Simulation for buffer energy model

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Dhawal Beohar
Dhawal Beohar 2023-7-18
回答： Yash 2025-2-16
I am trying to compare my montecarlo simulation result with the baseline model. I have the result and figure which I have got from the monte carlo simulation. And now I am trying to create a realistic energy model, calculate energy outage probability and compare the same.
I am not sure I am creating the energy model and then calculating enery outage correctly (last two line of the code). Can anyone please help me with this. Thank You!
clc
clear all;
%Parameters defined
NoB=1; % Noise Density
%rt=2*1.4426; % Rate converted into bits/sec/Hz---(1 nat = 1/ln
%2 bits) and (1 bits = 1/log e nats)
rt=3.0; % Rate in nats/sec/Hz
EH=0; % Initialising Energy Harvesting
Bmax=50; % Threshold level of battery in Joules
PmaxdB=[15];
Pmax=10.^(PmaxdB/10);
Qu=[-3:0.1:0];
QoS_Component_u=10.^(Qu);
for i=1:length(QoS_Component_u)
u=QoS_Component_u(i);
EHH(i)=0; %Initialising EHH equal to zero
muo_sum(i)=0; %Initialising mu_o equal to zero
for j=1:100000 %Monte-Carlo Simulations
hx=(1/sqrt(2))*randn(1,1);
hy=(1/sqrt(2))*randn(1,1);
ht_rayleigh=hx.^2+hy.^2;
muo(i)=min((((exp(rt)-1)*NoB)./ht_rayleigh),Pmax);
EHH(i)=EHH(i)+exp(u*muo(i));
end
EHHH(i)=(EHH(i))/100000; %Average for Expectation
mufix(i)=(1/u)*log(EHHH(i));
mu_o_avg(i)=(muo_sum(i))/100000; %Average of mu_o(i)
Epsilon(i)=(mu_o_avg(i))/(mufix(i));
Pout(i)= (Epsilon(i)*(exp(-u*(Bmax-Pmax)))); % Calculating Pout
energy_left(i) = (Bmax + mufix(i)) - muo(i); % creating energy model
el(i)= energy_left(i)/100000; % calculate energy outage probability
end

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回答（1 个）

Yash
Yash 2025-2-16
There's missing logic in the Monte Carlo simulation. muo_sum is initialized but never updated.
muo_sum(i)=0; %Initialising mu_o equal to zero
...
mu_o_avg(i)=(muo_sum(i))/100000; %Average of mu_o(i)
...
mu_o_avg(i) will always be evaluated as 0 and hence subsequent calculations of Epsilon, Pout and other parameters will also be incorrect.

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