### SYD Function Dictionary

Function Dictionary | Function Examples | Function Categories |

SYD | |||||||||

Purchase Value Of A New Car | £20,000 | ||||||||

Second Hand Value | £8,000 | ||||||||

Number Of Years Ownership | 6 | ||||||||

Deprecation in year 1 | £3,429 | =SYD(F3,F4,F5,1) | |||||||

Deprecation in year 2 | £2,857 | =SYD(F3,F4,F5,2) | |||||||

Deprecation in year 3 | £2,286 | =SYD(F3,F4,F5,3) | |||||||

Deprecation in year 4 | £1,714 | =SYD(F3,F4,F5,4) | |||||||

Deprecation in year 5 | £1,143 | =SYD(F3,F4,F5,5) | |||||||

Deprecation in year 6 | £571 | =SYD(F3,F4,F5,6) | |||||||

Total Depreciation : | £12,000 | =SUM(F7:F12) | |||||||

What Does It Do ? | |||||||||

This function calculates the depreciation of an item throughout its life, using the sum of the | |||||||||

years digits. | |||||||||

The depreciation is greatest in the earlier part of the items life. | |||||||||

What is the Sum Of The Years Digits ? | |||||||||

The sum of the years digits adds together the each of the years of the life. | |||||||||

A life of 3 years has a sum of 1+2+3 equalling 6. | |||||||||

Each of the years is then calculated as a percentage of the sum of the years. | |||||||||

Year 3 is 50% of 6, year 2 is 33% of 6, year 1 is 17% 6. | |||||||||

The total depreciation of the item is then allocated on the basis of these percentages. | |||||||||

A depreciation of £9000 is allocated as 50% being £4500, 33% being £3000, 17% being £1500. | |||||||||

£9,000 | |||||||||

1 | 17% | £1,500 | |||||||

2 | 33% | £3,000 | |||||||

3 | 50% | £4,500 | |||||||

As the greater part of the depreciation is allocated to the earliest years the values are | |||||||||

inverted, year 1 is $4500, year 2 is £3000 and year 1 is £1500. | |||||||||

Example 1 | |||||||||

Purchase Price Of A Car : | £10,000 | ||||||||

Salvage Value : | £1,000 | ||||||||

Expected Life in Years : | 3 | ||||||||

As % Of Total Depreciation | |||||||||

Depreciation in Year 1 : | £4,500 | ===> | 0.5 | ||||||

Depreciation in Year 2 : | £3,000 | ===> | 0.333333 | ||||||

Depreciation in Year 3 : | £1,500 | ===> | 0.166667 | ||||||

=SYD(E39,E40,E41,3) | |||||||||

1. Add together the digits of the Life to get the SumOfTheYearsDigits, 1+2+3=6. | |||||||||

2. Subtract the Salvage from the Purchase Price to get Total Deprectation, £10000-£1000=£9000. | |||||||||

3. Divide the Total Deprectation by the SumOfTheYearsDigits, £9000/6=£1500. | |||||||||

4. Invert the year digits, 1,2,3 becomes 3,2,1. | |||||||||

5. Multiply 3,2,1 by £1500 to get £4500, £3000, £1500, these values are the depreciation | |||||||||

values for each of the three years in the life of the item. | |||||||||

Example 2 | |||||||||

The same example using 4 years. | |||||||||

Purchase Price Of A Car : | £10,000 | ||||||||

Salvage Value : | £1,000 | ||||||||

Expected Life in Years : | 4 | ||||||||

As % Of Total Depriciation | |||||||||

Depreciation in Year 1 : | £3,600 | 0.4 | |||||||

Depreciation in Year 2 : | £2,700 | 0.3 | |||||||

Depreciation in Year 3 : | £1,800 | 0.2 | |||||||

Depreciation in Year 4 : | £900 | 0.1 | |||||||

Total Depreciation : | £9,000 | 100% | |||||||

Example 3 | |||||||||

This example will adjust itself to accommodate any number of years between 1 and 10. | |||||||||

Purchase Price Of A Car : | £10,000 | ||||||||

Salvage Value : | £1,000 | ||||||||

Expected Life in Years (1 to 10) : | 7 | ||||||||

As % Of Total Depriciation | |||||||||

Year | 1 | £2,250 | 25% | ||||||

Year | 2 | £1,929 | 21% | ||||||

Year | 3 | £1,607 | 18% | ||||||

Year | 4 | £1,286 | 14% | ||||||

Year | 5 | £964 | 11% | ||||||

Year | 6 | £643 | 7% | ||||||

Year | 7 | £321 | 4% | ||||||

Year | |||||||||

Year | |||||||||

Year | |||||||||

£9,000 | 100% | ||||||||

Syntax | |||||||||

=SYD(OriginalCost,SalvageValue,Life,PeriodToCalculate) | |||||||||

Formatting | |||||||||

No special formatting is needed. | |||||||||