The Mixed Integer Linear Programming Solver

Example 8.3 Facility Location

Consider the classic facility location problem. Given a set L of customer locations and a set F of candidate facility sites, you must decide on which sites to build facilities and assign coverage of customer demand to these sites so as to minimize cost. All customer demand $d_ i$ must be satisfied, and each facility has a demand capacity limit C. The total cost is the sum of the distances $c_{ij}$ between facility j and its assigned customer i, plus a fixed charge $f_ j$ for building a facility at site j. Let $y_ j = 1$ represent choosing site j to build a facility, and 0 otherwise. Also, let $x_{ij} = 1$ represent the assignment of customer i to facility j, and 0 otherwise. This model can be formulated as the following integer linear program:

\[ \begin{array}{llllll} \min & \displaystyle \sum _{i \in L} \displaystyle \sum _{j \in F} c_{ij} x_{ij} + \displaystyle \sum _{j \in F} f_ j y_ j \\ \mr{s.t.} & \displaystyle \sum _{j \in F} x_{ij} & = & 1 & \forall i \in L & \mr{(assign\_ def)} \\ & x_{ij} & \leq & y_ j & \forall i \in L, j \in F & \mr{(link)}\\ & \displaystyle \sum _{i \in L} d_ i x_{ij} & \leq & Cy_ j & \forall j \in F & \mr{(capacity)} \\ & x_{ij} \in \{ 0,1\} & & & \forall i \in L, j \in F \\ & y_{j} \in \{ 0,1\} & & & \forall j \in F \end{array} \]

Constraint (assign_def) ensures that each customer is assigned to exactly one site. Constraint (link) forces a facility to be built if any customer has been assigned to that facility. Finally, constraint (capacity) enforces the capacity limit at each site.

Consider also a variation of this same problem where there is no cost for building a facility. This problem is typically easier to solve than the original problem. For this variant, let the objective be

\[ \begin{array}{lllll} \min & \displaystyle \sum _{i \in L} \displaystyle \sum _{j \in F} c_{ij} x_{ij} \end{array} \]

First, construct a random instance of this problem by using the following DATA steps:

title 'Facility Location Problem';

%let NumCustomers  = 50;
%let NumSites      = 10;
%let SiteCapacity  = 35;
%let MaxDemand     = 10;
%let xmax          = 200;
%let ymax          = 100;
%let seed          = 938;

/* generate random customer locations */
data cdata(drop=i);
   length name $8;
   do i = 1 to &NumCustomers;
      name = compress('C'||put(i,best.));
      x = ranuni(&seed) * &xmax;
      y = ranuni(&seed) * &ymax;
      demand = ranuni(&seed) * &MaxDemand;
      output;
   end;
run;

/* generate random site locations and fixed charge */
data sdata(drop=i);
   length name $8;
   do i = 1 to &NumSites;
      name = compress('SITE'||put(i,best.));
      x = ranuni(&seed) * &xmax;
      y = ranuni(&seed) * &ymax;
      fixed_charge = 30 * (abs(&xmax/2-x) + abs(&ymax/2-y));
      output;
   end;
run;

The following PROC OPTMODEL statements first generate and solve the model with the no-fixed-charge variant of the cost function. Next, they solve the fixed-charge model. Note that the solution to the model with no fixed charge is feasible for the fixed-charge model and should provide a good starting point for the MILP solver. Use the PRIMALIN option to provide an incumbent solution (warm start).


