Optimal Resource Allocation for Multi-User MEC with Arbitrary Task Arrival Times and Deadlines

Optimal Resource Allocation for Multi-User MEC with Arbitrary Task Arrival Times and Deadlines Xinyun Wang, Ying Cui, Zhi Liu, Junfeng Guo, Mingyu Yang Abstract In this paper, we would like to investigate optimal resource allocation for a more practical multiuser mobile edge computing (MEC) system. First, we consider a computation task model with nonnegligible sizes of computation results and arbitrary task arrival times and deadlines. Based on it, we further establish a computation offloading model considering non-negligible execution durations and allowing parallel transmissions and executions for different tasks. Then, we formulate the weighted sum energy consumption minimization problem to optimize the task operation sequences and starting times for uploading, executing and downloading as well as uploading and downloading time durations. The problem is a challenging mixed discrete-continuous optimization problem. By analyzing its structural properties, we develop an algorithm to obtain an optimal solution. In addition, using several optimization techniques, we transform the problem to an equivalent Difference of Convex (DC) problem, and develop a low-complexity algorithm to obtain a suboptimal solution using Penalty Convex Concave Procedure (CCP). Both analytical and numerical results demonstrate the importance of the optimization of operation sequences in multi-user MEC systems, which is usually neglected in existing studies based on simplified models for multi-user MEC systems. I. INTRODUCTION With the support of megapixel on-device cameras and hi-precision built-in sensors, many advanced mobile applications, e.g., augmented reality, interactive online gaming and multimedia transformation are emerging. These desktop-level applications are typically computationintensive and latency-sensitive, posing increasingly high burdens on resource-limited mobile X. Wang, Y. Cui, J. Guo and M. Yang are with Shanghai Jiao Tong University, China. Z. Liu is with Shizuoka University, Japan.

2 devices. Mobile edge computing (MEC) is a promising technology that provides powerful computing capability at the wireless edge to improve the quality of experience of mobile users for these advanced applications. In an MEC system, computation tasks of mobile users can be offloaded to a serving node (e.g., base station and access point). Designing efficient MEC systems requires a joint optimization of communication and computation resources among distributed mobiles and MEC servers. A significant amount of research effort has been devoted to energy-efficient resource allocation for MEC systems [1] [8]. For example, [2] [4] consider energy-efficient resource allocation, such as offloading control and transmission power and time duration allocation, for computation tasks with the same arrival time and deadline. However, the synchronous assumption for computation tasks is suitable only for a limited number of applications. To obtain more practical energy-efficient resource allocation, [5], [6], [8] consider different task arrival times and deadlines. Note that [4], [5], [8] assume that the sizes of computation results are negligible, and fail to take account of the resource consumption for transmitting computation results from a serving node to mobiles. This assumption may not hold for many applications with computation results of large sizes, such as augmented reality, interactive online gaming and multi-media transformation. In addition, [2], [6], [8] assume that task execution durations are negligible. This assumption may not be suitable for several applications with heavy computation loads and long execution durations, such as multi-media transformation and 3D modeling/rendering. Although [1], [3] consider non-negligible task execution durations and non-negligible computation result sizes, they ignore the fact that the execution of one task can be conducted during the transmission of another task (i.e, transmissions and executions for different tasks can be performed in parallel). Under the simplifying assumptions, [1] [6] do not consider the optimization of operation sequences, which significantly reduces the opportunities for parallel processing. Notice that these opportunities may lead to great delay reduction under power constraints or energy consumption reduction under deadline constraints, especially for tasks with large sizes of computation results and long execution durations. In our previous work [7], we consider non-negligible computation result sizes and non-negligible task execution durations, and allow transmissions and executions of different tasks to be conducted in parallel. However, the synchronous assumption for task arrivals and completions in [7] limits the applications of the proposed solution in practical MEC systems. Therefore, further studies are required to obtain efficient resource allocation for more practical MEC systems.

