Solution of multi-layer ann with sigmoid units:
Assume here that we input the values 10, 30, 20 with the three input units and from top to bottom. So after then the weighted sum coming into H_{1} will be as:
S_{H1} = (0.2 * 10) + (-0.1 * 30) + (0.4 * 20) = 2 -3 + 8 = 7.
After then the σ function is applied to S_{H1} to give as:
σ(S_{H1}) = 1/(1+e-7) = 1/(1+0.000912) = 0.999
Here don't forget to negate S. Also the weighted sum coming into H_{2} will be as:
S_{H2} = (0.7 * 10) + (-1.2 * 30) + (1.2 * 20) = 7 - 36 + 24 = -5
Now next σ applied to S_{H2} gives as:
σ(S_{H2}) = 1/(1+e5) = 1/(1+148.4) = 0.0067
Furthermore from this we can see there that H_{1} has fired, but H_{2} has not. So now we can calculate the weights sum going in to output in unit O1 will be as:
SO_{1} = (1.1 * 0.999) + (0.1*0.0067) = 1.0996
And here next the weighted sum going in to output in unit O_{2} will be as:
SO_{2} = (3.1 * 0.999) + (1.17*0.0067) = 3.1047
However the output sigmoid unit in O_{1} will now calculate the output values from the network for O_{1 }as:
σ(SO_{1}) = 1/(1+e^{-1.0996}) = 1/(1+0.333) = 0.750
and the output from the network for O2:
σ(SO_{2}) = 1/(1+e^{-3.1047}) = 1/(1+0.045) = 0.957
However therefore if this network represented the learned rules for a categorisation problem then the input triple (10,30,20) would be categorised into the category associated with O2 it means it has the larger output.