3.8 Analysis of the Polynomial-Time Algorithm
3.8.3 Bounding the cost of clients
In this section we derive inequalities that are used to bound the service cost of each (α, z) produced during the algorithm. Consider some solution α produced by the algorithm,
and define
B = {j ∈ D : j is undecided and 2 ¯αj < d(j, j0) + 6 ¯α(0)j0 for all clients j0} . (3.8.1)
The set B is defined to contain those clients that are (potentially) bad, i.e., have worse connection cost than our target guarantee. Specifically, we now show that all clients
j ∈ D \ B, satisfy the first inequality of Property 2 in Definition 3.5.1 (with τi replaced
by ti), while all clients (in particular those in B) satisfy a slightly weaker inequality.
Lemma 3.8.7. Consider any (a, z) produced by RaisePrice. For every client j the
following holds:
• If j ∈ D\B, then there exists a tight facility i such that (1+√δ+) ¯αj ≥ d(j, i)+
√
δti.
• There exists a tight facility i such that 6 ¯α(0)j ≥ d(j, i) +√δti.
Proof. The proof is by induction on the well-ordered set (with respect to the natural
order ≤)
R = {0} ∪ {αj}j∈D\B∪ {(1 + )α (0) j }j∈D.
Specifically, we prove the following induction hypothesis: for r ∈ R,
(a) each client j ∈ D \ B with αj ≤ r has a tight facility i such that (1 +
√
δ + ) ¯αj ≥
d(j, i) +√δti;
(b) each client j ∈ D with (1 + )α(0)j ≤ r has a tight facility i such that 6 ¯α(0)j ≥
d(j, i) +√δti.
The statement then follows from the above with r = arg maxr∈Rr.
For the base case (when r = 0), the claim is vacuous since there is no client j such that
αj ≤ 0 or (1 + )α(0)j ≤ 0 (because every α-value is at least 1 by Invariant 2). For the
induction step, we assume that each client j ∈ D \ B with αj < r satisfies (a) and each
client j ∈ D with (1 + )α(0)j < r satisfies (b). We need to prove that any client j0∈ D \ B
with αj0 = r (respectively, j0∈ D with (1 + )α(0)j
0 = r) satisfies (a) (respectively, (b)). We divide the proof into two cases.
Case 1: j0 ∈ D \ B with αj0 = r. We prove that in this case j0 satisfies (a). Since
j0 6∈ B, either j0 has a witness, j0 is currently stopped, or there is another client j such
that 2 ¯αj0 ≥ d(j0, j) + 6 ¯α
(0) j .
Suppose first that j0 has a witness i. Then, i is a tight facility and, since j0 has a tight edge to i, d(j0, i) ≤ ¯αj0. Moreover, B(αj0) ≥ B(ti) which implies that (1 +
2) ¯αj0 ≥
p
(1 + ) ¯αj0 ≥ √
ti. Therefore, using that
√ δ ≤ 2, d(j0, i) + p δti ≤ (1 + √ δ + ) ¯αj0.
Now suppose that j0 is stopped by another client j. Then αj ≤ αj0/32 = r/9. On the one hand, if j ∈ D \ B, we have d(j0, i) +
√
δti ≤ (1 +
√
δ + ) ¯αj ≤ 6 ¯αj for some tight
facility i by the induction hypothesis (a). On the other hand, if j ∈ B then j is undecided so by Lemma 3.8.4, α(0)j ≤ αj. This in turn implies that α(0)j ≤ αj ≤ r/9 < r/(1 + ). We can thus apply the induction hypothesis (b) to j, to conclude that there is a tight
facility i such that d(j, i) +√δti≤ 6 ¯α(0)j ≤ 6 ¯αj. From above we have that, whether j is
in B or not, there is a tight facility i such that
d(j0, i) + p δti ≤ d(j0, j) + d(j, i) + p δti ≤ d(j0, j) + 6 ¯αj ≤ 2 ¯αj0 ≤ (1 + √ δ + ) ¯αj0,
where the penultimate inequality uses the fact that j0 is stopped by j and thus 2 ¯αj0 ≥
d(j, j0) + 6 ¯αj.
