FAR - Depreciation methods (Becker FAR 4) Flashcards Preview

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Flashcards in FAR - Depreciation methods (Becker FAR 4) Deck (12):
1

Straight-line depreciation formula

Straight line depreciation =

Cost - Salvage value   x   N = Depreciation expense
Estimated useful life        12

(1) Depreciation (put into service On Jan 1 Year 20XX):

$30,000 cost - $5,000 salvage value x  12 
              10 years                                        12 

$25,000 x  12     =  2,500 depreciation expense
    10 yrs       12

(2) Depreciation (put into service On March 1 Year 20XX)

$30,000 cost - $5,000 salvage value x  10 
              10 years                                        12 

$25,000 x  10     
    10 yrs       12

= 2,500  x 10
                  12

= $2,083 depreciation for March 1 to Dec 31.

(3) Depreciation (put into service On July  31 Year 20XX)

$30,000 cost - $5,000 salvage value x  5
              10 years                                        12 

$25,000 x  10     
    10 yrs       12

= 2,500  x  5
                  12

= $1,042 depreciation for July 31 to Dec 31.

 

2

Sum of the years digits formula

Sum of years digits Depreciation expense =

[Cost - salvage value] x  Remaining life of asset x  N
                                          Sum of years digits          12

Example:
$5,000 cost;    $2,000 salvage value;   5 years
Put into service on Jan 1 20X1

Year 1
[$5,000 - $2,000] x   5                           x  12
                                    [1 + 2 + 3 + 4 + 5]      12

= $3,000 x 5 / 15 x (12/12)
= $1,000 (year 1 depreciation expense)

Year 2
[$5,000 - $2,000] x   4                           x  12
                                    [1 + 2 + 3 + 4 + 5]      12

= $3,000 x 4 / 15 x (12/12)
= $800 (year 1 depreciation expense)

Year 3
[$5,000 - $2,000] x   3                           x 12
                                    [1 + 2 + 3 + 4 + 5]     12

= $3,000 x 3 / 15 x (12/12)
= $600 (year 1 depreciation expense)

Year 4
[$5,000 - $2,000] x   2                          x 12
                                    [1 + 2 + 3 + 4 + 5]    12

= $3,000 x 2 / 15 x (12/12)
= $400 (year 1 depreciation expense)

Year 5
[$5,000 - $2,000] x   1                            x 12
                                    [1 + 2 + 3 + 4 + 5]    12

= $3,000 x 1 / 15 x (12/12)
= $200 (year 1 depreciation expense)

 

Example 2: 

$5,000 cost;    $2,000 salvage value;   5 years
Put into service on Aug 1 20X1

Year 1 (August 1 to Dec 31)
[$5,000 - $2,000] x   5                           x  5
                                    [1 + 2 + 3 + 4 + 5]      12

= $3,000 x 5 / 15 x   (5/12 months)
= $417 (year 1 depreciation expense)

Year 2
Jan 1 to July 31: 

[$5,000 - $2,000] x   5                           x  7
                                    [1 + 2 + 3 + 4 + 5]     12

= $3,000 x 4 / 15 x (7/12 months)
= $800 

Aug 1 to Dec 31:
[$5,000 - $2,000] x   4                           x  5
                                    [1 + 2 + 3 + 4 + 5]      12

= $3,000 x 4 / 15 x (5/12 months)
= $333

Total year 2 depreciation expense:
Jan 1 to July 31:    800
Aug 1 to Dec 31:   333
                              1,133 year depreciation expense

 

3

Formula on  Sum of years digits over Long life

[Cost - salvage] x S   or    [Cost - Salv.] x (N x (N+1))/2

**S = N x (N + 1)
             2

Example: asset with 50-year life
cost = 10,000
salvage = 3,000

First:  
S = [50 x (50 + 1)] / 2
S = 2,550 / 2
S = 1,275 sum of years digits

Then: 
Year 1:
[10,000 cost - 3,000 salv] x  (50/1275)
= 7,000 x 0.039
= 273

Year 2:
[10,000 cost - 3,000 salv] x  (49/1275)
= 7,000 x 0.038
= 266

Year 3:
[10,000 cost - 3,000 salv] x  (48/1275)
= 7,000 x 0.037
= 259

Year 3:
[10,000 cost - 3,000 salv] x  (47/1275)
= 7,000 x 0.036
= 252

Year 4:
[10,000 cost - 3,000 salv] x  (46/1275)
= 7,000 x 0.035
= 245
 

Year 5:
[10,000 cost - 3,000 salv] x  (45/1275)
= 7,000 x 0.034
= 238
 

4

Formula for:

