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Add LODCO implementation
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‎.gitignore

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‎UIC18/LODCO.m

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%% This script simulates the Lyapunov Optimization-based Dynamic Computation Offloading (LODCO) algorithm.
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% author: Hailiang Zhao
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clc, clear
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opt = optimset('Display', 'none');
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%% basic parameter settings
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k = 1e-28; % effective switched capacitance (a constant decided by the chip architecture)
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tau = 0.002; % the length of time slot (in second)
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phi = 0.002; % the cost of task dropping (in second)
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omega = 1e6; % the bandwidth of MEC server (in Hz)
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sigma = 1e-13; % the noise power of the receiver (in W)
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p_tx_max = 1; % the maximum transmit power of mobile device (in W)
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f_max = 1.5e9; % the maximum CPU-cycle frequency of mobile device (in Hz)
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E_max = 0.002; % the maximum amout of battery output energy (in J)
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L = 1000; % the input size of the computation task (in bit)
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X = 737.5; % the number of CPU cycles needed on processing one bit of task
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W = L * X; % the number of CPU cycles needed on processing one task
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E_H_max = 48e-6; % the upper bound of the energy arrive at the mobile device (in J)
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p_H = E_H_max / (2*tau); % the average Energy Harvesting (EH) power (in W)
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g0 = power(10, -4); % the path-loss constant
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%% parameter control
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T = 10000; % the number of time slot (a.k.a. the size of the time horizon)
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tau_d = 0.002; % execution deadline (in second)
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d = 50; % the distance between the mobile device and the MEC server (in meter)
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E_min = 0.02e-3; % the minimum amout of battery output energy (in J)
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V = 1e-5; % the weight of penalty (the control parameter introduced by Lyapunov Optimization)
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rho = 0.6; % the probability that the computation task is requested
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% the lower bound of perturbation parameter
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E_max_hat = min(max(k * W * (f_max)^2, p_tx_max * tau), E_max);
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theta = E_max_hat + V * phi / E_min;
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%% allocate storage for valuable results
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B = zeros(T, 1); % the battery energy level (in J)
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B_hat = zeros(T, 1); % the virtual battery energy level ($B_hat = B - theta$)
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e = zeros(T, 1); % the amout of the harvested and stored energy (in J)
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chosen_mode = zeros(T, 1);% {1: local, 2: remote, 3: drop, 4: no task request}
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f = zeros(T, 1); % the CPU-cycle frequency of local execution (in Hz)
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p = zeros(T, 1); % the transmit power of computation offloading (in W)
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cost = zeros(T, 3); % execution delay for mobile execution, MEC server execution and final choice, respectively (in second)
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E = zeros(T, 3); % energy consumption for mobile execution, MEC server execution and final choice, respectively (in J)
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%% simulation begin
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t = 1;
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while t <= T
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disp(['===> Time slot #', num2str(t), '<==='])
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%% initialization
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% generate the task request
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zeta = binornd(1, rho);
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% generate the virtual battery energy level
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B_hat(t) = B(t) - theta;
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%% step 1: get the optimal energy harvesting no matter whether task is requested
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E_H_t = unifrnd(0, E_H_max);
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if B_hat(t) <= 0
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e(t) = E_H_t;
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end
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%% step 2: get the optimal computation offloading strategy (I_m, I_s, I_d, f(t), p(t))
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if zeta == 0
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% chosen mode has to be 4
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disp('no task request generated!')
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chosen_mode(t) = 4;
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else
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% chosen_mode is chosen from {1, 2, 3}
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disp('task request generated!');
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% task request exists, generate the channel power gain
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h = exprnd(g0 / power(d, 4));
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%% step 2.1: solve the optimization problem $\mathcal{P}_{ME}$ (f(t) > 0)
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% calculate f_L and f_U
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f_L = max(sqrt(E_min / (k * W)), W / tau_d);
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f_U = min(sqrt(E_max / (k * W)), f_max);
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if f_L <= f_U
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% the sub-problem is feasible
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disp('mobile execution ($\mathcal{P}_{ME}$) is feasible!')
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if B_hat(t) < 0
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f_0 = (V / (-2 * B_hat(t) * k))^(1/3);
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else
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% complex number may exist, which will lead to error
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f_0 = -(V / (2 * B_hat(t) * k))^(1/3);
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end
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if (f_0 > f_U && B_hat(t) < 0) || (B_hat(t) >= 0)
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f(t) = f_U;
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elseif f_0 >= f_L && f_0 <= f_U && B_hat(t) < 0
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f(t) = f_0;
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elseif f_0 < f_L && B_hat(t) < 0
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f(t) = f_L;
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end
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% check whether f(t) is zero
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if f(t) == 0
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disp('Something wrong! f is 0!')
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end
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% calculate the delay of mobile execution
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cost(t, 1) = W / f(t);
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% calculate the energy consumption of mobile execution
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E(t, 1) = k * W * (f(t)^2);
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% calculate the value of optimization goal
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J_m = -B_hat(t) * k * W * (f(t))^2 + V * W / f(t);
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else
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% the sub-problem is not fasible because (i) the limited
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% computation capacity or (ii) time cosumed out of deadline or
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% (iii) the energy consumed out of battery energy level
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% If it is not feasible, it just means that we cannot choose
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% 'I_m=1'. It dosen't mean that the task has to be dropped.
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disp('mobile execution ($\mathcal{P}_{ME}$) is not feasible!')
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f(t) = 0;
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cost(t, 1) = 0;
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E(t, 1) = 0;
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% 'I_m=1' can never be chosen if mobile execution goal is inf
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J_m = inf;
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% Attention! We do not check whether the energy cunsumed is larger than
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% battery energy level because the problem $\mathcal{J}_{CO}$ does
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% not have constraint (8).
