我正在寻找一个简单的javascript财务RATE函数,我找到了这个.但似乎很难理解.我想简化这个功能,我需要你的帮助.
如果有人有最简单的功能,请回答. (这是一个excel RATE函数等效.)
var rate = function(nper, pmt, pv, fv, type, guess) {
if (guess == null) guess = 0.01;
if (fv == null) fv = 0;
if (type == null) type = 0;
var FINANCIAL_MAX_ITERATIONS = 128;//Bet accuracy with 128
var FINANCIAL_PRECISION = 0.0000001;//1.0e-8
var y, y0, y1, x0, x1 = 0, f = 0, i = 0;
var rate = guess;
if (Math.abs(rate) < FINANCIAL_PRECISION) {
y = pv * (1 + nper * rate) + pmt * (1 + rate * type) * nper + fv;
} else {
f = Math.exp(nper * Math.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 Newton secant method
i = x0 = 0.0;
x1 = rate;
while ((Math.abs(y0 - y1) > FINANCIAL_PRECISION) && (i < FINANCIAL_MAX_ITERATIONS)) {
rate = (y1 * x0 - y0 * x1) / (y1 - y0);
x0 = x1;
x1 = rate;
if (Math.abs(rate) < FINANCIAL_PRECISION) {
y = pv * (1 + nper * rate) + pmt * (1 + rate * type) * nper + fv;
} else {
f = Math.exp(nper * Math.log(1 + rate));
y = pv * f + pmt * (1 / rate + type) * (f - 1) + fv;
}
y0 = y1;
y1 = y;
++i;
}
return rate;}
谢谢!
解决方法:
数学太复杂,我无法理解,但这可能更容易阅读.一些变量已被重命名为更有意义,并且它的格式更容易在您的眼睛
function rate(paymentsPerYear, paymentAmount, presentValue, futureValue, dueEndOrBeginning, interest)
{
//If interest, futureValue, dueEndorBeginning was not set, set now
if (interest == null)
interest = 0.01;
if (futureValue == null)
futureValue = 0;
if (dueEndOrBeginning == null)
dueEndOrBeginning = 0;
var FINANCIAL_MAX_ITERATIONS = 128;//Bet accuracy with 128
var FINANCIAL_PRECISION = 0.0000001;//1.0e-8
var y, y0, y1, x0, x1 = 0, f = 0, i = 0;
var rate = interest;
if (Math.abs(rate) < FINANCIAL_PRECISION)
{
y = presentValue * (1 + paymentsPerYear * rate) + paymentAmount * (1 + rate * dueEndOrBeginning) * paymentsPerYear + futureValue;
}
else
{
f = Math.exp(paymentsPerYear * Math.log(1 + rate));
y = presentValue * f + paymentAmount * (1 / rate + dueEndOrBeginning) * (f - 1) + futureValue;
}
y0 = presentValue + paymentAmount * paymentsPerYear + futureValue;
y1 = presentValue * f + paymentAmount * (1 / rate + dueEndOrBeginning) * (f - 1) + futureValue;
// find root by Newton secant method
i = x0 = 0.0;
x1 = rate;
while ((Math.abs(y0 - y1) > FINANCIAL_PRECISION)
&& (i < FINANCIAL_MAX_ITERATIONS))
{
rate = (y1 * x0 - y0 * x1) / (y1 - y0);
x0 = x1;
x1 = rate;
if (Math.abs(rate) < FINANCIAL_PRECISION)
{
y = presentValue * (1 + paymentsPerYear * rate) + paymentAmount * (1 + rate * dueEndOrBeginning) * paymentsPerYear + futureValue;
}
else
{
f = Math.exp(paymentsPerYear * Math.log(1 + rate));
y = presentValue * f + paymentAmount * (1 / rate + dueEndOrBeginning) * (f - 1) + futureValue;
}
y0 = y1;
y1 = y;
++i;
}
return rate;
}