Facility Location with Service Installation Costs Chaitanya Swamy Joint work with David Shmoys and Retsef Levi Cornell University

SODA Talk, 01/2004 Facility Location F : set of facilities. D : set of clients. Facility i has facility cost f i. c ij : distance between points i and j. client facility

SODA Talk, 01/2004 Facility Location with Services F : set of facilities. D : set of clients. S : set of services={tan, green, pink}. Facility i has facility cost f i. c ij : distance between points i and j. Client j requires service s(j). f i,s : cost of installing service s at fac. i. facility client services

SODA Talk, 01/2004 2)Install a set of services S(i) on each open facility i. We want to: Cost = ∑ i  A f i + ∑ i  A, s  S(i) f i,s + ∑ j  D c i(j)j = facility opening cost + service installation cost + client assignment cost facility client services 3)Assign each client j to an open facility i(j) on which service s(j) is installed. 1)Choose a set A of facilities to open. open facility

SODA Talk, 01/2004 Related Work Only one service  uncapacitated facility location LP rounding: Shmoys, Tardos & Aardal; Chudak & Shmoys; Sviridenko. Primal-dual algorithms: Jain & Vazirani; Jain, Mahdian, Markakis, Saberi & Vazirani. Best approx. - Mahdian, Ye & Zhang: 1.52 Baev & Rajaraman consider related caching problem: No facility costs, capacity on the number of services that may be installed at a facility. Gave a approx. algorithm, improved to 10 by (S). Ravi & Sinha consider similar model – multicommodity facility location. Give a log(| S |)-approx. algorithm.

SODA Talk, 01/2004 Our Results Give a 6-approx. primal-dual algorithm. Assume facilities can be ordered s.t. if i ≤ i’, then f i,s ≤ f i’,s for all services s. Special cases:  f i,s = h s – depends only on s  f i,s = g i – depends only on i General problem is set-cover hard. When f i,s = h s primal-dual gives a 5- approx. Improve ratio to using LP rounding. Consider k-median variant: give a 11.6-approx. when f i,s = h s using primal-dual+Lagrangian relaxation.

SODA Talk, 01/2004 LP formulation Minimize ∑ i f i y i +∑ i,s f i,s y i,s +∑ j,i c ij x ij (Primal) subject to∑ i x ij ≥ 1  j x ij ≤ y i,s(j)  i, j x ij ≤ y i  i, j x ij, y i ≥ 0  i, j y i : indicates if facility i is open. y i,s :indicates if service s is installed on facility i. x ij : indicates if client j is assigned to facility i. s(j):service required by j.

SODA Talk, 01/2004 ∑ i,j  D(s) z ij ≤ f i,s  i, s The Dual D(s) = clients requiring service s ∑ i,j w ij ≤ f i  i c ij z ij w ij v j, w ij, z ij ≥ 0  i, j Maximize ∑ j v j subject to v j ≤ c ij + w ij + z ij  i, j vjvj j

SODA Talk, 01/2004 Algorithm Outline Based on the primal-dual method. Any feasible dual solution is a lower bound on OPT - Weak Duality. 1)Construct primal solution and dual solution simultaneously. 2)Bound cost of primal by c ● (dual value)  get a c-approx. Algorithm strongly motivated by the Jain-Vazirani algorithm. Notion of time, t. Initially, t=0, all dual variables v j, z ij, w ij are 0. No facility is open, no service is installed.

SODA Talk, 01/2004 f d,s = 2 for all services s. f c,green = 2. t=0 unfrozen client not open facility a c d b

SODA Talk, 01/2004 Raise all v j at rate 1. t=0 unfrozen client not open facility a c d b

SODA Talk, 01/2004 t=1 unfrozen client not open facility f d,s = 2 for all services s. f c,green = 2.

SODA Talk, 01/2004 t=2 unfrozen client not open facility f d,s = 2 for all services s. f c,green = 2. If v j ≥ c ij say that j is tight with i. j becomes tight with i and s(j) is not installed on i: start increasing z ij.

SODA Talk, 01/2004 t=3 unfrozen client not open facility f d,s = 2 for all services s. f c,green = 2. For some i, service s, ∑ i,j  D(s) z ij = f i,s : tentatively install s on i, start raising w ij for all j in D(s) tight with i.

SODA Talk, 01/2004 unfrozen client not open facility t=4 unfrozen client not open facility tentatively open facility frozen client f d,s = 2 for all services s. f c,green = 2. For some i, ∑ i,j w ij = f i : tentatively open i, if j is tight with i and s(j) is tentatively installed at i, freeze j.

SODA Talk, 01/2004 t=5 unfrozen client not open facility tentatively open facility frozen client f d,s = 2 for all services s. f c,green = 2. When all demands are frozen, process stops.

