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- Question 3. Choose the best piping systems as shown in the table below using Annual Worth method (MARR: 6%). Concrete piping UPVC piping Initial cost (OR) Ca Initial cost (OR) Annual 0&M (OR/y)" 10000 Annual O&M (OR/yr) 5000 Rehabilitation cost at the 120000 5th year Useful life (years)¤ 10 Useful life (years)= 20 Ca 1720102 1136000 1828000:A company that manufactures amplified pressure transducers wishes to decide between the machines shown - variable speed (VS) and dual speed (DS). Determine the range of Ai*, incremental ROR. (i.e. 5%-6%). First cost, $ AOC, $ per year Overhaul in year 3, $ Overhaul in year 4, $ Salvage value, $ Life, years Vs -260,000 -230,000 -40,000 55,000 6 a. Ai*vs-ds 19% - 20% O b. Ai* vs.Ds = 10%-11% OC.Ai* *VS-Ds = 4%-5% O d. Ai* vs-Ds = 17% -18% VS-DS DS -215,000 -235,000 -24,000 14,000 6For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10% First cost, $ Annual cost, $/year Salvage value, $ Life, years Machine A 15000 B- AW for machine B= 8,342 4,000 Answer the below questions : Machine B 6 20,447 6,000 5,000 Machine C 10000 4,000 1,000
- A solid-waste recycling plant is considering two types of storage bins using an MARR of 10% per year. (a) Use ROR evaluation to determine which should be selected. (b) Confirm the selection using the regular AW method at MARR = 10% per year. Storage Bin P Q First cost, $ −18,000 −35,000 AOC, $ per year −4000 −3600 Salvage value, $ 1000 2700 Life, years 3 6A company that manufactures amplified pressure transducers wishes to decide between the machines shown - variable speed (VS) and dual speed (DS). Determine the range of Ai*, incremental ROR. (i.e. 5%-6%). First cost, $ AOC, $ per year Overhaul in year 3, $ Overhaul in year 4, $ Salvage value, $ Life, years VS -260,000 -230,000 -40,000 55,000 6 O a Ai* vs-ps = 19% - 20% Ob.Ai* vs os = 4%-5% OcAi* vs os = 10% -11% Ⓒd. Ai vs Ds = 17%-18% DS -215,000 -235,000 -24,000 14,000 6For the following alternatives compute the Delta B/C ratio of Alternative D minus Alternative A. Use 11% as MARR. (Remember for our convention, salvage value is a minus cost.) Project Initial Investment A -1500 -2000 -2500 -5200 Annual Benefit 350 500 600 850 Salvage Value Useful Life 320 610 820 2300 O1.27 1.18 O 1.46 O 1.73 O 0.92
- 6. Compare the following three alternatives by the IRR method, given MARR of 6%/year. First find if they are feasible and then compare them with the incremental rate of retum method (AROR). Construction cost S Benefits S/yr Salvage S Alt. Service Life (yrs) 510,000 145,000 -10,000 775.000 155,000 15,000 1,075,000 165,000 20,000 6.For the following alternatives compute the Delta B/C ratio of Alternative D minus Alternative A. Use 11% as MARR. (Remember for our convention, salvage value is a minus cost.) Project A B D Initial Investment -1500 -2000 -2500 -5200 Annual Benefit 350 500 600 850 Salvage Value Useful Life 320 610 820 2300 5 6 7 9Locations under consideration for a border patrol station have their costs estimated by the federal government. Use the B/C ratio method at an interest rate of 8% per year to determine which location to select, if any. Location North N South S Initial Cost. $ 960,000 2,900,000 Annual Cost, $ per Year 480,000 450,000 Disbenefits, $ per Year 70,000 60,000 00 Life, Years 00 The AB/C ratio is Select location (Click to select) v
- a- At 10%, find PW for the following Table Warehouse cost equipment cost Installation cost Annual maintenance costs Annual Revenues Salvage Machine life 0 1 2 3 -90,000 b- A project with an IRR = 20%, it has the following NCF (S) X X 4000 (SUS) 50,000 46.000 Find X? 6,500 8,000 45,000 15,000 15 yearsFor the following alternatives compute the Delta B/C ratio of Alternative C minus Alternative B. Use 11% as MARR. (Remember for our convention, salvage value is a minus cost.) Project Initial Investment Annual Benefit Salvage Value Useful Life 2.13 2.98 1.18 0.99 1.28 A -1500 350 320 5 B -2000 500 610 6 -2500 600 820 7 D -5200 850 2300 9An engineer must choose between two technically equivalent alternatives, A and B. Based upon the data in the table below over the 12year analysis period and an interest rate of 9%, the PW of machine A is: A B First Cost $3,000 $6,000 Annual Maintenance $500 $300 End of Useful Life Salvage Value $700 $1,000 Useful Life 5 years |15 уears O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/A,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,10)-500(P/A,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/F,9%,15)+700(P/F,9%,5)+700(P/F,9%,10)+ 700(P/F,9%,15) O PWA=-3000-3000(P/F,9%,5)-3000(P/F,9%,10)-500(P/A,9%,10)+700(P/F,9%,5)+700(P/F,9%,10)+ 700(P/F,9%,15)