Swimmer distribution
A swimming team plans to select four swimmers A, B, C and D to form a 4 * 100 meter medley relay team to participate in this year's championship. Their achievements in 100 meter freestyle, breaststroke, butterfly and backstroke are shown in the table below. Ask the four swimmers A, B, C and D what position they can swim in order to get the best results.
Table: Scores of four players
Performance freestyle/breaststroke/butterfly/backstroke/breaststroke
A 56 74 6 1 63
B 63 69 65 7 1
C 57 77 63 67
Ding 55 76 62 62
Please establish a mathematical model and write a solution program with Lingo software.
Solution: Introduce the variable Xij of 0- 1 If player I is selected to participate in swimming stroke J, record Xij= 1, otherwise record Xij=0. According to the requirements of forming a relay team, Xij should meet two constraints:
None is the most, only one of the four strokes can be selected, that is, I =1234; xij = 1；
Each stroke must have one person and only one person can be selected, that is, for j =1234; Expected Xij= 1.
When player I is selected for swimming stroke J, CijXij indicates his achievement, otherwise CijXij=0. So the result of relay race can be expressed as Z=, which is the objective function to change the problem.
To sum up, the 0- 1 programming model of this problem can be written.
The minimum Z = Z =S.t.= 1, i= 1, 2, 3, 4; = 1，i= 1，2，3，4
Substitute the header data into the model and enter LIGDO:
min = 56 * x 1 1+74 * x 12+6 1 * x 13+63 * x 14
+63 * x 2 1+69 * x22+65 * x23+7 1 * x24
+57 * x 3 1+77 * x32+63 * x33+67 * x34
+55 * x 4 1+76 * x42+62 * x43+62 * x44；
x 1 1+x 12+x 13+x 14 = 1；
x 2 1+x22+x23+x24 = 1；
x 3 1+x32+x33+x34 = 1；
x 4 1+x42+x43+x44 = 1；
x 1 1+x 2 1+x 3 1+x 4 1 = 1；
x 12+x22+x32+x42 = 1；
x 13+x23+x33+x43 = 1；
x 14+x24+x34+x44 = 1；
@ bin(x 1 1)；
@ bin(x 12)；
@ bin(x 13)；
@ bin(x 14)；
@ bin(x 2 1)；
@ bin(x22)；
@ bin(x23)；
@ bin(x24)；
@ bin(x 3 1)；
@ bin(x32)；
@ bin(x33)；
@ bin(x34)；
@ bin(x 4 1)；
@ bin(x42)；
@ bin(x43)；
@ bin(x44)；
The optimal solution can be obtained as follows:
Steel bar cutting problem
Set the length of steel bar raw materials of each specification as 10m, and solve the following optimization problems:
1) Now we need 28 pieces of this kind of steel bars with a length of 4m and 33 pieces with a length of1.8m. How many raw materials do we need to buy at least? How to cut?
2) If necessary, 28 steel bars with a length of 4m, 33 steel bars with a length of 1.8m, and 79 steel bars with a length of 3.6m, with a length of
2.4m 46 yuan, how many pieces of raw materials do you need to buy at least? How to cut (no more than 3 cutting methods can be considered)?
Please establish a mathematical model to solve the above problems, and write a solution program with Lingo software.
(1) solution: Through analysis, the following three cutting modes can be obtained:
Mode 4m rebar number 1.8m rebar number allowance1210.438+030.438+0.
Xi is used to indicate the number of original steel pipes cut according to mode 1 (I = 1, 2,3). Obviously, they should be non-negative integers.
As a goal, the total number of original steel pipes after cutting is at least, and there are
min Z = x 1+x2+x3；
In order to eliminate the demand, the constraints are:
2x 1+x2 & gt； =28;
x 1+3 x2+5x 3 & gt； =33;
Enter the positive form of the integer linear programming formed above into LINDO, as shown below:
min = x 1+x2+x3； ;
2x 1+x2 & gt； =28;
x 1+3 x2+5x 3 & gt； =33;
@ gin(x 1)；
@ gin(x2)；
@ gin(x3)；
The optimal solution can be obtained as follows:
(2) Solution: Xi represents the number of original steel pipes cut by the first method; (i= 1，2，3)
Suppose that the number of steel pipes that produce 1.8m, .4m, 3.6m and 4m per original steel pipe under the I-type cutting mode is r 1i, R2i, R3i and R4i respectively.
Because the total number of original steel pipes cannot be less than (1.8 * 33+2.4 * 46+3.6 * 79+4 * 28)/10 = 57.
therefore
Enter the positive form of the integer linear programming formed above into LINDO, as shown below:
min = x 1+x2+x3；
x 1 * r 1 1+x2 * r 12+x3 * r 13 & gt； =33;
x 1 * r 2 1+x2 * R22+x3 * r23 & gt； =46;
x 1 * r3 1+x2 * R32+x3 * r33 & gt； =79;
x 1 * r4 1+x2 * r42+x3 * r43 & gt； =28;
1.8 * r 1 1+2.4 * r 2 1+3.6 * r 3 1+4 * r 4 1 & lt； = 10;
1.8 * r 12+2.4 * R22+3.6 * R32+4 * r42 & lt； = 10;
1.8 * r 13+2.4 * r23+3.6 * r33+4 * r43 & lt； = 10;
1.8 * r 1 1+2.4 * r 2 1+3.6 * r 3 1+4 * r 4 1 & gt； 8.2;
1.8 * r 12+2.4 * R22+3.6 * R32+4 * r42 & gt； 8.2;
1.8 * r 13+2.4 * r23+3.6 * r33+4 * r43 & gt； 8.2;
x 1+x2+x3 & gt； =57;
x 1+x2+x3 & lt； =64;
x 1 & gt； = x2
x2 & gt= x3
@ gin(x 1)；
@ gin(x2)；
@ gin(x3)；
@ gin(r 1 1)；
@ gin(r 12)；
@ gin(r 13)；
@ gin(r 2 1)；
@ gin(R22)；
@ gin(r23)；
@ gin(r 3 1)；
@ gin(R32)；
@ gin(r33)；
@ gin(r 4 1)；
@ gin(r42)；
@ gin(r43)；
The optimal solution can be obtained as follows:
3. Talk about your experience and understanding of learning mathematical models.
Through the study of mathematical model this semester, I feel a lot. It taught me not only some theoretical knowledge of mathematics, but also the cultivation, exercise and improvement of comprehensive ability. Mathematics is closely integrated with life. I learned to solve simple mathematical modeling problems. I also exercised my logical reasoning ability and analytical ability. I believe this will be of great help to me in the future.