In: Accounting
Equipment1 was purchased at the beginning of the year 2016 for $50,000 cash. No salvage/residual value. Straight-line depreciation is used over a 10-year life.
Equipment2 was also purchased at the beginning of the year for 550,000 (no salvage) 10 year life. We decided to use SL method. The equipment2 required a $5,000 repair by year-end.
Equipment3 was purchased on 6/1 for 100,000 (20,000 salvage value)., 10 year life. We decided to use SYD as a depreciation method. At 12/31/2016 it required a capital improvements of $40,000 which we signed a note to pay in 9 months.
Prepare Journal entries for all transactions
Solution:
Journal
| Date | Journal | Debit | Credit | 
| 1/1/2016 | Equipment 1 | $50,000 | |
| To cash | $50,000 | ||
| (Being equipment 1 purchases for cash) | |||
| 1/1/2016 | Equipment 2 | $550,000 | |
| To cash | $550,000 | ||
| (Being equipment 2 purchases for cash) | |||
| 6/1/2016 | Equipment 3 | $100,000 | |
| To cash | $100,000 | ||
| (Being equipment 3 purchases for cash) | |||
| 12/31/2016 | Depreciation | $5000 | |
| To Accumulated Depreciation-Equipment 1 | $5000 | ||
| (Being depreciation provided for equipment 1 on straight line method) | |||
| 12/31/2016 | Depreciation | $55,000 | |
| To Accumulated Depreciation-Equipment 2 | $55,000 | ||
| (Being depreciation provided for equipment 2 on straight line method) | |||
| 12/31/2016 | Repairs | $5000 | |
| To cash | $5000 | ||
| (Being repairs paid on equipment 2 ) | |||
| 12/31/2016 | Equipment 3 - Capital improvements | $40,000 | |
| To Notes Payable | $40,000 | ||
| (Being amount of capital improvement capitalized in equipment 3) | |||
| 12/31/2016 | Depreciation | $12,727 | |
| To Accumulated depreciation- Equipment 3 | $12,727 | ||
| (Being depreciation provided on equipment 3 using sum of years digit method) | 
Workings: Calculation of depreciation:
Equipment 1 = ($50,000/10) = $5,000
Equipment 2 = ($550,000/10) = $ 55,000
Equipment 3 = Sum of years = [n x (n+1) ] / 2 = (10 x 11)/2 = 55
= (years left/ sum of years) x (Original cost- residual value)
= (10/55) x ($100,000+40,000- 20,000) x (7 /12)
= $12,727