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In electronics I need a certain resistance, however there are limited standard values on the market. If I make a combination of two resistors in parallel, I could get a very close result to the target. The question is which 2 resistors could give the best result.

So far i have these elements:

  1. Excel column of 86 standard values from 0.25 to 1000000 in A2:A87 cells
  2. Target resistance that I enter in cell C3
  3. R1 and R2 - variables that can get 86 values from A2 to A87 cells
  4. equation giving the parallel resistance: Rtarget=R1*R2/(R1+R2)

I am a VBA ignorant and find this task as a good opportunity to learn more about VBA operations. Here I started with the variables, but I do not know how to proceed. Your help would be very much appreciated.

Sub ResCom()

Dim Rstd As Double 
Dim Rt As Double, R1 As Double, R2 As Double

Rstd = Range("a2:a87").Value

.
.
.

End Sub

Thanks in advance!


We can create a tool that can easily be reused.

First enter the resistance values in A2 through A87. Then run:

Sub tablemaker()
    Dim i As Long, j As Long, K As Long

    K = 2
    For i = 2 To 87
        For j = i To 87
            Cells(K, 2) = Cells(i, 1)
            Cells(K, 3) = Cells(j, 1)
            K = K + 1
        Next j
    Next i

    Range("D2:D" & K - 1).Formula = "=B2*C2/(C2+B2)"
End Sub

This will create a resistance table in B2 through D3742. The table lists combination pairs and the associated resistance of each pair.

(permutations are not need since the pair 4,2 is really the same as 2,4)

Then enter the target value in cell E2. In E3 enter the array formula:

=MATCH(MIN(ABS(D2:D3742-E2)),ABS(D2:D3742-E2),0)

Array formulas must be entered with Ctrl + Shift + Enter rather than just the Enter key. If this is done correctly, the formula will appear with curly braces around it in the Formula Bar.

This gives the row in the table with the closest match.Finally in E4 enter:

=INDEX(B2:B3741,$E3)

and copy this to F4

To re-run this, just change the target in E2

Given an array A[] and a number x, check for pair in A[] with sum as x , Let an array be {1, 4, 45, 6, 10, -8} and sum to find be 16 Initialize two index variables to find the candidate Function to check if array has 2 elements end) if(s.find(target_sum - arr[i]) == s.end) insert(arr[i] into s) else print arr[i], target-arr[i] . Given array of n integers and given a number X, find all the unique pairs of elements (a,b), whose summation is equal to X. The following is my solution, it is O(nLog(n)+n), but I am not sure whether or not it is optimal.


Just to get you started, use this to continue

Option Explicit

Sub ResCom()

    Dim Rstd As Double
    Dim Rt As Double, R1 As Double, R2 As Double

    Rstd = Range("a2:a87").Value

    ' step thru upper triangle of R1/R2 values
    ' (lower triangle gives same results)
    Dim r1Row As Long, r2Row As Long
    For r1Row = LBound(Rstd) To UBound(Rstd)
        For r2Row = r1Row To UBound(Rstd)
            ' here is where you solve for  Rtarget=R1*R2/(R1+R2)
            ' R1 = Rstd(r1Row)
        Next r2Row
    Next r1Row


End Sub

Dynamic Programming - Subset Sum Problem, Input: set[] = {3, 34, 4, 12, 5, 2}, sum = 30 Output: False There is no subset that add up to 30. Consider the last element and now the required sum = target sum – value of 'last' element and number of elements Following is the recursive formula for isSubsetSum() problem. printf ( "Found a subset with given sum" );. else. Count of Array elements to be divided by 2 to make at least K elements equal; Value to be subtracted from array elements to make sum of all elements equals K; Construct an Array of size N in which sum of odd elements is equal to sum of even elements; Merge first two minimum elements of the array until all the elements are greater than K


Try this

Sub ResCom()
    Dim Rt as Double, R1 as Double, R2 as Double

    Rt = Range("C3").Value

    For i = 2 to 87
       R1 = Range("A" & i).Value
       if R1 < Rt Then
          For j = 2 to 87
             R2 = Range("A" & j).Value
             if Round(Rt,2) = Round((R1*R2/(R1+R2)),2) then
                ' Success you have found a Resistance Combination that matches
             End If
          Next j
       End If
    Next i
End Sub

...

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