Ask Question
25 May, 07:35

You can choose between two purchases: Machine A or Machine B.

Machine A costs $24,000 and has a salvage value of $12,000 after 3 years. Machine B costs $30,000 and has a salvage value of $16,000 after 4 years. You can lease a Machine B equivalent for $6,000 per year, if you initially purchased Machine B. You need a machine for a total of 6 years, and can purchase a new machine in the future at the same price with the same salvage value.

Required:

a) If i is 9% annual rate compounded annually, which machine should be purchased?

+2
Answers (2)
  1. 25 May, 07:47
    0
    Answer: machine A should be purchased since the cost is lower than the cost of machine B

    Explanation:

    Cost - Salvage value / number of years

    Cost = $24,000, salvage value = $12,000, n = 3 years

    24,000 - 12,000 / 3

    = 12,000 : 3

    = 4,000

    The yearly depreciation = $,4000

    Machine B

    Cost = $30,000, Salvage value = $16,000, n = 4 years

    30,000 - 16,000 / 4

    = 14,000 : 4

    = 3,500

    The yearly depreciation = $3,500

    Since machine B can be lease for $6,000 per year

    6,000 * 4 = 24,000

    To determine which machine to be purchased

    . Amount = P (1 + r / 100) ∧n

    Machine A principal = $4,000, r = 9 : 100 = 0.09, n = 6

    4,000 (1 + 0.09) ∧6

    4,000 (1.677100110841)

    = 6,708.40 - 4,000

    = 2,708.4

    Machine B principal $3,500, r = 0.09, n = 6

    3,500 (1 + 0.09) ∧6

    = 5,869.85 - 3,500

    = 2,369.85

    Amount + the amount earned from the lease of machine B

    = 2,369.85 + 24,000

    = 26,369.85

    The machine A should be purchased, since machine B was initially purchased and the price is higher than machine A and since new machine can be purchased in the future at the same salvage value.
  2. 25 May, 08:03
    0
    The price of machine A is less than that of machine B so the machine A will be bought

    Explanation:

    For the machine A

    Total Duration required=6 years

    Cost of Machine A=$24000

    Operational Time=3 years

    Salvage Value=$12000

    So for the total time of 6 years

    The machine A is to be bought for two times and salvaged for two times thus

    Total Cost of Machine A for 6 years = (Cost of Machine-Salvage Value) * 2

    Total Cost of Machine A for 6 years = ($24000-$12000) * 2

    Total Cost of Machine A for 6 years = ($12000) * 2

    Total Cost of Machine A for 6 years=$24000

    For the machine B

    Total Duration required=6 years

    Cost of Machine B=$30000

    Operational Time=4 years

    Salvage Value=$16000

    So for the total time of 6 years

    The machine B is to be bought for 1st time, the machine is to be salvaged and further the machine B is leased for the next two years at 9% interest compounded annually for $6000 per year

    Total Cost of Machine B for 6 years = (Cost of Machine-Salvage Value) + (Leasing Cost)

    Leasing Cost=First Installment+1st Interest+2nd Installment+2nd Interest

    Leasing Cost=6000+2160+6000+1814=15974

    Total Cost of Machine B for 6 years = (30000-16000) + (15974)

    Total Cost of Machine B for 6 years=$29974

    As the price of machine A is less than that of machine B so the machine A will be bought.
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “You can choose between two purchases: Machine A or Machine B. Machine A costs $24,000 and has a salvage value of $12,000 after 3 years. ...” in 📗 Business if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers