Forum Discussion

sg2021's avatar
sg2021
Copper Contributor
Feb 12, 2021

Material Leftovers

Hello,

 

I have possibly an unusual Formula request, but necessary for our company.  We have material that is being cut and would like to see if there is a formula that I can use to figure out how many piece I would be able to gain out of one piece.

 

For example, I would need to enter in original size of 31x104 and will need to cut pieces to the size of 30.5x67.75.  Is there a formula that tells me how many pieces I can get out of this 31x104 piece?

 

Thanks for your help!

13 Replies

  • Riny_van_Eekelen's avatar
    Riny_van_Eekelen
    Platinum Contributor

    sg2021 Now, I'm not a mathematician, so there could be more intelligent ways to resolve this. But, the following formula will do the trick.

    =MAX(INT(MatW/ProdW)*INT(MatL/ProdL),INT(MatW/ProdL)*INT(MatL/ProdW))

     MatW, ProdW, MatL and ProdL are named ranges, each referring to the Material's and Product's Width and Length. Dividing Widths and Lengths in both "directions" will enable you to determine the maximum amount of Product that can be cut out the Material. A working example is attached.

    • mtarler's avatar
      mtarler
      Silver Contributor

      sg2021  I liked Riny_van_Eekelen approach but thought we could do a bit better in some cases where the left over has enough scrap to rotate is and cut additional product at a 90 degree from the originals.  Here is the updated formula and sheet.  You'll notice the formula gets a bit longer but it is all based on Riny's orginal formula and named ranges:

      =MAX(INT(MatW/ProdW)*INT(MatL/ProdL)+INT(MOD(MatW,ProdW)/ProdL)*INT(MatL/ProdW)+INT(MOD(MatL,ProdL)/ProdW)*INT(MatW/ProdL),INT(MatW/ProdL)*INT(MatL/ProdW)+INT(MOD(MatW,ProdL)/ProdW)*INT(MatL/ProdL)+INT(MOD(MatL,ProdW)/ProdL)*INT(MatW/ProdW))
      • sg2021's avatar
        sg2021
        Copper Contributor

        mtarler

         

        What about the below? Doesn't seem like I can really get 39 pieces out of this material. Can you please confirm this formula works for this scenario? Thanks!

                W   L
        Mat 47 156
        Prod 3.5 48.5
        Max 39
        Max/Rotation 39

         

    • sg2021's avatar
      sg2021
      Copper Contributor
      Thank you for your response! I will give this a shot and update if I cannot make it a go. Thanks!
  • mathetes's avatar
    mathetes
    Silver Contributor

    sg2021 

     

    Unless I'm missing something, the answer in your example is 1. Given those dimensions, there'll be some scrap, but certainly not enough for another piece of the same size.

     

    So maybe that's not a good example.

     

    Are there other examples that would make it worthwhile to come up with a formula? And, if possible, could you include a spreadsheet that would give an idea of the range of material sizes you're dealing with and how you've got it laid out.

    • sg2021's avatar
      sg2021
      Copper Contributor

      mathetes 

       

      Good morning,

       

      Thank you for taking the time to reply to my request.  Below are more for you to review.  Thanks for your time!

       

      25 x 280 full
      8 x 280 cuts
       
      47 x 156 full
      3.5 x 48.5 cuts
       
      31 x 104 full
      30.5 x 67.75 cuts
       
      38.5 x 182.5 full 
      19 x 90  cuts
       
      48.5 x 144.5
      31 x 79
      • mathetes's avatar
        mathetes
        Silver Contributor

        sg2021 

         

        I'm pretty sure that the solution you got from Riny_van_Eekelen fully meets your need, so unless you come back with a request for further help, I'll defer to him.

Resources