面向服務(wù)的車輛網(wǎng)絡(luò)切片協(xié)調(diào)智能體設(shè)計(jì)
doi: 10.11999/JEIT190635 cstr: 32379.14.JEIT190635
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重慶郵電大學(xué)通信與信息工程學(xué)院 重慶 400065
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重慶高校市級(jí)光通信與網(wǎng)絡(luò)重點(diǎn)實(shí)驗(yàn)室 重慶 400065
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泛在感知與互聯(lián)重慶市重點(diǎn)實(shí)驗(yàn)室 重慶 400065
Service-oriented Coordination agent Design for Network Slicing in Vehicular Networks
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School of Communication and Information Engineering, Chongqing University of Posts and Telecommunications,Chongqing 400065, China
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Optical Communication and Networks Key Laboratory of Chongqing, Chongqing 400065, China
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3.
Ubiquitous Sensing and Networking Key Laboratory of Chongqing, Chongqing 400065, China
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摘要:
針對(duì)現(xiàn)有研究中缺乏對(duì)車輛網(wǎng)絡(luò)切片的部署和管理,該文設(shè)計(jì)了車輛網(wǎng)絡(luò)切片架構(gòu)中的切片協(xié)調(diào)智能體。首先基于K-means++聚類算法將車聯(lián)網(wǎng)通信業(yè)務(wù)根據(jù)相似度進(jìn)行聚類并映射到對(duì)應(yīng)的切片中。其次,在考慮應(yīng)用場(chǎng)景間的時(shí)空差異導(dǎo)致的無線資源利用不均衡現(xiàn)象,提出了共享比例公平方案以實(shí)現(xiàn)對(duì)無線資源的高效及差異化利用。最后,為了保證切片服務(wù)需求,采用線性規(guī)劃障礙方法求解最優(yōu)的切片權(quán)重分配,使切片負(fù)載變化容忍度最大化。仿真結(jié)果表明,共享比例公平方案相比于靜態(tài)切片方案平均比特傳輸時(shí)延(BTD)更小,在每切片用戶數(shù)為30的情況下均勻分布用戶負(fù)載場(chǎng)景中二者的BTD增益為1.4038,且在不同的用戶負(fù)載分布場(chǎng)景下都能求出最優(yōu)的切片權(quán)重分配。
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關(guān)鍵詞:
- 車聯(lián)網(wǎng) /
- 5G網(wǎng)絡(luò)切片 /
- 聚類 /
- 資源分配
Abstract:In view of the lack of deployment and management of slicing in vehicular network, a slice coordination agent of vehicular network slicing structure is designed. Firstly, based on the K-means++ clustering algorithm, the vehicle network communication services are clustered according to the similarity and then mapped into different slices. Secondly, considering the imbalance of radio resource utilization caused by the space-time characteristic among application scenarios, a shared proportional fairness scheme is proposed to utilize radio resources efficiently and differently. Finally, in order to ensure the requirements of slicing service, linear programming obstacle method is used to solve the optimal slice weight distribution to maximize the slice load variation tolerance. Simulation results show that the shared proportional fairness scheme has smaller average Bit Transmission Delay (BTD) than the static slicing scheme, and the optimal slice weight distribution can be obtained under different user load distribution scenarios. The BTD gain achieves 1.4038 in the uniform user load scenario with 30 users per slice.
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Key words:
- Vehicular networks /
- 5G network slicing /
- Clustering /
- Resource allocation
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表 1 符號(hào)縮寫
符號(hào) 定義 含義 ${\rho ^v}$ ${n^v}$ 切片$v$的總負(fù)載 ${{{\rho}} ^v}$ $\left( {\rho _b^v \triangleq n_b^v:b \in {\cal{B}}} \right)$ 切片$v$的負(fù)載分布 ${{\widetilde {{\rho}}} ^v}$ $\left( {\widetilde \rho _b^v \triangleq \dfrac{ {\rho _b^v} }{ { {\rho ^v} } }:b \in {\cal{B} } } \right)$ 切片$v$的相對(duì)負(fù)載分布 ${\widetilde {{g}}}$ $\left( { { {\widetilde g}_b} \triangleq \displaystyle\sum\nolimits_{v \in {\cal{V} } } { {s^v}\widetilde \rho _b^v:b \in {\cal{B} } } } \right)$ 總體權(quán)重相對(duì)負(fù)載分布 ${{{\delta}} ^v}$ $\left( {\delta _b^v \triangleq \mathbb{E}\left[ {\dfrac{1}{ {c_b^v} } } \right]:b \in {\cal{B} } } \right)$ 切片$v$的平均容量倒數(shù) $ {{\varDelta}} _v $ ${\rm{diag}}\left( {{{{\delta}} ^v}} \right)$ 切片$v$的平均容量倒數(shù)的對(duì)角矩陣 下載: 導(dǎo)出CSV
表 2 基于線性規(guī)劃障礙的資源分配算法(算法1)
輸入:初始${x_0}$,初始確定近似的參數(shù)${t_0}$,比例因子$\mu $,誤差閾值$\varepsilon $ 輸出:最優(yōu)解${x^*}$ (1) $x \leftarrow {x_0},t \leftarrow {t_0},\mu \leftarrow 50,\varepsilon \leftarrow {10^{ - 3}}$ (2) ${\rm{while}}\;({\rm{true}}) \;{\rm{do}}$ (3) 執(zhí)行表3所示的算法2,從$x$開始,最小化$t{f_0} + \phi $,得到對(duì)偶
可行解${x^*}(t)$(4) $x \leftarrow {x^*}(t)$
(5) 計(jì)算當(dāng)前對(duì)偶間隔${\rm{dualityGap} } \leftarrow \dfrac{ {2V} }{t}$(6) ${\rm{If}}\;{\rm{dualityGap}} < \varepsilon\;{\rm{ then}}$ (7) break (8) End if (9) $t \leftarrow \mu t$ (10) Endwhile (11) return $x$ 下載: 導(dǎo)出CSV
表 3 K-means++服務(wù)聚類算法(算法2)
步驟 1 選擇$K$個(gè)聚類${C_1},{C_2}, ··· ,{C_k}$的聚類中心; (1) 從數(shù)據(jù)集中隨機(jī)選取一個(gè)樣本作為初始聚類中心${\mu _1}$; (2) 首先計(jì)算每個(gè)樣本與當(dāng)前已有聚類中心之間的最短距離$D(x)$,其次計(jì)算每個(gè)樣本被選為下一個(gè)聚類中心的概率
$p(x) \leftarrow { {D{ {(x)}^2} } \Bigr/ {\displaystyle\sum\nolimits_{x \in X} {D{ {(x)}^2} } } }$,最后根據(jù)輪盤法選出下一個(gè)聚類中心;(3) 重復(fù)(2)直到選出$K$個(gè)聚類中心${\rm{(} }{\mu _1},{\mu _2}, ··· ,{\mu _k})$。 步驟 2 對(duì)剩下的每個(gè)樣本${x_i}$,計(jì)算其到$K$個(gè)聚類中心的距離${\rm{dist(}}{x_i},{\mu _k})$并將其分到距離最小的聚類中心所對(duì)應(yīng)的類中;
步驟 3 根據(jù)公式${\mu _k} = \dfrac{1}{ {\left| { {C_k} } \right|} }\displaystyle\sum\nolimits_{i \in {C_k} } { {x_i} } $重新計(jì)算聚類中心;步驟 4 重復(fù)步驟2和步驟3,直到聚類中心不再變化。 下載: 導(dǎo)出CSV
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