How many lengths of stock do I need?

Add up the length of every part, add one blade width (kerf) for every cut, divide by the length of one bar and round up. That is the fewest bars you could possibly buy. Whether the job actually cuts from that many depends on how the parts fit together on each bar, and that is a packing problem, not a division. A cut list optimizer does the packing, and tells you when it has hit the floor.

Step 1: the floor

A real job: 6000 mm bars, a 3 mm blade, and this cut list.

18 parts means 18 cuts, so add 18 × 3 = 54 mm of kerf: 24754 mm. Divide by 6000 and you get 4.13 bars. Nobody sells 0.13 of a bar, so the floor is 5 bars. No plan, on paper or on a computer, cuts this job from 4.

Step 2: why dividing is not the answer

The floor assumes every millimetre of every bar gets used. It rarely does: parts come in fixed lengths and a bar ends where it ends. Cut each size from its own bars and this job takes 6. The 2100s go two to a bar, the 1450s four, the 950s six: 2 + 2 + 2. The second bar of 1450s holds only two parts and the second bar of 950s only two, and those four parts would fit on one bar together. That is the bar you didn't need.

Mix the sizes on the same bar and the gaps fill. The optimiser's plan for this job:

5 bars. That is the floor, so the plan is proved optimal: no plan uses fewer. And the spare material ends up on one bar as a 3094 mm offcut you can rack, instead of being spread across several bars.

When the floor is too low

The floor is a minimum, not a forecast. The usual trap is parts longer than half a bar: two of them never share one, so the division undercounts. Six 3100 mm parts and six 1400 mm parts divide out to 4.5 bars, a floor of 5, and need 6. More in cutting from 6 m and 20 ft lengths.

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