PPE Part 2. Chapter 16 - Lecture notes 1,2 PDF

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Chapter 16Property, Plant and Equipment (Part 2)PROBLEM 1: TRUE OR FALSE1. FALSE – PAS 16 defines depreciation as the“systematic allocation of the depreciable amount of anasset over its estimated useful life.”2. FALSE – revaluation model3. TRUE = (120K – 20K) ÷ 10 years = 10K annualdepreciation;10K ...

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Chapter 16 Property, Plant and Equipment (Part 2) PROBLEM 1: TRUE OR FALSE 1. FALSE – PAS 16 defines depreciation as the “systematic allocation of the depreciable amount of an asset over its estimated useful life.” 2. FALSE – revaluation model 3. TRUE = (120K – 20K) ÷ 10 years = 10K annual depreciation; 10K annual depreciation ÷ 100K depreciable amount =10% 4. FALSE – 800K 5. TRUE 6. FALSE 7. TRUE 8. TRUE 9. FALSE – recognized in OCI and accumulated in equity 10. FALSE – (180K – 10K) – 200K = 30K loss PROBLEM 2: MULTIPLE CHOICE – THEORY 1. D 2. B 3. A F Depreciation starts when the asset is available for use in the manner intended by management. F Costs incurred while an item capable of operating in the manner intended by management has yet to be brought into use are recognized as expenses. 4. 5. 6. 7.

D – see the word “not” in the problem D D D - PAS 16 encourages the note disclosure of the gross carrying amounts of fully depreciated assets. If the fully depreciated assets were removed from the

Page | 2 ledger, information on the gross carrying amounts to be disclosed in the notes would not be readily available. 8. B 9. D 10. D PROBLEM 3: EXERCISES 1. Solutions: Requirement (a): Straight line method Initial cost (Historical cost) of equipment Residual value Depreciable amount Divide by: Estimated useful life

1,000, 000 (100,0 00) 900,0 00 4 225,0 00

Annual depreciation

 Depreciation table: Date Jan. 1, 20x1 Dec. 31, 20x1 Dec. 31, 20x2 Dec. 31, 20x3 Dec. 31, 20x4

Depreciati on

Accumulated depreciation

Carrying amount 1,000,000

225,000

775,000

450,000

550,000

225,000

675,000

325,000

225,000

900,000

100,000

225,000 225,000

900,000

 Journal entries: Dec. 31, 20x1 Dec. 31, 20x2

Depreciation expense Accumulated depreciation Depreciation expense Accumulated

225,00 0 225,00 0

225,00 0 225,00

Page | 3 depreciation

0

Requirement (b): Sum-of-the-years’ digits method SYD denominator = Life x [(Life + 1) ÷ 2] SYD denominator = 4 x [(4 + 1) ÷ 2] = 10

 Depreciation table:

Date

Deprecia ble amount

SY D rat e

Depreciati on

Accumulat ed depreciati on

1/1/x1 12/31/ x1 12/31/ x2 12/31/ x3 12/31/ x4

900,000 900,000 900,000 900,000

4/1 0 3/1 0 2/1 0 1/1 0

360,000

360,000

270,000

630,000

180,000

810,000

90,000

900,000

Carryin g amount

1,000,0 00 640,00 0 370,00 0 190,00 0 100,00 0

900,000

 Journal entries: Dec. 31, 20x1 Dec. 31, 20x2

Depreciation expense Accumulated depreciation Depreciation expense Accumulated depreciation

360,00 0 270,00 0

360,00 0 270,00 0

Requirement (c): Double declining balance method Double declining rate = 2 ÷ Life Double declining rate = 2 ÷ 4 = 50% Year 20x1 20x2 20x3 20x4

(1M x 50%) (1M – 500K) x 50% (1M – 500K – 250K) x 50% (1M – 500K – 250K – 125K – 100K RV)

Depreciati on 500,000 250,000 125,000 25,000

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 Depreciation table: Date

Jan. 1, 20x1 Dec. 31, 20x1 Dec. 31, 20x2 Dec. 31, 20x3 Dec. 31, 20x4

Depreciati on

Accumulat ed depreciatio n

Carrying amount

1,000,000 500,000

500,000

500,000

250,000

750,000

250,000

125,000

875,000

125,000

25,000

900,000

100,000

900,000

 Journal entries: Dec. 31, 20x1 Dec. 31, 20x2

Depreciation expense Accumulated depreciation Depreciation expense Accumulated depreciation

