基于終端能耗和系統(tǒng)時(shí)延最小化的邊緣計(jì)算卸載及資源分配機(jī)制
doi: 10.11999/JEIT180970 cstr: 32379.14.JEIT180970
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北京郵電大學(xué)網(wǎng)絡(luò)與交換技術(shù)國(guó)家重點(diǎn)實(shí)驗(yàn)室 北京 100876
A Computation Offloading and Resource Allocation Mechanism Based on Minimizing Devices Energy Consumption and System Delay
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State Key Laboratory of Networking and Switching Technology, Beijing University of Posts and Telecommunications, Beijing 100876, China
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摘要: 通過移動(dòng)邊緣計(jì)算下移云端的應(yīng)用功能和處理能力支撐計(jì)算密集或時(shí)延敏感任務(wù)的執(zhí)行成為當(dāng)前的發(fā)展趨勢(shì)。但面對(duì)眾多移動(dòng)終端用戶時(shí),如何有效利用計(jì)算資源有限的邊緣節(jié)點(diǎn)來保障終端用戶服務(wù)質(zhì)量(QoS)成為關(guān)鍵問題。為此,該文融合邊緣云與遠(yuǎn)端云構(gòu)建了一種分層的邊緣云計(jì)算架構(gòu),以此架構(gòu)為基礎(chǔ),以最小化移動(dòng)設(shè)備能耗和任務(wù)執(zhí)行時(shí)間為目標(biāo),將問題形式化描述為資源約束下的最小化能耗和時(shí)延加權(quán)和的凸優(yōu)化問題,并提出基于乘子法的計(jì)算卸載及資源分配機(jī)制解決該問題。實(shí)驗(yàn)結(jié)果表明,在計(jì)算任務(wù)量很大的情況下,提出的計(jì)算卸載及資源分配機(jī)制能夠有效降低移動(dòng)終端能耗,并在任務(wù)執(zhí)行時(shí)延方面較局部計(jì)算與計(jì)算卸載機(jī)制分別降低最高60%與10%,提高系統(tǒng)性能。
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關(guān)鍵詞:
- 邊緣計(jì)算 /
- 計(jì)算卸載 /
- 資源分配 /
- 終端能耗 /
- 系統(tǒng)時(shí)延
Abstract: To support the execution of computation-intensive, delay-sensitive computing task by moving down the computing and processing capability in mobile edge computing becomes the current trend. However, when serving a large number of mobile users, how to use effectively the edge nodes with limited computing resources to ensure Quality of service (QoS) of end-user has become a key issue. To solve this problem, the edge cloud and remote cloud are combined to build a layered edge cloud computing architecture. Based on this architecture, with the goal of minimizing mobile device energy consumption and task execution time, the problem which is proved to be convex is formulated to minimize the weight sum of energy and delay. A computation offloading and resource allocation mechanism based on multiplier method is proposed. Simulations are conducted to evaluate the proposed mechanism. Compared with local computing and computation offloading mechanism, the proposed mechanism can effectively reduce the energy consumption of mobile device and the delay of system by up to 60% and 10%, respectively.-
Key words:
- Edge computing /
- Computing offloading /
- Resource allocation /
- Energy consumption /
- Delay of system
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表 1 多用戶計(jì)算卸載
初始化:各移動(dòng)終端數(shù)量$n$及計(jì)算能力${C_i}$,邊緣節(jié)點(diǎn)計(jì)算能力
