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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