500,00 0 250,00 0

500,00 0 250,00 0

2. Solutions: Requirement (a): Based on Input Depreciation rate = Depreciable amount ÷ Estimated total hours Depreciation rate = 900,000 ÷ 12,000 Depreciation rate = 75 per hour of input Dec. 31, 20x1 Dec. 31, 20x2

Depreciation expense (3,600 x 75)

Accumulated depreciation Depreciation expense (3,000 x 75)

Accumulated depreciation

270,00 0 225,00 0

270,00 0 225,00 0

Requirement (a): Based on Output Depreciation rate = Depreciable amount ÷ Estimated total units

Page | 5 Depreciation rate = (900,000 ÷ 720,000) Depreciation rate = 1.25 per unit of output Dec. 31, 20x1

x 1.25)

Dec. 31, 20x2

x 1.25)

Depreciation expense (240K Accumulated depreciation Depreciation expense (200K Accumulated depreciation

300,00 0 250,00 0

300,00 0 250,00 0

3. Solution: Yr. 1 2 3 4

Straight line (75,000* / 4) = 18,750

SYD

18,750

4/10 x 75,000* = 30,000 3/10 x 75,000 = 22,500

18,750

2/10 x 75,000 = 15,000

18,750

1/10 x 75,000 = 7,500

Double declining balance 50% x 80,000 = 40,000 50% x 40,000 = 20,000 50% x 20,000 = 10,000 50% x 10,000 = 5,000

* 80,000 - 5,000 = 75,000 depreciable amount

Yea r 20x 1

Straight line 18,750 x 9/12 = 14,062.50

20x 2

18,750

20x 3

18,750

20x 4

18,750

20x 5

18,750 x 3/12 = 4,687.50

SYD

Double declining balance

30,000 x 9/12 = 22,500

40,000 x 9/12 = 30,000

30,000 x 3/12 = 7,500 22,500 x 9/12 = 16,875 22,500 x 3/12 = 5,625 15,000 x 9/12 = 11,250 15,000 x 3/12 = 3,750 7,500 x 9/12 = 5,625

40,000 x 3/12 = 10,000 20,000 x 9/12 = 15,000

7,500 x 3/12 = 1,875

20,000 x 3/12 = 5,000 10,000 x 9/12 = 7,500 10,000 x 3/12 = 2,500 5,000 x 9/12 = 3,750

5,000 x 3/12 = 1,250

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The asset is acquired on Mar. 18, 20x1 (last half of the month). Accordingly, it is depreciated starting on Apr. 1, 20x1.

4. Solution: Step 1: Carrying amount as at the beg. of the period of change  Double declining balance rate (2 ÷ Life) or (2 ÷ 20% 10 yrs.)

20,480,00 0

 Carrying amt. on Jan. 1, 20x4 (40M x 80% x 80% x 80%)

Step 2: Apply the changes Carrying amount on Jan. 1, 20x4 Residual value Depreciable amount Divide by: Revised remaining useful life (12 yrs. – 3 yrs.)

Straight line depreciation

Journal entry: Dec. 31, Depreciation expense 20x4 Accumulated depreciation

20,480, 000 (2,000,000) 18,480,000 9 2,053,333

2,053,33 3

2,053,33 3

5. Solutions: Requirement (a): Replacement cost Less: Depreciation (21M x 20/60(a)) Fair value Carrying amount (10M – 5M) Revaluation surplus – gross of tax

21,000, 000 (7,000,00 0) 14,000, 000 (5,000,00 0) 9,000,0

Page | 7 00 (2,700,000) 6,300,000

Less: Deferred tax (9M x 30%) Revaluation surplus – net of tax

(a) Total economic life = Effective life + Remaining economic life (20 + 40 = 60)

Requirement (b):

 Proportional method Historical Cost Building Accum. depreciation CA/ DRC/ RS

(a)

Replacement cost

10,000,000

21,000,000

(5,000,000)

(7,000,000)

5,000,000

14,000,000

Increase 11,000,0 00 (2,000,0 00) 9,000,00 0

(a)

Carrying amount/ Depreciated replacement cost/ Revaluation surplus – gross of tax Date

