To compute the amortized cost of the Server Hardware cost driver, you can configure the depreciation method and the depreciation period. Cost Insight supports two yearly depreciation methods and you can set the depreciation period from two to seven years.

Note:

Cost Insight calculates the yearly depreciation values and then divides the value by 12 to arrive at the monthly depreciation.

Method

Calculation

Straight line

`Yearly straight line depreciation = [(original cost - accumulated depreciation) / number of remaining depreciation years]`

Max of Double or Straight

`Yearly max of Double or Straight = Maximum (yearly depreciation of double declining balance method, yearly depreciation of straight line method) `

`Yearly depreciation of double declining method= [(original cost - accumulated depreciation) * depreciation rate].`

`Depreciation rate = 2 / number of depreciation years. `

Note:

`Double declining depreciation for the last year = original cost - accumulated depreciation`

## Example for Straight Line Depreciation Method

Year

Original Cost

Accumulated Depreciation

Straight Line Depreciation Cost

Year 1

10000

0

`[(10000-0)/5] =	2000`

Year 2

10000

2000

`[(10000-2000)/4] = 2000`

Year 3

10000

4000

`[(10000-2000)/3] = 2000`

Year 4

10000

6000

`[(10000-2000)/2] = 2000`

Year 5

10000

8000

`[(10000-2000)/1] = 2000`

## Example for Max of Double and Straight Line Depreciation Method

Year

Original Cost

Depreciation Rate

Accumulated Depreciation

Straight Line Depreciation Cost

Year 1

10000

0.4

0

```Maximum([(10000-0)*0.4],[(10000-0)/5])
= Maximum(4000, 2000) = 4000```

which is 333.33 per month.

Year 2

10000

0.4

4000

```Maximum([(10000-4000)*0.4],[(10000-4000)/4])
= Maximum (2400, 1500) = 2400```

which is 200 per month.

Year 3

10000

0.4

6400

```Maximum([(10000-6400)*0.4],[(10000-6400)/3])
= Maximum (1440, 1200) = 1440```

which is 120 per month.

Year 4

10000

0.4

7840

```Maximum([(10000-7840)*0.4],[(10000-7840)/2])
= Maximum (864, 1080) = 1080```

which is 90 per month.

Year 5

10000

0.4

8920

```Maximum([(10000-8920)*0.4],[(10000-8920)/1])
= Maximum (432, 1080) = 1080```

which is 90 per month.