VBA中的递归代码 [英] Recursion code in VBA

问题描述

我正在尝试运行此代码以计算附图中公式 16.4 中不同 Tn 下的 Q(n).但它没有给我正确的输出.我将不胜感激任何帮助.注意:delta1=delta2 =...deltan = dt=1(我在这里取过)并进一步将 S 项除以 10000，因为在等式中它处于基点，即 1% 的第 100 部分.

I am trying to run this code to calculate Q(n) at different Tn in the Equation 16.4 in the attached picture.But its not giving me the correct output. I would appreciate any help. Note: delta1=delta2 =...deltan = dt=1 ( I have taken here ) and further divided S term by 10000 just because in the Equation it is in basis point i.e. 100th part of 1 %.

Function Bootstrap(S As Range, Z As Range, L As Double) As Double
Dim j As Integer
Dim a As Variant
Dim b As Variant
Dim n As Integer
Dim Q() As Double
Dim sum As Double
Dim P As Double
Dim dt As Double

n = Application.WorksheetFunction.Max(S.Columns.Count, Z.Columns.Count)
a = S.Value
b = Z.Value
dt = 1
sum = 0
ReDim Q(0 To n)
Q(0) = 1

For j = 1 To n - 1
    P = (b(1, j) * (L * Q(j - 1) - (L + dt * a(1, n) / 10000) * Q(j))) / (b(1, n) * (L + a(1, n) * dt / 10000)) + Q(n - 1) * L / (L + a(1, n) * dt / 10000)
    sum = sum + P
    Q(n) = sum
Next j

Bootstrap = sum
End Function

推荐答案

要解决递归函数，你可以这样写，例如

To solve a recursive function you can write it this way, for example

Function Factorial(n as long) as long
    If n = 1 Then
        Factorial = 1
    Else
        Factorial = n * Factorial(n-1)
    End If
End function

是的，您可以看到 For...Loop 也可以进行因子计算，但在您的情况下，使用递归解决方案更容易.

Yes, you can see For...Loop can also do the Factorial calculation, but in your case, its much easier to use recursive solution.

此外，等式 16.4 被有意编写为递归函数.它没有写成求和函数，因为这样做更难.如果给你的是求和函数，那么你可以应用 For...Loop 解决方案.

Besides Eq 16.4 is intentionally written as a recursive function. It is not written as a summation function because it is harder to do so. If given to you is a summation function, then you can apply the For...Loop solution.

希望这会有所帮助.

编辑

Function Q(n as long) as double

    If n = 1 Then
        Q = 5
    Else
        Q = Z * ( L * Q_t - (L + d * S) * Q(n-1) ) / ( Z * ( L + d * S ) )
    End If

End Function

注意函数Q在n>1时会在Q(n-1)中不断调用自己.这就是所谓的递归解决方案.

Notice that the function Q keep calling itself in Q(n-1) when n>1. That is called recursive solution.

(检查公式.我可能抄错了)

(Check the formula. I might copy it wrong)

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