使用python中的二分搜索计算最低的每月付款

Question

我正在尝试使用以下方法计算每月最低付款以偿还贷款:

balance = 999999
annualInterestRate = .18
monthlyInterestRate = annualInterestRate/12

balanceCOPY = balance

#Bisection search parameters

lo = balance/12
hi = (balance*(1+monthlyInterestRate**12))/12
epsilon = .01

guess = (lo + hi)/2

while True:
   for month in range(1,13):
      balance = balance - guess
      balance = balance + (monthlyInterestRate*balance)

   if balance > 0 and balance > epsilon:
      lo = guess
      balance = balanceCOPY
   elif balance < 0 and balance < -epsilon:
      hi = guess
      balance = balanceCOPY
   else:
      print('Lowest payment: ',str(round(guess,2)))
      break

   guess = (lo + hi)/2

但是,我似乎陷入某种无限循环,我的guess变量没有被更新.如何突破无限循环并guess更新变量？

问题出在我的数学上.我的意思是说

hi = (balance*(1+monthlyInterestRate)**12)/12

谢谢大家的帮助!

Answer 1

首先,如果您正在进行MITx练习并完成之前的测试(只是在猜测中增加10),那么您只需要迈出一小步.只需要对条件进行一些调整并检查年度结果.

关于二分法搜索我将尝试澄清这个概念.你总会有两个肢体,最小值和最大值.而且总是会在四肢中间开始猜测.

在第一次猜测之后,您需要根据年度结果调整四肢.如果在一年之后支付最低限度的饮料,女孩,程序书籍和其他你没有支付总余额的东西,你肯定需要增加最低限度.否则,例如,如果您在第10个月支付总余额,那么明年你需要多喝酒并结识新女孩!开玩笑......你需要减少最低限度.这是您在完成一年的硬支付后需要进行的检查

在练习中,我们有:

• balance和annualInterestRate =给出(我们不需要关心)
• 最小值(下限)=余额/ 12
• 最大值(上限)=(余额x(1 +月利率)**12)/ 12.0

第一个猜测是(最小+最大)/ 2我调用guessMinimum所以:

guessMinimum = (minimum + maximum)/2

所以你将开始使用第一个猜测(guessMinimum).一年后,你会检查剩余的.如果遗骸是负数,则表示您已经支付了太多.你需要减少每月付款.另外,如果一个月后剩余是积极的(例如,超过你的精确度(例如0.10))你需要减少每月付款,好吗？!

试图设计思路.....

 +------------------------------------------------+ 
 |   /\                  /\                   /\  | 
 |   \/------------------\/-------------------\/  | 
 |MINIMUM               guess              MAXIMUM| 
 |                     Minimum                    | 
 +------------------------------------------------+ 

如果一年后,"保持"是负面的(例如).意味着'guessMinimum'非常多!你需要......不是你,程序!! 程序需要调整它,降低最小值......

 +---------------------------------------------------+ 
 |                        Got negative 'remain'      | 
 |                   ++                              | 
 |    /\             ||   /\                   /\    | 
 |    \/-------------||---\/-------------------\/    | 
 | MINIMUM           ||  guess              MAXIMUM  | 
 |                   ++ Minimum-,                    | 
 |                               ',                  | 
 |                                 `.                | 
 |                                   `.,             | 
 |                                      ',           | 
 |                                        ',         | 
 |                                          `.       | 
 |                                            `      | 
 |    /\                  /\                   /\    | 
 |    \/------------------\/-------------------\/    | 
 | MINIMUM               guess              MAXIMUM  | 
 +---------------------------------------------------+ 

对不起大家.我试图插入图像,但作为新成员.我不能.需要至少10个声望....帮助我!!!! 使用角色的工作太多!!!!

并且CODE需要做这项艰苦的工作来调整最小值,直到'保持'可接受(在你的精度,epsilon,或任何字母或变量或......之内)好吧.:)

在理解了概念和图纸之后..让我们检查一下CODE.

balance = 999999; 
annualInterestRate = 0.18

monthlyInterestRate = annualInterestRate / 12

minimum = balance / 12
maximum = (balance * (1 + monthlyInterestRate)**12) / 12.0

guessMinimum = (minimum + maximum)/2

remain = balance #if you payed nothin, the remain is the balance!!!!

precision = 0.10  #you choose....

while (remain >= precision):

    guessMinimum = (minimum + maximum)/2


    for i in range (1,13):

        newBalance = remain - guessMinimum
        monthInterest = annualInterestRate/12*newBalance
        remain = newBalance+monthInterest

    # after one month, the CODE need to check about the remain

    if (remain < 0): #paying too much.... need to decrease the value

        maximum = guessMinimum      #remember my beautiful draw above!!
        remain = balance  # reset the remain to start again!!

    elif (remain > precision): #paying less .... need to increase the value
        minimum = guessMinimum
        remain = balance  # reset the remain to start again!!   

print "Lowest Payment: %.2f" %(guessMinimum)

而已.

Answer 2

我认为这个解决方案应该有效,

balance = 999999
annualInterestRate = 0.18

monthlyInterestRate = annualInterestRate / 12
lowerBound = balance / 12
upperBound = (balance * (1 + annualInterestRate / 12) ** 12) / 12
originalBalance = balance
lowestBalance = 0.01 # Error margin e.g. $0.01

# Keep testing new payment values until the balance is +/- lowestBalance
while abs(balance) > lowestBalance:
    # Reset the value of balance to its original value
    balance = originalBalance
    # Calculate a new monthly payment value from the bounds
    payment = (upperBound - lowerBound) / 2 + lowerBound

    # Test if this payment value is sufficient to pay off the entire balance in 12 months
    for month in range(12):
        balance -= payment
        balance *= 1 + monthlyInterestRate

    # Reset bounds based on the final value of balance
    if balance > 0:
        # If the balance is too big, need higher payment so we increase the lower bound
        lowerBound = payment
    else:
        # If the balance is too small, we need a lower payment, so we decrease the upper bound
        upperBound = payment

# When the while loop terminates, we know we have our answer!
print "Lowest Payment:", round(payment, 2)

Answer 3

要找出这样的错误，一个好方法就是添加一些打印内容，例如，我将以下内容添加到您的代码中：

print(balance, lo, hi, guess)

然后看看会发生什么，你就能弄清楚发生了什么。事实证明：

hi = (balance*(1+monthlyInterestRate**12))/12

计算的上限太低。也许你的意思是：

hi = (balance*(1+monthlyInterestRate*12))/12