3 In this paper, our primary goal is to extend the task model and computation offloading model for the synchronous task scenario in our previous work [7] to the general scenario with arbitrary task arrival times and deadlines, to make them more applicable. Note that this is a highly nontrivial extension, as the operation mechanisms of MEC systems will change dramatically in the general scenario. We consider a multi-user MEC system with one serving node. In particular, we consider a computation task model with non-negligible sizes of computation results and arbitrary task arrival times and deadlines. Based on it, we further establish a computation offloading model considering non-negligible execution durations and allowing parallel transmissions and executions for different tasks. Then, we formulate the weighted sum energy consumption minimization problem to optimize the task operation sequences and starting times for uploading, executing and downloading as well as uploading and downloading time durations. The problem is a challenging mixed discrete-continuous optimization problem and is NP-hard in general. By analyzing its structural properties, we obtain an equivalent formulation which separates the discrete part and the continuous part, and develop an algorithm to obtain an optimal solution. In addition, using several optimization techniques, we transform the problem to an equivalent Difference of Convex (DC) problem, and develop a low-complexity algorithm to obtain a suboptimal solution using Penalty Convex Concave Procedure (CCP) [9]. Finally, numerical results demonstrate the advantages of the proposed suboptimal solution over some optimized schemes with simple designs of operation sequences. Both analytical and numerical results in this paper demonstrate the importance of the optimization of operation sequences in designing efficient multi-user MEC systems. II. SYSTEM MODEL As illustrated in Fig. 1, we consider a multi-user MEC system consisting of one single-antenna serving node and K single-antenna users, denoted by set K {1, 2,..., K}. The serving node has powerful computing capability by running an MEC server of a constant CPU-cycle frequency (in number of CPU-cycles per second) at the network edge. Each user k K has one computationintensive and latency-sensitive (computation) task, which is offloaded to the serving node for executing. 1 We consider a computation task model with non-negligible sizes of computation results and arbitrary task arrival times and deadlines. We also establish a computation offloading model considering non-negligible task execution durations and allowing parallel transmissions 1 The optimization results obtained in this paper can be extended to study a more general scenario, where some tasks can be executed locally.

4 Fig. 1. System model. and executions for different tasks. The two models are more complicated but more practical than the existing ones with the same task arrival time and deadline [2] [4], [7], negligible computation result sizes [4], [5], [8], or negligible task execution durations [2], [6], [8]. A. Computation Task Model We extend the computation task model in [7] by allowing computation tasks to have different arrival times and deadlines. The computation task at user k K, referred to as task k, is characterized by five parameters, i.e., the size of the uploaded task before computation L u,k > 0 (in bits), workload L e,k > 0 (in number of CPU-cycles), size of the downloaded computation result L d,k > 0 (in bits), arrival time A k 0 and deadline D k > A k. 2 Note that the considered task model is applicable for applications with computation results of large sizes, such as augmented reality, interactive online gaming and multi-media transformation. In addition, note that for all k K, L u,k, L e,k and L d,k are determined by the nature of task k itself, and for some computation tasks, such as html2text and x264 video encoding, these three parameters can be estimated to certain extent based on some prior offline measurements. B. Computation Offloading Model We offload each computation-intensive and latency-sensitive task to the serving node for executing. Offloading task k to the serving node for executing comprises three sequential stages: uploading task k from user k to the serving node, executing task k at the serving node, and 2 In this paper, we assume A k, k K and D k, k K are known in advance, and focus on the offline scenario to obtain first-order design insights as in [8]. In our future work, we shall study the online scenario where A k, k K and D k, k K are not known, based on the results obtained for the offline scenario in this paper.

5 downloading the computation result from the serving node to user k. We consider Time Division Multiple Access (TDMA) in Time-Division Duplexing (TDD) mode for transmission [7], [8]. Let t u,k, t e,k and t d,k denote the uploading, execution and downloading durations (in seconds) in the three stages, respectively, where t u,k, t d,k 0, k K, (1) t e,k = L e,k, k K. (2) F Here, F denotes the fixed CPU-cycle frequency of the MEC server at the serving node. Note that t e,k is fixed, whereas t u,k and t d,k can be optimized. Denote t u (t u,k ) k K, t e (t e,k ) k K, t d (t d,k ) k K and t (t u, t d ). The considered computation offloading model is suitable for applications with heavy computation loads, such as multi-media transformation and 3D modeling/rendering. Then, we introduce new notations and constraints to mathematically specify this model. Let s u,k, s e,k and s d,k denote the starting times for uploading, executing and downloading task k, respectively. Denote s u (s u,k ) k K, s e (s e,k ) k K, s d (s d,k ) k K, s (s u, s e, s d ), and O {u, e, d}. As each of the three stages cannot be interrupted, the completion times for uploading, executing and downloading task k are given by s o,k + t o,k, o O, k K. To ensure that the uploading, execution and downloading operations of each task are conducted sequentially, we require: s e,k s u,k + t u,k, k K, (3a) s d,k s e,k + t e,k, k K. (3b) To guarantee that the uploading of each task starts after its arrival and that the downloading of each task is completed before its deadline, we have: s u,k A k, k K, (4) s d,k + t d,k D k, k K. (5) Let Q denote the set of the K! different permutations of 1, 2,..., K. Let vectors q u (q u,i ) i K, q e (q e,i ) i K and q d (q d,i ) i K denote the three operation sequences (orders) for uploading