(a) Double Declining Balance (200% declining balance)

(b) 150 Decling balance

(a) Double declining balance:

2 x 1/N x (Cost - Accumualted depreciation) x  N/12

(b) 150 Declining balance

0.15 x 1/n x (cost - accumulated depreciation) x N/12

*N means:  number of years For example:

(i) Double Declining Balance for:
$100,000 cost
8 years
Put into service on Jan 1, Year X

Year 1
2 x 1/8 x 100,000 
= 1/4 x 100,000
= 0.25 x 100,000
= 25,000 Dep exp.

Year 2
2 x 1/8 x 75,000  or  2 x  1/8 x (100,000 - 25,000)
= 0.25 x 75,000
= 18750 Dep exp
 

(ii) 150 Declining balance for the following:
$200,000 cost
7 years
Put into service on Jan 1, Year X

Year 1
1.5 x 1/7 x 200,000
= 0.21 x 200,000
= $42,000 Dep exp. (for year 1)

Year 2
1.5 x 1/7 x 158,000
= $33,857 dep ex. (for year 2)

 

5

Example on Formula for:

(a) Double Declining Balance

(b) 150 Decling balance

On the following data:
$10,000 asset;    $2,000 est. salvage value
Useful Life = 10 years
 

(a) Double declining balance

Year       Double %    NBV             Dep. exp
1                 0.20       10,000           2,000
2                0.20        8,000            1,600
3                0.20        6,400             1,280
4                0.20        5,120              1,024
5                0.20        4,096             819*
6                0.20        3,277              655
7                0.20        2,622              524
8                0.20        2098                98**
Salvage                     2000                  0

 

*Going to use year 5 as illustration:

0.20 x ($10,000 - [2000+1600+1280+1024]
0.20 x ($10,000 - 5,904)
0.20 x 4096
= $819 year 5 depreciation

** only limit up to 98 depreciation expense on year 8 because cannot go below 2,000 salvage value

 

(b) 150 declining balance

Year         150 %       NBV             Dep. exp
1                 0.15       10,000           1,500
2                0.15        8,500            1,275
3                0.15        7,225             1,084
4                0.15        6,141               921
5                0.15        5220             783*
6                0.15        4,437              665
7                0.15        3771                565
8                0.15        3025               480
9                0.15        2,545              381
10               0.15        2,163               163* 
Salvage                     2000                  0

 

*Going to use year 5 as illustration:

0.15 x ($10,000 - [1500 + 1275 + 1084 + 921]
0.15 x ($10,000 - 4,780)
0.15 x 5220
= $783 year 5 depreciation expense

6

Decling Balance Example:  Placed during year (not at beginning of year like Jan 1)

(a) Double Declining Balance

(b) 150 Decling balance

On the following data:
$10,000 asset;    $2,000 est. salvage value
Useful Life = 10 years
Placed in service on July 1, year 1.

(a) Double declining balance

                (2 x 1/10)
Year       Double %    NBV       N/12      Dep. exp
1                 0.20       10,000      6/12      1,000*
2                0.20        9,000      12/12      1,800
3                0.20        7,200       12/12     1,440

*Year 1 gets affected on calculating a (0.20 x 10,000 x 6/12 (months))

(b) 150 Decling balance

               (1.5 x 1/10)
Year       Double %    NBV       N/12    Dep. exp
1                 0.15       10,000      6/12      750*
2                0.15        9,250      12/12     1,388
3                0.15        7,862       12/12     1,179**

*Year 1 gets affected on calculating a
[0.15 x (10,000) x 6/12]
= 1,500 x {6/12]
= 750

** Illustration on Year 3
0.15 x (10,000 - [750 + 1388]) x 12/12
0.15 x (10,000 - 2,138) x 12/12
0.15 x 7862 x 12/12
= 1,179

 

 

7

Units of Production Formula

Units of Production formula:

Cost - salvage value        = rate per unit or hour
Estimates units OR hours


Rate per hour  x  # of units produced = Depreciation 
(or unit)                   (or hours worked)         Exp.
 

