使用二分查找法在Python中计算最低月付款额

Question

我正在尝试计算每月最低还款额，以偿还一笔贷款，使用的是以下代码：

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，我似乎陷入了一个无限循环，导致我的 guess 变量没有被更新。我该如何打破这个无限循环，并更新我的 guess 变量呢？

问题出在我的数学计算上。我本来想说的是：

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

谢谢大家的帮助！

Answer 1

要找出像这样的错误，一个好办法就是加一些打印输出，比如我在你的代码里加了以下内容：

print(balance, lo, hi, guess)

然后看看会发生什么，这样你就能弄清楚到底出了什么问题。结果发现：

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

计算出的上限太低了。也许你想要的是：

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

Answer 2

我觉得这个解决方案应该可以用。

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

首先，如果你在做MITx的练习，并且完成了之前的测试（只需在猜测中加10），那么你离成功又近了一步。只需要对循环条件做一些调整，并检查一下年度结果。

关于二分查找，我来简单解释一下这个概念。你总是会有两个极端，一个是最小值，一个是最大值。每次猜测时，都是从这两个极端的中间开始。

第一次猜测后，你需要根据年度结果来调整这两个极端。如果在一年后，你支付了喝饮料、约女孩、买程序书和其他东西的最低金额，但仍然没有还清总账，那你肯定需要提高最低支付额。反之，如果比如说你在第十个月就还清了总账，那你明年就得多喝点、认识新女孩！开玩笑的……你需要降低最低支付额。这就是在完成一年的支付后需要检查的内容。

---

在这个练习中，我们有：

• 余额和年利率 = 给定的（我们不需要关心）
• 最低支付额（下限） = 余额 / 12
• 最高支付额（上限） = (余额 x (1 + 月利率)**12) / 12.0

第一次猜测将是（最低 + 最高）/ 2，我称之为guessMinimum，所以：

guessMinimum = (minimum + maximum)/2

你将开始使用第一次猜测（guessMinimum）。一年后你会检查剩余金额。如果剩余金额是负数，说明你支付得太多了，需要降低每月的支付额。另一方面，如果一个月后剩余金额是正数（比如说，超过你的精度（例如0.10）），你也需要降低每月的支付额，明白了吗？

试着设计一下思路……

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

如果一年后，“剩余”是负数（例如）。这意味着'guessMinimum'太高了！！！你需要……不是你，程序需要！！程序需要调整它，降低最低支付额……

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

抱歉，大家。我试着插入一张图片，但作为新成员，我做不到。需要至少10个声望……帮帮我！！用字符工作太麻烦了！！

而且代码需要做这些繁重的工作，调整最低支付额，直到“剩余”是可以接受的（在你的精度范围内，或者说epsilon，或者任何字母或变量……好吧。:)

在理解了这个概念和图示后……让我们来看看代码。

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)

就是这样。