6 and executing the K tasks, and downloading their computation results, respectively, where q o Q, o O, (6) and q o,i represents the i-th element of q o for all o O and i K. Denote q (q u, q e, q d ) and K K \ {K}. Given the operation sequence q o, we have the following constraints: s o,qo,i+1 s o,qo,i + t o,qo,i, o O, i K. (7) As we consider TDMA in TDD mode, at any time, there is at most one task being uploaded or downloaded. Let x i denote the number of downloading operations between the i-th uploading operation and the (i + 1)-th uploading operation, for all i K, and let x K denote the number of downloading operations after the K-th uploading operation, where x i {0} K, i K. (8) Denote x (x i ) i K. As the total number of downloading operations is K, we have: K x i = K. (9) Note that q and x jointly determine the order of all operations. Note that unlike [7], we allow q o, o O to be different and consider x, so as not to lose optimality in the case of different task arrival times and deadlines, as illustrated in Fig. 2. To ensure that at any time, there is at most one task being transmitted, it is sufficient to require that any adjacent uploading and downloading operations do not overlap in time given the constraints in (7), (8) and (9), i.e., ) I [x i 0] (s d,qd,ij=1 + t xj d,qd,ij=1 s u,qu,i+1, i K, (10a) xj I [x i+1 0] ( s u,qu,i+1 + t u,qu,i+1 ) sd,qd, ij=1 x j +1, i K, s u,qu,1 + t u,qu,1 s d,qd,1, (10b) (10c) where I[ ] denotes the indicator function. Here, i j=1 x j represents the number of downloading operations before the (i + 1)-th uploading operation. To increase the transmission time for all tasks so as to reduce the transmission energy consumption for latency-sensitive tasks, we allow the execution of one task to be conducted in parallel with the uploading or downloading of

7 (a) A 1 < A 2 < D 2 < D 1, q u = (1, 2), q d = (2, 1) and x = (0, 2). (b) A 1 < A 2 < D 1 < D 2 and q u = q d = (1, 2). Fig. 2. Illustration example of the importance of the choices for q and x, under given (t u, t e, t d ) at K = 2. another task at any time, as illustrated in Fig. 2. C. Energy Consumption Model In this paper, we consider low CPU voltage at the serving node, and adopt the energy consumption model for task execution as in [7], [10]. In particular, the energy consumption for executing task k at the serving node is E e,k µl e,k F 2, where µ is a constant factor determined by the switched capacitance at the MEC server. We consider a narrow band system and study the block fading channel model. Let h k denote the channel power gain for user k, which is assumed to be constant within the duration [A k, D k ]. For simplicity, we consider capacity achieving codes, as in [7], [10]. It can be easily shown that the energy consumption at user k for uploading task ( k to the serving node is E u,k (t u,k ) t u,k Lu,k h k g 2 t u,k ), and the energy consumption at the serving

8 ( node for transmitting the computation result of task k to user k is E d,k (t d,k ) t d,k Ld,k h k g 2 t d,k ), where g(x) n 0 ( 2 x B 1 ), and B and n 0 denote the bandwidth (in Hz) and the power (in Watt) of the complex additive white Gaussian channel noise, respectively. Thus, the weighted sum energy consumption for serving task k is E k (t u,k, t d,k ) = E u,k (t u,k ) + β(e e,k + E d,k (t d,k )), where 0 < β 1 is the corresponding weight factor. Note that 0 < β < 1 means imposing a higher cost on the energy consumption for user devices due to their limited battery power. Therefore, the weighted sum energy consumption for serving all K tasks is given by: E(t) k K E k (t u,k, t d,k ). (11) Note that E(t) is a convex function of t. III. PROBLEM FORMULATION AND OPTIMAL SOLUTION In this section, we first formulate the energy minimization problem. Then, we develop an algorithm to obtain an optimal solution by exploiting structural properties of the problem. A. Problem Formulation In this part, we would like to minimize the weighted sum energy consumption by optimizing the operation sequences q o, o O and x, operation starting times s and transmission durations t. Problem 1 (Energy Minimization): E min q,x,s,t E(t) s.t. (1), (3), (4), (5), (6), (7), (8), (9), (10), where E(t) is given by (11). Problem 1 is a mixed discrete-continuous optimization problem with two types of variables, i.e., the operation sequences (discrete variables) as well as the uploading and downloading durations and starting times for uploading, executing and downloading (continuous variables). Note that the set of possible choices for the discrete variables, {(q, x) : (6), (8), (9)}, has cardinality (K!) 3( 2K 1 K 1 ), which is prohibitively large for large K. Thus, Problem 1 is very challenging, and is in general NP-hard.