Building Accumulated depreciation Revaluation surplus Deferred tax liability

11,000,0 00

2,000,0 00 6,300,00 0 2,700,0 00

 Elimination method Dat e

Accumulated depreciation (elimination) Building (balancing figure)

Revaluation surplus Deferred tax liability

5,000,00 0 4,000,00 6,300,00 0 0 2,700,0 00

The building’s carrying amount after the revaluation is analyzed as follows: Building (10M + 11M); (10M + 4M) A/D (5M + 2M); (5M - 5M) Carrying amount (equal to fair value)

Proportion al 21,000,000 (7,000,000) 14,000,000

Eliminati on 14,000,000 14,000,00 0

Page | 8 Requirement (c): Fair value Residual value Depreciable amount Divide by: Revised annual depreciation

14,000,000 14,000,000 40 350,000

6. Solutions: Requirement (a): 48,000,0 00 (30,000,0 00) 18,000,00 0 (5,400,0 00) 12,600,0 00

Fair value Less: Carrying amount (40M – 10M) Revaluation surplus - gross of tax Less: Deferred tax (18M x 30%) Revaluation surplus - net of tax

Requirement (b):  Proportional method Building Accum. depreciation Carrying amount

Historical Cost 40,000,000

Fair value

% change

48,000,0 00

160%*

(10,000,000) 30,000,000

* (48,000,000 ÷ 30,000,000) = 160% increase

Building Accum. depreciation Carrying amount Dat e

Historical Cost 40,000,000

% change 160%

Revalued amounts 64,000,000

(10,000,000)

160%

(16,000,000)

30,000,000

Building (64M – 40M) Accum. depreciation (16M – 10M)

Deferred tax liability Revaluation surplus

48,000,000

24,000,0 00

6,000,00 0 5,400,00 0

Page | 9 12,600,0 00

Requirement (b): Elimination method 10,000,0 Accumulated depreciation (elimination) 00 Building (balancing figure) 8,000,00 5,400,0 Deferred tax liability 00 0 Revaluation surplus 12,600,0 00

Date

The building’s carrying amount after the revaluation is analyzed as follows: Proportio nal Building (40M + 24M); (40M + 8M)

64,000,000

Accum. Depreciation (10M + 6M); (10M - 10M)

(16,000,00 0) 48,000,00 0

Carrying amount (equal to fair value)

Eliminati on 48,000,00 0 48,000,0 00

PROBLEM 4: MULTIPLE CHOICE – COMPUTATIONAL 1. D  SLM = (1M x 95%) ÷ 10 = 95,000  SYD denominator = {10 x [(10 + 1) ÷ 2]} = 55 SYD depreciation in 20x2 = 950,000 x 9/55 = 155,455  DDB rate = 2 ÷ 10 = 20% DDB depreciation in 20x2 = 1M x 80% x 20% = 160,000  UOPM (input) depreciation in 20x2 = 950,000 x (2,800/28,000) = 95,000  UOPM (output) depreciation in 20x2 = 950,000 x (9,800/84,000) = 110,833 2. C Purchase price Commission

480,000 20,000

P a g e | 10 Freight Installation and testing Total cost Residual value Depreciable amount

22,000 18,000 540,000 (40,000) 500,000

 SLM = 500,000 x 8/10 + 40,000 = 440,000  SYD denominator = {10 x [(10 + 1) ÷ 2]} = 55 SYD accumulated depreciation on Dec. 31, 20x2 = 500,000 x [(10 + 9) ÷ 55] = 172,727 SYD carrying amount on Dec. 31, 20x2 = 540,000 172,727 = 367,273  DDB rate = 2 ÷ 10 = 20% DDB carrying amount on Dec. 31, 20x2 = 540M x 80% x 80% = 345,600  UOPM (input) Accumulated depreciation on Dec. 31, 20x2 = 500,000 x [(2,000 + 2,700 ) ÷ 25,000] = 94,000 Carrying amount on Dec. 31, 20x2 = 540,000 – 94,000 = 446,000  UOPM (output) Accumulated depreciation on Dec. 31, 20x2 = 500,000 x [(8,000 + 10,000 ) ÷ 100,000] = 90,000 Carrying amount on Dec. 31, 20x2 = 540,000 – 90,000 = 450,000 3. C Solution: SYD denominator = Life x [(Life + 1) / 2] = 4 x [(4+1) / 2] = 10 Historical cost