proc optmodel;
   set <str> CUSTOMERS;
   set <str> SITES init {};
   /* x and y coordinates of CUSTOMERS and SITES */
   num x {CUSTOMERS union SITES};
   num y {CUSTOMERS union SITES};
   num demand {CUSTOMERS};
   num fixed_charge {SITES};
   /* distance from customer i to site j */
   num dist {i in CUSTOMERS, j in SITES}
       = sqrt((x[i] - x[j])^2 + (y[i] - y[j])^2);
   read data cdata into CUSTOMERS=[name] x y demand;
   read data sdata into SITES=[name] x y fixed_charge;
   var Assign {CUSTOMERS, SITES} binary;
   var Build {SITES} binary;
   min CostNoFixedCharge
       = sum {i in CUSTOMERS, j in SITES} dist[i,j] * Assign[i,j];
   min CostFixedCharge
       = CostNoFixedCharge + sum {j in SITES} fixed_charge[j] * Build[j];
   /* each customer assigned to exactly one site */
   con assign_def {i in CUSTOMERS}:
      sum {j in SITES} Assign[i,j] = 1;
   /* if customer i assigned to site j, then facility must be built at j */
   con link {i in CUSTOMERS, j in SITES}:
      Assign[i,j] <= Build[j];
   /* each site can handle at most &SiteCapacity demand */
   con capacity {j in SITES}:
      sum {i in CUSTOMERS} demand[i] * Assign[i,j] <=
         &SiteCapacity * Build[j];
   /* solve the MILP with no fixed charges */
   solve obj CostNoFixedCharge with milp / logfreq = 500;
   /* clean up the solution */
   for {i in CUSTOMERS, j in SITES} Assign[i,j] = round(Assign[i,j]);
   for {j in SITES} Build[j] = round(Build[j]);
   call symput('varcostNo',put(CostNoFixedCharge,6.1));
   /* create a data set for use by GPLOT */
   create data CostNoFixedCharge_Data from
      [customer site]={i in CUSTOMERS, j in SITES: Assign[i,j] = 1}
      xi=x[i] yi=y[i] xj=x[j] yj=y[j];
   /* solve the MILP, with fixed charges with warm start */
   solve obj CostFixedCharge with milp / primalin logfreq = 500;
   /* clean up the solution */
   for {i in CUSTOMERS, j in SITES} Assign[i,j] = round(Assign[i,j]);
   for {j in SITES} Build[j] = round(Build[j]);
   num varcost = sum {i in CUSTOMERS, j in SITES} dist[i,j] * Assign[i,j].sol;
   num fixcost = sum {j in SITES} fixed_charge[j] * Build[j].sol;
   call symput('varcost', put(varcost,6.1));
   call symput('fixcost', put(fixcost,5.1));
   call symput('totalcost', put(CostFixedCharge,6.1));
   /* create a data set for use by GPLOT */
   create data CostFixedCharge_Data from
      [customer site]={i in CUSTOMERS, j in SITES: Assign[i,j] = 1}
      xi=x[i] yi=y[i] xj=x[j] yj=y[j];
quit;

The information printed in the log for the no-fixed-charge model is displayed in Output 8.3.1.

Output 8.3.1: OPTMODEL Log for Facility Location with No Fixed Charges

NOTE: Problem generation will use 4 threads.                                    
NOTE: The problem has 510 variables (0 free, 0 fixed).                          
NOTE: The problem has 510 binary and 0 integer variables.                       
NOTE: The problem has 560 linear constraints (510 LE, 50 EQ, 0 GE, 0 range).    
NOTE: The problem has 2010 linear constraint coefficients.                      
NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range).      
NOTE: The MILP presolver value AUTOMATIC is applied.                            
NOTE: The MILP presolver removed 10 variables and 500 constraints.              
NOTE: The MILP presolver removed 1010 constraint coefficients.                  
NOTE: The MILP presolver modified 0 constraint coefficients.                    
NOTE: The presolved problem has 500 variables, 60 constraints, and 1000         
      constraint coefficients.                                                  
NOTE: The MILP solver is called.                                                
NOTE: The parallel Branch and Cut algorithm is used.                            
NOTE: The Branch and Cut algorithm is using up to 4 threads.                    
          Node  Active    Sols    BestInteger      BestBound      Gap    Time   
             0       1       2    972.1737321              0    972.2       0   
             0       1       2    972.1737321    961.2403449    1.14%       0   
             0       1       2    972.1737321    961.2403449    1.14%       0   
             0       1       2    972.1737321    961.2403449    1.14%       0   
NOTE: The MILP presolver is applied again.                                      
             0       1       4    966.4832160    962.9120804    0.37%       0   
             0       1       5    966.4832160    966.4832160    0.00%       0   
             0       0       5    966.4832160    966.4832160    0.00%       0   
NOTE: The MILP solver added 0 cuts with 0 cut coefficients at the root.         
NOTE: Optimal.                                                                  
NOTE: Objective = 966.48321599.                                                 



The results from the warm start approach are shown in Output 8.3.2.

