Question

An asset for drilling was purchased and placed in service by a petroleum production company. Its cost basis is $157280, and it has an estimated market value of $20010 at the end of an estimated useful life of 5 years.

a) Compute the depreciation payments using SL (straight line) method.

b) Use DDB (Double declining balance) method and calculate the depreciation payments. (If it is necessary switch to SL)

c) If the market value at end of the n years is estimated as $0, calculate the DDB payments. (If it is necessary switch to SL)

Answer 1

a.

SL depreciation = (Cost – MV) / Estimated years

                           = (157,280 – 20,010) / 5

                            = 137,270 / 5

                            = 27,454

Depreciation payments

Year	Depreciation payments
1	27,454
2	27,454
3	27,454
4	27,454
5	27,454

b.

SL rate = (1 / Estimated years) × 100

            = (1 / 5) × 100

            = 100 / 5

            = 20%

DDB rate = SL rate × 2 [since it is “double”, means 2]

                 = 20% × 2

                 = 40%

Depreciation payments

Year	Beginning BV	Depreciation payments	Ending BV
1	157,280	157,280 × 40% = 62,912	157280-62912 = 94368
2	94368	94368 × 40% = 37,747	94368 – 37747 = 56,621
3	56,621	56621 × 40% = 22,648	56,621 – 22648 = 33,973
4	33,973	33973 × 40% = 13,589	33,973 – 13,589 = 20,384
5	20,384	20,384 – 20,010 = 374	20,010 = MV

Note: depreciation should be adjusted in 5^th year – the MV has to be kept in the “ending BV”; therefore, depreciation amount would be a difference.

c.

If MV = 0

There would be no adjustment in the 5^th year – ending BV would be 0; therefore, the whole beginning BV would be the depreciation amount.

Depreciation payments

Year	Beginning BV	Depreciation payments	Ending BV
1	157,280	157,280 × 40% = 62,912	157280-62912 = 94368
2	94368	94368 × 40% = 37,747	94368 – 37747 = 56,621
3	56,621	56621 × 40% = 22,648	56,621 – 22648 = 33,973
4	33,973	33973 × 40% = 13,589	33,973 – 13,589 = 20,384
5	20,384	20,384	0