V2X Task Offloading Scheme Based on Mobile Edge Computing
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摘要: 移动边缘计算(MEC)通过在移动网络边缘提供IT服务环境和云计算能力带来高带宽、低时延优势，从而在下一代移动网络的研究中引起了广泛的关注。该文研究车载网络中车辆卸载请求任务时搜寻服务节点为其服务的匹配问题，构建一个基于MEC的卸载框架，任务既可以卸载到MEC服务器以车辆到基础设施(V2I)形式通信，也可以卸载到邻近车辆进行车辆到车辆(V2V)通信。考虑到资源有限性、异构性，任务多样性，建模该框架为组合拍卖模式，提出一种多轮顺序组合拍卖机制，由层次分析法(AHP)排序、任务投标、获胜者决策3个阶段组成。仿真结果表明，所提机制可以在时延和容量约束下，使请求车辆效益提高的同时最大化服务节点的效益。Abstract: Mobile Edge Computing (MEC) draws much attention in the next generation of mobile networks with high bandwidth and low latency by enabling the IT and cloud computation capacity at the Radio Access Network (RAN). Matching problem between requesting nodes and servicing nodes is studied when a vehicle wants to offload tasks, a MEC-based offloading framework in vehicular networks is proposed, Vehicle can either offload task to MEC sever as V2I link or neighboring vehicle as V2V link. Taking into account the limited and heterogeneous resources, and the diversity of tasks, offloading framework is established as combination auction model, and a multi-round sequential combination auction mechanism is proposed, which consists of Analytic Hierarchy Process (AHP) ranking, task bidding and winners decision. Simulation results show that the proposed mechanism can maximize the efficiency of service nodes while increasing the efficiency of requesting vehicles under the constraints of the delay and the capacity.
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表 2 平均随机一致性指标RI
$n$ 1 2 3 4 5 … RI 0 0.58 0.90 1.12 1.24 … 注：当判断矩阵小于等于15阶时，平均随机一致性指标RI可由查表得到；当阶数大于15时，RI可由参考文献[12]中乘幂法得到。 
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表 3 多轮顺序组合拍卖卸载机制
算法1 多轮顺序组合拍卖卸载机制 (1)输入：车辆 $i$的请求信息为 $ \left\{ {{c_{ij}},{b_{ij}},p_{ij}^c,p_{ij}^b} \right\}$，服务节点 $j$的状态
$\left\{ {{C_j},{B_j}} \right\}$, $i \in \left\{ {1,2, ·\!·\!· ,L} \right\}$, $j \in \left\{ {1,2, ·\!·\!· ,M} \right\}$，
满意度序值 ${W^i} = \left\{ {w_{_1}^i,w_{_2}^i, ·\!·\!· ,w_M^i} \right\}$(2)输出：连接矩阵X，效益 ${U_M}(j)$，未投标矩阵 ${{F}}\!=\!\left\{ {{v_1},{v_2}, ·\!·\!· ,{v_f}} \right\}$
且 $0 \le f \le L$(3) for $r = 1:M$ (4) for $i = 1:f$ do (5) for $j = 1:M$ do (6) AHP计算F中各元素对 ${g_j}$排序，根据排序顺序地发布请求信息；
服务节点权衡与各请求车辆通信距离确定服务顺序(7) end for (8) end for (9) for $j = 1:M$ do
(10) ${U_{\rm temp}} = \sum\limits_{j = 1}^M {U(j)} $;(11) for $i = 1:f$ do (12) $(U(j),F,X) = $ $ {\rm Knapsack}(\left\{ {{c_{ij}},{b_{ij}},p_{ij}^c,p_{ij}^b} \right\},\left\{ {{C_j},{B_j}} \right\})$; (13) ${U_M} = {U_{{\text{tp}}}} + U(j)$; (14) end for (15) end for (16) for $k = 1:f$ do (17) if $p_{{kj}}^c \le z_i^c$ then (18) $p_{{kj}}^c = p_{{kj}}^c + \eta p_{{kj}}^c$ (19) end if (20) if $p_{{kj}}^b \le z_i^b$ then (21) $p_{{kj}}^b = p_{{kj}}^b + \eta p_{{kj}}^b$ (22) end if (23) end for
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表 4 多维分组背包算法
算法2 多维分组背包算法 (1)输入：车辆 $i$的请求信息为 $\left\{ {{c_{ij}},{b_{ij}},p_{ij}^c,p_{ij}^b} \right\}$，服务节点 $j$状态函
数 $\left\{ {{C_j},{B_j}} \right\}$(2)输出：选择矩阵X，效益 ${U_j}(j)$，未投标矩阵F (3) for $c = 1:{C_j}$ do (4) for $b = 1:{B_i}$ do (5) $ {U_j}(i,j,c,b) =\max \left\{ {U_j}(i,j - 1,c,b),\right. $
$ \left. {U_j}(i,j - 1,c - {c_{ij}},b - {b_{ij}}) + {p_{ij}} \right\}$;(6) If ${U_j}(i,j,c,b) \le {U_j}(i,j - 1,c,b)$ (7) ${x_{ij}} = 0$; (8) ${{F}} = {{F}} \cup \left\{ j \right\}$; (9) else (10) ${x_{ij}} = 1$; (11) ${{F}} = {{F}} \cup \left\{ {j{-}1} \right\}$; (12) end if (13) end for (14) end for
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