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同時透射反射可重構(gòu)智能表面賦能移動邊緣計算任務(wù)卸載研究

李斌 楊冬東

李斌, 楊冬東. 同時透射反射可重構(gòu)智能表面賦能移動邊緣計算任務(wù)卸載研究[J]. 電子與信息學(xué)報, 2025, 47(2): 418-426. doi: 10.11999/JEIT240733
引用本文: 李斌, 楊冬東. 同時透射反射可重構(gòu)智能表面賦能移動邊緣計算任務(wù)卸載研究[J]. 電子與信息學(xué)報, 2025, 47(2): 418-426. doi: 10.11999/JEIT240733
LI Bin, YANG Dongdong. Task Offloading for Simultaneously Transmitting and Reflecting Reconfigurable Intelligent Surface-assisted Mobile Edge Computing[J]. Journal of Electronics & Information Technology, 2025, 47(2): 418-426. doi: 10.11999/JEIT240733
Citation: LI Bin, YANG Dongdong. Task Offloading for Simultaneously Transmitting and Reflecting Reconfigurable Intelligent Surface-assisted Mobile Edge Computing[J]. Journal of Electronics & Information Technology, 2025, 47(2): 418-426. doi: 10.11999/JEIT240733

同時透射反射可重構(gòu)智能表面賦能移動邊緣計算任務(wù)卸載研究

doi: 10.11999/JEIT240733 cstr: 32379.14.JEIT240733
基金項目: 國家自然科學(xué)基金(62101277)
詳細(xì)信息
    作者簡介:

    李斌:男,副教授,研究方向為邊緣計算、無人機(jī)通信

    楊冬東:男,碩士生,研究方向為智能反射面、移動邊緣計算

    通訊作者:

    李斌 bin.li@nuist.edu.cn

  • 中圖分類號: TN929.5

Task Offloading for Simultaneously Transmitting and Reflecting Reconfigurable Intelligent Surface-assisted Mobile Edge Computing

Funds: The National Natural Science Foundation of China (62101277)
  • 摘要: 為彌補(bǔ)可重構(gòu)智能表面(RIS)半空間覆蓋和“乘性衰落”等不足,該文提出一種有源同時透射和反射可重構(gòu)智能表面(aSTAR-RIS)技術(shù)用于提升移動邊緣計算(MEC)卸載性能增益。首先,考慮MEC服務(wù)器計算資源、aSTAR-RIS能耗以及相移耦合約束,聯(lián)合設(shè)計任務(wù)卸載比例、計算資源配置、多用戶檢測矩陣(MUD)、aSTAR-RIS相移以及用戶上傳功率,建立一個多變量耦合的加權(quán)總時延最小化問題。然后,借助塊坐標(biāo)下降法(BCD)將原問題分解為兩個子問題,使用拉格朗日乘子法和罰項對偶分解法(PDD)交替優(yōu)化子問題。仿真結(jié)果表明,相較于無源STAR-RIS方案,所提aSTAR-RIS輔助MEC方案加權(quán)總時延降低了12.66%。
  • 圖  1  aSTAR-RIS輔助MEC系統(tǒng)模型

    圖  2  加權(quán)總時延迭代收斂圖

    圖  3  各方案時延與反射單元數(shù)量關(guān)系

    圖  4  不同功率約束下加權(quán)總時延與反射單元數(shù)量關(guān)系

    1  求解最優(yōu)任務(wù)卸載比和MEC計算資源分配算法

     初始化優(yōu)化變量,${n_1} = 0$,收斂閾值$ {\varepsilon }_{1}={10}^{-4} $
     步驟1 利用式(3)計算${R_{\tau ,i}}$,根據(jù)式(9)對${{\boldsymbol{\alpha}} ^{({n_1})}}$進(jìn)行更新;
     步驟2 利用二分法求得${\mu ^{({n_1})}}$,根據(jù)式(12)計算${{\boldsymbol{f}}^{\text{e}}}^{({n_1})}$;
     步驟3 計算$ {\varepsilon }^{({n}_{1})} $,若$ {\varepsilon }^{({n}_{1})}\ge {\varepsilon }_{1} $且${n_1} \le n_1^{\max }$,令${n_1} = {n_1} + 1$,
     回到步驟1;
     步驟4 輸出$({{\boldsymbol{\alpha}} ^*},{{\boldsymbol{f}}^{\text{e}}}^*)$
    下載: 導(dǎo)出CSV

