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Quickly solve the optimization problem of mathematical modeling scheme
In order to get the best benefit from the problem analysis, the total ore with an average grade of 45% should be the largest.

The model is established from the first mining point to the fourteenth mining point, and the ore blending amount of each mining point is 10000 tons (I stands for the number of mining points), and the average iron grade of each mining point is 1. Given some conditions in the topic, the following linear programming model can be obtained:

Max=∑ (Yi Xi), 1≤i≤ 14.

max = 0.37 16x 1.5 125 x2+0.4x 3+0.47 x4+0.42 X5+0.4996 X6+0.5 14 1x 7+0.4838 x8+0.4908 x9+0.4022 x 10+0.527 1x

Constraints are the average grade limit after ore mixing and the ore content limit every time:

∑ (Yi Xi) /∑xi≥0.45, 1≤i≤ 14.

That is, 0.3716x1+0.5125x2+0.4x3+0.47x4+0.42x5+0.4996x6+0.51x7+0.4838x8+0.4908x8.

Simplify:

0.0784 x 1-0.0625 x2+0.05 x3-0.02 x4+0.03 X5-0.0496 X6-0.064 1x 7-0.0338 x8-0.0408 x9+0.0478 x 10-0.077 1x 1 1-0.65438

0≤X 1≤70

0≤X2≤7

0≤X3≤ 17

0≤X4≤23

0≤X5≤3

0≤X6≤9.5

0≤X7≤ 1

0≤X8≤ 15.4

0≤X9≤2.7

0≤X 10≤7.6

0≤X 1 1≤ 13.5

0≤X 12≤2.7

0≤X 13≤ 1.2

0≤X 14≤7.2

Solve the model with matlab, create a new pk.m file, and input:

c =[0.37 16 0.5 125 0.4 0.47 0.42 0.4996 0.5 14 1 0.4908 0.4022 0.527 1.5692 0.4072 0.5020]*(- 1);

a=[0.0784,-0.0625,0.05,-0.02,0.03,-0.0496,-0.064 1,-0.0338,-0.0408,0.0478,-0.077 1,-0. 1 192,

0.0428,-0.052];

b = 0;

lb =[0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0];

ub =[70; 7; 17; 23; 3; 9.5; 1; 15.4; 2.7; 7.6; 13.5; 2.7; 1.2; 7.2; ];

[x,z]=linprog(c,a,b,[],[],lb,ub)

Save and run, the result is as follows:

& gt& gt park

Optimization terminated.

x =

3 1. 198 1

7.0000

17.0000

23.0000

3.0000

9.5000

1.0000

15.4000

2.7000

7.6000

13.5000

2.7000

1.2000

7.2000

z =

-63.899 1

Therefore, when x1= 31.1981,x2=7, x3= 17, x4=23, x5=3, x6=9.5, X7 =/kloc.

That is to say, the ore blending amount extracted from the first occurrence is 3 1. 1.98 1 1,000 tons, and all other occurrences are extracted.

Results According to the above results, all 13 ores can be extracted except the first one, because the first one has the lowest iron taste. Moreover, it can be seen that the average taste of iron extracted from other 13 ores is higher than 0.45, and the blending amount of the first ore has the greatest influence on the average taste after blending. Therefore, if the ore blending amount of the first ore spot is less than 31.1.981.000 tons, the average iron grade of the whole ore spot must be above 0.45.