We can do the analysis of best alternative based on present worth analysis. We calculate the present worth of alternative A. PW of A = -10,000 - 2000(P/A,i, n) + 3000(P/A, i, n) +2000(P/F, i,n) . Here I=7% and n=infinity for alternative A. Therefore PW of A = -10,000 -2000( P/A, 7%, infinity) + 3000(P/A, 7%, infinity) + 2000(P/F, 7%, infinity) = -10000 -2000/0.07 +3000/0.07 +0. As n=infinity salvage value at present period is negligible. So PW of A = -10,000 -2000/0.07+3000/0.07 = $4285.71 =$ 4286.
Similarly we can calculate PW of B is = -22000 +6000(P/A, 7%, 2) +10000(P/F, 7%, 2) = -10000 + 6000*1.808 +10000*0.8734 = $9582. (P/A, i, n) = {(1+0.07)^2 -0.07}/{0.07(1.07)^2}=1.808, (P/F,I,n) = (1.07)^-2=0.8734
So by comparing alternative A and B we can say present worth of A is less than PW of B. As B's PW is more ($9582) than A($4286) so we can say B is best alternative.
6. Analyze the two economic alternatives described below and seleot the annual interest rate of 7%...
Alternatives X and Y are described below. X involves using a CISC microprocessor while alternative Y is about using ASIC microprocessor. Wh you select based on their Future Worth? Use an annual interest rate of 5% for both alternatives. All values are in $ 7. ich microprocessor would Alternative Initial Cost Yearly Operating Expenses 6,000 500 2,500 1,500 6 4,000 Annual Revenues Salvage Value Life (years 1,000 900
ANSWER THE FOLLOWING QUESTIONS:-
Three mutually exclusive design alternatives are being considered. The estimated cash flows for each alternative are given. The interest rate is 20% per year. At the conclusion of the useful life, the investment will be sold A C Investment cost $28,000 $55,000 $13,000 $28,000 $8,000 $40,000 Annual expenses Annual revenues $15,000 $23,000 $6,000 10 years $22,000 $32,000 13 $10,000 Salvage value Useful life 10 years 10 years A decision-maker can select one of these alternatives or...
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II First Cost 160,000 25,000 Annual Operating Cost 15,000 3,000 Salvage Value 1,000,000 4,000 Life, Years
3. Compare the alternatives shown below on the basis of their Annual Worth, using an interest rate of 12% per year. Alternative I Alternative II 160.000 25,000 First Cost 15.000 3,000 Annual Operating Cost 1,000,000 4,000 Salvage Value Life. Years
Question 7 (15 points) Consider 2 alternatives with the following cash flows below at interest rate of 10%. Which is the more desirable alternative? A B Investment Cost $100,000 $120,000 Year 1: $1,000 Year 1: $1,500 Yearly Costs Year 2: $1,000+ $400 Year 2: $1,500+ $300 (From Year 1 to Year 3: $1,000 + 2 x $400 Year 3: S1,500 + 2 x $300 End of Useful Life)... Year 15: $1,000 + 14 x $400 Year 30: $1,500 + 29...
4. You are to choose between two competing alternatives using Annual Cash Flow Analysis. Altemative A has an expected life of ten (10) years. After making the appropriate calculations for Alternative A, it was found that EUAW- $4,450. Given below are the characteristics of Alternative B. Based on these characteristics, determine which option, trany, you should choose. MARR-10%. (S5233) Alternative B $75,000 3,000 each year S30,000 the first year, dropping by $2,000 each year after $10,000 Initial Cost O&M Costs...
2. Given the following data, if the interest rate is 10%, use present worth analysis to find the best alternative, A, B, or C. А в Initial cost, $ 10,000 15,000 12.000 Annual benefit. S 6,000 10,000 5.000 Salvage value, S 1,000 -2.000 3,000 Useful life, years 2
USE
ANNUAL WORTH Analysis
Two mutually exclusive design alternatives are being considered. The estimated cash flows for each alternative are given in the following table with MARR = 10% per year. Suggest your recommendation by using multiple attribute annual worth AW analysis. Design A Design B Investment cost (RM) 28,000 55,000 One-off expenses 10,000 13,000 In year 4 (RM) Annual revenues (RM) 22,000 28,000 Market value (RM) 6,000 8,000 Useful life 10 years 6 years
Question 1 The cash flows given in table below are for two different alternatives. MARR =10% Data IN Initial Cost Annual Benefits Salvage Value Useful Life in years M $20,000 $6,000 $5,000 $80,000 $10,000 $20,000 a) Determine the annual worth of alternative M b) Determine the annual worth of alternative N
5-73 Given the following data, use present worth analysis to find the best alternative, A, B, or C $10,000 15,000 $12,000 Initial cost Annual benefit 6,000 10,000 5,000 Salvage value 1,0002,000 3,000 Useful life 4 years 3 years 2 years Use an analysis period of 12 years and 15% interest.
5-73 Given the following data, use present worth analysis to find the best alternative, A, B, or C $10,000 15,000 $12,000 Initial cost Annual benefit 6,000 10,000 5,000 Salvage value...