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Outer approximation for quadratic facility location: solving different problem types in sequence Description A quadratic facility location problem is formulated and solved as an MIQP problem (quadfacloc.mos) and via an outer approximation algorithm that iterates over MIP and QP subproblems (quadfaclocoa.mos).
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quadfacloc.mos (!********************************************************************* Mosel NL examples ================= file quadfacloc.mos ``````````````````` Quadratic Facility Location model Model of quadratic facility location problem. The goal is to assign facilities to customers to minimize cost. Basic version (formulation as MIQP) (c) 2018 Fair Isaac Corporation author: Z.Csizmadia, Sven Leyffer, Aug. 2018 *********************************************************************!) model QuadFacLoc uses "mmxprs", "mmnl", "mmsystem" declarations Iter: integer end-declarations declarations I = 1..5 ! Set of facilities J = 1..10 ! Set of customers FacX,FacY: array(I) of real ! X/Y coordinates of facility location CustX,CustY: array(J) of real ! X/Y coordinates of customer location C: array(I) of real ! Fixed cost of opening facility at location Q: array(I,J) of real ! Quadr. cost of meeting demand j from facility i end-declarations ! **** Generate random data: ! **** Cost coefficients and coordinates of all facility + customer locations setrandseed(42) SCALE:=100 forall(i in I) do C(i) := SCALE*100*random FacX(i) := SCALE*random FacY(i) := SCALE*random end-do forall(j in J) do CustX(j) := SCALE*random CustY(j) := SCALE*random end-do forall(i in I, j in J) ! Formula used by Lee to create instances Q(i,j) := 50 * sqrt( (FacX(i) - CustX(j))*(FacX(i) - CustX(j)) + (FacY(i) - CustY(j))*(FacY(i) - CustY(j))) ! **** Optimization model formulation (MIQP) **** declarations x: array(I,J) of mpvar ! Percentage of customer j served by facility i z: array(I) of mpvar ! =1, iff facility i built QuadCost: nlctr BasicNodes: integer end-declarations ! Quadratic objective function QuadCost:= sum(i in I) C(i)*z(i) + sum(i in I, j in J) Q(i,j)*x(i,j)^2; ! Variable upper bound (only serve customers from open facilities) forall(i in I, j in J) x(i,j) <= z(i) ! Meet all customers' demands forall(j in J) sum(i in I) x(i,j) = 1 forall(i in I) z(i) is_binary setparam("xprs_verbose",true) starttime:=gettime minimize(QuadCost) BasicNodes:= getparam("XPRS_NODES") writeln(gettime-starttime, "sec Solution: ", getobjval, ". Basic formulation - nodes: ", BasicNodes) ! Display the solution write("Open Customer locations\nfacility") forall(j in J) write(formattext("%6d",j)) forall(i in I | z(i).sol>0) do write("\n ", textfmt(i,-8)) forall(j in J) write(formattext("%5.1f%%",x(i,j).sol*100)) end-do writeln end-model | |||||||||||||

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