A16364. Deterministic Scheduling for Extended Reality over 5G and Beyond

The input of a single test has $(4 + R \cdot K \cdot T + N \cdot R \cdot K + 1 + J)$ lines, which contains user number $N$ , cell number $K$ , TTI number $T$ , RBG number $R$ , initial SINRs $s_{0, r n t}^{(k)}$ , interference factors $d_{mrn}$ , frame number $J$ and information about $J$ frames.

The details are as follows:

- Line $1$ : User number $N$ , integer, $1 \leq N \leq 100$ . Users are numbered from $0$ to $N - 1$ .
- Line $2$ : Cell number $K$ , integer, $1 \leq K \leq 10$ . Cells are numbered from $0$ to $K - 1$ .
- Line $3$ : TTI number $T$ , integer, $1 \leq T \leq 1000$ . TTIs are numbered from $0$ to $T - 1$ .
- Line $4$ : RBG number $R$ , integer, $1 \leq R \leq 10$ . RBGs are numbered from $0$ to $R - 1$ .
- Line $5$ to $(4+R \cdot K\cdot T)$ : Initial SINRs $s_{0, r n t}^{(k)}$ , float, $0 < s_{0, r n t}^{(k)} < 10\,000$ . Each line has $N$ elements, corresponding to $N$ users. $s_{0, r n t}^{(k)}$ is the $(n+1)$ -th element of line $(5+r+k \cdot R+t \cdot K \cdot R)$ .
- Line $(5+R \cdot K \cdot T)$ to $(4+R \cdot K \cdot T + N \cdot R \cdot K)$ : Interference factors $d^{(k)}_{mrn}$ , float, $-2 \leq d^{(k)}_{mrn} \leq 0$ . Each line has $N$ elements, corresponding to $N$ users. $d^{(k)}_{mrn}$ is the $(n+1)$ -th element of line $(5+R \cdot K \cdot T+m+r \cdot N + k \cdot R \cdot N)$ .
- Line $(5+R \cdot K \cdot T + N \cdot R \cdot K)$ : Frame number $J$ , integer, $1 \leq J \leq 5000$ .
- Last $J$ lines: Frame information. Each line contains $5$ integers corresponding to a frame, which are, in order: frame ID $j\in\{0,\ldots,J-1\}$ in increasing order, size $TBS_j$ ( $0 < TBS_j \leq 100\,000$ ), user ID it belongs to, first TTI $t_{0,j} \in\{0,\ldots,T-1\}$ , and number of TTIs $t_{d,j} \in \left[ {1,100} \right]$ . Last TTI for frame $j$ can be found as $t_{1,j}=t_{0,j}+t_{d,j}-1$ ; it is guaranteed that $t_{1,j} \le T - 1$ .

It is guaranteed that each user has at most one frame at each TTI.

Output for a certain input is the optimization result of $p_{r n t}^{(k)}$ (float), which has $R \cdot K \cdot T$ lines. Each line has $N$ elements, corresponding to $N$ users. $p_{r n t}^{(k)}$ is the $(n+1)$ -th element of line $(1+r+k \cdot R+t \cdot K \cdot R)$ .

Note that the optimization result of $b_{r n t}^{(k)}$ does not need to be output, because $p_{r n t}^{(k)}>0$ and $p_{r n t}^{(k)} = 0$ means $b_{r n t}^{(k)}=1$ and $b_{r n t}^{(k)} = 0$ , respectively.

Please note that if the outputs do not meet the constraint (4), it will be judged as an incorrect answer and get score $0$ . Besides, transmit on some TTIs out of time window is valid, but usually results in a lower score due to resources waste.

Scoring

The goal is to maximize the number of successfully scheduled frames. When these numbers are tied, we will compare who used less power. To achieve that, $Score = X - 10^{-6}\times p$ , where $X$ and $p$ represent the number of successfully scheduled frames and the total power used for transmission, respectively.

The total score for a submission is the sum of scores on each test.

Two sets of tests are prepared in this problem. For the duration of the competition, each submission is tested on the preliminary set of tests. When the competition is finished, for each contestant:

The jury takes the latest submission with non-zero score on preliminary tests;

This submission is tested on the final set of tests for the final rank;

The two sets of tests are generated from the same pool of data, based on the real word data.

Two sets of tests are prepared in this problem. For the duration of the competition, each submission is tested on the preliminary set of tests. When the competition is finished, for each contestant:

The jury takes the latest submission with non-zero score on preliminary tests;

This submission is tested on the final set of tests for the final rank;

The two sets of tests are generated from the same pool of data, based on the real word data.