8

Disposals - Journal Entries:

(1) Sale of asset during its useful life

(2) Write-off Fully Deprecaited Asset  (remove from Balance sheet)

(3) Total and Permanment Impairment

(4) Partial Impairment

(1) Sale of asset during its useful life:

If gain,

Dr. Cash received on sale
Dr. Acc. Dep. sold old asset
      Cr. Sold asset (at purchase cost)
      Cr. Gain

If Loss,

Dr. Cash received on sale
Dr. Acc. Dep. sold old asset
Dr. Loss on sale
      Cr. Sold asset (at purchase cost)
     

(2) Write-off Fully Deprecaited Asset (remove from Balance sheet)
Dr. Accumulated Depreciation (at 100%)
      Cr. Old asset at full cost (at 100%)

 

(3) Total and Permanment Impairment

Dr. Accumulated Depreciation
Dr. Loss due to Impairment (difference)
      Cr. Old asset at full cost (at 100%)

Example:
Dr.  Accumualted Depreciation  2,000
Dr.  Loss due to impairment       3,000
       Cr.  Old asset                                    5,000

(4) Partial Impairment

Dr. Loss due to impariment  $100
           Accumulated Depreciation    $100

Example of Partial impairment (due to technological change).

 

9

Unit Depletion Rate (US GAAP) - What's the formula for:

(a) Depletion base

(b) Unit depreciation rate

(c) Depletion for year X

(d) Depletion amount in Cost of goods sold on units sold for year X

Use the following info:
Mining company purchase a mine for $5,500,000. Estimate ore amount to be removed is 4,000,000 tons.   Its estimated value after all ore is removed is $300,000.  Its restoration costs is $150,000 and development costs is $800,000. 

In year X, 400,000 units extracted. And 375,000 units sold. 

(a) Depletion base (Formula):

Depletion base = Land cost + Development cost + restoration cost - Residual value

Ex:

$5,500,000 purchase cost (land)
+ 800,000 development costs
+ 150,000 restoration cost
- 300,000 (residual value)
6,150,000 (depletion base)

(b) Unit depletion rate

Depletion base / estimate recoverable units
= 6,150,000 / 4,000,000 tons

= $1.53 per ton unit depletion rate

(c) Depletion for year X

$1.53 depletion rate x 400,000 units extracted
= $612,000 depletion for year x

(d) Depletion amount in Cost of goods sold on units sold  for year X


= $1.53 depletion rate x 375,000 units sold
= $573,750 cost of goods sold in Year X.

10

Advantages and Disadvantages for:

(1) Straight line depreciation

(2) Machine Hours and Units of Production Method

(3) Decling balance method

(1) Straight-line Depreciation

Advantages
a. Simple to compute.
b. Applies to virtually all assets.
c. Consistent from year to year.
d. Wide acceptability.
e. Similar to treatment of prepaid items.

Disadvantages
a. Does not refiect difference in usage of asset from year to year.
b. Does not accurately match costs with revenue.

(2) Machine Hours and Units of Production Method

Advantages
a. Matches costs with revenues.
b. Reflects activity of the enterprise.

Disadvantages
a. If no activity, no depreciation expensed; however, in reality, all assets depreciate.
b. Cannot be used for all assets (e.g., buildings).
c. Can be complex because it requires clerical work and records.
 

(3) Decling balance method

Advantages
a. Matches costs to revenues since greater utility is reflected in greater depreciation during earlier years. 
b. As the amount of depreciation decreases, repairs and maintenance charges increase thereby tending to balance out one another. 

Disadvantages
a. Does not reflect changes in the activity of the asset.
b. Computation can be complex.
c. Greater disparity in amount of depreciation between earlier years and later years.
d. Possibility that with decreasing depreciation and increasing repairs and maintenance, income is artificially smoothed over the years.

11

Disclosure on Depreciation:

What 4 things to disclose about in Financial Statements or Footnotes?

 

Disclosure​ on n Financial Statements or Footnotes on the following:

1. Depreciation expense for the period.

2. Balance of major classes of depreciable assets by nature or function.

3. Accumulated depreciation allowances by classes or in total.

4. The methods used, by major classes, in computing depreciation.

12

Depreciation expense methods (Straight line (SL), Sum-of-years-digits (SYD), and Production):

 

If no salvage value then _____ Dep exp. and ____ Net income.

If no salvage value then

INCREASE Deprec. expense

and

DECREASE net income.

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