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This category contains the mathematical finance functions of Office Calc.

Calculates the accrued interest of a security in the case of periodic payments.

`ACCRINT(Issue; FirstInterest; Settlement; Rate; [Par]; Frequency [; Basis])`

**Issue** (required) is the issue date of the security.

**FirstInterest** (required) is the first interest date of the security.

**Settlement** (required) is the date at which the interest accrued up until then is to be calculated.

**Rate** (required) is the annual nominal rate of interest (coupon interest rate)

**Par** (optional) is the par value of the security. If omitted, a default value of 1000 is used.

We recommend that you always specify the value that you require for ACCRINTâ€™s

Parargument, rather than allowing Calc to apply an arbitrary default. This will make your formula easier to understand and easier to maintain.

**Frequency** (required) is the number of interest payments per year (1, 2 or 4).

**Basis** (optional) is chosen from a list of options and indicates how the year is to be calculated.

Basis | Calculation |
---|---|

0 or missing | US method (NASD), 12 months of 30 days each |

1 | Exact number of days in months, exact number of days in year |

2 | Exact number of days in month, year has 360 days |

3 | Exact number of days in month, year has 365 days |

4 | European method, 12 months of 30 days each |

A security is issued on 28/02/2001. First interest is set for 31/08/2001. The settlement date is 01/05/2001. The Rate is 0.1 or 10% and Par is 1000 currency units. Interest is paid half-yearly (frequency is 2). The basis is the US method (0). How much interest has accrued?

=ACCRINT("28/02/2001";"31/08/2001";"01/05/2001";0.1;1000;2;0) returns 16.94444.

Calculates the accrued interest of a security in the case of one-off payment at the settlement date.

`ACCRINTM(Issue; Settlement; Rate [; Par [; Basis]])`

**Issue** (required) is the issue date of the security.

**Settlement** (required) is the date at which the interest accrued up until then is to be calculated.

**Rate** (required) is the annual nominal rate of interest (coupon interest rate).

**Par** (optional) is the par value of the security. If omitted, a default value of 1000 is used.

We recommend that you always specify the value that you require for ACCRINTMâ€™s

Parargument, rather than allowing Calc to apply an arbitrary default. This will make your formula easier to understand and easier to maintain.

**Basis** (optional) is chosen from a list of options and indicates how the year is to be calculated.

Basis | Calculation |
---|---|

0 or missing | US method (NASD), 12 months of 30 days each |

1 | Exact number of days in months, exact number of days in year |

2 | Exact number of days in month, year has 360 days |

3 | Exact number of days in month, year has 365 days |

4 | European method, 12 months of 30 days each |

A security is issued on 01/04/2001. The maturity date is set for 15/06/2001. The Rate is 0.1 or 10% and Par is 1000 currency units. The basis of the daily/annual calculation is the daily balance (3). How much interest has accrued?

=ACCRINTM("01/04/2001";"15/06/2001";0.1;1000;3) returns 20.54795.

Calculates the amount of depreciation for a settlement period as degressive amortisation. Unlike AMORLINC, a depreciation coefficient that is independent of the depreciable life is used here.

`AMORDEGRC(Cost; DatePurchased; FirstPeriod; Salvage; Period; Rate [; Basis])`

**Cost** is the acquisition costs.

**DatePurchased** is the date of acquisition.

**FirstPeriod** is the end date of the first settlement period.

**Salvage** is the salvage value of the capital asset at the end of the depreciable life.

**Period** is the settlement period to be considered.

**Rate** is the rate of depreciation.

**Basis** (optional) is chosen from a list of options and indicates how the year is to be calculated.

Basis | Calculation |
---|---|

0 or missing | US method (NASD), 12 months of 30 days each |

1 | Exact number of days in months, exact number of days in year |

2 | Exact number of days in month, year has 360 days |

3 | Exact number of days in month, year has 365 days |

4 | European method, 12 months of 30 days each |

An asset was acquired on 2020-02-01 at a cost of 2000 currency units. The end date of the first settlement period was 2020-12-31. The salvage value of the asset at the end of its depreciable life will be 10 currency units. The rate of depreciation is 0.1 (10%) and the year is calculated using the US method (Basis 0). Assuming degressive depreciation, what is the amount of depreciation in the fourth depreciation period?

=AMORDEGRC(2000; "2020-02-01"; "2020-12-31"; 10; 4; 0.1; 0) returns a depreciation amount of 163 currency units.

Beware that Basis 2 is not supported by Microsoft Excel. Hence, if you use Basis 2 and export your document to XLSX format, it will return an error when opened in Excel.

Calculates the amount of depreciation for a settlement period as linear amortisation. If the capital asset is purchased during the settlement period, the proportional amount of depreciation is considered.

