考虑到初始贷款金额,还款数量和还款金额,我正在尝试计算贷款利息。我似乎无法使用基本公式获得足够接近的数字,所以我一直在尝试使用Newton Raphson方法,这可以在这里使用:http://www.efunda.com/formulae/finance/loan_calculator.cfm(这是我的确切功能希望实施)
我已经尝试过使用PHPExcel的RATE()函数,但是我没有为输入获得正确的输出,即利率回到0.1%或类似(实际上它更像是5.75%)
这是PHPExcel的相关代码
/** FINANCIAL_MAX_ITERATIONS */
define('FINANCIAL_MAX_ITERATIONS', 128);
/** FINANCIAL_PRECISION */
define('FINANCIAL_PRECISION', 1.0e-08);
/**
* Convert an array to a single scalar value by extracting the first element
*
* @param mixed $value Array or scalar value
* @return mixed
*/
function flattenSingleValue($value = '') {
while (is_array($value)) {
$value = array_pop($value);
}
return $value;
}
/**
* RATE
*
* Returns the interest rate per period of an annuity.
* RATE is calculated by iteration and can have zero or more solutions.
* If the successive results of RATE do not converge to within 0.0000001 after 20 iterations,
* RATE returns the #NUM! error value.
*
* Excel Function:
* RATE(nper,pmt,pv[,fv[,type[,guess]]])
*
* @access public
* @category Financial Functions
* @param float nper The total number of payment periods in an annuity.
* @param float pmt The payment made each period and cannot change over the life
* of the annuity.
* Typically, pmt includes principal and interest but no other
* fees or taxes.
* @param float pv The present value - the total amount that a series of future
* payments is worth now.
* @param float fv The future value, or a cash balance you want to attain after
* the last payment is made. If fv is omitted, it is assumed
* to be 0 (the future value of a loan, for example, is 0).
* @param integer type A number 0 or 1 and indicates when payments are due:
* 0 or omitted At the end of the period.
* 1 At the beginning of the period.
* @param float guess Your guess for what the rate will be.
* If you omit guess, it is assumed to be 10 percent.
* @return float
**/
function RATE($nper, $pmt, $pv, $fv = 0.0, $type = 0, $guess = 0.1) {
$nper = (int) flattenSingleValue($nper);
$pmt = flattenSingleValue($pmt);
$pv = flattenSingleValue($pv);
$fv = (is_null($fv)) ? 0.0 : flattenSingleValue($fv);
$type = (is_null($type)) ? 0 : (int) flattenSingleValue($type);
$guess = (is_null($guess)) ? 0.1 : flattenSingleValue($guess);
$rate = $guess;
if (abs($rate) < FINANCIAL_PRECISION) {
$y = $pv * (1 + $nper * $rate) + $pmt * (1 + $rate * $type) * $nper + $fv;
} else {
$f = exp($nper * log(1 + $rate));
$y = $pv * $f + $pmt * (1 / $rate + $type) * ($f - 1) + $fv;
}
$y0 = $pv + $pmt * $nper + $fv;
$y1 = $pv * $f + $pmt * (1 / $rate + $type) * ($f - 1) + $fv;
// find root by secant method
$i = $x0 = 0.0;
$x1 = $rate;
while ((abs($y0 - $y1) > FINANCIAL_PRECISION) && ($i < FINANCIAL_MAX_ITERATIONS)) {
$rate = ($y1 * $x0 - $y0 * $x1) / ($y1 - $y0);
$x0 = $x1;
$x1 = $rate;
if (($nper * abs($pmt)) > ($pv - $fv))
$x1 = abs($x1);
if (abs($rate) < FINANCIAL_PRECISION) {
$y = $pv * (1 + $nper * $rate) + $pmt * (1 + $rate * $type) * $nper + $fv;
} else {
$f = exp($nper * log(1 + $rate));
$y = $pv * $f + $pmt * (1 / $rate + $type) * ($f - 1) + $fv;
}
$y0 = $y1;
$y1 = $y;
++$i;
}
return $rate;
} // function RATE()
我对该功能的输入是:
RATE(60, 1100, 50000); // Outputs 0.00420298759161
RATE(60, -1100, 50000); // Outputs 0.00959560344752
RATE(60, 1100, 66000); // Outputs -1.05036370955
如果60是5年的月数,则1100或-1100是每月偿还的金额,50,000是借款总额。
我不是数学家,上面的函数对我来说没有多大意义,但我的阅读说这是计算速率的最佳方法。希望我只是犯了一个愚蠢的错误......
答案 0 :(得分:3)
您可以使用“二进制搜索”代替“Newton Raphson方法”。
function rate($month, $payment, $amount)
{
// make an initial guess
$error = 0.0000001; $high = 1.00; $low = 0.00;
$rate = (2.0 * ($month * $payment - $amount)) / ($amount * $month);
while(true) {
// check for error margin
$calc = pow(1 + $rate, $month);
$calc = ($rate * $calc) / ($calc - 1.0);
$calc -= $payment / $amount;
if ($calc > $error) {
// guess too high, lower the guess
$high = $rate;
$rate = ($high + $low) / 2;
} elseif ($calc < -$error) {
// guess too low, higher the guess
$low = $rate;
$rate = ($high + $low) / 2;
} else {
// acceptable guess
break;
}
}
return $rate * 12;
}
var_dump(rate(60, 1000, 20000));
// Return 0.56138305664063, which means 56.1383%
“二元搜索”和“牛顿拉斐逊方法”基本上是一种猜测方法。它会进行初步猜测并随着时间的推移改进猜测,直到达到可接受的猜测。 “Newton Raphson方法”通常比“二分搜索”更快,因为它有更好的“改进猜测”策略。
概念很简单:
我们想知道r
哪个是利率。我们知道,N,C和P.我们不知道什么是r
,但是让我们猜测0.00到1.00之间的任何数字。 (在这种情况下,我们假设利率不能超过100%)。
r
是每月费率。多次乘12来获得年率。