20,000

Estimated residual value

(2,000)

Depreciable amount

18,000

Depreciation - 20x1 (18,000 x 4/10)

7,200

Depreciation - 20x2 (18,000 x 3/10)

5,400

P a g e | 11

Depreciation - 20x2 (18,000 x 2/10)

3,600

Accumulated depreciation - 12/31/20x3

16,200

Historical cost

20,000

Accumulated depreciation - 12/31/20x3

(16,200)

Carrying amount - 12/31/20x3

3,800

4. B Yr.

2

Straight line (100,000 – 10,000) ÷ 5 = 18,000 18,000

3

18,000

4 5

18,000 18,000

1

SYD 90,000 x 5/15 = 30,000 90,000 x 4/15 = 24,000 90,000 x 3/15 = 18,000 90,000 x 2/15 = 12,000 90,000 x 1/15 = 6,000

5. A (110,000 – 5,000) ÷ 10 yrs. = 10,500 6. A Solution: 150% declining balance rate = 1.5/Life = 1.5/5 = 30% Depreciation - 20x1 (200,000 x 30%) Depreciation - 20x2 (200,000 - 60,000) x 30%

60,000

Accumulated depreciation - 12/31/x2

102,000

7. D Solution:

42,000

P a g e | 12  Composite life = Depreciable amount ÷ Annual depreciation Composite life = 280,000 ÷ 70,000 = 4 years  Composite rate = Annual depreciation ÷ Total cost Composite rate = 70,000 ÷ 290,000 = 24.14%

 Depreciation in current year: Total depreciable amount Depreciable amount of old tools Depreciable amount of new tools Revised depreciable amount Divide by: Original composite life Revised annual depreciation

280,000 (8,000) 12,000 284,000 4 71,000

Total cost Cost of old tools Cost of new tools Revised total cost Multiply by: Original composite rate Revised annual depreciation

290,000 (8,000) 12,000 294,000 24.14% 70,972*

* Answer rounded-off to 71,000.

8. C Solutions:  Retirement method: Cost of disposals (12,000 + 24,000 + 36,000) Net disposal proceeds (1,000 + 1,600 + 2,000) Depreciation expense

72,000 (4,600) 67,400

 Replacement method: Cost of additions as replacements (20,000 + 44,000)

Cost of disposals but not replaced Proceeds from sale of old tools (1,000 + 1,600 + 2,000)

Depreciation expense

 Inventory method: Tools beg. bal.

300,000

4,600

Proceeds from asset disposals

64,000 24,000 (4,600) 83,400

P a g e | 13

Additions 124,000

67,40 0 352,00 0

Depreciation (squeeze) end. bal. (per physical count)

9. C  Useful life = 20 years  Remaining lease term as of 12/31/01 = (9* + 5 renewal) = 14 * Dec. 31, 2001 completion date of improvements to Dec. 31, 2010 end of original lease term = 9 yrs.

 Shorter = 14 years  480,000 x 13/14 = 445,714 10. C Solution: Step 1: Carrying amount as at the beg. of the period of change  Double declining balance rate (2 ÷ Life) or (2 ÷ 10% 20 yrs.)  Carrying amt. on Jan. 1, 20x8

(5M x 90% x 90% x 90% x 90% x 90% x 90% x 90%)

2,391,485

Step 2: Apply the changes  Remaining life = 20 yrs. – 7 yrs. = 13 years  SYD denominator = {13 x [(13 + 1) ÷ 2]} = 91  Carrying amount on Jan. 1, 20x8 2,391,485 Revised residual value (200,000 – 20,000) (180,000) Depreciable amount 2,211,485 Multiply by: 13/91 SYD depreciation in 20x8 315,926 11. D 264, 000

Historical cost Original estimated useful life

8 33,0 00

Original depreciation per year

Historical cost Accumulated depreciation - 1/1/x3 (33,000

264,000

P a g e | 14 x 3 yrs.)

(99,000)

Carrying amount - 1/1/x3

165,000

Revised residual value

(24,000)

Revised depreciable amount Divide by: Revised useful life (6 yrs. - 3 yrs.)

