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?

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.

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 .

its ii) i did this test

explanation: