Financial Functions Part One
This category contains the mathematical finance functions of LibreOffice Calc.
\<bookmark_value\>ACCRINT function\</bookmark_value\>ACCRINT
\<bookmark_value\>accrued interests;periodic payments\</bookmark_value\>Calculates the accrued interest of a security in the case of periodic payments.
Syntax
ACCRINT(Issue;First interest;Settlement;Rate;Par;Frequency;Basis)
\<emph\>Issue\</emph\>: the issue date of the security.
\<emph\>First interest\</emph\>: the first interest date of the security.
\<emph\>Settlement\</emph\>: the date at which the interest accrued up until then is to be calculated.
\<emph\>Rate\</emph\>: the annual nominal rate of interest (coupon interest rate).
\<emph\>Par\</emph\>: the par value of the security.
Frequency: number of interest payments per year (1, 2 or 4).
Example:
A security is issued on 2.28.2001. First interest is set for 8.31.2001. The settlement date is 5.1.2001. The Rate is 0.1 or 10% and Par is 1000 currency units. Interest is paid halfyearly (frequency is 2). The basis is the US method (0). How much interest has accrued?
=ACCRINT("2.28.2001"; "8.31.2001"; "5.1.2001"; 0.1; 1000; 2; 0) returns 16.94444.
\<bookmark_value\>ACCRINTM function\</bookmark_value\>\<bookmark_value\>accrued interests;oneoff payments\</bookmark_value\>ACCRINTM
Calculates the accrued interest of a security in the case of oneoff payment at the settlement date.
Syntax
ACCRINTM(Issue;Settlement;Rate;Par;Basis)
\<emph\>Issue\</emph\>: the issue date of the security.
\<emph\>Settlement\</emph\>: the date at which the interest accrued up until then is to be calculated.
\<emph\>Rate\</emph\>: the annual nominal rate of interest (coupon interest rate).
\<emph\>Par\</emph\>: the par value of the security.
Example:
A security is issued on 4.1.2001. The maturity date is set for 6.15.2001. 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("4.1.2001"; "6.15.2001"; 0.1; 1000; 3) returns 20.54795.
\<bookmark_value\>AMORDEGRC function\</bookmark_value\>\<bookmark_value\>depreciations;degressive amortizations\</bookmark_value\>AMORDEGRC
Calculates the amount of depreciation for a settlement period as degressive amortization. Unlike AMORLINC, a depreciation coefficient that is independent of the depreciable life is used here.
Syntax
AMORDEGRC(Cost;Date purchased;First period;Salvage;Period;Rate;Basis)
\<emph\>Cost\</emph\>: the acquisition costs.
\<emph\>Date purchased\</emph\>: the date of acquisition.
\<emph\>First period\</emph\>: the end date of the first settlement period.
\<emph\>Salvage\</emph\>: The salvage value of the capital asset at the end of the depreciable life.
\<emph\>Period\</emph\>: the settlement period to be considered.
\<emph\>Rate\</emph\>: the rate of depreciation.
\<bookmark_value\>AMORLINC function\</bookmark_value\>\<bookmark_value\>depreciations;linear amortizations\</bookmark_value\>AMORLINC
Calculates the amount of depreciation for a settlement period as linear amortization. If the capital asset is purchased during the settlement period, the proportional amount of depreciation is considered.
Syntax
AMORLINC(Cost;Date purchased;First period;Salvage;Period;Rate;Basis)
\<emph\>Cost\</emph\>: the acquisition costs.
\<emph\>Date purchased\</emph\>: the date of acquisition.
\<emph\>First period\</emph\>: the end date of the first settlement period.
\<emph\>Salvage\</emph\>: The salvage value of the capital asset at the end of the depreciable life.
\<emph\>Period\</emph\>: the settlement period to be considered.
\<emph\>Rate\</emph\>: the rate of depreciation.
\<bookmark_value\>calculating; geometricdegressive depreciations\</bookmark_value\>\<bookmark_value\>geometricdegressive depreciations\</bookmark_value\>\<bookmark_value\>depreciations;geometricdegressive\</bookmark_value\>\<bookmark_value\>DB function\</bookmark_value\>DB
Returns the depreciation of an asset for a specified period using the fixeddeclining 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.
Syntax
DB(Cost;Salvage;Life;Period;Month)
\<emph\>Cost\</emph\> fixes the initial cost of an asset.
\<emph\>Salvage\</emph\> is the value of an asset after depreciation.
\<emph\>Life\</emph\> Life defines the period over which an asset is depreciated.
\<emph\>Period\</emph\> is the length of each period. The length must be entered in the same date unit as the depreciation period.
\<emph\>Month\</emph\> (optional) denotes the number of months for the first year of depreciation. If an entry is not defined, 12 is used as the default.
Example:
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. One period is 30 days.
