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

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Sum of years digits example 1331389

Sum-of-Years’-Digits Example

  • Assume depreciable asset is a car with:

  • 4 year useful life

  • Original cost of $22,000

  • 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


Sum of years digits example 1331389

Sum-of-Years’-Digits Example

Sum of Year’s Digits


Sum of years digits example 1331389

Sum-of-Years’-Digits Example

This was corrected on July 16, 2002.

Depr. Fraction = Remaining Life/Sum of Years’ Digits


Sum of years digits example 1331389

Sum-of-Years’-Digits Example

This was corrected on July 16, 2002.


Sum of years digits example 1331389

Sum-of-Years’-Digits Example

This was corrected on July 16, 2002.


Sum of years digits example 1331389

Sum-of-Years’-Digits Example

This was corrected on July 16, 2002.


Sum of years digits example 1331389

Sum-of-Years’-Digits Example

This was corrected on July 16, 2002.


Sum of years digits example 1331389

Double Declining Balance Example

  • Assume depreciable asset is a car with:

  • 4 year useful life

  • Original cost of $22,000

  • Salvage Value of $7,000


Sum of years digits example 1331389

Double Declining Balance Example

Straight Line %age = 100%/Useful Life



Sum of years digits example 1331389

Double Declining Balance Example

Start with cost (not cost – salvage value)




Sum of years digits example 1331389

Double Declining Balance Example

Too much depreciation—below salvage value!


Sum of years digits example 1331389

Double Declining Balance Example

Throw out these final year computed values.


Sum of years digits example 1331389

Double Declining Balance Example

Make this enough to arrive exactly at ending salvage value.


Sum of years digits example 1331389

Double Declining Balance Example

Note that depreciation is complete after two years even

though asset has four year useful life.


Sum of years digits example 1331389

Partial Period Depreciation

Note that the prior examples assumed that the assets were put in use on January 1st 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.


Sum of years digits example 1331389

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


Sum of years digits example 1331389

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


Sum of years digits example 1331389

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


Sum of years digits example 1331389

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


Sum of years digits example 1331389

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


Sum of years digits example 1331389

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


Sum of years digits example 1331389

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


Sum of years digits example 1331389

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.


Sum of years digits example 1331389

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.

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.


Sum of years digits example 1331389

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.


Sum of years digits example 1331389

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.


Sum of years digits example 1331389

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.


Sum of years digits example 1331389

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.


Sum of years digits example 1331389

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.

Normal Schedule

½ Use First Year Schedule


Sum of years digits example 1331389

Partial Period Depreciation

First, take ½ of the first year’s normal depreciation.

Normal Schedule

½ Use First Year Schedule

x 1/2


Sum of years digits example 1331389

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.

Normal Schedule

½ Use First Year Schedule

x 1/2


Sum of years digits example 1331389

Partial Period Depreciation

Then add ½ of the second year’s normal depreciation to the roll forward amount.

Normal Schedule

½ Use First Year Schedule

x 1/2


Sum of years digits example 1331389

Partial Period Depreciation

Then add ½ of the second year’s normal depreciation to the roll forward amount.

Normal Schedule

½ Use First Year Schedule


Sum of years digits example 1331389

Partial Period Depreciation

Then roll forward ½ of the second year’s normal depreciation to the third year schedule.

Normal Schedule

½ Use First Year Schedule

x 1/2


Sum of years digits example 1331389

Partial Period Depreciation

Then add ½ of the third year’s normal depreciation to the roll forward amount.

Normal Schedule

½ Use First Year Schedule

x 1/2


Sum of years digits example 1331389

Partial Period Depreciation

Then add ½ of the third year’s normal depreciation to the roll forward amount.

Normal Schedule

½ Use First Year Schedule


Sum of years digits example 1331389

Partial Period Depreciation

Finally, roll forward ½ of the third year’s normal depreciation to the fourth year schedule.

Normal Schedule

½ Use First Year Schedule

x 1/2


Sum of years digits example 1331389

Yr 1

Yr 2

Yr 3

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.


Sum of years digits example 1331389

Yr 1

Yr 2

Yr 3

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000


Sum of years digits example 1331389

Yr 1

Yr 2

Yr 3

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$5,000

$5,000


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$5,000

$5,000


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$5,000

$5,000

$5,000


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$5,000

$5,000

$5,000

$5,000


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$2,500

$2,500

$5,000

$5,000


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$2,500

$2,500

$5,000

$5,000

$2,500

$2,500


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$2,500

$2,500

$5,000

$5,000

$2,500

$2,500

$7,500


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$500

$500

$5,000

$7,500

$2,500


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$500

$500

$5,000

$7,500

$2,500

$500

$500


Sum of years digits example 1331389

Yr 1

Yr 1

Yr 2

Yr 2

Yr 3

Yr 3

Yr 4

Yr 4

Partial Period Depreciation

Another way to look at it conceptually is with a timeline.

