ii) Select which alternative using AW and i= (10) 12% - per year. First Cost, $ -200,000 80,000 Annual operating cost, S/year 7,000 -30,000 Salvage value, $ 5,000 Life, years
Homework Answers
(i) Calculate the Annual Worth AW of Material X:-
First Cost = $15000
Maintenance Cost = $9000
Salvage = $2000
Life = 5 Years
PVAF @15% for 5 years = 3.3522
PVIF @15% for 5 years = 0.4972
[-First cost – Maintainance cost per year (PVAF @15% for 5 years) + Salvage (PVIF @15% for 5 years)]/ PVAF @15% for 5 years
[-$15000 - $9000(3.3522) + $2000(0.4972)]/3.3522
[-$15000 – $30169.8 + $994.4]/3.3522
-$44175.4/3.3522
= -$13178
Calculate the Annual Worth AW of Material Y:-
First Cost = $35000
Maintenance Cost = $7000
Salvage = $20000
Life = 5 Years
PVAF @15% for 5 years = 3.3522
PVIF @15% for 5 years = 0.4972
[-First cost – Maintainance cost per year (PVAF @15% for 5 years) + Salvage (PVIF @15% for 5 years)]/ PVAF @15% for 5 years
[-$35000 - $7000(3.3522) + $20000(0.4972)]/3.3522
[-$35000 – $23465.4 + $9944]/3.3522
-$48521.4/3.3522
= -$14474
On the basis of above analysis, Material X should be selected
(ii) Calculate the Annual Worth AW of Alternative C:-
First Cost = $80000
Annual Operating Cost = $30000
Salvage = $5000
Life = 8 Years
PVAF @12% for 8 years = 4.9676
PVIF @12% for 8 years = 0.4039
[-First cost – Annual Operating cost per year (PVAF @12% for 8 years) + Salvage (PVIF @12% for 8 years)]/ PVAF @12% for 8 years
[-$80000 - $30000(4.9676) + $5000(0.4039)]/4.9676
[-$80000 – $149028 + $2019.5]/4.9676
-$227008.5/4.9676
= -$45698
Calculate the Annual Worth AW of Alternative D:-
First Cost = $200000
Annual Operating cost = $7000
(-First cost * Int Rate) – Maintainance cost per year
(-$200000 * 0.12) - $7000
= - $31000
On the basis of above analysis, Alternative D should be selected
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