Dual Fitting Analysis of Mettu-Plaxton algorithm:
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(1) Dual LP. 

Recall that the dual LP is the following:

maximize Sum_i  v_i 
s.t.

v_i - w_{ij} <= c_{ij} for all i,j
sum_j w_{ij} <= f_j for all j

(2) Dual solution.

We can obtain a dual solution from the Mettu-Plaxton algorithm as
follows.

For each client i, facility j, set w_{ij} = r_j - c_{ij}, if i in B_j,
and 0 otherwise.

For each client i, set v_i = min_j (w_{ij} + c_{ij})

(3) Feasibility.

-- For each facility j, Sum_{i} w_{ij} = Sum_{i in B_j} r_j - c_{ij} =
f_j

-- For each client i, facility k, v_i = min_j (w_{ij} + c_{ij}) <=
w_{ik} + c_{ik}

(4) Cost bound

Let F denote the set of facilities selected by the Mettu-Plaxton
algorithm.  The cost of solution F can be written as the sum of two
terms:

(a) Sum, over all facilities j in F, of the sum over all clients i in
B_j, of w_{ij} + c_{ij}

(b) Sum, over clients i that do not belong to any B_j, j in F, over
c_{i s(i)}, where s(i) is the facility in F nearest to i.

Note that the clients that contribute to (a) are disjoint from the
clients that contribute to (b).

Consider any client i.  Let k be the facility such that v_i = w_{ik} +
c_{ik}.  We will show that the client's contribution to the cost of F
is at most 3v_i.

We will prove this by considering cases.  First, by the construction
of our solution, there exists p in F such that r_p <= r_k and c_{pk}
<= 2r_k.  We thus have

(*) c_{ip} <= c_{ik} + c_{kl} <= c_{ik} + 2r_k 

Case A: There exists j in F such that i is in B_j, then r_j <= 3v_i.

Case A.1: If j = p, then r_j = r_p <= r_k <= v_i.

Otherwise, we have from (*)

    r_j <= c_{ip} <= c_{ik} + 2r_k 
 
Case A.2: If i is in B_k, then
   
    r_j <= c_{ik} + 2r_k <= 3r_k = 3v_i

Case A.3: If i is not in B_k:

    r_j <= c_{ik} + 2r_k <= 3c_{ik} = 3v_i

Case B: There does not exist j in F such that i is in B_j.  So in this
case, i's contribution to the cost of F is c_{i s(i)}. 

Case B.1: If i is in B_k, then using (*)

    c_{i s(i)} <= c_{ip} <= c_{ik} + 2r_k <= 3r_k = 3v_i

Case B.2: If i is not in B_k, then using (*)

    c_{i s(i)} <= c_{ip} <= c_{ik} + 2r_k <= 3c_{ik} = 3v_i