`AMORLINC(Cost; DatePurchased; FirstPeriod; Salvage; Period; Rate [; Basis])`

**Cost** is the acquisition costs.

**DatePurchased** is the date of acquisition.

**FirstPeriod** is the end date of the first settlement period.

**Salvage** is the salvage value of the capital asset at the end of the depreciable life.

**Period** is the settlement period to be considered.

**Rate** is the rate of depreciation.

**Basis** (optional) is chosen from a list of options and indicates how the year is to be calculated.

Basis | Calculation |
---|---|

0 or missing | US method (NASD), 12 months of 30 days each |

1 | Exact number of days in months, exact number of days in year |

2 | Exact number of days in month, year has 360 days |

3 | Exact number of days in month, year has 365 days |

4 | European method, 12 months of 30 days each |

An asset was acquired on 2020-02-01 at a cost of 2000 currency units. The end date of the first settlement period was 2020-12-31. The salvage value of the asset at the end of its depreciable life will be 10 currency units. The rate of depreciation is 0.1 (10%) and the year is calculated using the US method (Basis 0). Assuming linear depreciation, what is the amount of depreciation in the fourth depreciation period?

=AMORLINC(2000; "2020-02-01"; "2020-12-31"; 10; 4; 0.1; 0) returns a depreciation amount of 200 currency units.

Beware that Basis 2 is not supported by Microsoft Excel. Hence, if you use Basis 2 and export your document to XLSX format, it will return an error when opened in Excel.

Returns the depreciation of an asset for a specified period using the fixed-declining balance method.

This form of depreciation is used if you want to get a higher depreciation value at the beginning of the depreciation (as opposed to linear depreciation). The depreciation value is reduced with every depreciation period by the depreciation already deducted from the initial cost.

`DB(Cost; Salvage; Life; Period [; Month])`

**Cost** is the initial cost of an asset.

**Salvage** is the value of an asset at the end of the depreciation.

**Life** defines the period over which an asset is depreciated.

**Period** is the length of each period. The length must be entered in the same date unit as the depreciation period.

**Month** (optional) denotes the number of months for the first year of depreciation. If an entry is not defined, 12 is used as the default.

A computer system with an initial cost of 25,000 currency units is to be depreciated over a three-year period. The salvage value is to be 1,000 currency units. The first period of depreciation comprises 6 months. What is the fixed-declining balance depreciation of the computer system in the second period, which is a full year starting from the end of the first six-month period?

=DB(25000; 1000; 3; 2; 6) returns 11,037.95 currency units.

Returns the depreciation of an asset for a specified period using the arithmetic-declining method.

Use this form of depreciation if you require a higher initial depreciation value as opposed to linear depreciation. The depreciation value gets less with each period and is usually used for assets whose value loss is higher shortly after purchase (for example, vehicles, computers). Please note that the book value will never reach zero under this calculation type.

`DDB(Cost; Salvage; Life; Period [; Factor])`

**Cost** fixes the initial cost of an asset.

**Salvage** fixes the value of an asset at the end of its life.

**Life** is the number of periods (for example, years or months) defining how long the asset is to be used.

**Period** defines the length of the period over which the value is to be calculated. The period must be entered in the same time unit as Life.

**Factor** (optional) is the factor by which depreciation decreases. If a value is not entered, the default is factor 2.

A computer system with an initial cost of 75,000 currency units is to be depreciated monthly over 5 years. The value at the end of the depreciation is to be 1 currency unit. The factor is 2.

=DDB(75000;1;60;12;2) = 1,721.81 currency units. Therefore, the double-declining depreciation during the twelfth month after purchase is 1,721.81 currency units.

Calculates the allowance (discount) of a security as a percentage.

`DISC(Settlement; Maturity; Price; Redemption [; Basis])`

**Settlement** is the date of purchase of the security.

**Maturity** is the date on which the security matures (expires).

**Price** is the price of the security per 100 currency units of par value.

**Redemption** is the redemption value of the security per 100 currency units of par value.

**Basis** (optional) is chosen from a list of options and indicates how the year is to be calculated.

Basis | Calculation |
---|---|

0 or missing | US method (NASD), 12 months of 30 days each |

1 | Exact number of days in months, exact number of days in year |

2 | Exact number of days in month, year has 360 days |

3 | Exact number of days in month, year has 365 days |

4 | European method, 12 months of 30 days each |

A security is purchased on 2001-01-25; the maturity date is 2001-11-15. The price (purchase price) is 97, the redemption value is 100. Using daily balance calculation (basis 3) how high is the settlement (discount)?

