<p class="Paragraph">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.</p>
<p class="Paragraph">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.</p>
<p class="Paragraph"><help:key-word value="accrued interest" tag="kw66877_33" xmlns:help="http://openoffice.org/2000/help"/>Calculates the accrued interest of a security in the case of periodic payments.</p>
<p class="Paragraph">A security is issued on 2.28.2001. The first interest date is set for 8.31.2001. The settlement date is 5.1.2001. The nominal rate is 0.1 or 10%, the par value is 1000 currency units. Interest is paid half-yearly (frequency is 2). The basis is the US method (0). How much interest has accrued?</p>
<p class="Paragraph">A security is issued on 4.1.2001. The maturity date is set for 6.15.2001. The nominal rate is 0.1 or 10%, the par value is 1000 currency units. The basis of the daily/annual calculation is the daily balance (3). How much interest has accrued?</p>
<p class="Paragraph"><help:help-text value="visible" xmlns:help="http://openoffice.org/2000/help">Returns the present value of an investment resulting from a series of regular payments.</help:help-text></p>
<p class="Paragraph">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.</p>
<p class="Paragraph">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 <span class="T1">Rate</span> and <help:productname xmlns:help="http://openoffice.org/2000/help">%PRODUCTNAME</help:productname> Calc with automatically calculate the correct factor.</p>
<p class="Paragraph"><span class="T1">Rate</span> defines the interest rate per period.</p>
<p class="Paragraph"><span class="T1">NPER</span> is the total number of periods (payment period).</p>
<p class="Paragraph"><span class="T1">PMT</span> is the regular payment made per period.</p>
<p class="Paragraph"><span class="T1">FV</span> (optional) defines the future value remaining after the final installment has been made.</p>
<p class="Paragraph"><span class="T1">Type</span> (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.</p>
<p class="Head3">Example</p>
<p class="Paragraph">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.</p>
<p class="Paragraph">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.</p>
<p class="Paragraph">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.</p>
<p class="Paragraph"><help:help-text value="visible" xmlns:help="http://openoffice.org/2000/help">Returns the arithmetic-declining depreciation rate.</help:help-text></p>
<p class="Paragraph">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.</p>
<p class="Paragraph"><span class="T1">Cost</span> is the initial cost of an asset.</p>
<p class="Paragraph"><span class="T1">Salvage</span> is the value of an asset after depreciation.</p>
<p class="Paragraph"><span class="T1">Life</span> is the period fixing the time span over which an asset is depreciated.</p>
<p class="Paragraph"><span class="T1">Period</span> defines the period for which the depreciation is to be calculated.</p>
<p class="Head3">Example</p>
<p class="Paragraph">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.</p>
<p class="Paragraph">SYD(50000;10000;5;1)=13,333.33 currency units. The depreciation amount for the first year is 13,333.33 currency units.</p>
<p class="Paragraph">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 <help:productname xmlns:help="http://openoffice.org/2000/help">%PRODUCTNAME</help:productname> Calc next to each other, you can see which depreciation form is the most appropriate. Enter the table as follows:</p>
<p class="Paragraph">The formula in E2 is as follows:</p>
<p class="Paragraph">=SYD($A$2;$B$2;$C$2;D2)</p>
<p class="Paragraph">This formula is duplicated in column E down to E11 (select E2, then drag down the lower right corner with the mouse).</p>
<p class="Paragraph">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:</p>
<p class="Paragraph">=SUMIF(E2:E11;A13)</p>
<p class="Paragraph">Now view the depreciation for a 10 year period, or at a salvage value of 1 currency unit, or enter a different initial cost, etc.</p>
<p class="Paragraph">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 allowance (discount)?</p>
<p class="Paragraph">=DISCOUNT("1.25.2001"; "11.15.2001"; 97; 100; 3) returns 0.03840 or 3.84 per cent.</p>
<p class="Paragraph">A security is purchased on 1.1.2001; the maturity date is 1.1.2006. The nominal rate of interest is 8%. <text:s text:c="" xmlns:text="http://openoffice.org/2000/text"/>The yield is 9.0%. Interest is paid half-yearly (frequency is 2). Using daily balance interest calculation (basis 3) how long is the duration?</p>
<p class="Paragraph"><help:help-text value="visible" xmlns:help="http://openoffice.org/2000/help">Returns the net annual interest rate for a nominal interest rate.</help:help-text></p>
<p class="Paragraph">As nominal interest refers to the interest amount due at the end of a calculation period, the effective interest increases with the number of payments made. That is to say interest is often paid in advance with the monthly, quarterly, etc. installments prior to the end of calculation period.</p>
<p class="Head3">Syntax</p>
<p class="Paragraph">EFFECTIVE(NOM;P)</p>
<p class="Paragraph"><span class="T1">NOM</span> is the nominal interest.</p>
