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Sum-of-Years’-Digits Example Assume depreciable asset is a car with: o 4 year useful life o Original cost of $22,000 o Salvage Value of $7,000 First, compute.

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Presentation on theme: "Sum-of-Years’-Digits Example Assume depreciable asset is a car with: o 4 year useful life o Original cost of $22,000 o Salvage Value of $7,000 First, compute."— Presentation transcript:

1 Sum-of-Years’-Digits Example Assume depreciable asset is a car with: o 4 year useful life o Original cost of $22,000 o Salvage Value of $7,000 First, compute Depreciable Base = Cost – Salvage Value = $22,000 - $7,000 = $15,000 Then, depreciate base x Sum of Years’ Digits Multiplier

2 Sum-of-Years’-Digits Example YearDepreciable Base = Cost - SV Remaining Life (Years) Depreciation Fraction Depr. Expense 14 23 32 41 10 Sum of Year’s Digits

3 Sum-of-Years’-Digits Example YearDepreciable Base = Cost - SV Remaining Life (Years) Depreciation Fraction Depr. Expense 144/10 233/10 322/10 411/10 1010/10 Depr. Fraction = Remaining Life/Sum of Years’ Digits This was corrected on July 16, 2002.

4 Sum-of-Years’-Digits Example YearDepreciable Base = Cost - SV Remaining Life (Years) Depreciation Fraction Depr. Expense 115,00044/106,000 233/10 322/10 411/10 1010/10 This was corrected on July 16, 2002.

5 Sum-of-Years’-Digits Example YearDepreciable Base = Cost - SV Remaining Life (Years) Depreciation Fraction Depr. Expense 115,00044/106,000 215,00033/104,500 322/10 411/10 1010/10 This was corrected on July 16, 2002.

6 Sum-of-Years’-Digits Example YearDepreciable Base = Cost - SV Remaining Life (Years) Depreciation Fraction Depr. Expense 115,00044/106,000 215,00033/104,500 315,00022/103,000 411/10 1010/10 This was corrected on July 16, 2002.

7 Sum-of-Years’-Digits Example YearDepreciable Base = Cost - SV Remaining Life (Years) Depreciation Fraction Depr. Expense 115,00044/106,000 215,00033/104,500 315,00022/103,000 415,00011/101,500 1010/10 This was corrected on July 16, 2002.

8 Double Declining Balance Example Assume depreciable asset is a car with: o 4 year useful life o Original cost of $22,000 o Salvage Value of $7,000

9 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 125% 2 3 4 Double Declining Balance Example Straight Line %age = 100%/Useful Life

10 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 125%50% 225%50% 325%50% 425%50% Double Declining Balance Example

11 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 122,00025%50%11,000 225%50% 325%50% 425%50% Double Declining Balance Example Start with cost (not cost – salvage value)

12 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 122,00025%50%11,000 2 25%50% 325%50% 425%50% Double Declining Balance Example

13 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 122,00025%50%11,000 2 25%50%5,500 325%50% 425%50% Double Declining Balance Example

14 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 122,00025%50%11,000 2 25%50%5,500 325%50% 425%50% Double Declining Balance Example Too much depreciation—below salvage value!

15 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 122,00025%50%11,000 2 25%50%5,500 325%50% 425%50% Double Declining Balance Example Throw out these final year computed values.

16 Make this enough to arrive exactly at ending salvage value. YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 122,00025%50%11,000 2 25%50%4,0007,000 325%50% 425%50% Double Declining Balance Example

17 Note that depreciation is complete after two years even though asset has four year useful life. YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 122,00025%50%11,000 2 25%50%4,0007,000 325%50% 425%50% Double Declining Balance Example

18 Partial Period Depreciation Note that the prior examples assumed that the assets were put in use on January 1 st of the year they were bought for use. Therefore, we took a full first year of depreciation expense. In reality, assets are usually put into use at all times throughout the year. So, we need to prorate the first year’s depreciation expense and adjust the following years’ depreciation expense accordingly.

19 Partial Period Depreciation This is easy to do with straight-line: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. Normal annual depreciation = $16,000 / 4 = $4,000 per year Year 1 use = 6 months/12 months = ½ year

20 Partial Period Depreciation This is easy to do with straight-line: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. Normal annual depreciation = $16,000 / 4 = $4,000 per year YearUsageNormal Depr.Annual Depr 11/2 2 3 4 5

21 Partial Period Depreciation This is easy to do with straight-line: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. Normal annual depreciation = $16,000 / 4 = $4,000 per year YearUsageNormal Depr.Annual Depr 11/24,000 2 3 4 5

22 Partial Period Depreciation This is easy to do with straight-line: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. Normal annual depreciation = $16,000 / 4 = $4,000 per year YearUsageNormal Depr.Annual Depr 11/24,0002,000 2 3 4 5