=DISC("2001-01-25";"2001-11-15";97;100;3) returns about 0.0372 or 3.72 per cent.

Calculates the duration of a fixed interest security in years.

`DURATION(Settlement; Maturity; Coupon; Yield; Frequency [; Basis])`

**Settlement** is the date of purchase of the security.

**Maturity** is the date on which the security matures (expires).

**Coupon** is the annual coupon interest rate (nominal rate of interest)

**Yield** is the annual yield of the security.

**Frequency** is the number of interest payments per year (1, 2 or 4).

**Basis** (optional) is chosen from a list of options and indicates how the year is to be calculated.

Basis | Calculation |
---|---|

0 or missing | US method (NASD), 12 months of 30 days each |

1 | Exact number of days in months, exact number of days in year |

2 | Exact number of days in month, year has 360 days |

3 | Exact number of days in month, year has 365 days |

4 | European method, 12 months of 30 days each |

A security is purchased on 2001-01-01; the maturity date is 2006-01-01. The Coupon rate of interest is 8%. The yield is 9.0%. Interest is paid half-yearly (frequency is 2). Using daily balance interest calculation (basis 3), what is the duration?

=DURATION("2001-01-01";"2006-01-01";0.08;0.09;2;3) returns 4.2 years.

Returns the net annual interest rate for a nominal interest rate.

Nominal interest refers to the amount of interest due at the end of a calculation period. Effective interest increases with the number of payments made. In other words, interest is often paid in instalments (for example, monthly or quarterly) before the end of the calculation period.

`EFFECT(Nom; P)`

**Nom** is the nominal interest.

**P** is the number of interest payment periods per year.

If the annual nominal interest rate is 9.75% and four interest calculation periods are defined, what is the actual interest rate (effective rate)?

=EFFECT(9.75%;4) = 10.11% The annual effective rate is therefore 10.11%.

Calculates the effective annual rate of interest on the basis of the nominal interest rate and the number of interest payments per annum.

The functions whose names end with -ADD or -EXCEL2003 return the same results as the corresponding Microsoft Excel 2003 functions without the suffix. Use the functions without suffix to get results based on international standards.

`EFFECT-ADD(NominalRate; NumPerYear)`

**NominalRate** is the annual nominal rate of interest.

**NumPerYear** is the number of interest payments per year.

What is the effective annual rate of interest for a 5.25% nominal rate and quarterly payment.

=EFFECT-ADD(0.0525;4) returns 0.053543 or 5.3543%.

Calculates the internal rate of return for an investment. The values represent cash flow values at regular intervals, at least one value must be negative (payments) and at least one value must be positive (income).

If the payments take place at irregular intervals, use the XIRR function.

`IRR(Values [; Guess])`

**Values** is an array containing the values.

**Guess** (optional) is the estimated value. An iterative method is used to calculate the internal rate of return. If you can provide only few values, you should provide an initial guess to enable the iteration.

This function ignores any text or empty cell within a data range. If you suspect wrong results from this function, look for text in the data ranges. To highlight text contents in a data range, use the value highlighting feature.

Under the assumption that the cell contents are A1=-10000, A2=3500, A3=7600 and A4=1000, the formula =IRR(A1:A4) gives a result of 11.33%.

Because of the iterative method used, it is possible for IRR to fail and return Error 523, with "Error: Calculation does not converge" in the status bar. In that case, try another value for Guess.

Calculates the level of interest for unchanged amortisation instalments.

`ISPMT(Rate; Period; TotalPeriods; Invest)`

**Rate** sets the periodic interest rate.

**Period** is the number of instalments for calculation of interest.

**TotalPeriods** is the total number of instalment periods.

**Invest** is the amount of the investment.

For a credit amount of 120,000 currency units with a two-year term and monthly instalments, at a yearly interest rate of 12% the level of interest after 1.5 years is required.

=ISPMT(1%;18;24;120000) = -300 currency units. The monthly interest after 1.5 years amounts to 300 currency units.

Returns the present (current) value of an investment resulting from a series of regular payments.

Use this function to calculate the amount of money needed to be invested at a fixed rate today, to receive a specific amount, an annuity, over a specified number of periods. You can also determine how much money is to remain after the elapse of the period. Specify as well if the amount is to be paid out at the beginning or at the end of each period.

Enter these values either as numbers, expressions or references. If, for example, interest is paid annually at 8%, but you want to use month as your period, enter 8%/12 under **Rate** and Office Calc with automatically calculate the correct factor.

`PV(Rate; NPer; Pmt [; FV [; Type]])`

**Rate** defines the interest rate per period.

**NumPeriods** is the total number of periods (payment period).

**Payment** is the constant payment made each period.

**FV** (optional) defines the future value remaining after the final instalment has been made.

**Type** (optional) denotes due date for payments. Type = 1 means due at the beginning of a period and Type = 0 (default) means due at the end of the period.

