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concern the semantics and operation of the commands. Also, PrK
=
=
“{Basic,
Pascal}” indicates that Kostas knows the programming languages 'Basic' and
'Pascal', already.
He is examining in C
6.5
: “processing arrays per row” and is succeeding 86 %.
So, the quintet, which describes Kostas' knowledge level on C
6.5
, is (0, 0, 0, 0.8,
0.2). However, according to the “strength of impact” of the knowledge dependen-
C
6.5
affects 99 % the concept C
6.4
and 77 % the concept C
6.6
. According to the
fuzzy rules (Figs.
3.6
and
3.7
) the following occur (Table
3.4
, column 'after'):
• According to R5 and R6 (h) are
μ
L
(C
6.4
)
=
0.79 and
μ
A
(C
6.4
)
=
0.5. However,
due to limitation L1 is
μ
L
(C
6.4
)
=
0.5. So, the quintet for C
6.4
is (0, 0, 0, 0.5,
0.5).
• According to R1 and R2 (h) are
μ
L
(C
6.6
)
=
0.62,
μ
A
(C
6.4
)
=
0.17 and it
remains 11 % 'Known' (
μ
Kn
(C
6.6
)
=
0.11). So, the quintet for C
6.6
is (0, 0,
0.11, 0.62, 0.17).
Consequently, Kostas remains to the same knowledge level (KL
=
6), but the system
infers that he does not need to read the domain concepts C
6.4.
Example 3
Stella's current student model has the following values: KL
=
3,
ErrTyp
=
“prone to logical errors”, PrK
=
“none”. The value KL
=
3 comes off
her current overlay model (Table
3.5
, column 'before'). ErrTyp is “prone to logical
errors” due to the fact that she had made usually errors that concern the semantics
and operation of the commands. PrK
=
“none” indicates that Stella does not have
previous knowledge on computer programming.
She is examining in C
4.2
: “calculating sum in a 'for' loop” and is succeeding 95 %.
So, the quintet, which describes Stella's knowledge level on C
4.2
, is (0, 0, 0, 0, 1).
However, according to the “strength of impact” of the knowledge dependencies that
45 % the concept C
4.3
, 81 % the concept C
4.4
, 100 % the concept C
5.2
, 45 % the con-
cept C
5.3
, and 39 % the concept C
5.4
.
According to the fuzzy rules (Figs.
3.6
and
3.7
) the following occur (Table
3.5
,
column 'after'):
• According to R2 (j) is are
μ
A
(C
4.3
)
=
0.45. However, the current value
of
μ
A
(C
4.3
) is 0.75. Therefore, according to R1
μ
A
(C
4.3
)
=
0.75 and
μ
L
(C
4.3
)
=
0.25 So, the quintet for C
4.3
is (0, 0, 0, 0.25, 0.75).
• According to R2 (j) IS
μ
A
(C
4.4
)
=
0.81 and it remains 19 % 'Learned'
(
μ
L
(C
4.4
)
=
0.19). So, the quintet for C
4.4
is (0, 0, 0, 0.19, 0.81).
• According to R2 (j) is
μ
A
(C
5.2
)
=
1. So, the quintet for C
5.2
is (0, 0, 0, 0, 1).
• According to R2 (j)
μ
A
(C
5.3
)
=
0.45 and it remains 35 % 'Learned'
(
μ
L
(C
5.3
)
=
0.35) So, the quintet for C
5.3
is (0, 0, 0, 0.35, 0.45).
• According to R2 (j)
μ
A
(C
5.4
)
=
0.39 and it remains 61 % 'Learned'
(
μ
Un
(C
5.4
)
=
0.61). So, the quintet for C
5.4
is (0, 0, 0, 0.61, 0.39).
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