Steel cutting problem
Steel cutting problem in building engineering are widely used, to solve the problem of the steel cutting, Can improve the utilization ratio of steel, saving production cost, improve efficiency, thus the enterprise to improve efficiency and enhance its competitiveness. So about the study of steel cutting is intentioned more and more by enterprise.
In order to solve some practical problems，we can plan some feasible treatment schemes according to the requirement of the problem, and then select one or several schemes by comparing their pros and cons, to make up the optimum combination, and use mathematical method to get the best solution. At the following, we can take an example to illustrate the application of this method.
After the rebar material list out, optimization can be done at cutting material scene in order to reduce the condition of lack of material or waste. On the basis of overall planning method and intelligent selection optimization techniques, we can roundly integrated rebars in the material list to reduce the waste to a minimum. We should cut a piece of rebar according to its number, diameter, length, quantity and length, from meterial which has a large number of quantity and the longer length, and as far as possible to reduce and shorten the short steel head, to save the steel.
Question: In order to make 20 rectangular steel frames, each frame is made up of two pieces of 2.2 meters rebar and two pieces of 1.5 meters rebar, and the rebars we have are 4.6 meters long. How to use the rebars is most efficient?
Analyze and answer: In order to make 20 rectangular steel frames, we need 40 pieces of 2.2 meters rebar and 40 pieces of 1.5meters rebar. There is a simple idea that we can cut out a piece of 2.2 meters rebar and a piece of 1.5 meters rebar from a piece of raw material, so each root of the raw material has 0.9 meters remnant. We need 40 pieces of rebars to make these 20 rectangular steel frames, and the remaining raw material is 0.9×40=36meters.
Obviously, the above idea is not very reasonable because of waste. Therefore, we should think about suit cut to save the material. In the following, we have three schemes available.
Table 28.2
We need to combine all kinds of schemes to save the material and get 20 steel frames. With x1 represents the number of raw materials in scheme I we need，x2 represents the number of raw materials in schemeⅡ, x3 represents the number of raw materials in scheme Ⅲ. The so-called using a minimum of raw material is to make the total remnants achieve the minimum. For this purpose, we can list the following mathematical model according to table 28.2,
y=0.1x1+0.2x2＋0.9x3，
The answer is
In the above results, 0≤x3≤40．Take x1, x2 into the equations of y to get
It is observed that, the greater value of x3 the greater value of y, so the value of x3 should be small as far as possible.
When x3 = 0, we can determine x1 = 14, x2 = 20.
When x3 =1，x1=13，x2=20，both of them expect to use 34 root raw materials, and the remnants is
y=34×4.6-(2.2+1.5)×40=8.4(meters)．
Therefore, the scheme which uses a minimum of raw material is: using 13 or 14 roots in scheme I, 20 roots in schemeⅡ, 1 or 0 root in scheme Ⅲ. So 20 steel frames can be made by just 34 roots raw materials.
The above problem is a simple case in using the rolled steel. At the following, there is a more complicated example which need more knowledge of mathematics to slove.
Question: Some factory accepted a batch of orders for processing, to produce 100 sets of steel frames and each steel frame should be made up of three pieces of rolled steels, each is 2.9m, 2.1m and 1.5m, and the factory only has a group of 7.4m long rolled steels now. How to design the reasonable scheme to make sure that the total length of the leftover material is the shortest?
There is a simple method to deal with this kind of problem: cutting out three roots rolled steels which are 2.9m,2.1m and 1.5m respectively from each raw rolled steel, and the remaining leftover material for each is 0.9m long. To meet the processing, we take 100 roots rolled steels into accout, and the total length of the leftover material is 90m. Obviously, it’s not the best method, because the reasonable suit cut is bound to save rolled steels. If you consider to get reasonable suit cut, you should list various suit cut plans first. The suit cut schemes of this case are listed in the following table.
suit cut scheme table
We can see from the table that schemeⅠ could save the most material with no tailing, but this scheme is not able to meet the regulation of suit cut. That is to say, it can’t meet the quantity requirements of steels which have various kinds of length. The rest four schemes also can not meet the regulation of suit cut alone. Therefore, we should not use only one kind of method to cut the material, and we should use a combination of several schemes instead, to get the purpose of meeting the regulation of suit cut and getting the shortest total length of the tailing.
The determination of objective function
Take the shortest total length of the tailing as the objective function, and suppose x1, x2, x3, x4, x5 represents the number of the rebar for each scheme. So we can write directly the expression about the total length of the remaining leftover material by the data in the last line in the table.
f(x)= 0.1x2+0.2x3+0.3x4+0.8x5
The determination of constraint function
According to the request for quantity of the rolled steel in various schemes, constraint function can build these functions:
The determination of mathematical model
Solution procedure and optimization result of simplex algorithm
In the mathematical model, the objective function and the constraint function are linear functions, so this design belongs to the linear programming problems and also the multi-dimensional optimization problem with constraints. We can get the optimal solution and optimum value with simplex algorithm(The concrete solving process is omitted).
