Example 5. A factory needs to make 100 sets of steel frames, each set of steel frames has a round steel, and the lengths are 2.9 m, 2. 1 m and 1.5 m respectively. It is known that each raw material is 7.4 m long. Q: How to cut materials to save raw materials to the maximum extent?

Solution: * * * The following five blanking schemes can be designed, as shown in the following table.

Let x 1, x2, x3, x4 and X5 be the number of raw materials to be cut in the above five schemes respectively. In this way, we establish the following mathematical model.

Objective function: Min x 1+x2+x3+x4+x5.

Constraint: s.t.x1+2x2+x4 ≥100.

2x3 + 2x4 + x5 ≥ 100

3x 1 + x2 + 2x3 + 3x5 ≥ 100

x 1，x2，x3，x4，x5 ≥ 0

The optimal blanking scheme is calculated by the software of "Management Operations Research": 30 pieces are cut according to the scheme 1; According to Scheme 2, cut 10 block; Cut 50 pieces according to Scheme 4.

X1= 30;

x2 = 10；

x3 = 0；

x4 = 50

X5 = 0；

Only 90 kinds of raw materials are needed to manufacture 100 steel frame.

Note: when establishing this kind of mathematical model, it is better to use the sign greater than or equal to as the constraint condition than the equal sign. Because sometimes when applying some blanking schemes, there may be an extra round steel with a certain specification, but it may be the best scheme. If the equal sign is used, this scheme is not feasible.

This is an example. Just draw a tiger according to the cat.