Hardware Reference
In-Depth Information
C
i
Major
Tasks
C
i+1
ε
ε
ε
ε
Associated
Tasks
ε
ε
ε
Figure 9.9
Task sequence generated by the adversary.
Note that the sequence of tasks generated by the adversary is constructed in such a
way that the player can schedule at most one task within its deadline (either a major
task or an associated task). Clearly, since task values are equal to their computation
times, the player never abandons a major task for an associated task because it would
accumulate a negligible value; that is,
. On the other hand, the values of the major
tasks (that is, their computation times) are chosen by the adversary to minimize the
resulting competitive factor. To find the worst-case sequence of values for the major
tasks, let
J
0
,J
1
,J
2
,...,J
i
,...,J
m
be the longest sequence of major tasks that can be generated by the adversary and,
without loss of generality, assume that the first task has a computation time equal to
C
0
=1. Now, consider the following three cases.
Case 0
. If the player decides to schedule
J
0
, the sequence terminates with
J
1
. In this
case, the cumulative value gained by the player is
C
0
, whereas the one obtained by the
adversary is (
C
0
+
C
1
−
). Note that this value can be accumulated by the adversary
either by executing all the associated tasks, or by executing
J
0
and all associated tasks
started after the release of
J
1
. Being
arbitrarily small, it can be neglected in the
cumulative value. Hence, the ratio among the two cumulative values is
C
0
C
0
+
C
1
1
1+
C
1
1
k
.
ϕ
0
=
=
=
If 1
/k
is the value of this ratio (
k>
0), then
C
1
=
k
−
1.
Case 1
. If the player decides to schedule
J
1
, the sequence terminates with
J
2
. In this
case, the cumulative value gained by the player is
C
1
, whereas the one obtained by the
adversary is (
C
0
+
C
1
+
C
2
). Hence, the ratio among the two cumulative values is
C
1
C
0
+
C
1
+
C
2
1
k
+
C
2
k
−
ϕ
1
=
=
.
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