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We use basic unit to refer to the minimum unit of resource in A. Block Assignment the time-frequency domain in the problem formulation. The We denote by matrix u of size |B|×|K| the utility matrix for set of basic units is denoted by I. We let a = 1 if block b,i all pairs of blocks and services. An element u represents the b includes basic unit i, otherwise a = 0. Recall that a b,k b,i utility of a block-service pair (b, k) (b ∈B and k ∈ K). Block block refers to a rectangular shape located at some specific assignments for K(ℓ) and K(c) are treated separately. The location in the resource grid, see Figure 1. Taking the figure former is performed first because of the latency requirement. as an example, there are 20 × 4 = 80 basic units. Hence I = {1,..., 80}. Block 1 is of shape 1×4, and for this block, Algorithm 1 BA(S,u). The input consists of a block-service a = 1 for four elements of I. Placing this 1 × 4 shape at 1,i assignment S which may be empty, and a utility matrix u. Set all possible positions of the grid, we obtain all blocks and the S is augmented and then returned by the algorithm. We remark corresponding a-values for this particular shape. Doing so for that b overlaps with b′ if and only if ∃i ∈ I a + a ′ > 1. all shapes generates the set of all candidate blocks B. As it can b,i b ,i be seen from Figure 1, for each block, the a-values are fully 1: repeat determined by the position of the block and the numerical 2: Remove from B the blocks in S and those overlapping indexing of the basic units, and the complexity of doing with the blocks in S. ′ ′ ′ ′ one mapping equals the size of I, i.e., |I|. The complexity 3: (b , k ) ← argmaxb∈B,k∈K(ℓ) ub,k, S ←S∪{(b , k )}. (ℓ) (ℓ) ′ for obtaining all the mappings is O(|B||I|). One possible 4: if qk′ is met: K ← K \{k }. (ℓ) implementation of performing the mapping is detailed in IEEE 5: until K = φ or B = φ (ℓ) DataPort [18]. This mapping is done in the pre-processing 6: if K 6= φ: (ℓ) stage once (not every TTI), and it is totally decoupled from 7: The demand of the users left in K cannot be met. any service to be scheduled, as it only concerns the two sets 8: repeat (ℓ) (c) B and I. 9: Lines 2–3 with the notation K replaced by K . The optimization task is to select blocks for each service, 10: until B = φ such that the latency and demand requirements are met for K(ℓ), without overlapping among chosen blocks. We use The operations in Lines 2 and 4 cost O(1) by hash-map optimization variable xb,k ∈{1, 0} to indicate whether block implementations. By sorting the utilities in advance (which b (b ∈ B) is assigned to the service k (k ∈ K). A block b is costs O |B||K(ℓ)| log |B||K(ℓ)| ), along with inspecting the (ℓ) infeasible for k (k ∈ K ), if the ending time of b exceeds τk. other algorithm operations, one can conclude that the overall

This is modeled by setting rb,k =0. The problem is formulated complexity is of O (|B||K| log (|B||K|)). below. The two sets of constraints impose the demand for K(ℓ) and block non-overlapping, respectively. B. Utility Estimation by LP Relaxation [P0] max rb,kxb,k (1a) x∈{0,1} b (c) One way to compute the utility matrix u is to solve the LP X∈B k∈KX (ℓ) relaxation of P0 and to use the LP optimum xLP. s.t. rb,kxb,k ≥ qk, k ∈ K (1b) b∈B X [P0-LP] max rb,kxb,k s.t. (1b), (1c) (2a) (1c) 0≤x≤1 ab,ixb,k ≤ 1, i ∈ I b∈B k∈K(c) k b X X X∈K X∈B We denote by uLP = xLP the LP-based utility. Also, xLP can Theorem 1: P0 is NP-hard. be used for initialization: S = {(b, k): uLP,b,k ≥ ρ,b ∈B, k ∈ Proof: We construct a polynomial-time reduction from K} with ρ being a threshold. the PARTITION PROBLEM (PP) for a set of integers {d1,...,dn}. The task is to determine whether or not there is a partition such that the two subsets have equal sum of C. Utility Estimation by LD n