## 3 Proposed Approach
### Discussion of Simple Heuristic Approaches that seem Obvious  
Let us study the possibilities of using simple heuristics to create an optimization algorithm. Therefore, the use of a simple example (Figure 1) shows that many \"logical\" heuristics lead away from the optimum. A network with a pair of nodes (Figure 1) is under study. There are two channels for transmitting information from the first node to the second node. The first channel has parameters \(\mu_{1}\) and \(N_{1}\), while \(x_{1}\) units of information per second are delivered through them. The second channel has parameters \(\mu_{2}\) and, similarly, \(x_{2}\) units of information per second are delivered through them. It is necessary to increase the amount of information delivered by \(\Delta x\) information units per second. One must determine which of the two channels should be increased in its load so that losses in the network are minimal. To solve this problem in a general way, one should analyze three hypotheses:  
* the most efficient decision is to increase the channel load for which \(x\) is less, because it delivers less information;
* the most efficient decision is to increase the channel load for which the loss probability is less, due to the assumption that it may have more free capacity than another;
* the most efficient decision is to to increase the channel load for which \(\mu-x\) is greater, since it has a larger bandwidth capacity. Using this simple example, it is possible to show that solving the flow optimization problem, even in a simple network, is not always intuitive. It is worth considering the hypotheses one by one. The first hypothesis is not true in general. Let us demonstrate it using the following example. Let network parameters be as follows: \(x_{1}=2\) units of information/second, \(\mu_{1}=1000\) units of information/second, \(N_{1}=5\) and \(x_{2}=1\) unit of information/second, \(\mu_{2}=2\) units of information/second, \(N_{2}=5\). According to the hypothesis, the first channel delivers twice as much information, so it is worth loading the second channel, although it is not correct, since its bandwidth capacity is too small. It is possible to prove that the second hypothesis is also wrong. Let us examine the graph in Figure 2. Axis \(X\) indicates the channel bandwidth capacity resource (information units/second), and axis \(Y\) indicates the change in the intensity of information loss (information units/second). In our calculations, the following network parameters are used: \(\mu_{1}=30\) units of information/s, \(N_{1}=3\) (blue curve in Figure 2) and \(\mu_{2}=60\) units of information/sec., \(N_{2}=9\) (red curve in Figure 2). The figure shows that when the useful information capacity resource reaches two units, it is preferable to load the second channel, and after four units it is preferable to load the first one. Therefore, it is impossible to determine which route is preferable for loading according to the remaining supply of useful information. According to the third hypothesis, it is necessary to load the channel with a lower probability of losses. An example will demonstrate this. Let the first channel have the following parameters: \(\mu_{1}=30\), \(N_{1}=3\), and \(x_{1}=13\). The parameters of the second channel are \(\mu_{2}=60\), \(N_{2}=9\), and \(x_{2}=50\). The calculated loss probability in the first and second channels are \(p_{1}=5.57\%\) and \(p_{2}=5.25\%\), respectively. According to the third hypothesis, it is necessary to \"load\" the second channel, and the losses will be equal to 3.57 units of information per second. However, when \"loading\" the first channel, the losses will be  
Figure 1: Example of a simple network. equal to 3.54 units of information per second, i.e., it is better to use the first channel, even considering the higher probability of losses. Therefore, the third hypothesis is also wrong. Thus, the use of obvious heuristics does not guarantee correct redistribution of information flows between two channels.