Output 8.3.2: OPTMODEL Log for Facility Location with Fixed Charges, Using Warm Start

NOTE: Problem generation will use 4 threads.                                    
NOTE: The problem has 510 variables (0 free, 0 fixed).                          
NOTE: The problem uses 1 implicit variables.                                    
NOTE: The problem has 510 binary and 0 integer variables.                       
NOTE: The problem has 560 linear constraints (510 LE, 50 EQ, 0 GE, 0 range).    
NOTE: The problem has 2010 linear constraint coefficients.                      
NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range).      
NOTE: The MILP presolver value AUTOMATIC is applied.                            
NOTE: The MILP presolver removed 0 variables and 0 constraints.                 
NOTE: The MILP presolver removed 0 constraint coefficients.                     
NOTE: The MILP presolver modified 0 constraint coefficients.                    
NOTE: The presolved problem has 510 variables, 560 constraints, and 2010        
      constraint coefficients.                                                  
NOTE: The MILP solver is called.                                                
NOTE: The parallel Branch and Cut algorithm is used.                            
NOTE: The Branch and Cut algorithm is using up to 4 threads.                    
          Node  Active    Sols    BestInteger      BestBound      Gap    Time   
             0       1       3  16070.0150023              0    16070       0   
             0       1       3  16070.0150023   9946.2514269   61.57%       0   
             0       1       3  16070.0150023   9962.4849932   61.31%       0   
             0       1       3  16070.0150023   9971.5893075   61.16%       0   
             0       1       3  16070.0150023   9974.5588580   61.11%       0   
             0       1       3  16070.0150023   9978.1322942   61.05%       0   
             0       1       3  16070.0150023   9978.3312183   61.05%       0   
             0       1       3  16070.0150023   9980.4930282   61.01%       0   
             0       1       3  16070.0150023   9981.2701907   61.00%       0   
             0       1       4  16009.9864088   9981.2701907   60.40%       0   
NOTE: The MILP solver added 20 cuts with 631 cut coefficients at the root.      
           207     181       5  14940.8468978  10254.0277196   45.71%       0   
           244     203       6  11574.7346191  10366.4971368   11.66%       0   
           304     250       7  11200.4907322  10384.6282929    7.86%       0   
           464     318       8  10952.9510709  10474.3603655    4.57%       0   
           500     339       8  10952.9510709  10548.0238535    3.84%       0   
           548     368       9  10951.0805543  10548.0238535    3.82%       0   
           560     374      10  10951.0805542  10548.0238535    3.82%       0   
           642       6      11  10950.0345546  10946.4429359    0.03%       0   
           660       0      12  10948.4603464  10948.4603464    0.00%       0   
NOTE: Optimal.                                                                  
NOTE: Objective = 10948.460346.                                                 



The following two SAS programs produce a plot of the solutions for both variants of the model, using data sets produced by PROC OPTMODEL:

title1 h=1.5 "Facility Location Problem";
title2 "TotalCost = &varcostNo (Variable = &varcostNo, Fixed = 0)";

data csdata;
   set cdata(rename=(y=cy)) sdata(rename=(y=sy));
run;

/* create Annotate data set to draw line between customer and assigned site */
%annomac;
data anno(drop=xi yi xj yj);
   %SYSTEM(2, 2, 2);
   set CostNoFixedCharge_Data(keep=xi yi xj yj);
   %LINE(xi, yi, xj, yj, *, 1, 1);
run;

proc gplot data=csdata anno=anno;
   axis1 label=none order=(0 to &xmax by 10);
   axis2 label=none order=(0 to &ymax by 10);
   symbol1 value=dot interpol=none
      pointlabel=("#name" nodropcollisions height=1) cv=black;
   symbol2 value=diamond interpol=none
      pointlabel=("#name" nodropcollisions color=blue height=1) cv=blue;
   plot cy*x sy*x / overlay haxis=axis1 vaxis=axis2;
run;
quit;

The output of the first program is shown in Output 8.3.3.