DB(25000;1000;36;1;6) = 1,075.00 currency units
The fixeddeclining depreciation of the computer system is 1,075.00 currency units.
\<bookmark_value\>calculating; arithmeticdegressive depreciations\</bookmark_value\>\<bookmark_value\>arithmeticdegressive depreciations\</bookmark_value\>\<bookmark_value\>depreciations;arithmeticdegressive\</bookmark_value\>\<bookmark_value\>DDB function\</bookmark_value\>DDB
Returns the depreciation of an asset for a specified period using the arithmeticdeclining 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.
Syntax
DDB(Cost;Salvage;Life;Period;Factor)
\<emph\>Cost\</emph\> fixes the initial cost of an asset.
\<emph\>Salvage\</emph\> fixes the value of an asset at the end of its life.
\<emph\>Life\</emph\> is the number of periods defining how long the asset is to be used.
\<emph\>Period\</emph\> defines the length of the period. The length must be entered in the same time unit as life.
\<emph\>Factor\</emph\> (optional) is the factor by which depreciation decreases. If a value is not entered, the default is factor 2.
Example:
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 doubledeclining depreciation during the first month after purchase is 1,721.81 currency units.
\<bookmark_value\>DISC function\</bookmark_value\>\<bookmark_value\>allowances\</bookmark_value\>\<bookmark_value\>discounts\</bookmark_value\>DISC
Calculates the allowance (discount) of a security as a percentage.
Syntax
DISC(Settlement;Maturity;Price;Redemption;Basis)
Issue: the date of issue of the security.
Maturity: the date on which the security is sold.
\<emph\>Price\</emph\>: The price of the security per 100 currency units of par value.
\<emph\>Redemption\</emph\>: the redemption value of the security per 100 currency units of par value.
Example:
A security is purchased on 1.25.2001; the maturity date is 11.15.2001. The price (purchase price) is 97, the redemption value is 100. Using daily balance calculation (basis 3) how high is the settlement (discount)?
=DISC("1.25.2001"; "11.15.2001"; 97; 100; 3) returns 0.03840 or 3.84 per cent.
\<bookmark_value\>DURATION_ADD function\</bookmark_value\>\<bookmark_value\>Microsoft Excel functions\</bookmark_value\>\<bookmark_value\>durations;fixed interest securities\</bookmark_value\>DURATION_ADD
Calculates the duration of a fixed interest security in years.
Syntax
DURATION_ADD(Settlement;Maturity;Coupon;Yield;Frequency;Basis)
Issue: the date of issue of the security.
Maturity: the date on which the security is sold.
\<emph\>Coupon\</emph\>: the annual coupon interest rate (nominal rate of interest)
\<emph\>Par\</emph\>: the par value of the security.
Frequency: number of interest payments per year (1, 2 or 4).
Example:
A security is purchased on 1.1.2001; the maturity date is 1.1.2006. The Coupon rate of interest is 8%. The yield is 9.0%. Interest is paid halfyearly (frequency is 2). Using daily balance interest calculation (basis 3) how long is the duration?
=DURATION_ADD("1.1.2001"; "1.1.2006"; 0.08; 0.09; 2; 3)
\<bookmark_value\>annual net interest rates\</bookmark_value\>\<bookmark_value\>calculating; annual net interest rates\</bookmark_value\>\<bookmark_value\>net annual interest rates\</bookmark_value\>\<bookmark_value\>EFFECTIVE function\</bookmark_value\>EFFECTIVE
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 installments (for example, monthly or quarterly) before the end of the calculation period.
Syntax
EFFECTIVE(NOM;P)
\<emph\>NOM\</emph\> is the nominal interest.
\<emph\>P\</emph\> is the number of interest payment periods per year.
Example:
If the annual nominal interest rate is 9.75% and four interest calculation periods are defined, what is the actual interest rate (effective rate)?
EFFECTIVE(9.75%;4) = 10.11% The annual effective rate is therefore 10.11%.
\<bookmark_value\>effective interest rates\</bookmark_value\>\<bookmark_value\>EFFECT_ADD function\</bookmark_value\>EFFECT_ADD
Calculates the effective annual rate of interest on the basis of the nominal interest rate and the number of interest payments per annum.
Syntax
EFFECT_ADD(Nominal rate;Npery)
Nominal rate: the annual nominal rate of interest.
Npery: the number of interest payments per year.
Example:
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 5.3534%.
\<bookmark_value\>IRR function\</bookmark_value\>\<bookmark_value\>calculating;internal rates of return, regular payments\</bookmark_value\>\<bookmark_value\>internal rates of return;regular payments\</bookmark_value\>IRR
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.
Syntax
IRR(Values;Guess)