9 B. Optimal Solution In this part, we develop an algorithm to obtain an optimal solution of Problem 1. 1) Structural Properties: First, we analyze structural properties of Problem 1. For all q o Q and o O, let idx(q o, i) denote the index of element i in q o, i.e., q o,idx(qo,i) = i. 3 For all feasible (q, x) of Problem 1, we have the following results. Lemma 1 (Structural Properties of Problem 1): (i) For all k K, idx(q u,k) 1 x i < idx(q d, k). (ii) For all k, k K with idx(q d, k) idx(q u,k ) 1 x i, idx(q e, k) < idx(q e, k ). (iii) For all k, k K with D k < A k, idx(q u, k) < idx(q u, k ), idx(q d, k) < idx(q d, k ) and idx(q d, k) idx(qu,k ) 1 x i. Proof 1: Please refer to Appendix. A. Property (i) indicates that the uploading of task k should be conducted before its downloading. Property (ii) indicates that if the downloading of task k is conducted before the uploading of task k, the executing of task k should be conducted before the executing of task k. Property (iii) indicates that if the deadline of task k is before the arrival time of task k, the uploading of task k should be conducted before the uploading of task k, the downloading of task k should be conducted before the downloading of task k, and the downloading of task k should be conducted before the uploading of task k. Lemma 1 offers important design insights for choosing appropriate operation sequences. Later, we shall see that Lemma 1 also facilitates the process of solving Problem 1. 2) Algorithm: It can be seen that the structural properties for discrete variables (q, x) in Lemma 1 reflect the constraints on continuous variables (s, t) in (3), (4) and (5) to certain extent. Based on Lemma 1, we first propose an equivalent formulation of Problem 1 to facilitate the optimization. Problem 2 (Operation Sequences): E = min Eseq(q, x), (q,x) S where S {(q, x) : (6), (8), (9), Properties (i), (ii) and (iii)}. Let (q, x ) denote an optimal solution. E seq(q, x) is given by the following problem. 3 We assume 0 xi = 0.

10 Algorithm 1 : Algorithm for Obtaining An Optimal Solution Output: (q, x, s, t ). 1: Set E = 2: for each (q, x) S do 3: Obtain (s (q, x), t (q, x)) and E seq(q, x) by solving Problem 3 4: if E seq(q, x) E then 5: Set E = E seq(q, x), (q, x, s, t ) = (q, x, s (q, x), t (q, x)) 6: end if 7: end for Problem 3 (Starting Times and Transmission Durations): For any (q, x) S, E seq(q, x) = min s,t E(t), s.t. (1), (3), (4), (5), (7), (10). If the problem is infeasible, we have Eseq(q, x) = ; Otherwise, we have Eseq(q, x) <, and denote an optimal solution by (s (q, x), t (q, x)). Due to the equivalence between Problem 1 and Problems 2-3, we know that (q, x, s (q, x ), t (q, x )) is an optimal solution of Problem 1. This equivalent formulation separates the discrete and continuous variables, and enables solving a mixed discrete-continuous optimization problem (i.e., Problem 1) by solving a discrete optimization problem (i.e., Problem 2) and a continuous optimization problem for each (q, x) S (i.e., Problem 3). Note that for any given (q, x) S, Problem 3 is a convex optimization problem with 5K variables, and thus can be solved using standard convex optimization techniques. Problem 2 is a discrete optimization problem with 4K variables, and can be solved by exhaustive search over set S. Note that the structural properties in Lemma 1 enable a great reduction of the search space for the values of the discrete variables of Problem 2 (from {(q, x) : (6), (8), (9)} to S). The details for solving Problem 1 based on the equivalent formulation in Problem 2 and Problem 3 are summarized in Algorithm 1. IV. LOW-COMPLEXITY SUBOPTIMAL SOLUTION Although the complexity for obtaining an optimal solution of Problem 1 has been reduced based on Lemma 1, the computation complexity of Algorithm 1 may not be acceptable when K is large. In this section, we propose a low-complexity algorithm to obtain a suboptimal solution of Problem 1.