Output 8.3.3: Solution Plot for Facility Location with No Fixed Charges

Solution Plot for Facility Location with No Fixed Charges


The output of the second program is shown in Output 8.3.4.

title1 "Facility Location Problem";
title2 "TotalCost = &totalcost (Variable = &varcost, Fixed = &fixcost)";

/* create Annotate data set to draw line between customer and assigned site */
data anno(drop=xi yi xj yj);
   %SYSTEM(2, 2, 2);
   set CostFixedCharge_Data(keep=xi yi xj yj);
   %LINE(xi, yi, xj, yj, *, 1, 1);
run;

proc gplot data=csdata anno=anno;
   axis1 label=none order=(0 to &xmax by 10);
   axis2 label=none order=(0 to &ymax by 10);
   symbol1 value=dot interpol=none
      pointlabel=("#name" nodropcollisions height=1) cv=black;
   symbol2 value=diamond interpol=none
      pointlabel=("#name" nodropcollisions color=blue height=1) cv=blue;
   plot cy*x sy*x / overlay haxis=axis1 vaxis=axis2;
run;
quit;

Output 8.3.4: Solution Plot for Facility Location with Fixed Charges

Solution Plot for Facility Location with Fixed Charges


The economic trade-off for the fixed-charge model forces you to build fewer sites and push more demand to each site.

It is possible to expedite the solution of the fixed-charge facility location problem by choosing appropriate branching priorities for the decision variables. Recall that for each site j, the value of the variable $y_ j$ determines whether or not a facility is built on that site. Suppose you decide to branch on the variables $y_ j$ before the variables $x_{ij}$. You can set a higher branching priority for $y_ j$ by using the .priority suffix for the Build variables in PROC OPTMODEL, as follows:

   for{j in SITES} Build[j].priority=10;

Setting higher branching priorities for certain variables is not guaranteed to speed up the MILP solver, but it can be helpful in some instances. The following program creates and solves an instance of the facility location problem, giving higher priority to the variables $y_ j$. The LOGFREQ= option is used to abbreviate the node log.


%let NumCustomers  = 45;
%let NumSites      = 8;
%let SiteCapacity  = 35;
%let MaxDemand     = 10;
%let xmax          = 200;
%let ymax          = 100;
%let seed          = 2345;

/* generate random customer locations */
data cdata(drop=i);
   length name $8;
   do i = 1 to &NumCustomers;
      name = compress('C'||put(i,best.));
      x = ranuni(&seed) * &xmax;
      y = ranuni(&seed) * &ymax;
      demand = ranuni(&seed) * &MaxDemand;
      output;
   end;
run;

/* generate random site locations and fixed charge */
data sdata(drop=i);
length name $8;
   do i = 1 to &NumSites;
      name = compress('SITE'||put(i,best.));
      x = ranuni(&seed) * &xmax;
      y = ranuni(&seed) * &ymax;
      fixed_charge = (abs(&xmax/2-x) + abs(&ymax/2-y)) / 2;
      output;
   end;
run;
proc optmodel;
   set <str> CUSTOMERS;
   set <str> SITES init {};

   /* x and y coordinates of CUSTOMERS and SITES */
   num x {CUSTOMERS union SITES};
   num y {CUSTOMERS union SITES};
   num demand {CUSTOMERS};
   num fixed_charge {SITES};

   /* distance from customer i to site j */
   num dist {i in CUSTOMERS, j in SITES}
       = sqrt((x[i] - x[j])^2 + (y[i] - y[j])^2);

   read data cdata into CUSTOMERS=[name] x y demand;
   read data sdata into SITES=[name] x y fixed_charge;

   var Assign {CUSTOMERS, SITES} binary;
   var Build {SITES} binary;

   min CostFixedCharge
       = sum {i in CUSTOMERS, j in SITES} dist[i,j] * Assign[i,j]
         + sum {j in SITES} fixed_charge[j] * Build[j];

   /* each customer assigned to exactly one site */
   con assign_def {i in CUSTOMERS}:
      sum {j in SITES} Assign[i,j] = 1;