23 Partial Period Depreciation This is easy to do with straight-line: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. Normal annual depreciation = $16,000 / 4 = $4,000 per year YearUsageNormal Depr.Annual Depr 11/24,0002,000 214,000 3 4 5

24 Partial Period Depreciation This is easy to do with straight-line: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. Normal annual depreciation = $16,000 / 4 = $4,000 per year YearUsageNormal Depr.Annual Depr 11/24,0002,000 214,000 31 41 5

25 Partial Period Depreciation This is easy to do with straight-line: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. Normal annual depreciation = $16,000 / 4 = $4,000 per year YearUsageNormal Depr.Annual Depr 11/24,0002,000 214,000 31 41 51/24,0002,000

26 Partial Period Depreciation This is harder to do with accelerated (Sum-of-Year’s Digits or Double-Declining Balance): The idea for prorating in a partial period asset placement is the same regardless of the method used for accelerated depreciation.

27 Example: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. He uses double-declining balance method. Partial Period Depreciation This is harder to do with accelerated (Sum-of-Year’s Digits or Double-Declining Balance): First, compute normal annual depreciation as if the asset were used the entire year.

28 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 120,00025%50%10,000 225%50% 325%50% 425%50% Partial Period Depreciation Example: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. He uses double-declining balance method.

29 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 120,00025%50%10,000 2 25%50%5,000 325%50% 425%50% Partial Period Depreciation Example: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. He uses double-declining balance method.

30 YearBook Value, Beg. of Year Straight- Line %age Double SL %age Depr. Expense Book Value, End of Year 120,00025%50%10,000 2 25%50%5,000 3 25%50%1,0004,000 4---25%50%0--- Partial Period Depreciation Example: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. He uses double-declining balance method.

31 Partial Period Depreciation Example: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. He uses double-declining balance method. Since the car was place in service for only ½ the first year, we need to prorate and adjust the depreciation schedule. We effectively do this by taking ½ the first year’s depreciation, and then rolling the rest of the depreciation schedule forward.

32 Partial Period Depreciation Example: JoePa bought a car for $20,000, 4 yr. life, $4,000 salvage value. He started driving the car on July 1. He uses double-declining balance method. YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 1 2 3 4 ½ Use First Year Schedule

33 Partial Period Depreciation First, take ½ of the first year’s normal depreciation. YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 15,000 2 3 4 ½ Use First Year Schedule x 1/2

34 Partial Period Depreciation First, take ½ of the first year’s normal depreciation. Then roll forward the second ½ of the first year’s normal depreciation. YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 15,000 2 3 4 ½ Use First Year Schedule x 1/2

35 Partial Period Depreciation Then add ½ of the second year’s normal depreciation to the roll forward amount. YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 15,000 2 5,000 + 2,500 3 4 ½ Use First Year Schedule x 1/2

36 Partial Period Depreciation Then add ½ of the second year’s normal depreciation to the roll forward amount. YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 15,000 27,500 3 4 ½ Use First Year Schedule

37 Partial Period Depreciation Then roll forward ½ of the second year’s normal depreciation to the third year schedule. YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 15,000 27,500 32,500 4 ½ Use First Year Schedule x 1/2

38 Partial Period Depreciation YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 15,000 27,500 3 2,500 + 500 4 ½ Use First Year Schedule x 1/2 Then add ½ of the third year’s normal depreciation to the roll forward amount.

39 Partial Period Depreciation YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 15,000 27,500 33,000 4 ½ Use First Year Schedule Then add ½ of the third year’s normal depreciation to the roll forward amount.

40 Partial Period Depreciation YearDepr. Expense 110,000 25,000 31,000 40 Normal Schedule YearDepr. Expense 15,000 27,500 33,000 4500 ½ Use First Year Schedule Finally, roll forward ½ of the third year’s normal depreciation to the fourth year schedule. x 1/2

41 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4

42 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000

43 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 $5,000

44 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 $5,000 Yr 1Yr 2Yr 3Yr 4

45 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 $5,000 Yr 1Yr 2Yr 3Yr 4 $5,000

46 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 Yr 1Yr 2Yr 3Yr 4 $5,000

47 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 Yr 1Yr 2Yr 3Yr 4 $5,000 $2,500

48 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 Yr 1Yr 2Yr 3Yr 4 $5,000 $2,500

49 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 Yr 1Yr 2Yr 3Yr 4 $5,000 $2,500 $7,500

50 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 Yr 1Yr 2Yr 3Yr 4 $5,000 $500 $2,500$7,500

51 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 Yr 1Yr 2Yr 3Yr 4 $5,000$2,500$7,500 $500

52 Partial Period Depreciation Another way to look at it conceptually is with a timeline. Yr 1Yr 2Yr 3Yr 4 $10,000$5,000$1,000 Yr 1Yr 2Yr 3Yr 4 $5,000$7,500$500$3,000

53 Fixed Asset Impairment Example Assume Company A has equipment: o Original cost of $120,000 o Accumulated depreciation of $20,000 o Market value of $97,000 o Market interest rate of 8% o Expected cash flows: $24,000 for four years (paid at end of yr.)

