Please help me, solve this question to me Q3: A piece of research equipment is e
ID: 2654323 • Letter: P
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Please help me, solve this question to meQ3: A piece of research equipment is expected to require an investment of $18,000, with $6,000 committed now and next year $7000 while the remaining $6K will be pay out at the end of year 2. Annual operating costs for the system are expected to start in the first year and continue at $800 per year. The life of the equipment is 8 years with a salvage value of $500. Calculate the CR and AW values for the system, if the corporate MARR is 14.18% per year. Explain the results of CR and AW.
Please help me, solve this question to me
Q3: A piece of research equipment is expected to require an investment of $18,000, with $6,000 committed now and next year $7000 while the remaining $6K will be pay out at the end of year 2. Annual operating costs for the system are expected to start in the first year and continue at $800 per year. The life of the equipment is 8 years with a salvage value of $500. Calculate the CR and AW values for the system, if the corporate MARR is 14.18% per year. Explain the results of CR and AW.
Q3: A piece of research equipment is expected to require an investment of $18,000, with $6,000 committed now and next year $7000 while the remaining $6K will be pay out at the end of year 2. Annual operating costs for the system are expected to start in the first year and continue at $800 per year. The life of the equipment is 8 years with a salvage value of $500. Calculate the CR and AW values for the system, if the corporate MARR is 14.18% per year. Explain the results of CR and AW.
Q3: A piece of research equipment is expected to require an investment of $18,000, with $6,000 committed now and next year $7000 while the remaining $6K will be pay out at the end of year 2. Annual operating costs for the system are expected to start in the first year and continue at $800 per year. The life of the equipment is 8 years with a salvage value of $500. Calculate the CR and AW values for the system, if the corporate MARR is 14.18% per year. Explain the results of CR and AW.
Explanation / Answer
Year
investment cost
Operating cost
total cost
Salvage value
0
6000
6000
1
7000
800
7800
2
6000
800
6800
3
800
800
4
800
800
5
800
800
6
800
800
7
800
800
8
800
800
500
MARR (R) = 14.18%
Present value of all cost = cost0 + cost1/(1+R) + cost2/(1+R)^2 + cost3/(1+R)^3 + cost4/(1+R)^4 + cost5/(1+R)^5 + cost6/(1+R)^6 + cost7/(1+R)^7 + cost8/(1+R)^8 - Salvage value / (1+R)^8
Present value of all cost = 6000 + 7800/1.1418 + 6800/1.1418^2 + 6800/1.1418^3 + 6800/1.1418^4 + 6800/1.1418^5 + 6800/1.1418^6 + 6800/1.1418^7 + 6800/1.1418^8 - 500/1.1418^8
Present value of all cost = $38057.44
Equivalent annual worth or cost = Present value of all cost / (( 1-1/(1+R)^n) / R)
Equivalent annual worth or cost =38057.44/ ((1-1/1.1418^8)/.1418) = $8253.65
Now, the equipment has to produce the min value of work of $8253.65 as cost recovery to justify the investment.
Year
investment cost
Operating cost
total cost
Salvage value
0
6000
6000
1
7000
800
7800
2
6000
800
6800
3
800
800
4
800
800
5
800
800
6
800
800
7
800
800
8
800
800
500
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