二分搜索

Question

  

可能重复：
  Using bisection search to determine

我已发布其他帖子，但没有收到答案，因此我试图提供一些工作以使其更清晰。

我需要用二分法确定每月付款，以便在一年内偿还债务。

这里有一些代码：

originalBalance = 320000
annualInterestRate = 0.2
monthly_interest = annualInterestRate / 12
low = originalBalance/12
high = (originalBalance*(1 + monthly_interest)**12)/12
epsilon = 0.01
min_payment = (high + low)/2.0

while min_payment*12 - originalBalance >= epsilon:
    for month in range(0, 12):
        balance = (originalBalance - min_payment) * (1+monthly_interest)

    if balance < 0:
        low = min_payment
    elif balance > 0:
        high = min_payment
        min_payment = (high + low)/2.0
print "lowest payment: " + str(balance)

然而，我收到的答案很简单：298222.173851

我的朋友告诉我，正确答案是：29157.09

这比我的要低很多......我想问题是四舍五入（我还没有做）并在每次循环后保持平衡，如果余额超过0则重置它。我无法弄清楚如何尝试这个问题，请帮助别人：）

Answer 1

这是正确的。

originalBalance = 320000
annualInterestRate = 0.2
monthly_interest = annualInterestRate / 12
low = originalBalance/12
high = (originalBalance*(1 + monthly_interest)**12)/12
epsilon = 0.01
min_payment = (high + low)/2.0

while min_payment*12 - originalBalance >= epsilon:
    for month in range(0, 12):
        balance = (originalBalance - min_payment)/10 * (1+monthly_interest)

    if balance < 0:
        low = min_payment
    elif balance > 0:
        high = min_payment
        min_payment = (high + low)/2.0
print "lowest payment: " + str(balance)

Answer 2

这就是你需要的

while abs(x) > epsilon:
    x = balance
    for month in range(0, 12):
        x = (x - ans) * (1+monthly_interest)