AW of alternative A=AW1=-25000*(A/P,0.10,3)-3000+1100*(A/F,0.10,3)
Let us calculate the interest factors
So,
AW of alternative A=AW1=-25000*0.402115-3000+1100*0.302115=-$12720.55
Now take the case of alternative B
AW of alternative B=AW2=-17000*(A/P,0.10,5)-3200+2100*(A/F,0.10,5)
Let us calculate the interest factors
So,
AW of alternative B=AW2=-17000*0.263797-3200+2100*0.163797=-$7340.58
AW of costs is lower in case of alternative B. Alternative B should be selected.
So, correct option is
B
The cash flows for two alternatives are shown. Determine which should be selected on the basis...
The cash flows of two alternatives for an electronic machine are given below. Which one should be selected on the basis of AW-Based rate of return analysis? MARR is 20% per year. A First Cost ($) AOC ($/year) Salvage Value ($) Life (years) -270,000 -135,000 75,000 3 -245,000 - 139,000 35,000 Both O None OB
Either of the cost alternatives shown below can be used in a chemical refining process. If the company’s MARR is 15% per year, determine which should be selected on the basis of an incremental ROR analysis. A B First cost ,$ − 40,000 − 61,000 Annual cost, $/year − 25,000 − 19,000 Salvage value, $ 8,000 11,000 Life, years 5 5 5 - A. B. C. D. E. F. The incremental rate of return computed using a present worth analysis...
Question 1 A. Using Ms Excel, find out which alternative should be selected on the basis of the Present Worth method, if the rate of interest is 8% per year. • Alternative 1: Initial purchase price = $2500000, Annual operating cost $45000 at the end of 1st year and increasing by $3000 in the subsequent years till the end of useful life, Annual income = $120000, Salvage value = $550000, Useful life = 3 years. Alternative 2: Initial purchase price...
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
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
show the formula
Compare the following alternatives on the basis of Present worth Analysis at an interest rate of 6% per year.|| First Cost, $ Annual Operating cost, $ /year Annual Revenues, $/year Salvage Value, $ Life years Petroleum Based Feedstock -350,000 -130,000 300,000 45,000 Inorganic Based Feedstock -120,000 -60,000 290,000 34,000 You should consider Least Common Multiple of 6 and 4) which is 12 years in your analysis instead of 4 or 6 years! Petroleum Based Feedstock Inorganic Based...
Determine whether either of the alternatives below should be
selected. Use an MARR of 15% per year. Incremental Rate of Return
Analysis must be used. (PLEASE DO NOT USE EXCEL).
First Cost Annual Operating Cost Annual Repair Cost Annual Increase in Repair Cost Salvage Value Life (years) Project A -$60,000 -$15,000 -$5,000 -$1,000 $8,000 15 Project B -$90,000 -$8,000 -$2,000 -$1,500 $12,000 15
Consider the following EOY cash flows for two mutually
exclusive alternatives (one must be chosen). The MARR is 5% per
year.
I need the PW of the Lead Acid and Lithium Ion.
Problem 6-28 (algorithmic) EQuestion Help Consider the following EOY cash flows for two mutually exclusive alternatives (one must be chosen) The MARR is 5% per year ead Acid $7,000 thium lon Capital investment Annual expenses Useful life Market value at end of useful life $13,000 $2.500 $2,750 12...
For the below Me alternatives, which machine should be selected based on the future worth analysis. MARR-10% First costs Annual cost, s/year Salvage value, $ Life, years Machine A Machine B 15000 36,202 10000 4,808 4,000 5,000 Machine C 10000 4,000 1,000 Answer the below questions: B. Future worth for machine B, FW B-
Compare two alternatives, A and B. on the basis of a present worth evaluation using /= 10% per year and a study period of 8 years. Alternative A B First Cost $-19,000 $-46,000 Annual Operating Cost $-6,000 $-10,000 Overhaul in Year 4 $0 $-3,850 Salvage Value $1,200 $6,200 Life 8 years 4 years The present worth of alternative A is $ and that of alternative B is $ Alternative (Click to select) is selected.