task2.m

% 2.a For each value of p, determine the probability of the link being in the interference state
% and in the normal state when one control frame is received with errors (fulfil the
% following table). What do you conclude?

% |             | p(normal) | p(interference) |
% |-------------|-----------|-----------------|
% | p = 99%     |           |                 |
% | p = 99.9%   |           |                 |
% | p = 99.99%  |           |                 |
% | p = 99.999% |           |                 |

pAll = [0.99, 0.999, 0.9999, 0.99999];

pEF1 = 0;
pEF2 = 0;
pF1 = 0;
pF2 = 0;

%pF1E = (pEF1*pF1) / (pEF1*pF1 + pEF2*pF2)

results = [];

for i=1:size(pAll,2)
    pEF1 = 0.0001;
    pEF2 = 0.5;
    pF1 = pAll(i);
    pF2 = 1-pAll(i);
    
    results(i,1) = (pEF1*pF1) / (pEF1*pF1 + pEF2*pF2);
    results(i,2) = (pEF2*pF2) / (pEF1*pF1 + pEF2*pF2);
    
end
format SHORTE; % em probabilidade
fprintf('2a')
results
format shorte;

% 2.b. For each value of p and for n = 2, 3, 4 and 5, determine the probability of false positives
% and fulfil the follow table:

% |             | Probability of false positives |
% |             | n=2    | n=3   | n=4   | n=5   |
% |-------------|--------|-------|-------|-------|
% | p = 99%     |        |       |       |       |
% | p = 99.9%   |        |       |       |       |
% | p = 99.99%  |        |       |       |       |
% | p = 99.999% |        |       |       |       |

results = [];
for n=2:5
    for i=1:size(pAll,2)
        pEF1 = 0.0001^n;
        pEF2 = 0.5^n;
        pF1 = pAll(i);
        pF2 = 1 - pAll(i);
        
        results(i,n-1) = (pEF1*pF1) / (pEF1*pF1 + pEF2*pF2);
    end
end

fprintf('2b')
results
results = [];

% 2.c. For each value of p and for n = 2, 3, 4 and 5, determine the probability of false
% negatives and fulfil the follow table:

% |             | Probability of false negatives |
% |             | n=2    | n=3   | n=4   | n=5   |
% |-------------|--------|-------|-------|-------|
% | p = 99%     |        |       |       |       |
% | p = 99.9%   |        |       |       |       |
% | p = 99.99%  |        |       |       |       |
% | p = 99.999% |        |       |       |       |

results = [];
for n=2:5
    for i=1:size(pAll,2)
        pEF1 = 1- 0.0001^n;
        pEF2 = 1- 0.5^n;
        pF1 = pAll(i);
        pF2 = 1 - pAll(i);
        
        results(i,n-1) = (pEF2*pF2) / (pEF1*pF1 + pEF2*pF2);
    end
end
fprintf('2c')
results