In the Office Calc functions, parameters marked as "optional" can be left out only when no parameter follows. For example, in a function with four parameters, where the last two parameters are marked as "optional", you can leave out parameter 4 or parameters 3 and 4, but you cannot leave out parameter 3 alone.

What is the present value of an investment, if 500 currency units are paid out monthly and the annual interest rate is 8%? The payment period is 48 months and 20,000 currency units are to remain at the end of the payment period.

=PV(8%/12;48;500;20000) = -35,019.37 currency units. Under the named conditions, you must deposit 35,019.37 currency units today, if you want to receive 500 currency units per month for 48 months and have 20,000 currency units left over at the end. Cross-checking shows that 48 x 500 currency units + 20,000 currency units = 44,000 currency units. The difference between this amount and the 35,000 currency units deposited represents the interest paid.

If you enter references instead of these values into the formula, you can calculate any number of "If-then" scenarios. Please note: references to constants must be defined as absolute references. Examples of this type of application are found under the depreciation functions.

Calculates the amount received that is paid for a fixed-interest security at a given point in time.

`RECEIVED(Settlement; Maturity; Investment; Discount [; Basis])`

**Settlement** is the date of purchase of the security.

**Maturity** is the date on which the security matures (expires).

**Investment** is the purchase sum.

**Discount** is the percentage discount on acquisition of the security.

**Basis** (optional) is chosen from a list of options and indicates how the year is to be calculated.

Basis | Calculation |
---|---|

0 or missing | US method (NASD), 12 months of 30 days each |

1 | Exact number of days in months, exact number of days in year |

2 | Exact number of days in month, year has 360 days |

3 | Exact number of days in month, year has 365 days |

4 | European method, 12 months of 30 days each |

Settlement date: 15th February 1999, maturity date: 15th May 1999, investment sum: 1000 currency units, discount: 5.75 per cent, basis: Daily balance/360 = 2.

The amount received on the maturity date is calculated as follows:

=RECEIVED("1999-02-15";"1999-05-15";1000;0.0575;2) returns 1014.420266.

Returns the arithmetic-declining depreciation rate.

Use this function to calculate the depreciation amount for one period of the total depreciation span of an object. Arithmetic declining depreciation reduces the depreciation amount from period to period by a fixed sum.

`SYD(Cost; Salvage; Life; Period)`

**Cost** is the initial cost of an asset.

**Salvage** is the value of an asset after depreciation.

**Life** is the period fixing the time span over which an asset is depreciated.

**Period** defines the period for which the depreciation is to be calculated.

A video system initially costing 50,000 currency units is to be depreciated annually for the next 5 years. The salvage value is to be 10,000 currency units. You want to calculate depreciation for the first year.

=SYD(50000;10000;5;1)=13,333.33 currency units. The depreciation amount for the first year is 13,333.33 currency units.

To have an overview of depreciation rates per period, it is best to define a depreciation table. By entering the different depreciation formulae available in Office Calc next to each other, you can see which depreciation form is the most appropriate. Enter the table as follows:

| | **A** | **B** | **C** | **D** | **E** | |
| -------- | --------------------- | --------------------- | ----------- | ----------- | ------------------------ |
| 1 | Initial Cost | Salvage Value | Useful Life | Time Period | Deprec. SYD |
| 2 | 50,000 currency units | 10,000 currency units | 5 | 1 | 13,333.33 currency units |
| 3 | | | | 2 | 10,666.67 currency units |
| 4 | | | | 3 | 8,000.00 currency units |
| 5 | | | | 4 | 5,333.33 currency units |
| 6 | | | | 5 | 2,666.67 currency units |
| 7 | | | | 6 | 0.00 currency units |
| 8 | | | | 7 | |
| 9 | | | | 8 | |
| 10 | | | | 9 | |
| 11 | | | | 10 | |
| 12 | | | | | |
| 13 | >0 | | | Total | 40,000.00 currency units |

The formula in E2 is as follows:

=SYD($A$2;$B$2;$C$2;D2)

This formula is duplicated in column E down to E11 (select E2, then drag down the lower right corner with the mouse).

Cell E13 contains the formula used to check the total of the depreciation amounts. It uses the SUMIF function as the negative values in E8:E11 must not be considered. The condition >0 is contained in cell A13. The formula in E13 is as follows:

=SUMIF(E2:E11;A13)

Now view the depreciation for a 10 year period, or at a salvage value of 1 currency unit, or enter a different initial cost and so on.