Operations Research Goal Planning Questions 30

Summary. Operations research is an important professional basic course in modern management. It is a new discipline developed in the early 30s of the 20th century, and its main purpose is to provide scientific basis for managers when making decisions, and is one of the important methods to achieve effective management, correct decision-making and modern management.

This discipline is applied to the interdisciplinary study of mathematics and formal sciences, using statistics, mathematical models and algorithms to find the best or near-best solutions to complex problems.

Any questions.

Operations research is an important professional basic course in modern management. It is a new discipline developed in the early 30s of the 20th century, and its main purpose is to provide scientific basis for managers when making decisions, and it is one of the important methods to achieve effective management, correct decision-making and modern management. This discipline is applied to the interdisciplinary study of mathematics and formal sciences, using statistics, mathematical models and algorithms to find the best or near-best solutions to complex problems.

Operations research is often used to solve complex real-life problems, specifically to improve or optimize the efficiency of existing systems. The fundamentals of operations research include real analysis, matrix theory, stochastic routines, discrete mathematics, and algorithmic foundations. In terms of application, the number of oranges is mostly related to warehousing, logistics, algorithms and other fields.

Therefore, operations research is related to majors such as applied mathematics, industrial engineering, computer science, and economic management [1].

Take a look. Can you ask one more question.

OK. Similar.

In the sexual programming, because the constraints are all linear functions, the feasible domain is the convex set. Referring to the ** method of the two-dimensional problem, the feasible domain is an area enclosed by several lines, so it must be a convex set. Then, the optimal solution is searched in this convex set.

If the solution is searched by the movement of the contour of the objective function, the optimal solution must reach the optimal value at the edge of its convex set, and the edge of the convex set is either a line segment or a vertex, so the optimal solution of the linear programming problem must be at the vertex of the feasible domain.

In fact, these vertices are the basic feasible solutions to the linear programming problem.

So how do you find these vertices (the base feasible solution) from the model?

The key to solving the model is to solve ax=b.

Since the a matrix is a m n matrix, it is not possible to derive a unique solution to the above constraint equations. The nonsingular submatrix b of m m must be found in the a matrix, i.e., satisfying b is not equal to zero (the determinant is not zero), so that the unique solution of bx b can be obtained. At this point, the decision variables corresponding to matrix b are called base variables, and the rest are non-basis variables.

If the value of the base variable in x is the solution of bx b, and the value of the non-fundamental variable is zero, then x is the basic (feasible) solution of the problem, that is, the solution corresponding to the vertices of the feasible domain.

This is written as I understand it, and I hope it helps.

First, let's take a look at the knowledge of solving systems of linear equations related to advanced algebra.

(1) The convex set in linear programming means that its feasible domain (the set of all feasible solutions) is a convex set (convex plane polygon in 2ary linear programming), that is, if x1 and x2 are any two feasible solutions in the feasible domain, then x=1 2(x1+x2) is still a feasible solution, and it still falls in the feasible domain x1 and x2;

2) The linear basic feasible solution is a special set of feasible solutions: it divides the variables into two categories, one is the basic variable (the number of variables is the number of independent equations in the constraints), and the other is the non-fundamental variable (the number of variables is the difference between the number of decision variables and the number of basic variables), so that the value of all non-fundamental variables is 0, if the basic variables correspond to the only set of solutions and meet the constraints of the variables, then the group of solutions corresponding to all the decision variables is called the basic feasible solution of the problem about this basic variable group;

3) The basic feasible solution, which geometrically corresponds to the vertices of the feasible domain, also known as the angular top feasible solution.

4) When solving a linear programming problem, the group of fundamental variables corresponding to the first basic feasible solution is called the initial group of fundamental variables.

8x1+x2-4x3=2x5=10

There is a problem with this constraint It should be 8x1+x2-4x3+2x5=10

Yes, if yes, all the basis solutions are: x1=(0,16 3,-7 6,0,0).

x2=(0,10,0,-7,0,0) x3=(0,3,0,0,7/3,0) x4=(7/4,-4,0,0,0,21/4) x5=0,16/3,-7/6,0,0,0)

x6=0,10,0,-7,0,0) x7=(0,3,0,0,7/3,0) x8=(3/4,0,0,0,4/3,9/4) x9=（5/4,0,0,-2,0,15/4)

x10=(0,0,0,3,10/3,0) x11=(1,0,-1/2,0,0,3) x12=(0,0,3/2,,0,16/3,0)

x13=(0,0,-5/2,8,0,0) x14=0,0,0,310/3,0) x15=(0,0,3/2,0,16/3,0) x16=(0,0,-5/2,8,0,0)

All the fundamental solutions that satisfy the non-negative are the basic feasible solutions, and the optimal solutions are the fundamental feasible solutions that make the objective function the largest.

