1. 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.
Prove the bases case (k<=K=3) hold (k=2 and k=3).
Suppose that k<=K hold.
Prove that k=K+1 hold

2. 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.
D >> d d
Fig.1 Initial Configuration
WEST
EAST

3. For K=3
D >> d d
Fig.1 Initial Configuration
WEST
EAST
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.

4. Notations Now, we define following notations.
D >> d d
Fig.1 Initial Configuration
WEST
EAST
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

5. 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.
D >> d d
Fig. 2: ALG moves its second server to EAST
EAST
WEST

6. 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.
D >> d d
Fig. 2: ALG moves its second server to EAST
EAST
WEST
D >> d d
Fig. 4: ALG moves its third server to EAST
EAST
WEST

7. 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.
D >> d d
Fig. 5: final action of the phase
EAST
WEST
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.

8. 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.

9. 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.
D >> d
Fig.6 Initial Configuration
WEST
EAST

10. 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.

11. 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.
D >> d
Fig. 7: ALG moves its one server to EAST
EAST
WEST

12. 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.
D >> d d
Fig. 9: final action of the phase
EAST
WEST
Fig. 8: ALG moves another third server to EAST
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.

13. Questions ?