输入参数:Lre为第i辆车的剩余里程,E 为第i辆车的电池容量,PE为第i辆车的能耗,S0 为第i辆车的初始剩余电量。定义充电方案Project,包括3个参数:1.充电类型(Ch_type):Ch_no,Ch_slow,Ch_fast;2.充电成本(Ch_cost):Ch_cost=E*Ch_charge/Pt_price;3.充电量(Ch_charge):当Ch_type为Ch_no时Ch_charge为0;当Ch_type为Ch_fast时Ch_charge取值为4个:10,20,50,80;当Ch_type为Ch_slow时Ch_charge取值为4个:10,20,50,80。其中,当Ch_type为Ch_no时Pt_price为0,当Ch_type为Ch_slow时Pt_price为2.37,当Ch_type为Ch_fast时Pt_price为5.52;生成J个方案的m个参数,包括初始剩余电量(S0),电费成本(Ch_cost),未来出行满足度(Travel_sat):未来出行满足度等于(S0+Ch_charge)E/PELre。然后编写函数计算方案i的后悔值R,计算公式1为:Ri⟶jm=ln(1+exp(βm(xjm-xim))),其中,S0的β值为0.2,Ch_cost的β值为0.3,Travel_sat的β值为-0.1;计算公式2为:Ri=∑j≠i,j=1J∑m=1MRi⟶jm,计算公式3为Pi=exp(-Rj)/∑j=1Jexp(-Rj),输出内容为(i,Ch_type,Ch_charge,Ch_cost,Travel_sat, Ri,Pi)。用matlab代码描述这一过程。
时间: 2024-02-12 20:08:42 浏览: 128
通信与网络中的推动以太网接入的LRE技术说明
以下是用Matlab编写的代码,可以实现上述功能:
```matlab
% 输入参数
Lre = [100, 80, 120]; % 第i辆车的剩余里程
E = [50, 60, 70]; % 第i辆车的电池容量
PE = [0.2, 0.25, 0.3]; % 第i辆车的能耗
S0 = [30, 40, 50]; % 第i辆车的初始剩余电量
% 定义常量
Pt_price = [0, 2.37, 5.52]; % 充电价格,分别对应Ch_no、Ch_slow、Ch_fast
Ch_slow_charge = [10, 20, 50, 80]; % 慢充电量
Ch_fast_charge = [10, 20, 50, 80]; % 快充电量
beta_S0 = 0.2; % S0的β值
beta_Ch_cost = 0.3; % Ch_cost的β值
beta_Travel_sat = -0.1; % Travel_sat的β值
% 生成J个方案的m个参数
Project = []; % 存储所有方案的参数
for i = 1:length(Lre)
for j = 1:3 % 充电类型
if j == 1 % Ch_no
Ch_type = "Ch_no";
Ch_charge = 0;
Ch_cost = 0;
elseif j == 2 % Ch_slow
Ch_type = "Ch_slow";
for k = 1:length(Ch_slow_charge)
Ch_charge = Ch_slow_charge(k);
Ch_cost = E(i) * Ch_charge / Pt_price(j);
Travel_sat = (S0(i) + Ch_charge) * E(i) / (PE(i) * Lre(i));
Project = [Project; S0(i), Ch_cost, Travel_sat];
end
elseif j == 3 % Ch_fast
Ch_type = "Ch_fast";
for k = 1:length(Ch_fast_charge)
Ch_charge = Ch_fast_charge(k);
Ch_cost = E(i) * Ch_charge / Pt_price(j);
Travel_sat = (S0(i) + Ch_charge) * E(i) / (PE(i) * Lre(i));
Project = [Project; S0(i), Ch_cost, Travel_sat];
end
end
end
end
% 计算方案的后悔值R
R = []; % 存储所有方案的后悔值
for i = 1:size(Project, 1)
Rijm = []; % 存储方案i与其他方案j的后悔值
for j = 1:size(Project, 1)
if i ~= j % 不计算方案i与自身的后悔值
Rijm_j = log(1 + exp(beta_S0 * (Project(j, 1) - Project(i, 1))) ...
+ exp(beta_Ch_cost * (Project(j, 2) - Project(i, 2))) ...
+ exp(beta_Travel_sat * (Project(j, 3) - Project(i, 3))));
Rijm = [Rijm, Rijm_j];
end
end
Ri = sum(Rijm);
R = [R; Ri];
end
% 计算方案的概率值Pi
Pi = exp(-R) ./ sum(exp(-R));
% 输出结果
for i = 1:size(Project, 1)
fprintf("(%d, %s, %d, %.2f, %.2f, %.2f, %.2f)\n", i, Project(i, 1), ...
Project(i, 2), Project(i, 3), R(i), Pi(i));
end
```
注意:由于题目中未说明J和M的具体值,此处假设J和M均为方案总数。如果需要指定J和M的值,可以在代码中进行相应修改。
阅读全文