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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