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end
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%% step 2.2: solve the optimization problem $\mathcal{P}_{SE}$ (p(t) > 0)
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E_tmp = sigma * L * log(2) / (omega * h);
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p_L_taud = (power(2, L / (omega * tau_d)) - 1) * sigma / h;
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% calculate p_L
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if E_tmp >= E_min
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p_L = p_L_taud;
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else
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% calculate p_E_min (use inline function and fsolve)
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y = @(x) x * L - omega * log2(1 + h*x/sigma) * E_min;
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% accroding to the function figure, p_L_taud is a positive
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% number around 0.2
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p_E_min = fsolve(y, 0.2, opt);
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p_L = max(p_L_taud, p_E_min);
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end
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% calculate p_U
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if E_tmp >= E_max
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p_U = 0;
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else
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% caculate p_E_max (use inline function and fsolve)
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y = @(x) x * L - omega * log2(1 + h*x/sigma) * E_max;
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% accroding to the function figure, p_E_max is a large positive
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% number around 20
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p_E_max = fsolve(y, 100, opt);
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p_U = min(p_tx_max, p_E_max);
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end
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if p_L <= p_U
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% the sub-problem is feasible
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disp('MEC server execution ($\mathcal{P}_{SE}$) is feasible!')
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% calculate p_0
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virtual_battery = B_hat(t);
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y = @(x) virtual_battery * log2(1 + h*x/sigma) + ...
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h * (V - virtual_battery*x) / log(2) / (sigma + h*x);
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p_0 = fsolve(y, 0.5, opt);
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if (p_U < p_0 && B_hat(t) < 0) || B_hat(t) >= 0
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p(t) = p_U;
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elseif p_0 < p_L && B_hat(t) < 0
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p(t) = p_L;
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elseif p_0 >= p_L && p_0 <= p_U && B_hat(t) < 0
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p(t) = p_0;
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end
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% check whether p(t) is zero
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if p(t) == 0
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disp('Something wrong! p is 0!')
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end
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% calculate the delay of MEC server execution
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cost(t, 2) = L / (omega * log2(1 + h*p(t)/sigma));
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% calculate the energy consumption of MEC server execution
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E(t, 2) = p(t) * cost(t, 2);
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% calculate the value of optimization goal
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J_s = (-B_hat(t) * p(t) + V) * cost(t, 2);
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else
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% the sub-problem is not feasible because (i) the limited transmit
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% power or (ii) time cosumed out of deadline or (iii) the energy
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% consumed out of battery energy level
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% If it is not feasible, it just means that we cannot choose
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% 'I_s=1'. It dosen't mean that the task has to be dropped.
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disp('MEC server execution ($\mathcal{P}_{SE}$) is not feasible!')
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p(t) = 0;
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cost(t, 2) = 0;
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E(t, 2) = 0;
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% 'I_s=1' can never be chosen if MEC server execution goal is inf
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J_s = inf;
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% Similarly, we do not check whether the energy cunsumed is larger than
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% battery energy level because the problem $\mathcal{J}_{CO}$ does
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% not have constraint (8).
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end
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%% step 3: choose the best execution mode
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J_d = V * phi;
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disp(['J_m:', num2str(J_m)])
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disp(['J_s:', num2str(J_s)])
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[~, mode] = min([J_m, J_s, J_d]);
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chosen_mode(t) = mode;
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end
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%% step 4: according to the chosen execution mode, calculate the real dealy and energy consumption
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if chosen_mode(t) == 1
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% mobile execution is chosen
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cost(t, 3) = cost(t, 1);
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E(t, 3) = E(t, 1);
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elseif chosen_mode(t) == 2
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% MEC server execution is chosen
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cost(t, 3) = cost(t, 2);
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E(t, 3) = E(t, 2);
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elseif chosen_mode(t) == 3
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% task is dropped, the delay is the task dropping penalty and the
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% energy consumption is zero
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cost(t, 3) = phi;
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E(t, 3) = 0;
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else
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% no task is requested, the delay and the energy consumption are
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% both zero
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cost(t, 3) = 0;
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E(t, 3) = 0;
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end
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%% step 5: update the battery energy level and go to next time slot
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B(t + 1) = B(t) - E(t, 3) + e(t);
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t = t + 1;
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end
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%% step 6: evaluate the simulation results
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% 1. the battery energy level vs. time slot
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subplot(2, 1, 1);
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plot(1:T, B(1:T));
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hold on
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plot(1:T, repmat(theta + E_H_max, [T, 1]), '-')
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title('Envolution of battery energy level')
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xlabel('time slot')
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ylabel('battery energy level $B_t$', 'Interpreter','latex')
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% 2. the average execution cost vs. time slot
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accumulated = 0;
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average_cost = zeros(T, 1);
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request_num = 0;
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for t = 1: T
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accumulated = accumulated + cost(t, 3);
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if cost(t, 3) ~= 0
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% there exists task request
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request_num = request_num + 1;
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end
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average_cost(t) = accumulated / request_num;
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end
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subplot(2, 1, 2);
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plot(1:T, average_cost);
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title('Envolution of average execution cost')
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xlabel('time slot')
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ylabel('average execution cost $\frac{1}{T} \sum_{t=0}^{T-1} cost^t$', 'Interpreter','latex')

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