   /* if customer i assigned to site j, then facility must be built at j */
   con link {i in CUSTOMERS, j in SITES}:
      Assign[i,j] <= Build[j];

   /* each site can handle at most &SiteCapacity demand */
   con capacity {j in SITES}:
      sum {i in CUSTOMERS} demand[i] * Assign[i,j] <= &SiteCapacity * Build[j];

   /* assign priority to Build variables (y) */
   for{j in SITES} Build[j].priority=10;

   /* solve the MILP with fixed charges, using branching priorities */
   solve obj CostFixedCharge with milp / logfreq=1000;
quit;

The resulting output is shown in Output 8.3.5.

Output 8.3.5: PROC OPTMODEL Log for Facility Location with Branching Priorities

NOTE: There were 45 observations read from the data set WORK.CDATA.             
NOTE: There were 8 observations read from the data set WORK.SDATA.              
NOTE: Problem generation will use 4 threads.                                    
NOTE: The problem has 368 variables (0 free, 0 fixed).                          
NOTE: The problem has 368 binary and 0 integer variables.                       
NOTE: The problem has 413 linear constraints (368 LE, 45 EQ, 0 GE, 0 range).    
NOTE: The problem has 1448 linear constraint coefficients.                      
NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range).      
NOTE: The MILP presolver value AUTOMATIC is applied.                            
NOTE: The MILP presolver removed 0 variables and 0 constraints.                 
NOTE: The MILP presolver removed 0 constraint coefficients.                     
NOTE: The MILP presolver modified 0 constraint coefficients.                    
NOTE: The presolved problem has 368 variables, 413 constraints, and 1448        
      constraint coefficients.                                                  
NOTE: The MILP solver is called.                                                
NOTE: The parallel Branch and Cut algorithm is used.                            
NOTE: The Branch and Cut algorithm is using up to 4 threads.                    
          Node  Active    Sols    BestInteger      BestBound      Gap    Time   
             0       1       3   2823.1827978              0   2823.2       0   
             0       1       3   2823.1827978   1727.0208789   63.47%       0   
             0       1       3   2823.1827978   1756.0637224   60.77%       0   
             0       1       5   1906.4633474   1764.3961986    8.05%       0   
             0       1       5   1906.4633474   1772.0991932    7.58%       0   
             0       1       5   1906.4633474   1784.2955379    6.85%       0   
             0       1       5   1906.4633474   1788.0570704    6.62%       0   
             0       1       5   1906.4633474   1788.0570704    6.62%       0   
NOTE: The MILP presolver is applied again.                                      
             0       1       5   1906.4633474   1788.0570704    6.62%       0   
             0       1       5   1906.4633474   1788.0570704    6.62%       0   
             0       1       5   1906.4633474   1788.0570704    6.62%       0   
             0       1       5   1906.4633474   1788.0570704    6.62%       0   
             0       1       5   1906.4633474   1788.0570704    6.62%       0   
             0       1       5   1906.4633474   1788.1857838    6.61%       0   
             0       1       5   1906.4633474   1793.0963734    6.32%       0   
             0       1       5   1906.4633474   1793.5540695    6.30%       0   
             0       1       5   1906.4633474   1793.6034154    6.29%       0   
             0       1       5   1906.4633474   1793.7200171    6.29%       0   
             0       1       5   1906.4633474   1793.7651419    6.28%       0   
             0       1       6   1903.8518326   1793.7651419    6.14%       0   
NOTE: The MILP solver added 28 cuts with 666 cut coefficients at the root.      
           209     182       7   1826.3615927   1800.9334244    1.41%       0   
           816      46       8   1823.4483963   1805.5532343    0.99%       0   
          1000     195       8   1823.4483963   1808.1519690    0.85%       0   
          1390     397       9   1823.4483963   1811.0458067    0.68%       0   
          1845     352      10   1819.9124343   1814.9424888    0.27%       1   
          2000     153      10   1819.9124343   1817.0347069    0.16%       1   
          2167       7      10   1819.9124343   1819.7338302    0.01%       1   
NOTE: Optimal within relative gap.                                              
NOTE: Objective = 1819.9124343.