11 First, using several optimization techniques, we transform the mixed discrete-continuous problem (i.e., Problem 1) to a DC problem. Denote Q o (Q o,i,j ) i,j K, o O, Q (Q o ) o O, X (X i,j ) i,j K, Z (Z i,j ) i,j K, α o (α o,i,j ) i,j K, o O, α (α o ) o O, φ o (φ o,i,j ) i,j K, o O, φ (φ o ) o O, λ (λ i,n,j ) j K,i,n K and γ (γ i,m,j ) m,j K,i K. 4 Problem 4 (DC Problem of Problem 1): Ẽ min 0 Q,X,Z 1, s,t,α,φ,λ,γ 0 E(t) s.t. (1), (3), (4), (5), P (Q, X, Z) 0, (12) Q o,i,j = Q o,j,i = 1, o O, j K (13) i K i K X i,j = K, (14) i,j K Z i,j = K 1, i,j K j K X m,n (i 1) K (1 Z i,j ), i, j K (16) m=1 n=1 j K α o,i,j j K (15) φ o,i+1,j, o O, i K (17) φ o,i,j min{ sq o,i,j, s o,j }, o O, i, j K s o,j (1 Q o,i,j ) s φ o,i,j, o O, i, j K α o,i,j min{( s + t)q o,i,j, s o,j + t o,j }, o O, i, j K s o,j + t o,j (1 Q o,i,j )( s + t) α o,i,j, o O, i, j K (18a) (18b) (18c) (18d) 4 In this paper, and represent componentwise inequities.

12 λ i,m,j min {( s + t)q d,m,j, ( s + t)z m+1,i, s d,j + t d,j }, j K, i, m K s d,j + t d,j (2 Q d,m,j Z m+1,i ) ( s + t) λ i,m,j, j K, i, m K γ i,n,j min{ sq d,n,j, sz n,i, s d,j }, n, j K, i K s d,j (2 Q d,n,j Z n,i ) s γ i,n,j, n, j K, i K (19a) (19b) (19c) (19d) λ i,m,j φ u,i+1,j, i K j K m K j K α u,i+1,j γ i,n,j, i K j K j K n K α u,1,j φ d,1,j, j K j K (20a) (20b) (20c) where s max k K D k, t max k K D k min k K A k, P (Q, X, Z) is given by P (Q, X, Z) i,j K (X i,j (1 X i,j ) + Z i,j (1 Z i,j )) + o O i,j K Let (Q, X, Z, s, t, α, φ, λ, γ ) denote an optimal solution of Problem 4. Q o,i,j (1 Q o,i,j ). (21) Note that the constraints in (12), (13) and (14) correspond to the constraints in (6), (8) and (9), the constraints in (12), (13), (17) and (18) correspond to the constraints in (7), the constraints in (12), (13), (14), (15), (16), (18), (19) and (20) correspond to the constraints in (10). The relationship between Problem 1 and Problem 4 is shown below. Theorem 1 (Equivalence between Problem 1 and Problem 4): Problem 4 is equivalent to Problem 1, i.e., q o,i = K j=1 jq o,i,j, o O, i K, x i = K j=1 X i,j, i K, s o,k = s o,k, o O, k K, t u,k = t u,k, k K, t d,k = t d,k, k K, and E = Ẽ. Proof 2: Firstly, we equivalently convert sequences q and x satisfying (6), (8) and (9) to binary variables Q and X satisfying (13), (14) and X i,j, Q o,i,j {0, 1}, o O, i, j K. (22) The equivalence can be seen from Q o,i,j = I[q o,i = j], q o,i = j K ji[q o,i,j = 1], X i,j = I[j x i ] and x i = j K X i,j. Then, we introduce new variables Z i,j I[ j m=1 K n=1 X m,n = (i

