On Jun 21, 11:48 am, "Dana" wrote:
I would cut
1 pattern of P6 (4-3')
1 pattern of P4 (1-3', 2-4')
and
2 patterns of P1 (2-5')

Nice work. But I believe your solution results in 5 feet of
"scrap" (by that, I mean 1- and 2-foot pieces, which are unusable)
plus the extra 7-foot length that you mention. That includes the
unavoidable 2 feet of scrap from the 10-foot length.

I found another solution by trial-and-error. Coincidentally, it is
the same solution that Solver finds, using the design and set-up that
I describe in my response to "brownti". That is:

1 of the 5+3+3 pattern
2 of the 5+4+3 pattern
1 of the 9+3 pattern

That results in only 3 feet of "scrap" plus an extra 9-foot length.

After finding that solution, then asking Solver to minimize scrap,
Solver finds:

3 of the 5+4+3 pattern
1 of the 6+3+3 pattern

Although that results in only 2 feet of scrap (the minimum), it is a
dubious optimal solution. it results in extra 4- and 6-foot lengths,
instead of the one 7- or 9-foot length in yours and my solutions. It
also requires 9 cuts, compared to the 8 cuts required for all of the
above solutions, including yours.

(All of these solutions require 5 of the 12-foot raw material.)

Oh well.... Fun problem!