Change from random Distribution to Normal/Poisson Distribution

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Brave A
Brave A 2021-8-18
评论： Brave A 2021-8-19
Hello there ,
I have this code which distrbuted values of C and f_i as randome distrbutuons.
I need to change this distrbutuins to :
1- Normal Distrbutions
2- Poission Distrbutions
but any time I tried those it gave me errors.
s_input = struct('V_POSITION_X_INTERVAL',[0 1],... %(km)
'V_POSITION_Y_INTERVAL',[0 1],... %(km)
'V_SPEED_INTERVAL',[0.01 0.1],... %(km/s) originally set to [0.2 2.2]?
'V_PAUSE_INTERVAL',[0 1],... % pause time (s) originally [1 4]
'V_WALK_INTERVAL',[1 5],... % walk time (s)
'V_DIRECTION_INTERVAL',[-180 180],... % (degrees)
'SIMULATION_TIME',20,... %(s) % originally 20
'NB_NODES', 200); % originally 300
global BaseX BaseY
BaseX = 0.5; %(km)
BaseY = 0.5; %(km)
num_users = s_input.NB_NODES;
s_mobility = Generate_Mobility(s_input);
% timeStep is used for interpolating positions of nodes
% within the positions specified in the s_mobility data generated above
timeStep = 0.5; % (s) % originally 1; shouldn't really affect anything
% except to increase number of node time-positions
% generate and display an animated version of the nodes
% using the timeStep to interpolate position data available
% in the s_mobility position data
test_Animate(s_mobility,s_input,timeStep);
%%
% ======================== %
% PARAMETERS %
% ======================== %
% Loss function parameters
alpha = 128.1;
beta = 37.6;
% Computational Latency:
C_i = 5000; % (eventually want C to be) Between C_i and 3*C_i
C_t=69;
% D_i = 1; % Between 5000-10000 bytes
D_i=500; % modified by wdc 01/09/2021 for use below in calc of D
z)
f_i = randi([1500000000, 3000000000], num_users, 1); % give each user her own f value
%Y = mu + sqrt(2)*sigma*erfinv(2*X-1);
eta = 1/10; % Got value from notes (was 1/1000)
Delta = 0.1;
L = 4; % From notes but it should came From hessian
ggamma = 2; % From notes but it should came From hessian
% Uploading Latency:
global B P_i N_0 S
B = 11000000; % bandwidth is 10 MHz = 10,000,000 HZ; trying .1 MHz?
P_i = 10; % dBm; Transmission power
% No idea what the N_0 is supposed to be here. Is ?noise? = N_0?
% Is the following really in bytes and not bits?
S = 1000000;
N_0 = 10^((-104-30)/10); % noise=-104dbm. Noise
% transmit data size, trying 1 Mbit = 1,000,000 bits
% Define Tau
% Tau = 0.00009563
Tau = 1; % originally 0.00009563
Selected = s_mobility;
% Create a vector of all desired time values/steps, in the form
% [0, timeStep, 2*timeStep, ..., SIMULATION_TIME]
v_t = 0:timeStep:s_input.SIMULATION_TIME;
for nodeIndex = 1:Selected.NB_NODES
Selected.VS_NODE2(nodeIndex).v_t = v_t;
Selected.VS_NODE2(nodeIndex).v_x = interp1(Selected.VS_NODE(nodeIndex).V_TIME,Selected.VS_NODE(nodeIndex).V_POSITION_X,v_t);
Selected.VS_NODE2(nodeIndex).v_y = interp1(Selected.VS_NODE(nodeIndex).V_TIME,Selected.VS_NODE(nodeIndex).V_POSITION_Y,v_t);
end
for N = 1:Selected.NB_NODES
% this tlen should actually be the same each time, being the
% number of (x,y) positions created in the previous for loop above
% for each node
tlen = length(Selected.VS_NODE2(N).v_t);
for t = 1:tlen
Selected.VS_NODE2(N).DISTANCE(t) = disFn(Selected.VS_NODE2(N).v_x(t),Selected.VS_NODE2(N).v_y(t));
Selected.VS_NODE2(N).LOSS(t) = lossFn(alpha,beta,Selected.VS_NODE2(N).DISTANCE(t));
Selected.VS_NODE2(N).ULAT(t) = UlatFn(Selected.VS_NODE2(N).LOSS(t));
end
end
C_i = 5000; % (eventually want C to be) Between C_i and 3*C_i
C_t=69;
C = C_i + (C_t*C_i)*rand(Selected.NB_NODES,1);
f_i = randi([1500000000, 3000000000], num_users, 1); % give each user her own f value
function distance=disFn(x,y)
global BaseX BaseY
% distance=sqrt(sum((x-BaseX).^2)+sum((y-BaseY)^2));
% cleaned this up a bit -- was working but the sums were not necessary
distance=sqrt((x-BaseX).^2 + (y-BaseY)^2);
end
function loss=lossFn(alpha,beta,d)
loss=alpha+beta*log10(d);
end
function Ulatency=UlatFn(loss)
global B P_i N_0 S
h2=10.^(-loss/10);
Ulatency = S/(B*log2(1+P_i.*h2./N_0)) ;
end
function RateLat=RatelatFn(loss)
global B P_i N_0
h2=10.^(loss/10);
RateLat = B*log2(1+P_i.*h2./N_0);
end
Any suggestions willl be helpful.

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Image Analyst
Image Analyst 2021-8-18
编辑：Image Analyst 2021-8-18
Well you probably solved it by now, assuming you looked up random in the help, but for what it's worth, here is a solution:
randomNumbers = random('poisson', 5, 1, 100000);
histogram(randomNumbers);
grid on;
fontSize = 15;
title('Poisson Distribution', 'FontSize', fontSize);
xlabel('Value', 'FontSize', fontSize);
ylabel('Count', 'FontSize', fontSize);
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Walter Roberson
Walter Roberson 2021-8-19
lb = 1500000000; ub = 3000000000;
mu = (ub+lb)/2;
normalized_sigma = 3; %prob would have been in bounds is 99.7%
sigma = (ub-lb)/normalized_sigma;
pdn = makedist('Normal', 'mu', mu, 'sigma', sigma);
tpdn = truncate(pdn, lb, ub);
pdp = makedist('Poisson', 'lambda', mu);
tpdp = truncate(pdp, lb, ub);
f_in = random(tpdn, num_users, 1); %normal
f_ip = random(tpdp, num_users, 1); %poisson
In both cases, the distribution is arranged to peak half way between the upper bound and the lower bound. In both cases, the distribution is truncated in a way that only values between the lower bound and the upper bound can be generated.
Brave A
Brave A 2021-8-19
@Image Analyst Thank you so much for your help, I read all replies and I got sense how is my result will be from your previous reply.
@Walter Roberson Thanks a lot Walter for all comments, I think I got some resluts. Thank you both, I really appriciated :)

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