Exercise 10.12
Exercise 10.12 Proof that
for any k, there exists a metric space (e.g., the line) in
which 2 is the lower bound for the (2,k) server
problem, where 2 is the total number of servers for
offline algorithm(ADV), and k is the total number of
servers for online algorithm (ALG).
Theorem 10.1 Let M be any metric space with at least k+1
points. For any 1<=h<=k , k/(k-h+1) is a lower bound on the
competitive ratio of any online (h, k) server algorithm for M.
Main idea: for any k<=K, there exits a metric space in which 2 is
the lower bound for any (2,k) server problem.
1.
2.
3.
Prove the bases case (k<=K=3) hold (k=2 and k=3).
Suppose that k<=K hold.
Prove that k=K+1 hold
Base Case k<=K and K=3
Base: prove k=2 and k=3, the results hold.
For k=2,
From theorem 10.1, let h =2 and k=2, then we have
the competitive ratio 2 for any metric space.
For k =3,
We consider h(2,3) in the line space illustrated in
Fig.1 and construct a request sequence that
consists of arbitrary number of phases such that
the competitive ratio for h(2,3) in a line space is
close 2.

WEST
EAST
d
D >> d
Fig.1 Initial Configuration
d
For K=3
WEST
EAST
d
D >> d
d
Fig.1 Initial Configuration


For this configuration, we place a sequence of
requests at the EAST side. ALG eventually will
move all three servers to the EAST side. If it is
not move all three servers to the EAST side, this
configuration is same as the h(2,2)
configuration. Therefore it has a lower bound 2.
Each phase starts with a configuration similar
with the initial configuration or the symmetric
configuration.
Notations
WEST
EAST
d
D >> d
d
Fig.1 Initial Configuration
Now, we define following notations.
 the cost e1 is the total cost incurred by ALG for
servicing requests within the EAST side before it
moves its second server to the EAST side,
 and e2 is the total cost incurred before it moves
its third server to the EAST side.
 The cost of moving servers across the long
distance D is not included either e1 or e2
Phase construction

If e1 <= D, then ADV keeps one server at each
front. ADV pays e1, ALG pays D+e1. ADV
continues place requests at EAST side until ALG
pays at least twice as much as ADV. Then the
phase is end. The end configuration is
symmetric to the initial one, so next phase can
start it.
WEST
EAST
d
D >> d
Fig. 2: ALG moves its second server to EAST
d
Phase construction

If e1 > D, ADV immediately moves it
second server to the EAST side, place
requests in favor of ADV until ALG moves
its second server to the EAST side. After
that, it is a h(2,2) situation, which
guarantees the ratio of 2 until the ALG
moves its third server to the EAST side.
WEST
EAST
d
D >> d
d
Fig. 2: ALG moves its second server to EAST
WEST
EAST
d
D >> d
Fig. 4: ALG moves its third server to EAST
d
Phase construction-Cont.

ADV moves one of its server to the WEST
side right after the ALG moves its third
server to the EAST. ADV makes a request
at WEST side. The phase is end. The next
phase starts from WEST side.
WEST
EAST
d
D >> d
d
Fig. 5: final action of the phase
The total cost paid by ALG is e1+e2+3D, while the total
cost paid by ADV Is 2D+e2/2. Since e1>D, the close (bigger)
than 2.
Induction Statement

For any k<=K, there exists a metric space
(I.e.,the line) in which 2 is the lower bound
for the (2,k)-server problem.
For k = K+1




It stars with the following initial configuration.
The ALG has K servers at WEST side, and one
server at the EAST side.
The ADV has one server at each side.
The ADV places requests at the EAST side initially.
Eventually, ALG will move all servers at the WEST
side to the EAST side. Otherwise, ADV moves its
server to the EAST, it is a h(2,k<=K) situation,
which guarantees the ratio of 2 according to our
induction.
WEST
EAST
D >> d
Fig.6 Initial Configuration
Notation
Let e2 be the total cost paid by ALG before
it moves its last servers in WEST side to
the EAST side.
 The costs of moving servers across the
long distance D are not included in e2.

Phase Construction

ADV immediately moves its second server
to the EAST side and places requests at
EAST side. before the ALG moves its last
server to the EAST side, the situation is
h(2, k<=K), According to the induction, it
guarantees the lower bound of 2.
WEST
EAST
D >> d
Fig. 7: ALG moves its one server to EAST
Phase Construction

ADV moves one of servers to the WEST
side and places a request there right after
the ALG moves its last server to the EAST
side. The next phase starts from the
WEST side.
WEST
EAST
D >> d
d
Fig. 8: ALG moves another third server to EAST
WEST
EAST
D >> d
d
Fig. 9: final action of the phase
The total cost paid by ADV, is at most 2D+e2/2, while the total cost paid
by ALG is e2+(K+1)D. Since K > =3, therefore, ADV pays less than half
as much as ALG does for this phase.
Questions ?