Facility Location with Service Installation Costs Chaitanya Swamy

Facility Location with Service Installation Costs Chaitanya Swamy Joint work with David Shmoys and Retsef Levi Cornell University SODA Talk, 01/2004

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

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

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

Related Work Only one service º 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 20. 5 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 fi, s ≤ fi’, s for all services s. Special cases: § fi, s = hs – depends only on s § fi, s = gi – depends only on i General problem is set-cover hard. • When fi, s = hs primal-dual gives a 5 - approx. Improve ratio to 2. 391 using LP rounding. • Consider k-median variant: give a 11. 6 -approx. when fi, s = hs using primal-dual+Lagrangian relaxation. SODA Talk, 01/2004

LP formulation Minimize ∑i fiyi+∑i, s fi, syi, s+∑j, i cijxij (Primal) subject to ∑i xij ≥ 1 yi j xij ≤ yi, s(j) i, j xij ≤ yi i, j xij, yi ≥ 0 i, j : indicates if facility i is open. yi, s : indicates if service s is installed on facility i. xij : indicates if client j is assigned to facility i. s(j) : service required by j. SODA Talk, 01/2004

The Dual vj wij zij cij j D(s) = clients requiring service s Maximize ∑j vj subject to vj ≤ cij + wij + zij ∑i, jÎD(s) zij ≤ fi, s i, j i, s ∑i, j wij ≤ fi i vj, wij, zij ≥ 0 i, 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) Þ get a c-approx. Algorithm strongly motivated by the Jain-Vazirani algorithm. Notion of time, t. Initially, t=0, all dual variables vj, zij, wij are 0. No facility is open, no service is installed. SODA Talk, 01/2004

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

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

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

If vj ≥ cij say that j is tight with i. fd, s = 2 for all services s. j becomes tight with i and s(j) is not fc, green = 2. installed on i: start increasing zij. t=2 not open facility unfrozen client SODA Talk, 01/2004

For some i, service s, ∑i, jÎD(s) zij = fi, s: ftentatively = 2 for all services s. d, s install s on i, start raising fwc, green = 2. j in D(s) tight with i. for all ij t=3 not open facility unfrozen client SODA Talk, 01/2004

For some i, ∑i, j wij = fi : tentatively fd, s = 2 for all services s. open i, if j is tight with i and s(j) is fc, green = 2. tentatively installed at i, freeze j. t=4 not open facility tentatively open facility unfrozen client SODA Talk, 01/2004

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

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

Opening facilities X a d Xc b Say i, i’ are dependent if there is client j s. t. wij, wi’j > 0. Let O: ordering on facilities. 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’. i tentatively open, not opened Þ there is i’ s. t. i’ ≤ i and i, i’ are dependent. Denote i’ = nbr(i). SODA Talk, 01/2004

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

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

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

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

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

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

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

Installing services X F’ = facilities opened. X Consider service s. Let Fs = tentatively open facilities where service s is tentatively installed. Say i, i’ are s-dependent if there is a client jÎD(s) s. t. zij, zi’j > 0. Carefully pick a maximal independent set F’s Í Fs. For each i in F’s, if iÎF’ install service s on i, otherwise install s on nbr(i). 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. wij > 0}. Facility opening cost ≤ ∑jÎD’ vj. Service installation cost ≤ ∑j vj. Can show that the assignment cost of j is at most 3 vj if jÎD’ and at most 5 vj otherwise. SODA Talk, 01/2004

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

Thank You. SODA Talk, 01/2004