i=1 di/2, where the numerator is assumed to be even. We define a single TTI size and multiple blocks that take the shape By relaxing the constraints (1c) of P0 with Lagrangian Pof one basic unit. There are two services, denoted by kℓ and multiplier λi (i ∈ I), the Lagrangian is defined as follows: kc, in K(ℓ) and K(c), respectively. The latency parameter of kℓ n equals that of the TTI size, and the demand equals i=1 di/2. L(x, λ)= rb,kxb,k+ λi 1 − ab,ixb,k Moreover, rb,1 = rb,2 = db,b = 1,...,n. By construction, b∈B k∈K(c) i∈I b∈B k∈K ! (1c) has no effect. Next, one can observe that partitioningP the n X X X X X basic units into two subsets, each providing a total throughput The LD function is defined in (3). n of i di/2, is equivalent to a feasible solution to PP. In =1 [P1] g(λ)= max L(x, λ) s.t.(1b) (3) addition, this can occur if and only if the objective function x∈{0,1} P c n defined for k reaches i=1 di/2. Hence the conclusion. Accordingly, we have the LD problem: IV. PROBLEM SOLVING [P0-LD] min g(λ). (4) We propose a sub-optimal but low-complexity algorithm, λ≥0 consisting in performing assignment of blocks to services, based on utility values generated from linear programming We define αb = i∈I λiab,i. Problem P1 decomposes for (LP) relaxation and the Lagrangian dual (LD). K(c) and K(ℓ): The constraints (1b) are only for K(ℓ). There- P 3 fore, solving P1 amounts to solving the two problems P2 and kHz, µ =0, 1, 2 ...) originate from Release 15 [20, Table 4.2- P3 for K(c) and K(ℓ) respectively, shown below. 1]. Note that Release 15 also specifies subcarrier spacing up to 240 kHz. However, by [21, Table I], a TTI of 0.125 ms [P2] max (r − α )x (5a) x b,k b b,k meets all the worst-case transmission latencies for the listed b (c) X∈B k∈KX 5G ultra-reliable low-latency communication configurations. s.t. xb,k ≤ 1, b ∈B (5b) Parameter settings are given in Table I. We test (c) our algorithm for a set of candidate thresholds ρ ∈ k∈KX (c) {0.05, 0.5,..., 0.95} among which the one achieves the best xb,k ∈{0, 1}, b ∈B, k ∈ K (5c) objective is selected. The maximum sub-gradient iterations is set to 200. While calculating the block rates, the impact on capacity due to guardband is included by following the model [P3] min αbxb,k (6a) x∈{0,1} in [14]. The rate reduction due to control overhead follows that b (ℓ) X∈B k∈KX in [22], where two symbols per TTI constitute the overhead. (ℓ) s.t. rb,kxb,k ≥ qk, k ∈ K (6b) We emphasize on accurate assessment in terms of optimality, b X∈B that is, how much does the proposed algorithm perform with respect to global optimum. We use the global optimum Note constraints (5b) are not present in P0, though these are obtained by solving the integer programming problem (1) via (c) implied by (1c) for services in K . Computing the optimum a solver. This is not a scalable method. The purpose here of P2 is straightforward. Each block b is allocated to the is for benchmarking, to demonstrate that our low-complexity service argmaxk rb,k − αb with rb,k − αb > 0. algorithm has little loss in optimality. The number of users as (ℓ) Problem P3 further decomposes to K problems P3[k] well as the bandwidth is chosen such that the global optimum (k ∈ K(ℓ)), each with objective max α x , constraints b∈B b b,k can be obtained with reasonable amount of computing effort. b∈B rb,kxb,k ≥ qk, and binary variables x b,k (b ∈B). Similarly, in view of the computational effort of obtaining P global optimum for benchmarking, we do not include all TTI P [P3[k]] min αbxb,k (7a) x∈{0,1} sizes that are permitted by 5G NR [20]. b X∈B s.t. rb,kxb,k ≥ qk (7b) Table I b SIMULATION PARAMETERS. X∈B Each P3[k] can be reformulated as a KNAPSACK PROBLEM, Parameter Value and optimally solved by dynamic programming. Number of users 10 with |K(ℓ)| = |K(c)| = 5 The dual problem P0-LD can be solved using a