    2  MUD矩陣、aSTAR-RIS相移和用戶上傳功率交替優(yōu)化算法

     初始化優(yōu)化變量,${n_2} = 0$,收斂閾值$ \zeta ={\varepsilon }_{2}={\varepsilon }_{3}={10}^{-4} $,
     $\rho = 10$
     步驟1 根據(jù)式(20)更新$ {{\boldsymbol{W}}^{({n_2})}} $;
     步驟2 解決問題式P2.5更新$\{ {{\boldsymbol{\theta}} _{\text{t}}}^{({n_2})},{{\boldsymbol{\theta}} _{\text{r}}}^{({n_2})}\} $;
     步驟3 更新$\left\{ {\tilde {\boldsymbol{\psi}} _{\text{t}}^{({n_2})},\tilde{\boldsymbol{ \psi}} _{\text{r}}^{({n_2})}} \right\}$和$\left\{ {{{\tilde {\boldsymbol{\beta}} }_{\text{t}}}^{({n_2})},{{\tilde {\boldsymbol{\beta }}}_{\text{r}}}^{({n_2})}} \right\}$;
     步驟4 解決問題式P2.8更新$ {{\boldsymbol{p}}^{({n_2})}} $;
     步驟5 更新輔助變量${{\boldsymbol{\varphi}} ^{({n_2})}}$;
     步驟6 計算$ {\varepsilon }^{\left({n}_{2}\right)} $,若$ {\varepsilon }^{\left({n}_{2}\right)} \gt {\varepsilon }_{2} $,且${n_2} \le n_2^{{\text{max}}}$,令
     ${n_2} = {n_2} + 1$,回到步驟1;
     步驟7 更新$ {{\boldsymbol{\lambda}} ^{({n_2})}} $, $ {{\boldsymbol{\xi}} ^{({n_2})}} $,若$ \left|{\lambda }_{k}^{({n}_{2})}{R}_{k}^{({n}_{2})}-1\right| \gt {\varepsilon }_{2} $或
     $ \left|{\xi }_{k}^{({n}_{2})}{R}_{k}^{({n}_{2})}-{\varpi }_{k}{\alpha }_{k}{L}_{k}\right| \gt {\varepsilon }_{2} $,令${n_2} = {n_2} + 1$,回到步驟1;
     步驟8 若$ \upsilon \le \zeta $, $ {{\boldsymbol{\eta}} _{\,\tau }} = {{\boldsymbol{\eta}} _{\,\tau }} + \dfrac{1}{\rho }({\tilde {\boldsymbol{\theta}} _\tau } - {{\boldsymbol{\theta}} _\tau }) $,否則設(shè)置$ \rho = c\rho $;
     步驟9 $ \zeta = 0.9\upsilon $,若$ \upsilon \gt {\varepsilon }_{3} $,令${n_2} = 0$,回到步驟1;
     輸出 $ \left( {{{\boldsymbol{W}}^*},{{\boldsymbol{\theta}} _{\text{t}}}^*,{{\boldsymbol{\theta}} _{\text{r}}}^*,{{\boldsymbol{p}}^*}} \right) $
    下載: 導(dǎo)出CSV

    3  整體算法

     初始化優(yōu)化變量,${n_3} = 0$,收斂閾值$ \varepsilon ={10}^{-4} $
     步驟1 根據(jù)算法1,給定$ {{\boldsymbol{W}}^{({n_3} - 1)}} $, ${{\boldsymbol{\theta}} _{\text{t}}}^{({n_3} - 1)}$, ${{\boldsymbol{\theta}} _{\text{r}}}^{({n_3} - 1)}$,
     $ {{\boldsymbol{p}}^{({n_3} - 1)}} $優(yōu)化${{\boldsymbol{\alpha}} ^{({n_3})}}$, ${{\boldsymbol{f}}^{\text{e}}}^{({n_3})}$;
     步驟2 根據(jù)算法2,給定${{\boldsymbol{\alpha}} ^{({n_3})}}$, ${{\boldsymbol{f}}^{\text{e}}}^{({n_3})}$優(yōu)化$ {{\boldsymbol{W}}^{({n_3})}} $, ${{\boldsymbol{\theta }}_{\text{t}}}^{({n_3})}$,
     ${{\boldsymbol{\theta}} _{\text{r}}}^{({n_3})}$, ${{\boldsymbol{p}}^{({n_3})}}$;
     步驟3 計算$ {\varepsilon }^{\left({n}_{3}\right)} $,若$ {\varepsilon }^{\left({n}_{3}\right)} \gt \varepsilon $且$ {n_3} \le n_3^{\max } $,令${n_3} = {n_3} + 1$,
     回到步驟1;
     輸出:$\left( {{{\boldsymbol{\alpha}} ^*},{{\boldsymbol{f}}^{\text{e}}}^*,{{\boldsymbol{W}}^*},{{\boldsymbol{\theta}} _{\text{t}}}^*,{{\boldsymbol{\theta}} _{\text{r}}}^*,{{\boldsymbol{p}}^*}} \right)$
    下載: 導(dǎo)出CSV
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  • 收稿日期:  2024-08-26
  • 修回日期:  2024-12-24
  • 網(wǎng)絡(luò)出版日期:  2024-12-31
  • 刊出日期:  2025-02-28

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