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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)

吳大鵬 鄭豪 崔亞平

吳大鵬, 鄭豪, 崔亞平. 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)[J]. 電子與信息學(xué)報(bào), 2020, 42(8): 1910-1917. doi: 10.11999/JEIT190635
引用本文: 吳大鵬, 鄭豪, 崔亞平. 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)[J]. 電子與信息學(xué)報(bào), 2020, 42(8): 1910-1917. doi: 10.11999/JEIT190635
Dapeng WU, Hao ZHENG, Yaping CUI. Service-oriented Coordination agent Design for Network Slicing in Vehicular Networks[J]. Journal of Electronics & Information Technology, 2020, 42(8): 1910-1917. doi: 10.11999/JEIT190635
Citation: Dapeng WU, Hao ZHENG, Yaping CUI. Service-oriented Coordination agent Design for Network Slicing in Vehicular Networks[J]. Journal of Electronics & Information Technology, 2020, 42(8): 1910-1917. doi: 10.11999/JEIT190635

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
基金項(xiàng)目: 國(guó)家自然科學(xué)基金(61871062, 61771082, 61801065),重慶市高校創(chuàng)新團(tuán)隊(duì)建設(shè)計(jì)劃資助項(xiàng)目(CXTDX201601020)
詳細(xì)信息
    作者簡(jiǎn)介:

    吳大鵬:男,1979年生,教授,研究方向?yàn)榉涸跓o線網(wǎng)絡(luò)、社會(huì)計(jì)算、無線網(wǎng)絡(luò)服務(wù)質(zhì)量控制等

    鄭豪:1995年生,碩士生,研究方向?yàn)檐嚶?lián)網(wǎng)、網(wǎng)絡(luò)切片與虛擬化

    崔亞平:1986年生,講師,研究方向?yàn)楹撩撞ㄍㄐ?、多天線技術(shù)、車聯(lián)網(wǎng)等

    通訊作者:

    鄭豪 547721540@qq.com

  • 中圖分類號(hào): TN915; TP393

Service-oriented Coordination agent Design for Network Slicing in Vehicular Networks

Funds: The National Natural Science Foundation of China (61871062, 61771082, 61801065), The Program for Innovation Team Building at Institutions of Higher Education in Chongqing (CXTDX201601020)
  • 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。

  • 圖  1  車輛網(wǎng)絡(luò)切片系統(tǒng)架構(gòu)

    圖  2  聚類前后數(shù)據(jù)點(diǎn)對(duì)比

    圖  3  3種用戶負(fù)載分布場(chǎng)景下的理論和仿真BTD增益對(duì)比

    圖  4  不同用戶負(fù)載分布場(chǎng)景下的最優(yōu)切片權(quán)重分配

    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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  • 收稿日期:  2019-08-26
  • 修回日期:  2020-03-10
  • 網(wǎng)絡(luò)出版日期:  2020-04-21
  • 刊出日期:  2020-08-18

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