Finally, suppose that j0 is not stopped or witnessed. Then, j0 is currently undecided
and, as j0 6∈ B, there is a client j such that 2 ¯αj0 ≥ d(j0, j) + 6 ¯α
(0)
j . This implies that
α(0)j ≤ αj0/9 = r/9 < r/(1 + ). We can thus apply the induction hypothesis (b) to j to
conclude, that there is a tight facility i such that d(j, i) +√δti ≤ 6 ¯α(0)j . Now, we have:
d(j0, i) + p δti ≤ d(j0, j) + d(j, i) + p δti ≤ d(j0, j) + 6 ¯α(0)j ≤ 2 ¯αj0 ≤ (1 + √ δ + ) ¯αj0.
Case 2: j0 ∈ D with (1 + )α(0)j0 = r. We prove that in this case j0 satisfies (b). Suppose first that αj0 < α
(0)
j0 . Then j0 is decided by Lemma 3.8.4. Therefore j0∈ D \ B with αj0 < r and so by the induction hypothesis (a) there is a tight facility i satisfying d(j0, i) + √ δti ≤ (1 + √ δ + ) ¯αj0 < 6 ¯α (0) j0 , as required. Similarly, if j0 ∈ U (0) then by
Corollary 3.8.6, j0 is decided and so
αj0 ≤ α
(0)
j0 + z< (1 + )α
(0) j0 = r ,
where the second inequality follows from z < and α(0)j0 ≥ 1 by Invariant 2. We can
i satisfying d(j0, i) + √ δti ≤ (1 + √ δ + ) ¯αj0 ≤ 6 ¯α (0)
j0 . Thus, from now on, we assume that αj0 ≥ α(0)j
0 and that j0 6∈ U
(0). We divide the remaining part of the analysis into
two sub-cases depending on whether j0 was stopped in α(0).
First, suppose that j0 was stopped in α(0) by another client j. Then α(0) j ≤ α
(0) j0 /9 <
r/(1 + ) and so by the induction hypothesis (b), there is a tight facility i satisfying d(j, i) +√δti≤ 6 ¯α(0)j . Hence, d(j0, i) + p δti ≤ d(j0, j) + d(j, i) + p δti ≤ d(j0, j) + 6 ¯α(0)j ≤ 2 ¯α(0)j 0 < 6 ¯α (0) j0 .
Finally, suppose that j0 was not stopped in α(0). Then since every client is decided in α(0) (Invariant 4) j0had a witness i in α(0). Moreover, as j0 6∈ U(0), we may assume that i 6= i+
and so zi = z(0)i . By the definition of a witness, α (0) j1 ≤ (1 + )α (0) j0 for all j1 ∈ N (0)(i). If αj1 ≥ α (0) j1 for all j1∈ N
(0)(i), then, since z
i= zi(0), our feasibility invariant (Invariant 2)
implies that in fact αj1 = α
(0)
j1 for all j1 ∈ N
(0)(i) and so N (i) = N(0)(i). Therefore,
in this case i is still a witness for j0 and d(j0, i) +
√
δti ≤ (1 +
√
δ + ) ¯α(0)j0 ≤ 6 ¯α(0)j0 . It remains to consider the case when αj1 < α
(0)
j1 for some j1 ∈ N
(0)(i) (note that
j1 6= j0, since αj0 ≥ α
(0)
j0 by assumption). Since αj1 < α
(0)
j1 , j1 must be decided (by Lemma 3.8.4) and so j1 ∈ D \ B. Moreover, αj1 < α(0)j
1 ≤ (1 + )α
(0)
j0 = r, and so we can apply the induction hypothesis (a) to conclude that there is a tight facility i1 satisfying
d(j1, i1) +pδti1 ≤ (1 + √ δ + ) ¯αj1 < (1 + √ δ + ) ¯α(0)j1 . Then, d(j0, i1) + q δti1 ≤ d(j0, i) + d(i, j1) + d(j1, i1) + q δti1 < ¯α(0)j 0 + ¯α (0) j1 + (1 + √ δ + ) ¯α(0)j 1 ≤ ¯α(0)j0 + (1 + ) ¯α(0)j0 + (1 + )(1 + √ δ + ) ¯α(0)j0 ≤ 6 ¯α(0)j0 , as required.
Lemma 3.8.7 shows that the clients in D \ B satisfy the first inequality of Property 2 in Definition 3.5.1 while the potentially bad clients j ∈ B satisfy a slightly weaker inequality. It remains to prove that the potentially bad clients will have a small contribution towards the total cost of our solution.