$10,000

$5,000

$1,000

$5,000

$7,500

$3,000

$500


Sum of years digits example 1331389

Fixed Asset Impairment Example

  • Assume Company A has equipment:

  • Original cost of $120,000

  • Accumulated depreciation of $20,000

  • Market value of $97,000

  • Market interest rate of 8%

  • Expected cash flows:

    • $24,000 for four years (paid at end of yr.)


Sum of years digits example 1331389

Fixed Asset Impairment Example

Recoverability Test (do we need to record an impairment?)

  • Book value = $120,000 – 20,000 = $100,000

Orig. Cost

Accum. Depr.


Sum of years digits example 1331389

Fixed Asset Impairment Example

Recoverability Test (do we need to record an impairment?)

  • Book value = $120,000 – 20,000 = $100,000

  • Net future cash flows = $24,000 x 4 = $96,000

Not discounted for interest rate


Sum of years digits example 1331389

Fixed Asset Impairment Example

Recoverability Test (do we need to record an impairment?)

  • Book value = $120,000 – 20,000 = $100,000

  • Net future cash flows = $24,000 x 4 = $96,000

  • NFCF < BV, so we need to record an impairment


Sum of years digits example 1331389

Fixed Asset Impairment Example

Recoverability Test (do we need to record an impairment?)

  • Book value = $120,000 – 20,000 = $100,000

  • Net future cash flows = $24,000 x 4 = $96,000

  • NFCF < BV, so we need to record an impairment

Amount of Impairment Loss

  • Market value is determinable, so use BV – FMV:

  • $100,000 - $97,000 = $3,000


Sum of years digits example 1331389

Fixed Asset Impairment Example

Recoverability Test (do we need to record an impairment?)

  • Book value = $120,000 – 20,000 = $100,000

  • Net future cash flows = $24,000 x 4 = $96,000

  • NFCF < BV, so we need to record an impairment

Amount of Impairment Loss

  • Market value is determinable, so use BV – FMV:

  • $100,000 - $97,000 = $3,000

3/31 Loss on Impairment 3,000

Accum. Depr, Equipment 3,000

Note: Record impairment to equipment


Sum of years digits example 1331389

Fixed Asset Impairment Example

Recoverability Test (do we need to record an impairment?)

  • Book value = $120,000 – 20,000 = $100,000

  • Net future cash flows = $24,000 x 4 = $96,000

  • NFCF < BV, so we need to record an impairment

Amount of Impairment Loss

  • Market value is determinable, so use BV – FMV:

  • $100,000 - $97,000 = $3,000

  • If FMV is undeterminable, use BV – Discounted CF


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

Fixed Asset Impairment Example

Discounted Cash Flows


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

Fixed Asset Impairment Example

Discounted Cash Flows

$24,000

$24,000

$24,000

$24,000


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

1

Each year’s discount rate =

(1 + int rate)year

Fixed Asset Impairment Example

Discounted Cash Flows

$24,000

$24,000

$24,000

$24,000


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

x

x

x

x

1

1

1

1

1

Each year’s discount rate =

(1.08)3

(1.08)1

(1.08)2

(1.08)4

(1 + int rate)year

Fixed Asset Impairment Example

Discounted Cash Flows

$24,000

$24,000

$24,000

$24,000


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

1

(1.08)1

Fixed Asset Impairment Example

Discounted Cash Flows

$24,000

$24,000

$24,000

$24,000

x 0.926

x 0.857

x 0.794

x 0.735


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

Fixed Asset Impairment Example

Discounted Cash Flows

$24,000

$24,000

$24,000

$24,000

x 0.926

x 0.857

x 0.794

x 0.735

= 22,224

= 20,568

= 19,056

= 17,640


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

Fixed Asset Impairment Example

Discounted Cash Flows

$24,000

$24,000

$24,000

$24,000

x 0.926

x 0.857

x 0.794

x 0.735

= 22,224

= 20,568

= 19,056

= 17,640

Net Discounted Cash Flows = 22,224 + 20,568 + 19,056 + 17,640 = $79,488


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

Fixed Asset Impairment Example

Discounted Cash Flows

$24,000

$24,000

$24,000

$24,000

x 0.926

x 0.857

x 0.794

x 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:


Sum of years digits example 1331389

Fixed Asset Impairment Example

Present Value of $1 Annuity


Sum of years digits example 1331389

Fixed Asset Impairment Example

Present Value of $1 Annuity


Sum of years digits example 1331389

1/1/01

12/31/01

12/31/02

12/31/03

12/31/04

Fixed Asset Impairment Example

Discounted Cash Flows

$24,000

$24,000

$24,000

$24,000

x 0.926

x 0.857

x 0.794

x 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


Sum of years digits example 1331389

Fixed Asset Impairment Example

Recoverability Test (do we need to record an impairment?)

  • Book value = $120,000 – 20,000 = $100,000

  • Net future cash flows = $24,000 x 4 = $96,000

  • NFCF < BV, so we need to record an impairment

Amount of Impairment Loss

  • Market value is determinable, so use BV – FMV:

  • $100,000 - $97,000 = $3,000

  • If FMV is undeterminable, use BV – Discounted CF:

  • $100,000 – 79,488 = $20,512