【发布时间】:2016-08-30 20:16:25
【问题描述】:
我正在创建一个在 javascript 中每月贡献的复利公式,用于填充 chartkick.js 图表,以显示每年年底本金增长到的总金额。这是我遵循的公式。
Compound Interest For Principal
P(1 + r/n) ^ nt
Future Value of a Series
PMT * (((1 + r/n)^nt - 1) / (r/n)) * (1+r/n)
A = [Compound Interest for Principal] + [Future Value of a Series]
A = the future value of the investment/loan, including interest
P = the principal investment amount (the initial deposit or loan amount)
PMT = the monthly payment
r = the annual interest rate (decimal)
n = the number of times that interest is compounded per year AND additional payment frequency
t = the number of years the money is invested or borrowed for
*/
这是我的 javascript:
P = $("#start-amount").val().replace(/[^\d\.]/g, '');
var t = $("#years").val();
var PMT = $("#contributions-amount").val().replace(/[^\d\.]/g, '');
var n = $("#contribution-rate").val();
//If you choose monthly it is equal to 12 and annually it is equal to 1
//Convert Interest Rate to Decimal
var r_percentage = $("#interest-rate").val().replace(/[^\d\.]/g, '');
var r = parseFloat(r_percentage) / 100.0;
//Create an arary with list of numbers less than number of years
var t_arr = [];
for (i = 0; i <= t; i++) {
if (i > 0 && i <= t) {
t_arr.push(i)
}
}
//For Chartkick Chart
data = [
{"name": "With interest", "data": {}
},
{"name": "No Interest", "data": {}
}];
/*
Needs to be like this!
data = [
{"name":"Workout", "data": {"2013-02-10 00:00:00 -0800": 3, "2013-02-17 00:00:00 -0800": 4}},
{"name":"Call parents", "data": {"2013-02-10 00:00:00 -0800": 5, "2013-02-17 00:00:00 -0800": 3}}
];*/
/*With Interest
J is equal to each individual year less than the total number of years*/
for (j = 1; j <= t_arr.length; j++) {
var compound_interest_for_principal = P * (Math.pow(1 + r / n, n * t));
var value_rounded1 = Math.round(compound_interest_for_principal * 100) /100;
var future_value_of_series = PMT * (Math.pow(1 + r / n, n * j) - 1) / (r / n) * (1 + r / n);
var value_rounded2 = Math.round(future_value_of_series * 100) / 100;
var A = value_rounded1 + value_rounded2;
var A = A.toFixed(2);
if (data[0]["data"][j] === undefined) {
data[0]["data"][j] = A;
}
}
/*Without Interest */
for (k = 1; k <= t_arr.length; k++) {
var r = 0;
var principal_no_interest= P;
var value_rounded1 = Math.round(principal_no_interest * 100) /100;
var monthly_no_interest = PMT * n * k;
var value_rounded2 = Math.round(monthly_no_interest * 100) / 100;
var A = value_rounded1 + value_rounded2;
var A = A.toFixed(2);
if (data[1]["data"][k] === undefined) {
data[1]["data"][k] = A;
}
}
new Chartkick.LineChart("savings-chart", data, {"discrete": true});
我正在测试的值是 P = 1,000 美元,r = 5%(年利率),PMT = 每月 100 美元,n = 12(每年复利的次数和额外的付款频率)。
问题出在我得到的值。
1 年的利息后,我得到 2,880 美元,我应该得到 2,277 美元。 10 年后,我应该得到 17,065.21 美元,但我得到了 17,239.94 美元。我不明白为什么它在原版的基础上增加了大约 1,800 美元,而应该增加了大约 1,200 美元。有什么建议吗?
【问题讨论】:
-
将任何
x ^ y替换为Math.pow(x, y)。有什么问题? -
问题在于将公式的前半部分嵌套在那里......答案将数百万。
-
不,不会的。你保留嵌套,所以
Math.pow(x, y)中的x是(1 + r/n)... -
可能是
var future_value = contributions_amount * (((1 + interest_rate/contribution_rate) ^ (contribution_rate * years) - 1 / (interest_rate/years) * (1 + (interest_rate/years)));末尾缺少右括号 -
@JasonBourne IEEE 754 可以处理高达 1.7976931348623157e+308 的双精度。有什么问题?
标签: javascript math chartkick