## 4 Results  
We will describe the work on finding the optimal time for students to complete their homework using a non-stationary multichannel queuing system with a limited waiting queue. In this optimization problem, we represent students of 6 classes doing their homework as service channels; applications are tasks in subjects; the input stream is formed by the teachers' requirements for students to complete their homework in accordance with SanPiN and Federal State Educational Standard (for example, the full employment of the student during the time allotted for homework). It is necessary to introduce the following designations: \(M\)--the number of students (the model takes into account the possibilities of performing both individual tasks and project tasks for a group of students); \(Q\)--the length of the queue of items with tasks; \(\lambda(t)\)--the intensity of receipt of a set of tasks in subjects, distributed according to the exponential distribution law; \(\mu(t)\)--the intensity of the student's homework, distributed according to the exponential distribution law; \(T\)--the considered operating time of the system, limited to a 2.5 h working day (for students of 6 grades). Since the software implementation of the optimization problem posed is solved in the MATLAB environment (in which the variable indices start from one, and not from zero), the mathematical model of the functioning of the homework process by students has the form (10-12). Let \(p_{k}\) the probability of the system being in a state in which it is necessary to complete tasks in \(k\) subjects, \(k=\overline{2,M+Q+1}\). \[ \begin{array}{l}\frac{dp_{1}(t)}{dt}=-\lambda p_{1}(t)+\mu p_{2}(t)\\ \frac{dp_{1}(t)}{dt}=-(\lambda+(i-1)*\mu)p_{i}(t)+\lambda p_{i-1}(t)+i*\mu p_{ i+1}(t),i=\overline{2,M};\\ \frac{dp_{i+M}(t)}{dt}=-(\lambda+M*\mu)p_{j+M}(t)+\lambda p_{j+M-1}(t)+M*\mu p_ {j+M+1}(t),j=\overline{1,Q};\\ \frac{dp_{Q+M+1}(t)}{dt}=\lambda p_{Q+M}(t)-M*\mu p_{Q+M+1}(t);\end{array}  \tag{10}\]  
\[p_{k}(0)=p_{k0}>0,\ \ \ \ \ k=\overline{1,M+Q+1} \tag{11}\]  
In addition, the normalization condition must be met. \[\sum_{k}p_{k}(t)=1,t\in[0,T] \tag{12}\]  
The parameter to be optimized is the average time for completing homework by one student, which is determined through the intensity and is equal to the reciprocal of the average time for making \(\overline{t}_{lesson}\) homework \(\mu=\frac{1}{t_{lesson}}\). We will use the following characteristics as optimality criteria:  
1. The average relative time for completing homework, taking into account the difficulty of a particular subject, is as follows: \[K_{1}=\frac{1}{M\times T}\times\overline{t}_{lesson}\to min\] (13) where \(\overline{t}_{lesson}\)--average time to complete homework (min.) for one lesson. The time limit for completing homework (min.) for one lesson is as follows: \[t_{lesson}=60\times D\frac{B_{i}}{\sum_{i}B_{i}P_{i}}\] where \(D\)--the time limit for completing homework during the week, according to hygiene requirements (hours); \(B_{i}\)--the difficulty of a separate \(i\) subject; \(P_{i}\)--the number of hours in the curriculum for the \(i\) subject. 2. Average relative efficiency of homework in terms of material assimilation: \[ \begin{array}{l}K_{2}=\frac{\overline{Ef}\times\overline{t}_{lesson}}{Ef_ {max}}\to max,\\ \overline{Ef}=a_{N}*kol,\end{array} \] (14) where \(\overline{t}_{lesson}\) is the average time for completing homework (min.) for one lesson; \(\overline{Ef}\)--the average efficiency of assignments in subjects assigned to the home; \(Ef_{max}\)--maximum efficiency, \(a_{N}\)--task efficiency coefficient obtained using a neural network, \(kol\)--number of tasks. It is necessary to impose the condition on the parameter to be optimized with a constant average time for completing homework:  
\[a\leq\mu(t)=\mu\leq b, \tag{15}\]  
where \(a\) and \(b\) are determined by the decision-maker.