<p class="Paragraph"><span class="T1">P</span> is the number of interest payment periods per year.</p>
<p class="Head3">Example</p>
<p class="Paragraph">If the annual nominal interest rate is 9.75% and four interest calculation periods are defined, what is the actual interest rate (effective rate)?</p>
<p class="Paragraph">EFFECTIVE(9.75%;4) = 10.11% The annual effective rate is therefore 10.11%.</p>
<p class="Paragraph">Calculates the effective annual rate of interest on the basis of the nominal interest rate and the number of interest payments per annum.</p>
<p class="Paragraph"><help:help-text value="visible" xmlns:help="http://openoffice.org/2000/help">Returns the depreciation of an asset for a specified period using the arithmetich-declining method.</help:help-text></p>
<p class="Paragraph">Use this form of depreciation if you require a higher initial depreciation value (as opposed to linear depreciation). The depreciation value get less with each period and is usually used for assets whose value loss is higher shortly after purchase (e.g. vehicles, computers). Please note that the book value will never reach zero under this calculation type.</p>
<p class="Paragraph"><span class="T1">Cost</span> fixes the initial cost of an asset.</p>
<p class="Paragraph"><span class="T1">Salvage</span> fixes the value of an asset at the end of its life.</p>
<p class="Paragraph"><span class="T1">Life</span> is the number of periods defining how long the asset is to be used.</p>
<p class="Paragraph"><span class="T1">Period</span> defines the length of the period. The length must be entered in the same time unit as life.</p>
<p class="Paragraph"><span class="T1">Factor</span> (optional) is the factor by which depreciation decreases. If a value is not entered, the default is factor 2.</p>
<p class="Head3">Example</p>
<p class="Paragraph">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.</p>
<p class="Paragraph">DDB(75000;1;60;12;2) = 1,721.81 currency units. Therefore, the double-declining depreciation during the first month after purchase is 1,721.81 currency units.</p>
<p class="Paragraph"><help:help-text value="visible" xmlns:help="http://openoffice.org/2000/help">Returns the depreciation of an asset for a specified period using the double-declining balance method.</help:help-text></p>
<p class="Paragraph">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.</p>
<p class="Paragraph"><span class="T1">Cost</span> is the initial cost of an asset.</p>
<p class="Paragraph"><span class="T1">Salvage</span> is the value of an asset at the end of the depreciation.</p>
<p class="Paragraph"><span class="T1">Life</span> Life defines the period over which an asset is depreciated.</p>
<p class="Paragraph"><span class="T1">Period</span> is the length of each period. The length must be entered in the same date unit as the depreciation period.</p>
<p class="Paragraph"><span class="T1">Month</span> (optional) denotes the number of months for the first year of depreciation. If an entry is not defined, 12 is used as the default.</p>
<p class="Head3">Example</p>
<p class="Paragraph">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.</p>
<p class="Paragraph"><help:help-text value="visible" xmlns:help="http://openoffice.org/2000/help">Use this function to calculate the internal interest rate for an investment without considering costs or profits. This allows you to verify the profitability of an investment.</help:help-text></p>
<p class="Head3">Syntax</p>
<p class="Paragraph">IRR(Values;Guess)</p>
<p class="Paragraph"><span class="T1">Values</span> represents a cell reference or an array containing the values of the payment amounts.</p>
<p class="Paragraph"><span class="T1">Guess</span> (optional) is the estimated value. The initial interest rate value.</p>
<p class="Head3">Example</p>
<p class="Paragraph">Under the assumption that cell contents are A1=-10000, A2=13500, A3=7600 and A4=1000, a result of 80.24% is shown.</p>
<p class="Paragraph"><help:help-text value="visible" xmlns:help="http://openoffice.org/2000/help">This function allows you to calculate the level of interest for unchanged amortization installments.</help:help-text></p>
<p class="Paragraph"><span class="T1">Interest</span> sets the periodic interest rate.</p>
<p class="Paragraph"><span class="T1">Period</span> is the number of installments for calculation of interest.</p>
<p class="Paragraph"><span class="T1">Total periods</span> is the total number of installment periods.</p>
<p class="Paragraph"><span class="T1">Invest</span> is the amount of the investment.</p>
<p class="Head3">Example</p>
<p class="Paragraph">For a credit amount of 120,000 currency units with a two-year term and monthly installments, at an interest rate of 12% the level of interest after 1.5 years is required.</p>
<p class="Paragraph">ISPMT(1;18;24;120000) = -30,000 currency units. The total interest after 1.5 years amounts to 30,000 currency units.</p>
<p class="Paragraph"><help:link Id="68336" xmlns:help="http://openoffice.org/2000/help">Forward to Financal Functions Part Two</help:link></p>
<p class="Paragraph"><help:link Id="68335" xmlns:help="http://openoffice.org/2000/help">Forward to Financal Functions Part Three</help:link></p>
