Gobe Productive User's Guide - Financial Functions

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# Financial Functions

DB | FV | PMT
DDB | NPER | PV

Financial functions are useful for computing complex financial information such as interest rates, payment and equity data.

## DB

> Returns the depreciation of an investment using the fixed-declining balance method for the specified period. The 'life' and 'period' arguments must use the same units of measure.

> ### Syntax

> DB( cost, salvage, life, period, {months} )

> ### Arguments

> cost | The initial cost to depreciate. Must be a positive number.
> salvage | The value of the item at the end of the depreciation period. Must be a positive number.
> life | The number of periods in the depreciation cycle. Must be a positive number.
> period | The period of time to compute the depreciation. Must be a positive number.
> months | The number of months in the first year of depreciation. If omitted, 12 months is used. Must be a positive number.

> ### Example

> DB( 10000, 1000, 5, 1 ) is $3,690

> A $10,000 investment depreciates $3,690 in the first year if it is depreciated over 5 years with a final salvage value of $1,000.

## DDB

> Returns the depreciation of an investment using the variable declining balance method for the specified period. The 'life' and 'period' arguments must use the same units of measure.

> ### Syntax

> DB( cost, salvage, life, period, {factor} )

> ### Arguments

> cost | The initial cost to depreciate. Must be a positive number.
> salvage | The value of the item at the end of the depreciation period. Must be a positive number.
> life | The number of periods in the depreciation cycle. Must be a positive number.
> period | The period of time to compute the depreciation. Must be a positive number.
> factor | The rate of balance decline. If omitted, the system assumes 2 (double declining balance method). Must be a positive number. |

> ### Example

> DDB( 10000, 1000, 5, 1, 3) is $6,000.

> A $10,000 investment depreciates $6,000 in the first year if it is depreciated over 5 years with a final salvage value of $1,000, and a depreciation factor of 3.

## FV

> Returns the future value of an investment based on constant periodic payments and a constant interest rate.

> Make sure that the rate corresponds to the number of payment periods. For a monthly payment period remember to divide an annual interest rate by 12 (months) to get a monthly interest rate. Enter cash paid out as negative values, and any cash received as positive values.

> ### Syntax

> FV( rate, periods, payment, {pv}, {atEnd} )

> ### Arguments

> rate | The interest rate for the period of the investment.
> periods | The number of payment periods.
> payment | The amount paid each period.
> pv | The present value of the investment (optional, default is 0).
> atEnd | If TRUE payments take place at end of a payment period. (optional, default is TRUE.)

> ### Example

> FV( .10/12, 5*12, -100 ) = $7,743.71

> You save $7,743.71 after 5 years depositing $100 a month with no initial deposit and a 10% interest rate.

## NPER

> Returns the number of periods for an investment with a constant cash flow and interest rate given a present and future value.

> Make sure that the rate corresponds to the number of payment periods. For a monthly payment period remember to divide an annual interest rate by 12 (months) to get a monthly interest rate. Enter cash paid out as negative values, and any cash received as positive values.

> ### Syntax

> NPER( rate, pmt, pv, {fv}, {atEnd} )

> ### Arguments

> rate | The interest rate for the period of the investment.
> payment | The amount paid each period.
> pv | The present value of the investment.
> fv | The future value of the investment (optional, default is 0).
> atEnd | If TRUE payments take place at end of a payment period. (optional, default is TRUE.)

> ### Example

> NPER( .08/12, -100, 2000 ) = 21.54.

> At $100.00 per month it takes 21.54 months to repay $2,000.00 at 8% interest.

> NPER( .08/12, -100, 0, 2000 ) = 18.84

> At $100.00 per month in an 8% interest account, it takes 18.84 months to save $2,000.00.

## PMT

> Returns the payment amount needed to finance an investment.

> Make sure that the rate corresponds to the number of payment periods. For a monthly payment period remember to divide an annual interest rate by 12 (months) to get a monthly interest rate. Enter cash paid out as negative values, and any cash received as positive values.

> ### Syntax

> PMT( rate, periods, pv, {fv}, {atEnd} )

> ### Arguments

> rate | The interest rate for the period of the investment.
> periods | The number of payment periods.
> pv | The present value of the investment.
> fv | The future value of the investment (optional, default is 0).
> atEnd | If TRUE payments take place at end of a payment period. (optional, default is TRUE.)

> ### Example

> PMT( .08/12, 12*3, 10000 ) = -$313.36

> $10,000 borrowed at 8% annual interest requires a monthly payment of $313.36 to pay off the loan in 3 years.

## PV

> Returns the present value of an investment based on constant periodic payments and a constant interest rate.

> Make sure that the rate corresponds to the number of payment periods. For a monthly payment period remember to divide an annual interest rate by 12 (months) to get a monthly interest rate. Enter cash paid out as negative values, and any cash received as positive values.

> ### Syntax

> PV( rate, periods, payment, {fv}, {atEnd} )

> ### Arguments

> rate | The interest rate for the period of the investment.
> periods | The number of payment periods.
> payment | The amount paid each period.
> fv | The future value of the investment (optional, default is 0).
> atEnd | If TRUE payments take place at end of a payment period. (optional, default is TRUE.)

> ### Example

> PV( .10/12, 5*12, -100, 10000 ) is -$1371.35.

> Depositing $100 a month into a 10% savings account requires an initial outlay of $1,371.35. to have $10,000 in the account in 5 years.

> PV( .08/12, 3*12, -300 ) is $9,573.54

> A $9,573.54 loan at 8% can be paid off in 3 years with monthly payments of $300.

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