## Posted By

mertnuhoglu on 11/14/07

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# Solving p-median location allocation problem with Lagrangian Relaxation heuristics

/ Published in: MatLab   --- solve_problems.m function report = solve_problems()% data_files={'Alberta';'Galvao100';'Galvao100';'Galvao150';'Galvao150';'Galvao150'};% p=[10 10 15 5 15 20];% data_files={'TestData'};% p=;data_files={'Alberta'};p=;for i=1:length(data_files)    data_file=char(data_files(i));    dist=distances([data_file '_distances.txt']);    demand=load([data_file '_demands.txt']);    [bestLB,iterations,debug]=solve_p_median(dist,demand,p(i));    report(i).bestLB=bestLB;    report(i).iterations=iterations;    report(i).debug=debug;    dlmwrite([data_file '_p' int2str(p(i)) '_iterations.txt'],iterations);    dlmwrite([data_file '_p' int2str(p(i)) '_debug.txt'],debug);end --- distances.m function distances=distances(data_file)distance_data=load(data_file);distances=zeros(distance_data(end,1)+1);for row=1:length(distance_data)    from=distance_data(row,1);    to=distance_data(row,2);    distance=distance_data(row,3);    distances(to,from)=distance;    distances(from,to)=distance;end --- solve_p_median.m function [bestLB,iterations,debug]=solve_p_median(dist,demand,p)bestLB=0;bestUB=inf;currentLB=0;currentUB=inf;iterations(1,:)=[0 currentLB bestLB currentUB bestUB];pi=2;n_c=length(demand); % number of customersn_s=length(dist(:,1));  % number of sitesu=ones(n_c,1); % LR (Lagrangean Relaxation) multiplierszero=zeros(n_c,1);x=zeros(n_s,n_c); % site/customer assignmentsi=1;piUpdateTime=1;improvementOccurred=0;debug=[0 0 2 zeros(1,p) u']; % parameters stored for debugging (iteration, step_size, pi, open facilities, u)while(~stoppingCondition(pi,bestLB,bestUB,i))    for s=1:n_s        cost=dist(:,s).*demand;        newCost=cost-u;        z_LR(s)=sum(min(zero,newCost));        x(s,find(newCost<0))=1; % x_{s,c} = 1 if cost negative    end    [z_sorted,order]=sort(z_LR);    currentLB=sum(z_sorted(1:p))+sum(u); % z_{LR}(u) value is current LB    facilities=order(1:p); % open p facilities where z is smallest    x(setdiff(1:n_s,facilities),:)=0; % x_{s,c} \leq y_s \forall s,c constraint    if(currentLB>bestLB)        bestLB=currentLB;    end    currentUB=findUB(facilities,dist,demand);    if(currentUB<bestUB)        bestUB=currentUB;    end    iterations(i+1,:)=[i currentLB bestLB currentUB bestUB];    normOfRelaxedCsts=sum((1-sum(x)).^2);    if(normOfRelaxedCsts == 0) % hit the lower bound         break    end    step=pi*(bestUB-currentLB)/normOfRelaxedCsts; % s^t = {\pi (UB* - z_{LR}(u^t)) \over \sum_c (1-\sum_s x_{sc})^2}    u=u+step*(1-sum(x))'; % u_c^{t+1} = u_c^t + s^t (1-\sum_s x_{sc}^t)    if(~improvementsOccur(iterations,piUpdateTime))        pi=pi/2;        piUpdateTime=i;    end    debug(i+1,:)=[i step pi facilities u'];    i=i+1end function result=stoppingCondition(pi,bestLB,bestUB,iterationNo)result = (pi < 0.005) | (bestUB == bestLB);% result = iterationNo > 15000 | (bestUB == bestLB); function result=improvementsOccur(iterations,piUpdateTime)n=30;currentTime=length(iterations(:,1));timeSinceLastPiUpdate=currentTime-piUpdateTime;if(currentTime <= n | timeSinceLastPiUpdate <= n)    result = 1;else    lastImpForLB=whenDidLastImprovementOccur(iterations(end-n:end,3));    lastImpForUB=whenDidLastImprovementOccur(iterations(end-n:end,5));    if(lastImpForLB <= n | lastImpForUB <= n)        result = 1;    else        result = 0;    endend function lastImp=whenDidLastImprovementOccur(iterations)lastValue=iterations(end);ixLastImp=length(find(iterations~=lastValue));lastImp=length(iterations)-ixLastImp; function currentUB=findUB(facilities,dist,demand)feasibleAssignments=assignCustomers(facilities,dist);currentUB=sum(feasibleAssignments.*dist)*demand; function customerAssignments=assignCustomers(facilities,dist)n_s=length(dist(:,1)); % number of sitesn_c=length(dist(1,:)); % number of customerscustomerAssignments=zeros(n_s,n_c);distOpenFacilities=dist(facilities,:);[minDist,order]=min(distOpenFacilities);for c=1:n_c    customerAssignments(facilities(order(c)),c)=1;end Subscribe to comments
Posted By: pastoral on November 16, 2007

Thank you for sharing! But could you tell me what the format of these ".txt" files, such as Albertadistances.txt and Albertademands.txt? Of course, an example will be better.... Thank you in advance!

Posted By: neb_sar on August 21, 2008

kardeÅŸ paylaÅŸÄ±m iÃ§in Ã§ok saÄŸol. Data dosyalarÄ±nÄ±n ne olduklarÄ±nÄ± anlatan bir not yazarsan ya da data dosyalarÄ±nÄ± da bizimle paylaÅŸÄ±rsan ya da e mailime gÃ¶nderirsen Ã§ok sevinirim. location problemi ile ilgili tez yapacaÄŸÄ±m da bu kodlar gerÃ§ekten iÅŸime yarayabilir. teÅŸekkÃ¼rler.

Posted By: mertnuhoglu on September 30, 2008

I put all the data files under http://sites.google.com/site/mertnuhoglu/programming/files/ie517-heuristics

Posted By: Michalis on November 4, 2008

Could you pls add some more comments within the code pls?

Posted By: majid_amirklabir on July 8, 2012

Hi, my friend do u have any code for a two echelon problem that u have solved it by Lagrangian relaxation, it would be grateful if u would/could send me yours. sincerely yours. [email protected]

Posted By: setienne24 on February 19, 2013

Hi, i am searching an algorithm for location allocation.

Your development seems to be what i am searching for. Could you give a little more explanations on the code. There are two files which are created and i don t really know the meaning of these two files : result and debug. do you have such explanations about these 2 files ? Thany you.

in my case, i have a city of N points with the matrix of distances also. I must place, each 8 cities, a depot. And at the end, each point of the city must be near a depot.

do your algorithm can resolve my problem?