# IEEE 802.11 ax OFDMA Resource Allocation with Frequency-Selective Fading
## 4 Problem Statement  
Consider a network with several APs. Let \(R\) be the maximum allowable distance between an AP and a STA associated with it. The STAs are uniformly distributed in an area consisting of circles of radius \(R\) around all APs, as shown in Figure 3. All key notations used in this paper are summarized in Table 2.  
Every STA is associated with the closest AP. At random time instances, the STAs generate data flows that should be transmitted to the corresponding APs. After generating a flow, a STA transmits a BSR to inform the AP that this STA has data for the transmission. The AP allocates resources to the associated STAs that have sent BSRs. Note that every AP makes resource allocation independently of the other APs. Let \(\{r_{j}\}\) be the set of all RUs, which differ from each other in size and location in the band. For the 40 MHz band, all \(r_{j}\) are numbered in Figure 1. Based on the RU structure, we determine the inclusion and intersection operations of different RUs. Inclusion: \(r_{i}\subset r_{j}\) if \(r_{j}\) contains all \(r_{i}\) frequencies,  
Intersection: \(r_{i}\cap r_{j}= \begin{cases}r_{i},\ r_{i}\subset r_{j},\\ r_{j},\ r_{j}\subset r_{i},\\ \emptyset,\ r_{i}\
ot\subset r_{j}\text{ and }r_{j}\
ot\subset\ r_{i}.\end{cases} \)  
Note that there are no partially overlapping RUs. If two RUs contain the same frequencies, it means that one of them fully contains the other one. For the RU structure from Figure 1: \(r_{2}\subset r_{26};r_{26}\subset r_{30};r_{19}\cap r_{26}=r_{19}\). Let \(S=\{s_{i}\}_{i=1}^{n}\) be the set of STAs that have data to transmit. The RU allocation to STAs is the set of pairs \(X=\{(s_{i},r_{j})\}\), where \((s_{i},r_{j})\) means that STA \(s_{i}\) was assigned to RU \(r_{j}\). To simplify the notation, we further write \(s\in X\), meaning that such a pair \((s,r)\) exists, that \((s,r)\in X\). Similarly, we write \(r\in X\), meaning that such a pair \((s,r)\) exists, that \((s,r)\in X\). Using the above notations, we determine the _restrictions_ on the RU allocations:  
\[ \begin{split}&\forall(s_{i},r_{i}),(s_{j},r_{j})\in X,r_{i}\
eq r _{j}\to s_{i}\
eq s_{j},\\ &\forall(s_{i},r_{i}),(s_{j},r_{j})\in X,s_{i}\
eq s_{j}\to r_{i} \cap r_{j}=\emptyset,\end{split}  \tag{1}\]  
where the first restriction means that a STA cannot be allocated two RUs at once, and the second restriction means that the RUs should not intersect with each other. In addition to RU allocation, the AP assigns MCSs and transmit powers. As noted in Section 2, parts of one OFDMA frame from different STAs have to reach AP with approximately equal power. To satisfy this requirement, we make the STAs involved in the same OFDMA transmission use the same MCS. In this case, power equalizing can be easily done. Suppose that the RU allocation is built, and one of the MCSs was chosen. If the AP knows channel conditions, it can compute the required transmit power and, consequently, the received power for each STA. The AP finds the minimum received power value, and then the STAs decrease their transmit power so that the received power would be equal to this minimum. If the STAs transmit at the different MCSs, a more complex equalizing procedure is required. Note that the STAs can change not only the transmit powers but also the MCSs. Figure 3: Studied network scheme: different kinds of hatching show different basic service sets. To state the allocation problem, we introduce a utility function \(U\) that indicates the quality of resource allocation. The scheduler should select the allocation that maximizes this function. Let \(C\) be the set of available MCSs. Let \(\lambda(s,r,c)\) be the gain of the utility function if the scheduler assigns a STA \(s\) to an RU \(r\) at an MCS \(c\in C\) (see Section 5.2 for specific examples of utility function). If the STA does not support transmission in the RU \(r\) at the MCS \(c\) (for example, due to the lack of power), then \(\lambda=0\). Let \(H\) be the set of all RUs to STAs allocations satisfying the conditions (1).