二分搜索

时间:2012-10-10 13:03:02

标签: python bisect

  

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:)

2 个答案:

答案 0 :(得分:4)

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)

答案 1 :(得分:3)

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)