QUESTION 3
For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%
Machine A | Machine B | Machine C | ||
---|---|---|---|---|
First cost, $ | 26,538 | 30000 | 10000 | |
Annual cost, $/year | 8,060 | 6,000 | 4,000 | |
Salvage value, $ | 4,000 | 5,000 | 1,000 | |
Life, years | 3 | 6 | 2 |
Answer the below questions:
A- AW for machine A=
QUESTION 4
For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%
Machine A | Machine B | Machine C | |||
---|---|---|---|---|---|
First cost, $ | 15000 | 21,667 | 10000 | ||
Annual cost, $/year | 8,870 | 6,000 | 4,000 | ||
Salvage value, $ | 4,000 | 5,000 | 1,000 | ||
Life, years | 3 | 6 | 2 |
Answer the below questions:
B- AW for machine B=
3)
Annual worth = Net present worth Capital recovery factor
Net present worth of machine A = $ 43576.76784
Annual worth of machine A = $ 43576.76784 ( A/P, 10%, 3 years)
( A/P, 10%, 3 years) = Capital recovery factor
Annual worth of machine A = $ 43576.76784 0.4021
Annual worth of machine A = $ 17522.22
Net present worth machine B = $ 53309.2
Annual worth of machine B = $ 53309.2 ( A/P, 10%, 6 years)
Annual worth of machine B = $ 53309.2 0.2296
Annual worth of machine B = $ 12239.79
Net present worth of machine C = $ 16115.7
Annual worth of machine C = $ 16115.7 ( A/P, 10%, 2 years)
Annual worth of machine C = $ 16115.7 0.5762
Annual worth of machine C = $ 9285.87
Based on the annual worth analysis, machine C must be selected because it has the lowest annual worth of costs compared to machine A and machine B.
---------------------------------------------------------------------------------------------------------------
4)
Annual worth = Net present worth Capital recovery factor
Net present worth of machine A = $ 34053.12
Annual worth = $ 34053.12 ( A/P, 10%, 3 years)
Annual worth machine A = $ 34053.12 0.4021 = $ 13692.76
Net present worth of machine B = $ 44976.2
Annual worth machine B = $ 44976.2 0.2296
Annual worth machine B = $ 10326.53
Annual worth of machine C = $ 16115.7 ( A/P, 10%, 2 years)
Annual worth of machine C = $ 16115.7 0.5762
Annual worth of machine C = $ 9285.87
Based on the annual worth analysis, machine C must be selected because it has the lowest annual worth of costs compared to machine A and machine B.
SOLUTION :
Q4 :
r = MARR = 10% = 0.1
=> (1 + r) = 1.1
PV of costs of Machine A :
= 26538 + 8060/ 1.1 + 8060/1.1^2 + 8060/1.1^3 - 4000/1.1^3
= 43576.77 ($)
PV of costs of Machine B :
= 30000 + 6000 (1.1^6 - 1)/(0.1*1.1^6) - 5000/1.1^6
= 53309.19 ($)
PV of costs of Machine C :
= 10000 + 4000/1.1 + 4000/1.1^2 - 1000/1.1^2
= 16115.70 ($)
Therefore, as per PV of costs , Machine C has lowest cost. So Machine C should be selected.
(ANSWER)
Q3 :
r = MARR = 10% = 0.1
=> (1 + r) = 1.1
PV of costs of Machine A :
= 15000 + 8870/ 1.1 + 8870/1.1^2 + 8870/1.1^3 - 4000/1.1^3
= 34053.12 ($)
PV of costs of Machine B :
= 21667 + 6000 (1.1^6 - 1)/(0.1*1.1^6) - 5000/1.1^6
= 44976.19 ($)
PV of costs of Machine C :
= 10000 + 4000/1.1 + 4000/1.1^2 - 1000/1.1^2
= 16115.70 ($)
Therefore, as per PV of costs , Machine C has lowest cost. So Machine C should be selected.
(ANSWER)
For the below ME alternatives, which machine should be selected based on the AW analysis. MARR=10%
For the below ME alternatives, which machine should be selected based on the PW analysis. MARR=10%.Machine AMachine BMachine CFirst cost, $ 15000 30000 10,360Annual cost, $/year 8,320 6,000 4,000Salvage value, $ 4,000 5,000 1,000Life, years 362Answer the below questions :C- PW for machine C =
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-
QUESTION 3 10 points For the below Me alternatives, which machine should be selected based on the PW analysis. MARR=10% First cost, $ Annual cost, $/year Salvage value, $ Life, years Machine A 28,822 9,821 4,000 Machine B Machine C 30000 10000 6,000 4,000 5,000 1,000 2 Answer the below questions: A-PW for machine A= Click Save and Submit to save and submit. Click Save All Answers to save all answers. Save All Answers Save an
For the below ME alternatives , which machine should be selected based on the PW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $ 24,425 30000 10000Annual cost, $/year 8,323 6,000 4,000Salvage value, $ 4,000 5,000 1,000Life, years 362Answer the below questions :A- PW for machine A=
For the below ME alternatives , which machine should be selected based on the PW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $ 27,272 30000 10000Annual cost, $/year 9,401 6,000 4,000Salvage value, $ 4,000 5,000 1,000Life, years 362Answer the below questions :A- PW for machine A=
For the below ME alternatives , which machine should be selected based on the PW analysis. MARR=10%.Machine AMachine BMachine CFirst cost, $ 15000 30000 10,423Annual cost, $/year 12,340 6,000 4,000Salvage value, $ 4,000 5,000 1,000Life, years 362Answer the below questions :C- PW for machine C =
or the below ME alternatives , which machine should be selected based on the PW analysis. MARR=10%Machine AMachine BMachine CFirst cost, $ 15000 27,898 10000Annual cost, $/year 8,415 6,000 4,000Salvage value, $ 4,000 5,000 1,000Life, years 362Answer the below questions :B- PW for machine B=
Moving to another question will save this response. Question 1 of 8 Question 1 10 points Save Ariswer A new production system for a factory is to be purchased and installed for $169,342. This system will save approximately 300,000 kWh of electric power each year for a 6 year period. Assume the cost of electricity is 50.10 per kWh, and factory MARR IS 15% per year, and the salvage value of the system will be $9,636 at year 6. Using...
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
1. Compare the ME alternatives below using an MARR of 10% per year. Assume that "do nothing" is NOT an option. Pick the better alternative using AW analysis Alternative Initial cost, $ Annual operating cost, $ Life, years 60,000 21,000 400,000 5,000 'forever