SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Knowing the sizes of jobs in  $I$ ,  $T_{i}^{\mathsf{RR}}(I)$  can be computed as
$$
T_{i}^{\mathsf{RR}}(I) = x_{i} + \sum_{\substack{j\in I\\ j\neq i}}D^{\mathsf{RR}}(j,i) = x_{i} + \sum_{\substack{j\in I\\ x_{j}\leq x_{i}}}x_{j} + \sum_{\substack{j\in I\\ x_{j} > x_{i}}}x_{i}.
$$
We deduce that PAR with SRR update rule can be implemented running concurrently OPT(I) and RR(J) with respective rates  $\alpha^{\mathbb{R}\mathbb{R}}$  and  $1 - \alpha^{\mathbb{R}\mathbb{R}}$ , where the processing rate updates as follows: initialize  $c(I) = B, c(J) = N - B$  and  $\alpha^{\mathbb{R}\mathbb{R}} = \frac{c(I)}{c(I) + c(J)} = \frac{B}{N}$ , and for any  $t > 0$ ,
(i) if a job in  $J$  terminates, then  $c(J)\gets c(J) - 1$
(ii) if  $T_i^{\mathsf{RR}}(I) = \int_0^t \alpha_u^{\mathsf{RR}} du$  for some  $i \in I$ , then  $c(I) \gets c(I) - 1$ ,
and in both cases update  $\alpha^{\mathsf{RR}} = \frac{c(I)}{c(I) + c(J)}$ .
Proof of Theorem 4.3. Using the processing rate  $(\alpha_{t}^{\mathsf{RR}})_{t > 0}$ , the delays generated by  $I$  on  $J$  are exactly the same as if RR was running, and the delays by  $J$  on  $I$  are smaller than RR's, because OPT(I) runs with the same processing rate as RR(I) but all the jobs finish earlier, thus using Lemma C.1 we deduce that
$$
\begin{array}{l} \mathbf {E} [ D ^ {\mathsf {P A R}} (I, J) + D ^ {\mathsf {P A R}} (J, I) ] \leq 2 \sum_ {i = 1} ^ {B} \sum_ {j = B + 1} ^ {N} \mathbf {E} [ \min  \{x _ {\sigma (i)}, x _ {\sigma (j)} \} ] \\ = \frac {4 B (N - B)}{N (N - 1)} \sum_ {k = 1} ^ {N} (N - k) x _ {k}. \\ \end{array}
$$
Using Lemma 4.2 then Equation 2 we obtain
$$
\begin{array}{l} \mathbf {E} [ \mathsf {P A R} (\mathcal {J}) ] \leq \sum_ {i = 1} ^ {N} x _ {i} + \left(2 - \frac {B}{N} \left(4 - 3 \frac {B - 1}{N - 1}\right) + \frac {4 B (N - B)}{N (N - 1)}\right) \sum_ {k = 1} ^ {N} (N - k) x _ {k} \\ = \sum_ {i = 1} ^ {N} x _ {i} + \left(2 - \frac {B (B - 1)}{N (N - 1)}\right) \sum_ {k = 1} ^ {N} (N - k) x _ {k} \\ \leq \left(2 - \frac {B (B - 1)}{N (N - 1)}\right) \left(\sum_ {i = 1} ^ {N} x _ {i} + \sum_ {k = 1} ^ {N} (N - k) x _ {k}\right) \\ = \left(2 - \frac {B (B - 1)}{N (N - 1)}\right) \mathsf {O P T} (\mathcal {J}). \\ \end{array}
$$
Which proves the theorem.