sub- Time-frequency domain 2 ms and 2 MHz x(h) SNR range [5, 30] (dB) gradient method [19]. Denote by LD the LD solution in Demand q {16, 32, 64, 128, 256, 512} (kbps) the hth iteration of the sub-gradient method. We let uLD = Latency tolerance τ {0.25, 0.5, 1.0, 1.5, 2} (ms) x(h) Threshold (Section IV-B) ρ ∈ {0.05, 0.1,..., 0.95} h LD to be the LD-based utility. P Figure 2 shows the average bit rate of services in K(c) with D. Algorithm Implementation respect to the latency tolerance τ of K(ℓ). For the non-flexible structures, Shape 1, Shape 2, and Shape 3 are used separately. In addition to BA(S,uLP ) and BA(S,uLD ), we consider al- Each of these structures is referred to in the format of “TTI- gorithm “LP+LD” that returns the best solution of BA(S,uLP ) SCS” (e.g. 0.25ms-30kHz means a shape of a fixed TTI of and BA(S,uLD ). We remark that BA(S,uLD ) is quite flexible in terms of computational effort, as one can use accumulated 0.25 ms and a fixed SCS of 30 kHz). The flexible structure significantly outperforms the non- uLD before full convergence. Overall, the algorithm scales well. Moreover, if necessary, the service sets can be decom- flexible ones. The system tends to benefit more from flexible posed into subsets, and the algorithm can be applies to one structure when the latency tolerance becomes more stringent. subset at a time to further reduce complexity. Note that our algorithm with flexible structure is near-optimal. Among the three non-flexible schemes, 0.25ms-30kHz out- performs the other two. This result is related to that, in the V. NUMERICAL RESULTS optimization problem, the throughput of the services in K(c), is The use of flexible numerology is expected to outperform subject to latency constraints of services in K(ℓ). For resource fixed numerology. The purpose of performance evaluation is to allocation, blocks of 0.5ms-15kHz and 0.125ms-60kHz have examine the amount of improvement, which is of significance low flexibility in the time and frequency domains, respectively. in particular as the control channel overhead for supporting The former does not have many choices in meeting the latency the flexible structure is accounted for. The result also tell how requirements of K(ℓ), leading to poor throughput for K(c). well the proposed algorithm is suited for the flexible structure. The latter is not efficient on the frequency domain (due to Comparing to LTE that applies a fixed SCS of 15 kHz and frequency selectivity), though this inefficiency is mitigated TTI of 1.0 ms, we consider four shapes, Shape 1, Shape 2, when the latency tolerance is high, as more choices become Shape 3, and Shape 4, with SCS being 15 kHz, 30 kHz, 60 available in the time domain. Blocks of 0.25ms-30kHz strikes kHz, and 60 kHz, CP 4.7 µs, 2.3 µs 1.2 µs, and 4.17 µs, and a balance between short TTI size (to meet latency-constrained the number of symbols 7, 7, 7, and 6, respectively. The TTI services) and flexibility in the frequency domain. durations of the four shapes are 0.5 ms, 0.25 ms, 0.125 ms, Without showing by the figures, we remark that the problem and 0.125 ms, respectively. The numerologies (∆f =2µ × 15 feasibility of the three non-flexible schemes is very sensitive 4

to the latency tolerance. This issue is alleviated by the flexible ACKNOWLEDGEMENT structure. In comparison to the related work [23] considering This work has been partially supported by European Union advantages of flexible numerology, our results emphasize the H2020 MSCA projects ACT5G (643002) and DECADE significance of block-service assignment optimization. (645705), and the Center for Industrial Information Tech- nology (CENIIT). The work of the first author was partly accomplished while he was at Link¨oping University, Sweden. 0.125ms-60kHz 0.5ms-15kHz 0.25ms-30kHz

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