54 Fixed Asset Impairment Example Recoverability Test (do we need to record an impairment?) 1.Book value = $120,000 – 20,000 = $100,000 Orig. Cost Accum. Depr.

55 Fixed Asset Impairment Example Recoverability Test (do we need to record an impairment?) 1.Book value = $120,000 – 20,000 = $100,000 2.Net future cash flows = $24,000 x 4 = $96,000 Not discounted for interest rate

56 Fixed Asset Impairment Example Recoverability Test (do we need to record an impairment?) 1.Book value = $120,000 – 20,000 = $100,000 2.Net future cash flows = $24,000 x 4 = $96,000 3.NFCF < BV, so we need to record an impairment

57 Fixed Asset Impairment Example Recoverability Test (do we need to record an impairment?) 1.Book value = $120,000 – 20,000 = $100,000 2.Net future cash flows = $24,000 x 4 = $96,000 3.NFCF < BV, so we need to record an impairment Amount of Impairment Loss 1.Market value is determinable, so use BV – FMV: $100,000 - $97,000 = $3,000

58 Fixed Asset Impairment Example Recoverability Test (do we need to record an impairment?) 1.Book value = $120,000 – 20,000 = $100,000 2.Net future cash flows = $24,000 x 4 = $96,000 3.NFCF < BV, so we need to record an impairment Amount of Impairment Loss 1.Market value is determinable, so use BV – FMV: $100,000 - $97,000 = $3,000 3/31Loss on Impairment3,000 Accum. Depr, Equipment 3,000 Note: Record impairment to equipment

59 Fixed Asset Impairment Example Recoverability Test (do we need to record an impairment?) 1.Book value = $120,000 – 20,000 = $100,000 2.Net future cash flows = $24,000 x 4 = $96,000 3.NFCF < BV, so we need to record an impairment Amount of Impairment Loss 1.Market value is determinable, so use BV – FMV: $100,000 - $97,000 = $3,000 2.If FMV is undeterminable, use BV – Discounted CF

60 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04

61 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04 $24,000

62 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04 $24,000 Each year’s discount rate = 1 (1 + int rate) year

63 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04 $24,000 Each year’s discount rate = 1 (1 + int rate) year 1 (1.08) 1 x 1 (1.08) 2 x 1 (1.08) 4 x 1 (1.08) 3 x

64 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04 $24,000 1 (1.08) 1 x 0.926x 0.857x 0.794x 0.735

65 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04 $24,000 x 0.926x 0.857x 0.794x 0.735 = 22,224= 20,568= 19,056= 17,640

66 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04 $24,000 x 0.926x 0.857x 0.794x 0.735 = 22,224= 20,568= 19,056= 17,640 Net Discounted Cash Flows = 22,224 + 20,568 + 19,056 + 17,640 = $79,488

67 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04 $24,000 x 0.926x 0.857x 0.794x 0.735 = 22,224= 20,568= 19,056= 17,640 Net Discounted Cash Flows = 22,224 + 20,568 + 19,056 + 17,640 = $79,488 Note: We can arrive at the same answer by using the Annuity formula:

68 Fixed Asset Impairment Example Present Value of $1 Annuity 7%8%9% 1 year0.93509260.917 2 years1.8081.7831.759 3 years2.6242.5772.531 4 years3.3873.3123.240 5 years4.1003.9933.890

69 Fixed Asset Impairment Example Present Value of $1 Annuity 7%8%9% 1 year0.93509260.917 2 years1.8081.7831.759 3 years2.6242.5772.531 4 years3.3873.3123.240 5 years4.1003.9933.890

70 Fixed Asset Impairment Example Discounted Cash Flows 1/1/0112/31/0112/31/0212/31/0312/31/04 $24,000 x 0.926x 0.857x 0.794x 0.735 = 22,224= 20,568= 19,056= 17,640 Net Discounted Cash Flows = 22,224 + 20,568 + 19,056 + 17,640 = $79,488 Note: We can arrive at the same answer by using the Annuity formula: $24,000 x 3.312 = $79,488

71 Fixed Asset Impairment Example Recoverability Test (do we need to record an impairment?) 1.Book value = $120,000 – 20,000 = $100,000 2.Net future cash flows = $24,000 x 4 = $96,000 3.NFCF < BV, so we need to record an impairment Amount of Impairment Loss 1.Market value is determinable, so use BV – FMV: $100,000 - $97,000 = $3,000 2.If FMV is undeterminable, use BV – Discounted CF: $100,000 – 79,488 = $20,512


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