|
Junction point
|
Hamburg
|
Köln
|
München
|
Leipzig
|
Berlin
|
|
Test vector
|
.5
|
.4
|
.6
|
.8
|
.1
|
is generated. Only the tolerance hypothesis associated with the junction points München, Leipzig and Berlin are confirmed by the acceptable operational states. Only in those factors the procedure adds 50 to the counter values. The value of 50 is calculated by subtracting the current counter value of 1 from the maximum value (preset to 100) and then dividing the difference by 2. This value of 2 is derived from the learning rule preset in the beginning of the procedere. There is stated, that confirmed tolerance options will be reinforced by adding half of the difference that is left up to the maximum counter value. All other tolerance options stay represented by the counter value of 1.
Why are the counter values for junction points Hamburg and Köln left unchanged? The candidate State 1 differs from reference example State 5 in the values of junction points Hamburg and Köln. Operational State 5 would confirm the tolerance test vector if only those two factors would differ between those operational states. In fact, the sum of all tolerance hypothesis of all differing values is 2.3 as candidate example and reference example differ additionally in the values of junction points München and Leipzig. Consequently, the tolerance test vector is not confirmed by the pair of operational states 1 and 5.
Counter values after adaption cycle 1
|
Junction point
|
Hamburg
|
Köln
|
München
|
Leipzig
|
Berlin
|
|
Tolerance option 0
|
1
|
1
|
1
|
1
|
1
|
|
Tolerance option .1
|
1
|
1
|
1
|
1
|
51
|
|
Tolerance option .2
|
1
|
1
|
1
|
1
|
1
|
|
Tolerance option .3
|
1
|
1
|
1
|
1
|
1
|
|
Tolerance option .4
|
1
|
1
|
1
|
1
|
1
|
|
Tolerance option .5
|
1
|
1
|
1
|
1
|
1
|
|
Tolerance option .6
|
1
|
1
|
51
|
1
|
1
|
|
Tolerance option .7
|
1
|
1
|
1
|
1
|
1
|
|
Tolerance option .8
|
1
|
1
|
1
|
51
|
1
|
|
Tolerance option .9
|
1
|
1
|
1
|
1
|
1
|
|
Tolerance option 1
|
1
|
1
|
1
|
1
|
1
|