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
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