| | Why use a cutting optimizer?
by Theodor Augustin Floare |
|
The role of cutting optimizer software is to save money by
reducing the scraps and the number of used bars. Let me give you an example. Suppose that
we have to cut from bars of 5000 mm length, the following items:
| Length | Quantity |
| 710 mm | 4 |
| 1068 mm | 4 |
| 1750 mm | 6 |
| 2540 mm | 4 |
| 960 mm | 2 |
| 1630 mm | 10 |
I will consider that
- stubs that are longer than 600 mm are reused and therefore they are not scraps;
- price of bar is 3.6 euros per meter;
- saw cut length is 0 mm;
Now, how do we cut the items?
I will present you three solutions: one "manual" solution and two solutions generated by
the Bar Cut Optimizer & Manager application.
|
| |
|
1. "manual" solution
Depending on your available time, you can come up with a better solution.
Time is the keyword.
| used bars |
cutting pattern |
stub |
| 1 |
2540 mm + 710 mm * 3 |
330 mm (waste) |
| 1 |
2540 mm + 1750 mm + 710 mm |
0 mm |
| 2 |
2540 mm + 1068 mm + 960 mm |
432 mm (waste) |
| 2 |
1750 mm + 1630 mm + 1068 mm |
552 mm (waste) |
| 2 |
1630 mm * 3 |
110 mm (waste) |
| 1 |
1750 mm + 1630 mm |
1620 mm (reused) |
| 1 |
1750 mm * 2 |
1500 mm (reused) |
| 1 |
1630 mm |
3370 mm (reused) |
|
This solution needs 11 raw bars of 5000 mm length.
| Total length |
55000 mm |
100.00 % |
198.00 euro |
| Used length |
45992 mm |
83.62 % |
165.57 euro |
| Reused length |
6490 mm |
11.8 % |
23.36 euro |
| Waste length |
2518 mm |
4.57 % |
9.06 euro |
|
| |
|
2. optimal solution A
| used bars |
cutting pattern |
stub |
| 4 |
2540 mm + 1750 mm + 710 mm |
0 mm |
| 1 |
1750 mm + 1068 mm * 3 |
46 mm (waste) |
| 3 |
1630 mm * 3 |
110 mm (waste) |
| 1 |
1750 mm + 1068 mm + 960 mm * 2 |
262 mm (waste) |
| 1 |
1630 |
3370 mm (reused) |
|
This solution needs 10 raw bars of 5000 mm length.
| Total length |
50000 mm |
100.00 % |
180.00 euro |
| Used length |
45992 mm |
91.98 % |
165.57 euro |
| Reused length |
3370 mm |
6.74 % |
12.13 euro |
| Waste length |
638 mm |
1.27 % |
2.29 euro |
|
| |
|
3. optimal solution B
There is no waste material in this solution because all resulted stubs can be reused (> 600 mm).
| used bars |
cutting pattern |
stub |
| 4 |
2540 mm + 1750 mm + 710 mm |
0 mm |
| 2 |
1750 mm + 1630 mm + 960 mm |
660 mm (reused) |
| 4 |
1630 mm * 2 + 1068 mm |
672 mm (reused) |
|
This solution needs 10 raw bars of 5000 mm length.
| Total length |
50000 mm |
100.00 % |
180.00 euro |
| Used length |
45992 mm |
91.98 % |
165.57 euro |
| Reused length |
4008 mm |
8.01 % |
14.42 euro |
| Waste length |
0 mm |
0.00 % |
0.00 euro |
|
| |
| So, from this simple example, the conclusion I can
draw is that the optimal solution uses less raw bars (10 instead of 11 for a manual
solution) and has less scraps. |
| In programming literature this problem is known
as "knapsack problem" or "cutting stock problem". Because an algorithm in polynomial
time for this problem is not discovered yet, the solution is to generate many
cases and to choose the best case. That is why a software program is
needed. |
