在VBA函数中调用Excel求解器*没有*直接引用单元格？

Question

如何在不引用现有单元的情况下在VBA函数中调用求解器？我不想为这些纯粹的中间参数保留单元格。

Public Function calcSomething(param1 As Double, param2 As Double) As Double
' do something
' need to invoke solver to obtain some intermediate params
    Dim intermediate_param As Double
'hopefully should be like intermediate_param = calcWithSolver(param1, param2)
'do something
End Function

Public Function calcWithSolver(param1 As Double, param2 As Double) As Double
' how to call solver here without direct reference to existing cells?
End Function

我在网上发现的使用SolverAdd函数的教程似乎都需要引用已经存在的单元格。有办法避免这种情况吗？

Answer 1

作为替代方案，您可以尝试使用这种非常简单的求解器，该求解器使用对分法，并且即使在大多数复杂情况下也能很好地工作。

它没有excel求解器的灵活性，并且不能处理跨越工作簿的深层嵌套引用（您的解决方案参数需要在单个单元格中唯一定义）。但是，只要您可以安排它，它具有自动计算表格的优势，而Excel Solver则无法实现。

还可以通过更改二等分条件（用户练习）来修改代码以找到最大值和最小值

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以excel为例，如果单元格A1中包含方程式

Function Bisect(F_ref As Range, y As Double, xname As String, xmin As Double, xmax As Double, Optional ytol As Double, Optional iter As Double)

'Finds the solution for x of the equation F(x) = y using the bisection method
'Requires a function F(x) and an interval [xmin, xmax] (in which F(x) is continuous) containing one and only one solution.

'F_ref     - reference to single cell containing the function f(x). NOTE : Do not specify the function directly as an input. Cell referenced needs to contain all the dependencies on x directly (no 
'y         - value of F(x) required
'xname     - reference text name for x in F_ref
'xmin,xmax - maximum and minimum values of xname that solution is known to lie within
'ytol      - tolerance to within which solution is needed [optional, default 1e-3]
'iter      - maximum number of iterations [optional, default 20]
'            returns #TOL! if tolerance not met. Increase ytol or increase iter

Dim fun As String, grad As Double, count As Integer

If ytol = 0 Then ytol = 0.001
If iter = 0 Then iter = 20

fun = F_ref.Formula & "-" & y
grad = (Evaluate(Replace(fun, xname, xmax)) - Evaluate(Replace(fun, xname, xmin))) / (xmax - xmin)

For count = 1 To iter
    x = (xmax + xmin) / 2
    y = Evaluate(Replace(fun, xname, x))
    If (y * y) < (ytol * ytol) Then Exit For
    If grad > 0 Then
        If y > 0 Then xmax = x
        If y < 0 Then xmin = x
    Else
        If y < 0 Then xmax = x
        If y > 0 Then xmin = x
    End If
Next count
If count > iter Then x = "#TOL! " & x
Bisect = x
End Function

然后调用函数

=2*LN(B1) + B1

在1和2的值之间求解方程2ln（x）+ x = 2，将x = 1.370的答案返回到默认的小数点后三位。