SODA Talk, 01/2004 The Primal-Dual process Keep raising all v j at rate 1 until: 1)j reaches facility i: if s(j) is not tentatively installed at i, increase z ij ; else, if i is not tentatively open, raise w ij ; otherwise freeze j. 2)For some i and s, ∑ i,j  D(s) z ij = f i,s : tentatively install s on i. If i is not tentatively open raise w ij for all j in D(s) tight with i, else freeze j. 3)For some i, ∑ i,j w ij = f i : tentatively open i. If j is tight with i and s(j) is tentatively installed at i, freeze j. Now only raise v j of unfrozen demands. Continue until all demands are frozen.

SODA Talk, 01/2004 Opening facilities a c d b X X Say i, i’ are dependent if there is client j s.t. w ij, w i’j > 0. Suppose O is a ≤ b ≤ c ≤ d. Consider tentatively open facilities in this order, pick a maximal independent subset. Open all these facilities – call this set F’. Let O : ordering on facilities. i tentatively open, not opened  there is i’ s.t. i’ ≤ i and i, i’ are dependent. Denote i’ = nbr(i).

SODA Talk, 01/2004 Installing services X X F’ = facilities opened.

SODA Talk, 01/2004 X X Installing services Consider service s. Let F s = tentatively open facilities where service s is tentatively installed. Say i, i’ are s-dependent if there is a client j  D(s) s.t. z ij, z i’j > 0. Carefully pick a maximal independent set F’ s  F s. F’ = facilities opened.

SODA Talk, 01/2004 X X Installing services For each i in F’ s, if i  F’ install service s on i, otherwise install s on nbr(i). F’ = facilities opened. Consider service s. Let F s = tentatively open facilities where service s is tentatively installed. Say i, i’ are s-dependent if there is a client j  D(s) s.t. z ij, z i’j > 0. Carefully pick a maximal independent set F’ s  F s.

SODA Talk, 01/2004 Installing services For each i in F’ s, if i  F’ install service s on i, otherwise install s on nbr(i). F’ = facilities opened. Consider service s. Let F s = tentatively open facilities where service s is tentatively installed. Say i, i’ are s-dependent if there is a client j  D(s) s.t. z ij, z i’j > 0. Carefully pick a maximal independent set F’ s  F s. X X

SODA Talk, 01/2004 Installing services For each i in F’ s, if i  F’ install service s on i, otherwise install s on nbr(i). F’ = facilities opened. Consider service s. Let F s = tentatively open facilities where service s is tentatively installed. Say i, i’ are s-dependent if there is a client j  D(s) s.t. z ij, z i’j > 0. Carefully pick a maximal independent set F’ s  F s. X X

SODA Talk, 01/2004 Installing services F’ = facilities opened. For each i in F’ s, if i  F’ install service s on i, otherwise install s on nbr(i). Consider service s. Let F s = tentatively open facilities where service s is tentatively installed. Say i, i’ are s-dependent if there is a client j  D(s) s.t. z ij, z i’j > 0. Carefully pick a maximal independent set F’ s  F s. X X

SODA Talk, 01/2004 X X Installing services For each i in F’ s, if i  F’ install service s on i, otherwise install s on nbr(i). F’ = facilities opened. Consider service s. Let F s = tentatively open facilities where service s is tentatively installed. Say i, i’ are s-dependent if there is a client j  D(s) s.t. z ij, z i’j > 0. Carefully pick a maximal independent set F’ s  F s.

SODA Talk, 01/2004 X X Installing services For each i in F’ s, if i  F’ install service s on i, otherwise install s on nbr(i). F’ = facilities opened. Consider service s. Let F s = tentatively open facilities where service s is tentatively installed. Say i, i’ are s-dependent if there is a client j  D(s) s.t. z ij, z i’j > 0. Carefully pick a maximal independent set F’ s  F s.

SODA Talk, 01/2004 Finally, Client j is assigned to the nearest open facility on which service s(j) is installed.

SODA Talk, 01/2004 Analysis sketch Let D’ = {j:  i in F’ s.t. w ij > 0}. Facility opening cost ≤ ∑ j  D’ v j. Service installation cost ≤ ∑ j v j. Can show that the assignment cost of j is at most 3v j if j  D’ and at most 5v j otherwise.

SODA Talk, 01/2004 Analysis sketch Let D’ = {j:  i in F’ s.t. w ij > 0}. Facility opening cost ≤ ∑ j  D’ v j. Service installation cost ≤ ∑ j v j. Can show that the assignment cost of j is at most 3v j if j  D’ and at most 5v j otherwise. Theorem: Total cost ≤ 6 ● ∑ j v j ≤ 6 ● OPT.

SODA Talk, 01/2004 Thank You.