${C_{{\rm{edge}}}}$,遠(yuǎn)端云節(jié)點(diǎn)計(jì)算能力${C_{{\rm{cloud}}}}$,無線帶寬資源$B$,權(quán)值$V\,$, $S = \varnothing $;輸入:各用戶終端計(jì)算任務(wù)請(qǐng)求REQ($\left[ {{\lambda _1}, {\lambda _2}, ·\!·\!· , {\lambda _n}} \right]$); 輸出:最優(yōu)卸載決策$S = {X^*}$; $C_i^{{\ \rm{edge}}} = {{{C_{{\rm{edge}}}}} / n}$; while TRUE do; 接收用戶計(jì)算卸載請(qǐng)求REQ,提取請(qǐng)求中的對(duì)應(yīng)任務(wù)信息: $B_i^{{\rm{in}}}, {V_i}, B_i^{{\rm{out}}}, P_i^{\rm{c}}, P_i^{{\rm{up}}}, {\lambda _i}$; for each $i \in \left\{ {1, 2, ·\!·\!· , n} \right\}$ do; 引入拉格朗日函數(shù),求得滿足KKT條件的最優(yōu)解
$ < {x_i}, x_i^{{\rm{edge}}}, x_i^{{\rm{cloud}}} > $;最優(yōu)解向下取整,得整數(shù)解$ < x' + {1_i}, x_i^{'{\rm{edge}}}, x_i^{'{\rm{cloud}}} > $, $ < {x'_i}, x_i^{'{\rm{edge}}} + 1, x_i^{'{\rm{cloud}}} > $, $ < {x'_i}, x_i^{'{\rm{edge}}}, x_i^{'{\rm{cloud}}} + 1 > $; 將整數(shù)可行解代入目標(biāo)函數(shù),取使目標(biāo)函數(shù)最小的整數(shù)解為最優(yōu) 整數(shù)解; end for; 回傳最優(yōu)解${X^*}$,移動(dòng)終端接收卸載決策,執(zhí)行任務(wù); end while. 下載: 導(dǎo)出CSV
表 2 多用戶計(jì)算卸載及資源分配機(jī)制
初始化:$n$, ${C_i}$, ${C_{{\rm{edge}}}}$, ${C_{{\rm{cloud}}}}$, $B$,權(quán)值$V\,$, $S = \varnothing $ 輸入:各用戶終端計(jì)算任務(wù)請(qǐng)求REQ($\left[ {{\lambda _1}, {\lambda _2}, ·\!·\!· , {\lambda _n}} \right]$) 輸出:最優(yōu)卸載決策$S = {X^*}$ $C_i^{{\ \rm{edge}}} = {{{C_{{\rm{edge}}}}} / n}$, ${C_0} = < C_1^{{\ \rm{edge}}}, C_2^{{\ \rm{edge}}}, ·\!·\!· , C_n^{{\ \rm{edge}}} > $; while TRUE do; 接收用戶計(jì)算卸載請(qǐng)求REQ,提取任務(wù)信息:
$B_i^{{\rm{in}}}, {V_i}, B_i^{{\rm{out}}}, P_i^{\rm{c}}, P_i^{{\rm{up}}}, {\lambda _i}$;for each $i \in \left\{ {1, 2, ·\!·\!· , n} \right\}$ do; 引入拉格朗日函數(shù),求得滿足KKT條件的最優(yōu)解
$ < {x_i}, x_i^{{\rm{edge}}}, x_i^{{\rm{cloud}}} > $;end for; 得到平均資源分配條件下的初始最優(yōu)解${X^*}$, ${X_0} = {X^*}$; ${S_0} = < {X_0}, {C_0} > $; ${\mu ^{\left( 1 \right)}} = \left( {1, 1, ·\!·\!· , 1} \right)$, ${\eta ^{\left( 1 \right)}} = \left( {1, 1, ·\!·\!· , 1} \right)$, $\varepsilon = {10^{ - 5}}$, $M = 2$,
$\theta = 0.8$, $\alpha = 2$;$k = k + 1$; ${S_1} = {\rm{BFGS}}\left( {\varphi \left( {S, \mu , \eta , M} \right)} \right)$; ${\beta _k} = {\left\{ {\sum\limits_{i = 1}^n {{h_i}^2\left( {{S_k}} \right)} + \sum\limits_{j = 1}^{4n + 1} {{{\left[ {\left( {\min {g_j}\left( {{S_k}} \right), \frac{{{{\left( {{\eta ^{\left( K \right)}}} \right)}_j}}}{M}} \right)} \right]}^2}} } \right\}^{{1 / 2}}}$; while ${\beta _k} > \varepsilon $ do; 更新罰函數(shù):若${\beta _k} > \theta \cdot {\beta _k}$,則$M = \alpha \cdot M$,否則$M$不變; 更新乘子向量${\mu ^{\left( k \right)}}$, ${\eta ^{\left( k \right)}}$; $k = k + 1$; ${S_k} = {\rm{BFGS}}\left( {\varphi \left( {S, \mu , \eta , M} \right)} \right)$; 依據(jù)上述公式計(jì)算${\beta _k}$值; end while; 對(duì)$ < {x_i}, x_i^{{\rm{edge}}}, x_i^{{\rm{cloud}}} > $求最優(yōu)整數(shù)解,返回${S_k}^* = < {X_k}^*, {C_k}^* > $,
按${X_k}^*$進(jìn)行計(jì)算卸載,按${C_k}^*$進(jìn)行計(jì)算資源分配;end while. 下載: 導(dǎo)出CSV
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