Problem 2. Compare the following mutually exclusive alternatives using the conventional incremental B/C ratio with annual worth to determine the most economical alternative with an interest rate of 15% per year. The Do Nothing option does not exist. Y 50K 75K 25K Initial Cost ($) Maintenance cost ($/yr) Income ($/yr) Life (years) 20K 25K 30K 120K 170K 150K 10 15 20

Problem 2. Compare the following mutually exclusive alternatives using the conventional incremental B/C ratio with annual worth to determine the most economical alternative with an interest rate of 15% per year. The Do Nothing option does not exist. Y 50K 75K 25K Initial Cost ($) Maintenance cost ($/yr) Income ($/yr) Life (years) 20K 25K 30K 120K 170K 150K 10 15 20

Essentials Of Investments

11th Edition

ISBN:9781260013924

Author:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Publisher:Bodie, Zvi, Kane, Alex, MARCUS, Alan J.

Chapter1: Investments: Background And Issues

Section: Chapter Questions

Problem 1PS

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Consider two mutually exclusive alternative. Construction a choice table using PW for interest rate from 0% to 20%. If the minimum attractive rate of return is 8%, which alternative should be selected? Alternative A B Fist cost $2,000 $3,000 Uniform annual benefit $450 $600 Salvage Value $200 700 Useful life 6 years 6 years Hint: Equation: PW_A = 450(P/A, i, 6) + 200(P/F, i, 6) – 2000 PW_B = 600(P/A, i, 6) + 700(P/F, i, 6) – 3000 or Spreadsheet: PW_A = -PV(i, nper, pmt, fv) – 2000 = -PV(i, 6, 450, 200) - 2000 PW_B = -PV(i, nper, pmt, fv) – 3000 = -PV(i, 6, 600, 700) - 3000
QUESTION-1 A defender and a challenger are considered in a replacement analysis. The following decision tree includes the data where FC: first cost, AB: annual benefit; C: Challenger and D: Defender. If annual compound interest rate is 10%, when must we replace the defender? Node no branch Cash flow FC=-45,000 $ Life=20 yrs AB= +8,500 $ 1 10 FC=-10,000 $ Life=5 yrs AB= +4,500 $ FC=-10,000 $ Life=5 yrs AB= +5,500 $ FC=-35,000 $ Life=15 yrs 2 D 2 1D 20 AB= +4,500 $ FC=-10,000 $ Life=5 yrs AB= +5,000 $ FC=-25,000 $ Life=10 yrs AB= +9,000 $ FC=-20,000 $ Life=5 yrs 1D 20 4 10 AB= +6,000 $ FC=-15,000 $ Life=5 yrs 2 D AB= +5,000 $ 3.
An alternative with an infinite life has a B/C ration of 1.5. The alternative has benefits of P50,000 per year and annual maintenance costs of P10,000 per year. The initial cost of the alternative at an interest rate of 10% per year is closes to which value? Select one: O a. P233,000 o b. P23,300 o c. P85,400 o d. P146,100
Given the financial data in the table below for two mutually exclusive alternatives, determine the value "X" for the two alternatives to be equally attractive. Use an interest rate of 10% per year. Initial cost $2,500 $4,000 Annual benefit 500 600 Life You do not need to interpolate, just answer the closest n that surrounds your number. For example between year so and so.
Two different alternatives are being compared. Using Net Present Worth Analysis which alternative is best? Use an interest rate of 9%. Alternative Initial Cost A В $ 1,425 $ 2,168 $ 1,325 $ $ Annual Benefit 620 Salvage Value Useful Life (years) 465 $ 897 3 9. A is the best alternative Choose neither alternative Choose either alternative B is the best alternative
Consider the following two investment alternatives. Determine the range of investment costs for Alternative B (i.e., min. value < X
Consider alternatives A and B as shown below. The interest rate is 9%. For parts a) and b), let X = $250. For parts a) and b), which alternative is preferred: a) by equivalent annual worth analysis? b) by payback period? c) Compute the value of X that makes the two alternatives equally desirable. Initial cost Uniform annual benefit Salvage value Useful life, in years A $500 200 100 5 B $900 X 120 5
4. If the terminal balance of a 5-year project at 10% s0 then which one below is true? A. The IRR of the project is less than 10% B. The future worth of the project is greater than 0 C. Discounted payback perlod is 5 years OD. The present worth of the project is less than 0 5 In an incremental IRR analysis, which analysis period should be used to compare 2 mutually altemative projects with 3 and 4 years of useful lives? (Assume an indefinite analynis perlod and that the projects are repeatable) O A. B years O B. 12 years OC. 4 years D. 3 years
Which alternative should be selected using the incremental rate of return analysis, if MARR =11.0%? Do- nothing A B C D First Cost 0 $10,000 $4000 $10,000 $7000 Annual benefit 0 1,806 828 1,880 1,067 Life 10 Years ROR 12.5% 16.0% 13.5% 8.5% a. B, because its ROR is the highest b. Something other than C, because C costs the most initially c. C, because the C-B increment has a ROR of 11.78% and the A-B increment has a ROR of 10.5% d. C because C has the highest annual benefit
Consider the following four alternatives. Three are do-something and one is Do -Nothing. Alternative A B D Cost $O $50 $30 $40 Net annual benefit Useful life (years) Which is the preferred alternative? If 10% interest rate is selected. Use PW analysis. $O $14 $5 $7 5 10 10
An alternative has the following cash flows: Benefits of $50,000 per year; Disbenefits of $27,000 per year; Initial cost of $250,000; M&O costs of $10,000 per.year. If the alternative has an infinite life and the interest rate is 5% per year. What is the B/C ratio for this alternative? O 0.91 O 1.02 O 1.16 O 0.58 73° Chi
Time Value Comparison of Single Amounts Assume you need to choose one of the alternatives shown in the following table in exchange for a payment of $20,000 today. Alternative 1: $28,500 at the end of 3 years Alternative 2: $54,000 at the end of 9 years Alternative 3: $160,000 at the end of 20 years Required: Find the value today of each alternative assuming your opportunity cost is 11%. Are all the alternatives acceptable, meaning are they worth $20,000 today? Which alternative, if any, will you choose?

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