Solver not giving me optimal result. Need help setting up for Solver.

Hello, I am trying to find the best width cut size that minimizes scrap loss. I have a Master Sheet that I need to cut into multiple sizes (various widths). I'm given a specific width that we need to buy that is cut from that Master Sheet but we also have other sizes we use out of that same Master sheet. I need Solver to tell me which of those cuts and how many of each cut I should make to minimize the scrap loss. First it must pick the cut part that I need then decide based on the highest usage what other sizes to cut to. The Master Sheets can be 48-58" wide and there is always a minimum scrap loss of 1".

I've tried running Solver but it does not give me the minimum scrap loss possible. It's selecting one size and one cut giving me a huge scrap loss. Not sure what I'm doing wrong but I also need help in incorporating the usage % to hopefully pick highest usage parts first after my initial size is selected. Any help is appreciated!

I'm struggling to understand the data.

One cut of each of the widths in M8:M14 gives a total 75.2 which is more than the maximum 58 width. So how could they all be satisfied from one sheet.

Does this boil down to trying different combinations of cuts with several sheets to minimise the waste. e.g.
Taking say M8:M11 & M14 which total 47, and cutting them from a 48 wide sheet, with a loss of 1, then accepting M12 & M13 will be cut from a second sheet.

Then checking all other combinations to see if the loss of 1 in this example could be further reduced.

What part if any do the usage %s in L8:L14 play?

I'm not sure I understand your request, anyhow, here is a workbook with a different setup.

You can use Simplex, and you don't need binary variables, just integers.

HTH,

Francesco

Is there any way to say if one size has the highest usage (besides the part selected, PartA6125) that the # of Cuts should be >=1 for that size?

You could add a helper column with the formula

=IF(E8=MAX($E$8:$E$14),1,0)

and then use it as a constraint on the number of cuts.

HTH,

Francesco