141,000

Depreciation - 20x3

47,000

3

Accumulated depreciation - 1/1/x3 (33,000 x 3 yrs.) Depreciation - 20x3 Accumulated depreciation - 12/31/x3

12. C Solutions: (1) Jan. 1, Cash 20x7 Accumulated depreciation

(1.8M x

5/15)

Loss on replacement (squeeze)

99, 000 47,0 00 146 ,000

100,000 600,000 1,100,0 00 1,800,0 00

Equipment (old part) to derecognize the old part Jan. 1, 20x7

Equipment (new part) Cash to recognize the new replacement part

2,100,0 00 2,100,0 00

On derecognition, the difference between the carrying amount of the derecognized PPE and the net disposal proceeds, if any, is recognized as gain or loss in profit or loss. (2) Jan. 1, 20x7

Cash Accumulated depreciation (2.1M x

100,000 700,000

P a g e | 15 5/15)

Loss on replacement (squeeze)

1,300,0 00 2,100,0 00

Equipment (old part) to derecognize the old part Jan. 1, 20x7

Equipment (new part) Cash to recognize the new replacement part

2,100,0 00 2,100,0 00

13. A Solution: (1) Replacement cost Less: Depreciation (50M x 8(a)/32(b)) Fair value (Depreciated replacement cost) Less: Carrying amount (40,000,000 – 16,000,000) Revaluation surplus, gross of tax Less: Deferred tax consequence (13.5M x 30%) Revaluation surplus, net of tax

50,000,00 0 (12,500,0 00) 37,500,00 0 (24,000,0 00) 13,500,00 0 (4,050,00 0) 9,450,00 0

(a)

Effective life (Effective age) Total economic life = Effective life + Remaining economic life = (8 + 24) = 32 (b)

Fair value (Depreciated replacement cost) Divide by: Remaining economic life Revised annual depreciation

14. B Solutions: Replacement cost

37,500,0 00 24 1,562,5 00

30,000,000

P a g e | 16

Less: Depreciation (30M – 3M) x 7(a)/28 Fair value Carrying amount (22M – 2M) x 19/25 + 2M Revaluation surplus, gross of tax Less: Deferred tax consequence (6.050M x 30%) Revaluation surplus, net of tax – 12/31/x6

(6,750,000 ) 23,250,00 0 (17,200,00 0) 6,050,000 (1,815,000 ) 4,235,000

(a)

28 yrs. total economic life – 21 yrs. remaining economic life = 7 yrs. effective life

(1) Carrying amount of building on 12/31/x7: Fair value on 12/31/x6 23,250,000 (3,000,000 Revised residual value ) 20,250,00 Revised depreciable amount 0 Divide by: Remaining economic life 21 Revised annual depreciation 964,286 Fair value on 12/31/x6 Less: Depreciation in 20x7 Carrying amount of building on 12/31/x7

23,250,000 (964,286) 22,285,71 4

(2) Carrying amount of revaluation surplus on 12/31/x7: Revaluation surplus, net of tax – 12/31/x6 4,235,000 Divide by: Remaining economic life 21 Annual transfer to retained earnings 201,667 Revaluation surplus, net of tax – 12/31/x6 Less: Amount transferred to R/E in 20x7 Revaluation surplus, net of tax – 12/31/x7

4,235,000 (201,667) 4,033,333

P a g e | 17 15. A Solution:  Building: Replacement cost Less: Depreciation (12M x 10/40*) Fair value Carrying amount [8M - (8M x 15**/25)] Revaluation surplus – gross of tax Multiply by: Revaluation surplus – net of tax (Building)

12,000,00 0 (3,000,000 ) 9,000,000 (3,200,000 ) 5,800,000 70 % 4,060,000

* 10 yrs. effective life + 30 yrs. remaining life = 40 total economic life **Actual life

 Patio: Replacement cost Less: Depreciation (4.2M x 10/25*) Fair value Carrying amount [3M – (3M x 10**/20)] Revaluation surplus – gross of tax Multiply by: Revaluation surplus – net of tax (Patio)

4,200,000 (1,680,000 ) 2,520,000 (1,500,000 ) 1,020,000 70 % 714,000

* 10 yrs. effective life + 15 yrs. remaining life = 25 total economic life **Actual life

Total Revaluation Surplus, net of tax: (4.06M + 714K) = 4,774,000

16. D Solution:

P a g e | 18

17. D Solution:  Changes in accounting estimates in 20x4: Step 1: Carrying amount as at...