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