\<emph\>Values\</emph\> represents an array containing the values.
\<emph\>Guess\</emph\> (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.
Example:
Under the assumption that cell contents are A1=10000, A2=3500, A3=7600 and A4=1000, the formula =IRR(A1:A4) gives a result of 80.24%.
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.
\<bookmark_value\>calculating; interests for unchanged amortization installments\</bookmark_value\>\<bookmark_value\>interests for unchanged amortization installments\</bookmark_value\>\<bookmark_value\>ISPMT function\</bookmark_value\>
ISPMT
Calculates the level of interest for unchanged amortization installments.
Syntax
ISPMT(Rate; Period; Total_periods; Invest)
\<emph\>Rate\</emph\> sets the periodic interest rate.
\<emph\>Period\</emph\> is the number of installments for calculation of interest.
\<emph\>Total_periods\</emph\> is the total number of installment periods.
\<emph\>Invest\</emph\> is the amount of the investment.
Example:
For a credit amount of 120,000 currency units with a twoyear term and monthly installments, 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.
\<bookmark_value\>PV function\</bookmark_value\>\<bookmark_value\>present values\</bookmark_value\>\<bookmark_value\>calculating; present values\</bookmark_value\>PV
Returns the present 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 \<emph\>Rate\</emph\> and \<item type=\"productname\"\>LibreOffice\</item\> Calc with automatically calculate the correct factor.
Syntax
PV(Rate; NPER; PMT; FV; Type)
\<emph\>Rate\</emph\> defines the interest rate per period.
\<emph\>Total_periods\</emph\> is the total number of installment periods.
\<emph\>PMT\</emph\> is the regular payment made per period.
\<emph\>FV\</emph\> (optional) defines the future value remaining after the final installment has been made.
\<emph\>Type\</emph\> (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 LibreOffice 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.
Example:
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. Crosschecking 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 "Ifthen" scenarios. Please note: references to constants must be defined as absolute references. Examples of this type of application are found under the depreciation functions.
\<bookmark_value\>RECEIVED function\</bookmark_value\>\<bookmark_value\>amount received for fixedinterest securities\</bookmark_value\>RECEIVED
Calculates the amount received that is paid for a fixedinterest security at a given point in time.
Syntax
RECEIVED(Settlement;Maturity;Investment;Discount;Basis)
Issue: the date of issue of the security.
Maturity: the date on which the security is sold.
\<emph\>Investment\</emph\>: the purchase sum.
Discount: the percentage discount upon acquisition of the security.
Example:
Settlement date: February 15 1999, maturity date: May 15 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("2.15.99";"5.15.99";1000; 0.0575;2) returns 1014.420266.
\<bookmark_value\>calculating; depreciations\</bookmark_value\>\<bookmark_value\>SYD function\</bookmark_value\>\<bookmark_value\>depreciations; arithmetic declining\</bookmark_value\>\<bookmark_value\>arithmetic declining depreciations\</bookmark_value\>SYD
Returns the arithmeticdeclining 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.
Syntax
SYD(Cost;Salvage;Life;Period)
\<emph\>Cost\</emph\> fixes the initial cost of an asset.
\<emph\>Salvage\</emph\> is the value of an asset after depreciation.
\<emph\>Life\</emph\> is the period fixing the time span over which an asset is depreciated.
\<emph\>Period\</emph\> defines the period for which the depreciation is to be calculated.
Example:
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 formulas available in \<item type=\"productname\"\>LibreOffice\</item\> Calc next to each other, you can see which depreciation form is the most appropriate. Enter the table as follows:
\<emph\>Mode\</emph\> 
\<emph\>Mode\</emph\> 
\<emph\>Mode\</emph\> 
\<emph\>Mode\</emph\> 
\<emph\>Mode\</emph\> 

1 
195 
195 
195 
195 
195 
2 
10,666.67 currency units 
10,666.67 currency units 
195 
195 
10,666.67 currency units 
3 
195 
10,666.67 currency units 

4 
183 
10,666.67 currency units 

5 
148 
10,666.67 currency units 

6 
195 
10,666.67 currency units 

7 
195 
10,666.67 currency units 

8 
170 

9 
148 

10 
195 

11 
170 

12 

13 
170 
195 
10,666.67 currency units 
The formula in E2 is as follows:
=TDIST(12; 5; 1)
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:
=TDIST(12; 5; 1)
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.