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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. |
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.
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.
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.
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.
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.
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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