For the problem of finding the maximum, the m objective function needs to multiply -m by the artificial variable xi (if there are several artificial variables, you need to subtract a few mxi): first, as with the simplex method, the constraint <=, plus the relaxation variable, and the constraint 1 plus x4 in this problem, I don't need to talk about this. The other two constraints are the same, > = minus a residual variable, because when we are listing a simplex table, we need to find a set of bases, which are generally coefficients of 1, that is, to form an identity matrix, which I don't need to say.

The second constraint is -x5, x5 is the residual variable, the previous coefficient is -1, and the identity matrix cannot be made, so in order to make an identity matrix, we need to add a variable ourselves, that is, the artificial variable x6, the coefficient is 1, and the third constraint also needs to add an artificial variable x7, which can be made into a base. The basis can be found intuitively in the initial simplex table. i.e. p4, p6, p7, that is, the column where the base variable x4, x6, x7 is located, and the three columns form an identity matrix.

The iterative process is similar, for the problem of finding the maximum, m is seen as infinity, which is a number. Do the same. The same goes for the optimal solution.

However, if at the end of the iteration, it is found that the artificial variable is the base variable and is not 0, then there is no solution, if the base variable does not contain an artificial variable or the artificial variable is 0, then the specific solution is determined according to the discriminant formula. This is a matter of maximum, and the question of minima is another matter. As for the others, the same.

x1 x2 x3 x4 x5 x6 x7

For the maximum value problem, when swapping into the base, the discriminant is: the column with the positive test number and the largest absolute value, not as good as m-2 and m-3 comparison, m is infinite, m-2 is larger, choose the column with the largest test number, when swapping out the base, then choose the smallest ratio and not a negative number, the intersecting variable into the base, as the main element, that is, the one that hits [], this you should be clear, because we are looking for the maximum value, we should make the target value tend to the maximum as soon as possible, Therefore, the large number of tests is selected as the base variable, and the optimal solution is obtained until all the test numbers are <=0. Minimum problem, in the objective function + mxi (there are several artificial variables, add a few), to determine whether the optimal solution, when swapping into the base, choose the smallest number of tests and negative, to approach the minimum value as soon as possible, out of the base is the same, choose the smaller ratio, and then the intersection of the variable is.

Hope it helps.

x5 is a relaxation variable that turns the inequality constraint into an equality constraint, while x6 is an artificial variable with the aim of obtaining an initial feasible basis of an identity matrix. Artificial variables are redundant, and if there is a feasible solution to the problem, it means that the artificial variable must be equal to zero. The large m method, i.e., the coefficient of the artificial variable is m{find the minimum problem}, or m{find the maximum problem}, and the purpose is to swap the artificial variable out of the feasible basis as soon as possible.

The basic idea of the big m method: for the standard linear programming with an objective function of max, the value coefficient of the artificial variable in the objective function is -m, and m is a large positive number. The purpose is to change the artificial variable from a basic variable to a non-fundamental variable as quickly as possible.

In the initial normalization process, the constraints become: (1)x1-2x2+x3+x4=11(2)-4x1+x2+2x3-x5=3(3)-2x1+x3=1

4）x1，x2，x3，x4，x5>=0。Then there's the process of adding artificial variables, and now we need to get an identity matrix that we can use as a viable basis to get an initial basic viable solution. Observing the technical coefficient matrix, we can't make up an identity matrix in the sub-matrix, so we add two artificial variables, x6 and x7, so that (x4, x6, x7) can be obtained as a feasible base.

So x5 and x6 in the second constraint are not added at the same time. Then iterate and know to swap the artificial variables from base to non-basis variables.

** method is also applicable to the goal planning problem of two variables, but it is simple to operate and the principle is clear at a glance. At the same time, it also helps to understand the principle and process of solving general goal planning.

**The steps to solve the problem are as follows:

1. Determine the feasible domain of each constraint, that is, all constraints (including target constraints and absolute constraints, regardless of positive and negative deviation variables).

Represented on the coordinate plane;

2. On the boundary line represented by the target constraint, the direction of the increase of the positive and negative deviation variable values is marked with arrows;

3. Find a solution that satisfies the highest priority goal;

4. Go to the next priority target, and find the solution of the priority target without destroying all the higher priority goals;

5. Repeat 4 until all priority objectives have been reviewed;

6. Determine the optimal solution and satisfactory solution.