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B PM-symbol : change of processing time of task 2 entity in PM at the beginning
and end of practice. At the beginning of practice in single task 2, RT of eight
alternative choice reaction time (Hick's law: 50 ms, slope: 170 ms/bit) is com-
posed of one average perception cycle (100 ms), ( A PM-symbol +
B PM-symbol ),
one motor cycle (70 ms). Thus, B PM-symbol =
+
×
(
)
150
170
Log2
8
100
443 ms.
α PM-symbol : learning rate of PM in processing the task 2 entity. The speed of
formation of the automatic process in PM is slower than Hicog because it receives
the entities from CE server via the indirect parallel learning mechanism with
the four incompatible alternatives [33]. Thus,
A PM-symbol
70
=
α PM-symbol =(
0
.
001
/
4
) /
4
=
1
/
16
,
000.
Appendix 2
Lemma 9.1. At any transition state t ( t
=
0 ), if 1
/ μ 2 , t ,
t
<
1
/ μ 3 , t , then Q t + 1 (
1
,
2
) >
Q t + 1 (
)
Proof . Using mathematic deduction method
(i) At t
1
,
3
0: Q 1
Q 1
Q 1
Q 1
=
(
1
,
3
)=
(
1
,
2
)=
(
2
,
4
)=
(
3
,
4
)=
0.
1: Using the online Q learning formula: Q 2
Q 1
(ii)At t
=
(
1
,
3
)=
(
1
,
3
)+ ε [
r t +
Q 1
Q 1
γ
(
3
,
4
)
(
1
,
3
)] = εμ 3 , 1 .
Note: because entity routes to only one server (server 4) max b Q t
(
S t +
1
,
b
)=
Q 2
Q 2
Q 2
Q
(
3
,
4
) ,
(
1
,
2
)= εμ
,
(
3
,
4
)= εμ
,
(
2
,
4
)= εμ
4 ;If1
/ μ
<
1
/ μ
1 then
2
,
1
4
2
,
1
3
,
1), i.e., Q 2
Q 2
< εμ
< ε <
(
,
) >
(
,
)
εμ
1 (given 0
1
2
1
3
. Thus, lemma is proved
3
,
1
2
,
=
1.
iii According to mathematic deduction method, Lemma 9.1 is correct: i.e., at tran-
sition state t
at t
k then Q k + 1
Q k + 1
=
k :if1
/ μ
<
1
/ μ
(
1
,
2
) >
(
1
,
3
)
. We want
2
,
k
3
,
to prove at transition state k
+
1, lemma is still correct: i.e., At transition state
t
=
k
+
1:
/ μ 3 , k + 1 , then Q k + 2
Q k + 2
1: Q k + 2
if 1
/ μ 2 , k + 1 <
1
(
1
,
2
) >
(
1
,
3
)
At t
=
k
+
(
1
,
2
)=
Q k + 1
Q k + 1
(
1
,
2
)+ ε [ μ 2 , k + 1 + γεμ 4
(
1
,
2
)]
Q k + 2
Q k + 1
Q k + 1
(
1
,
3
)=
(
1
,
3
)+ ε [ μ
+ γεμ
(
1
,
3
)] ,
(9.9)
, k +
4
3
1
Q k + 2
Q k + 2
(
1
,
2
)
(
1
,
3
)
Q k + 1
Q k + 1
=
(
1
,
2
)+ ε [ μ 2 , k + 1 + γεμ 4
(
1
,
2
)]
Q k + 1
Q k + 1
(
,
)+ ε [ μ 3 , k + 1 + γεμ
(
,
)]
1
3
1
3
4
Q k + 1
Q k + 1
=(
1
ε )[
(
1
,
2
)
(
1
,
3
)]+( εμ 2 , k + 1 εμ 3 , k + 1 )
(9.10)
With Equation (9.3) and 0
< ε <
1, we have
Q k + 1
Q k + 1
(
1
ε )[
(
1
,
2
)
(
1
,
3
)] >
0
.
(9.11)
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