Hardware Reference
In-Depth Information
PROOF OF THE BOUND
Whenever the player chooses to schedule a task J i , the sequence stops with J i +1
and
the ratio of the cumulative values is
C i
j =0 C j + C i +1
1
k .
ϕ i =
=
However, if the player chooses to schedule the last task J m , the ratio of the cumulative
values is
C m
j =0 C j
ϕ m =
.
Note that if k and m can be chosen such that ϕ m
1 /k ; that is,
C m
j =0 C j
1
k ,
(9.4)
then we can conclude that, in the worst case, a player cannot achieve a cumulative
value greater than 1 /k times the adversary's value. Note that
C m
j =0 C j
C m
m− 1
j =0
C m
C m
kC m− 1
=
=
=
.
m− 1
j =0
C j + kC m− 1 m− 1
C j + C m
C j
j =0
Hence, if there exists an m that satisfies Equation (9.4), it also satisfies the following
equation:
C m
C m− 1 .
(9.5)
Thus, (9.5) is satisfied if and only if (9.4) is satisfied.
From (9.4) we can also write
i +1
C i +2
=
kC i +1
C j
j =0
i
C i +1
=
kC i
C j ,
j =0
and subtracting the second equation from the first one, we obtain
C i +2
C i +1
= k ( C i +1
C i )
C i +1 ;
that is,
C i +2
= k ( C i +1
C i ) .
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