backpropagation - Neural network: How to calculate the error for a unit

Question

I am trying to work out question 26 from this exam paper (the exam is from 2002, not one I'm getting marked on!)

This is the exact question:

The answer is B.

Could someone point out where I'm going wrong?

I worked out I1 from the previous question on the paper to be 0.982.

The activation function is sigmoid. So should the sum be, for output 1:

d1 = f(Ik)[1-f(Ik)](Tk-Zk)

From the question:

T1 = 0.58
Z1 = 0.83
T1 - Z1 = -0.25
sigmoid(I1) = sigmoid(0.982) = 0.728
1-sigmoid(I1) = 1-0.728 = 0.272

So putting this all together:

d1 = (0.728)(0.272)(-0.25)
d1 = -0.049

But the answer should be d1 = -0.0353

Can anyone show me where I'm going wrong?

Edit 1: I tried to work backwards to understand the situation, but I still got stuck.

I said:

d1 = f(Ik)[1-f(Ik)](Tk-Zk)
-0.0353 = f'(Ik)(-0.25) (where I know -0.0353 is the right answer, and -0.25 is Tk - Zk)
0.1412 = f'(Ik)
0.1412 = f(Ik)[1-f(Ik)]
0.1412 = sigmoid(x).(1-sigmoid(x))

...but then I got stuck, if anyone has an idea

Answer 1

The problem is, that the I? you got from the previous question is not the same I? you need for this task.

The value of I? changes depending on the input values(which are different for this question)!

For the solution of this question you can instead use the fact that f(I?) = z?:

δ? = f(I?)·[1 - f(I?)]·(t? - z?)

= z?·[1 - z?]·(t? - z?)

→ δ? = 0.83·[1 - 0.83]·(-0.25) = -0.2075·0.17 = -0.035275 ≈ -0.0353

→ δ? = 0.26·[1 - 0.26]·(0.70 - 0.26) ≈ -0.0847

→ δ? = 0.56·[1 - 0.56]·(0.20 - 0.56) ≈ -0.0887