13 1)], i, j K and impose (15) and (16). Thus, we can equivalently convert (7) and (10) to m K Z m+1,i Q u,i+1,j s u,j, i K, (23) Q d,m,j (s d,j + t d,j ) j K j K Z n,i Q d,n,j s d,j, i K. (24) Q u,i+1,j (s u,j + t u,j ) j K n K j K Q o,i+1,j s o,j, o O, i K, (25) j K Q o,i,j (s o,j + t o,j ) j K j K Q u,1,j (s u,j + t u,j ) j K Q d,1,j s d,j. (26) Secondly, we adopt the big-m formulation to decompose the product terms Q o,i,j (s o,j + t o,j ), Q o,i+1,j s o,j, Z m+1,i Q d,m,j (s d,j +t d,j ), Z n,i Q d,n,j s d,j in (23), (24), (25) and (26) by introducing new variables α, φ, λ, γ and imposing the linear constraints in (18) and (19). Finally, we equivalently convert the discrete constraints in (22) to continuous constraints in (12) and 0 Q, X, Z 1. By noting that x a (a 0) can be rewritten as x a and a x, and x min{a 1, a 2,..., a n } can be rewritten as x a i, i = 1, 2,..., n, the constraints in (16), (18) and (19) can be viewed as linear constraints. In addition, note that P (Q, X, Z) given by (21) is concave, the other constraints are linear, and E(t) is convex. Thus, Problem 4 is a DC problem, which is non-convex. A classical goal of solving a non-convex problem is to obtain a stationary point. Next, we obtain a stationary point of Problem 4 using Penalty CCP [9]. Specifically, at the (l + 1)-th iteration, we solve the following convex approximation problem, which is obtained via linearizing the concave portion in P (Q, X, Z) at the solution obtained at the l-th iteration, and relaxing the constraint in (12) by introducing a slack variable y. Problem 5 (Convex Approximation Problem at (l + 1)-th iteration): arg min 0 Q,X,Z 1, s,t,α,φ,λ,γ 0,y 0 k K E(t) + ρ (l+1) y s.t. (1), (3), (4), (5), (13), (14), (15), (16), (17), (18), (19), (20) ˆP ( Q, X, Z; Q (l), X (l), Z (l)) y, where ρ (l+1) = min{θρ (l), ρ} for some θ > 1 and ρ > 0, and ˆP ( Q, X, Z; Q (l), X (l), Z (l)) is

14 Algorithm 2 : Algorithm for Obtaining A Low-complexity Suboptimal Solution ) 1: Initialization: choose a feasible initial point (Q (0), X (0), Z (0), s (0), t (0), α (0), φ (0), λ (0), γ (0) of Problem 4, and set ρ (0) > 0 and l = 0. 2: repeat ( 3: Obtain Q (l+1), X (l+1), Z (l+1), s (l+1), t (l+1), α (l+1), ) φ (l+1), λ (l+1), γ (l+1) by solving Problem 5 using standard convex optimization techniques. 4: Set ρ (l+1) = min{θρ (l), ρ} and l = l + 1 5: until convergence criteria is met given by ˆP ( Q, X, Z; Q (l), X (l), Z (l)) (( 1 2Z (l) i,j i,j K i,j K o O ) ) Z i,j + Z (l) 2 i,j + i,j K (( + ((( ) )) 1 2Q (l) o,i,j Q o,i,j + Q (l) 2 o,i,j 1 2X (l) i,j ) ) X i,j + X (l) 2 i,j ) Let (Q (l+1), X (l+1), Z (l+1), s (l+1), t (l+1), α (l+1), φ (l+1), λ (l+1), γ (l+1), y (l+1) denote an optimal solution of Problem 5 at the (l + 1)-th iteration. The details of the algorithm are summarized in Algorithm 2. By [9], we know that the sequence )} {(Q (l), X (l), Z (l), s (l), t (l), α (l), φ (l), λ (l), γ (l) generated by Algorithm 2 is convergent, )} and the limit point of {(Q (l), X (l), Z (l), s (l), t (l), α (l), φ (l), λ (l), γ (l) is a stationary point of Problem 4. We can run Algorithm 2 multiple times, each with a random initial feasible point of Problem 4, and select the stationary point with the minimum weighted sum energy, denoted by ( Q, X, Z, s, t ). Then, we can construct a suboptimal solution of Problem 1, denoted by ( q, x, s, t ), where q o,i = K j=1 jq o,i,j, o O, i K and x i = K j=1 X i,j, i K. V. NUMERICAL RESULTS In this section, we compare the proposed optimal and suboptimal solutions with two optimized baselines schemes in three different cases of task arrival times and deadlines as illustrated in Fig. 3 using numerical results. Both baseline schemes allow parallel transmissions and executions and consider the optimization of starting times and transmission durations but for two simple choices of operation sequences, i.e., adopt the solutions of Problem 2 for two typical (q, x) S. Our primary goal is to demonstrate the key impact of the operation sequences in designing efficient MEC systems.

