二等分搜索 [英] Bisection search

问题描述

可能重复:
使用对分搜索来确定

Possible Duplicate:
Using bisection search to determine

我发布了其他主题，但未收到答案，因此，我试图提供一些工作以使其更加清晰.

I have posted other thread but it did not receive answers thus i'm trying to provide some of my work to make more clear.

我需要使用二等分法来确定每月还款额，以便准确地在一年内还清债务.

I need to use bisection method to determine monthly payment in order to pay off debt in one year exactly.

以下是一些代码:

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)

但是，我得到的答案很远:298222.173851

However, I receive very way off answer: 298222.173851

我的朋友告诉我正确答案是:29157.09

My friend told me that correct answer is : 29157.09

哪个比我的低很多...我想问题出在四舍五入(我还没有这样做)，并且在每次循环后保留余额并在余额超过0时将其重置.我不知道该怎么办尝试此问题，请帮助某人:)

Which is a lot lower than my...I guess the problem is in rounding(which I did not do yet) and preserving the balance after every looping and resetting it if balance is over 0. I cannot figure out how to attempt this problem and please, help someone :)

推荐答案

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)

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