vba If within if - vba

Question

I want to create a function which give me what contribution an employee has to make to a pension plan in function of its salary (with given down and upper limit) and in function of its age (doesn't contribute before 25 and after 64).
With the following code, I try to achieve that. I may have spend too much time on front of my screen but i don't see the problem. The debugger runs fine but I keep getting a "#Value!" alert. It may be the If-statment within another If. If someone may see where the problem could be, it would be very helpful
Option Explicit
Constant M = 1175
Function coti_min_LPP(x As Integer, salaire_AVS As Double)
Dim salaire_coord As Double
'The basis for the calculation uses the salary in different layers
If salaire_AVS < 18 * M Then
salaire_coord = 0
ElseIf salaire_AVS > 18 * M And salaire_AVS < 24 * M Then
salaire_coord = 3 * M
ElseIf salaire_AVS > 24 * M Then
salaire_coord = salaire_AVS - 21 * M
End If
'Under a certain limit, the contribution is a percentage of the basis
'according to the age
If salaire_AVS < 72 * M Then
If x < 25 Then
coti_min_LPP = 0
ElseIf x > 24 And x < 35 Then
coti_min_LPP = salaire_coord * 0.07
ElseIf x > 34 And x < 45 Then
coti_min_LPP = salaire_coord * 0.1
ElseIf x > 44 And x < 55 Then
coti_min_LPP = salaire_coord * 0.15
ElseIf x > 54 And x < 65 Then
coti_min_LPP = salaire_coord * 0.18
ElseIf x > 64 Then
coti_min_LPP = 0
End If
Else
'Above the limit, it is the percentage of a fixed amount
If x < 25 Then
coti_min_LPP = 0
ElseIf x > 24 And x < 35 Then
coti_min_LPP = 51 * M * 0.07
ElseIf x > 34 And x < 45 Then
coti_min_LPP = 51 * M * 0.1
ElseIf x > 44 And x < 55 Then
coti_min_LPP = 51 * M * 0.15
ElseIf x > 54 And x < 65 Then
coti_min_LPP = 51 * M * 0.19
ElseIf x > 64 Then
coti_min_LPP = 0
End If
End If
End Function

Answer 1

There are two immediate things that result in an issue:
Instead of Constant M = 1175 you should use Private Const M As Long = 1175. Why Long? Please look at the line If salaire_AVS < 72 * M Then if you add watch and go through your code in the debbuger you will find out that, if M is declared as an integer, after this line (according to watch window) M has value "out of context". This is basically due to the fact that you have achieved integer overflow. Please consider following code:
Example 1:
Sub test2()
Dim a As Integer
Dim b As Double
a = 1175
b = 15
Debug.Print b < 72 * a
End Sub
Option Explicit is a good practice, however requires you to declare all of the variables in your code. Declare them. You may remove Option Explicit for testing purposes. Some lecture on variable declaration: http://www.excelfunctions.net/VBA-Variables-And-Constants.html
After above changes your function works for me (tested =coti_min_LPP(27,19*1175))
Hope that helps. Btw. Please notice that having so many nested ifs with hardcoded values is troublesome to maintain, control and usually is not considered as good coding practice.