15 Fig. 3. Illustration of the three cases of task arrival times and deadlines. Let v A Q and v D Q denote the orders of the arrival times and deadlines of the K tasks, where v A,i and v D,i represent the indices of the tasks corresponding to the i-th smallest arrival time and deadline, respectively. Both baseline schemes choose q u = q e = v A, which seems quite reasonable. In addition, Baseline 1 chooses x with x i = 0, i K, x K = K and q d = v D, and Baseline 2 chooses x with x i = 1, i K and q d = v A. That is, in Baseline 1, the uploading operations of all K tasks are completed before the downloading operation of any task, and in Baseline 2, the uploading and downloading operations of a task are completed before those of another task. Note that the two choices for q d and x are also typical. The considered three cases have different arrival times and deadlines. As illustrated in Fig. 3, in Case 1, the order of arrival times is the reverse of that of deadlines, i.e., v A,i = v D,K+1 i, i K; in Case 2, the order of arrival times is the same as that of deadlines, i.e., v A,i = v D,i, i K; in Case 3, the order of arrival times and the order of deadlines can be treated as a combination of those in Case 1 and Case 2, i.e., v A,i = v D,i I[i is odd] + v D,2 I[i is even], i K, where denotes the floor K 2 +2 i function. In the simulation, we consider the following settings. For those three cases, we set L u,k = k 10 4 bits, L d,k = 1.5k 10 4 bits, L e,k = 1 K 105 CPU-cycles, h k = 10 3, k K, β = 0.1, µ = 10 29, F = 5 10 8, B = 10 6 Hz and n 0 = k B BT 0, where k B = 1.38 10 23 Joule/Kelvin is the Boltzmann constant and T 0 = 300 Kelvin is the temperature. For ease of illustration, let

16 Energy consumption(j) 10-6 10-8 Optimal Suboptimal Baseline 1 Baseline 2 2 3 4 5 6 Number of users K (a) Energy consumption versus K. Computation Time (s) 1000 800 600 400 200 Optimal Suboptimal Baseline 1 Baseline 2 0 2 3 4 5 6 Number of users K (b) Computation time versus K. Fig. 4. Comparisions between proposed solutions and baseline schemes in Case 1, where A i = 15 K, D i = 30 K and D 1 A 1 = 15 + 6K 15 K. Energy consumption(j) 10-4 10-6 10-8 Optimal Suboptimal Baseline 1 Baseline 2 2 3 4 5 6 Number of users K (a) Energy consumption versus K. Computation Time (s) 1000 800 600 400 200 Optimal Suboptimal Baseline 1 Baseline 2 0 2 3 4 5 6 Number of users K (b) Computation time versus K. Fig. 5. Comparisions between proposed solutions and baseline schemes in Case 2, where A i = 15 K, D i = 45 K and D 1 A 1 = 6K + 30 K. A i A i+1 A i, D i D i+1 D i, i K, and we consider time in ms. Fig. 4, Fig. 5 and Fig. 6 illustrate the weighted sum energy consumption and computation time (reflecting computation complexity) of the proposed solutions and baseline schemes versus the number of users K in the three cases, respectively. Note that since the computation complexity for obtaining an optimal solution is prohibitively high when K is large, the proposed optimal solution is only evaluated at K = 2, 3. In addition, note that in Case 3, Baseline 1 is not feasible when K 5. From Fig. 4(a), Fig. 5(a) and Fig. 6(a), we can observe that the weighted sum

17 Energy consumption(j) 10-4 10-6 10-8 Optimal Suboptimal Baseline 1 Baseline 2 2 3 4 5 6 Number of users K (a) Energy consumption versus K. Computation Time (s) 2000 1500 1000 500 Optimal Suboptimal Baseline 1 Baseline 2 0 2 3 4 5 6 Number of users K (b) Computation time versus K. Fig. 6. Comparisions between proposed solutions and baseline schemes in Case 3, where A i = 12I[i is odd] + 3I[i is even], D i + D i+1 = 25I[i is odd] 11I[i is even], D 1 A 1 = 25 and D 2 A 2 = 27 + 15 K 2. energy consumption of the proposed suboptimal solution is close to that of the optimal solution when K = 2, 3 in the three cases. In addition, the proposed suboptimal solution achieves smaller weighted sum energy consumption than Baseline 2 in Case 1, Baseline 1 in Case 2, and both baseline schemes in Case 3 (e.g., 97% smaller than Baseline 1 and 66% smaller than Baseline 2 both at K = 4). This observation demonstrates that the typical choices for operation sequences may perform well in some special cases of task arrival times and deadlines, but cannot guarantee universally good performance in the general scenario. From Fig. 4(b), Fig. 5(b) and Fig. 6(b), we see that the computation complexity of the optimal solution is not acceptable when K 4, and the computation complexity of the proposed suboptimal solution is quite close to those of both baseline schemes. These numerical results demonstrate the applicability and effectiveness of the proposed suboptimal solution. VI. CONCLUSION In this paper, we considered a computation task model and a computation offloading model, which are more practical than the existing ones. Then, we formulated the weighted sum energy consumption minimization problem to optimize the task operation sequences and starting times for uploading, executing and downloading as well as uploading and downloading time durations. The problem is a challenging mixed discrete-continuous optimization problem. By carefully using several optimization techniques, we developed an algorithm to obtain an optimal solution and a

18 low-complexity algorithm to obtain a suboptimal solution. Both analytical and numerical results demonstrate the importance of the optimization of operation sequences in practical multi-user MEC systems. A. Proof of Lemma 1 APPENDIX First, we prove Property (i) by contradiction. Suppose that there exists k K satisfying idx(qu,k ) 1 x i idx(q d, k ). By (7) and idx(q u,k ) 1 x i idx(q d, k ), we have: s d,k + t d,k = s + t d,qd,idx(qd,k ) d,q s d,idx(qd,k ) d,q. (27) idx(qu,k ) 1 d, x i By (10), we have: By (27) and (28), we have: s d,qd, s idx(qu,k ) 1 u,qu,idx(qu,k ) = s u,k. (28) x i s d,k s u,k. (29) In addition, by (2) and (3), we have: s u,k s e,k < s d,k. (30) It is clear that (30) contradicts (29). Thus, by contradiction, we can prove Property (i). Next, we prove Property (ii) by contradiction. Suppose that there exist k, k K satisfying idx(q d, k) idx(q u,k ) 1 x i and idx(q e, k) idx(q e, k ). By (3), (7), (10) and idx(q d, k) idx(qu,k ) 1 x i, we have: s e,k < s d,k + t d,k = s + t d,qd,idx(qd,k) d,q s d,idx(qd,k) d,q s idx(qu,k ) 1 u,k s e,k. (31) d, x i In addition, by (7) and idx(q e, k) idx(q e, k ), we have: s e,k s e,k. (32) It is clear that (32) contradicts (31). Thus, by contradiction, we can prove Property (ii). Finally, we prove Property (iii) by contradiction. For all k, k K, by (2), (3) and (7), we

19 have: A k s u,k s e,k < s d,k D k, (33) A k s u,k s e,k < s d,k D k. (34) 1) Suppose that there exist k, k K with D k < A k satisfying idx(q u, k) idx(q u, k ). By (7) and idx(q u, k) idx(q u, k ), we have: s u,k s u,k (35) By (33), (34) and D k < A k, we have: s u,k < s u,k (36) It is clear that (35) contradicts (36). 2) Suppose that there exist k, k K with D k < A k satisfying idx(q d, k) idx(q d, k ). By (7) and idx(q d, k) idx(q d, k ), we have: s d,k s d,k. (37) By (33), (34) and D k < A k, we have: s d,k < s d,k. (38) It is clear that (37) contradicts (38). 3) Suppose that there exist k, k K with D k < A k satisfying idx(q d, k) > idx(q u,k ) 1 x i. By (7), (10) and idx(q d, k) > idx(q u,k ) 1 x i, we have: s u,k = s u,qu,idx(qu,k ) s d,qd, s = s idx(qu,k ) 1 d,qd,idx(qd,k) d,k. (39) x i +1 By (33), (34) and D k < A k, we have: s d,k < s u,k. (40) It is clear that (39) contradicts (40). Thus, by contradiction, we can prove Property (iii). Therefore, we complete the proof of Lemma 1.

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