The Decomposition Algorithm

Example 15.2 Generalized Assignment Problem

The generalized assignment problem (GAP) is that of finding a maximum profit assignment from n tasks to m machines such that each task is assigned to precisely one machine subject to capacity restrictions on the machines. With each possible assignment, associate a binary variable $x_{ij}$, which, if set to 1, indicates that machine i is assigned to task j. For ease of notation, define two index sets $M=\left\{ 1,\dots ,m\right\} $ and $N=\left\{ 1,\dots ,n\right\} $. A GAP can be formulated as a MILP as follows:

\begin{align*} & \text {maximize} &  \sum _{i \in M} \sum _{j \in N} p_{ij} x_{ij}\\ & \text {subject to} &  \sum _{i \in M} x_{ij} &  = 1 & &  j \in N & &  \text {(Assignment)}\\ & &  \sum _{j \in N} w_{ij} x_{ij} &  \leq b_ i & &  i \in M & &  \text {(Knapsack)} \\ & &  x_{ij} &  \in \left\{ 0,1\right\}  & &  i \in M, \  j \in N \end{align*}

In this formulation, Assignment constraints ensure that each task is assigned to exactly one machine. Knapsack constraints ensure that for each machine, the capacity restrictions are met.

Consider the following example taken from Koch et al. (2011) with $n=24$ tasks to be assigned to $m=8$ machines. The data set profit_data provides the profit for assigning a particular task to a particular machine:

%let NumTasks    = 24;
%let NumMachines = 8;

data profit_data;
   input p1-p&NumTasks;
   datalines;
25 23 20 16 19 22 20 16 15 22 15 21 20 23 20 22 19 25 25 24 21 17 23 17
16 19 22 22 19 23 17 24 15 24 18 19 20 24 25 25 19 24 18 21 16 25 15 20
20 18 23 23 23 17 19 16 24 24 17 23 19 22 23 25 23 18 19 24 20 17 23 23
16 16 15 23 15 15 25 22 17 20 19 16 17 17 20 17 17 18 16 18 15 25 22 17
17 23 21 20 24 22 25 17 22 20 16 22 21 23 24 15 22 25 18 19 19 17 22 23
24 21 23 17 21 19 19 17 18 24 15 15 17 18 15 24 19 21 23 24 17 20 16 21
18 21 22 23 22 15 18 15 21 22 15 23 21 25 25 23 20 16 25 17 15 15 18 16
19 24 18 17 21 18 24 25 18 23 21 15 24 23 18 18 23 23 16 20 20 19 25 21
;

The data set weight_data provides the amount of resources used by a particular task when assigned to a particular machine:

data weight_data;
   input w1-w&NumTasks;
   datalines;
 8 18 22  5 11 11 22 11 17 22 11 20 13 13  7 22 15 22 24  8  8 24 18  8
24 14 11 15 24  8 10 15 19 25  6 13 10 25 19 24 13 12  5 18 10 24  8  5
22 22 21 22 13 16 21  5 25 13 12  9 24  6 22 24 11 21 11 14 12 10 20  6
13  8 19 12 19 18 10 21  5  9 11  9 22  8 12 13  9 25 19 24 22  6 19 14
25 16 13  5 11  8  7  8 25 20 24 20 11  6 10 10  6 22 10 10 13 21  5 19
19 19  5 11 22 24 18 11  6 13 24 24 22  6 22  5 14  6 16 11  6  8 18 10
24 10  9 10  6 15  7 13 20  8  7  9 24  9 21  9 11 19 10  5 23 20  5 21
 6  9  9  5 12 10 16 15 19 18 20 18 16 21 11 12 22 16 21 25  7 14 16 10
;

Finally, the data set capacity_data provides the resource capacity for each machine:

data capacity_data;
   input b @@;
   datalines;
36 35 38 34 32 34 31 34
;

The following PROC OPTMODEL statements read in the data and define the necessary sets and parameters:

proc optmodel;
   /* declare index sets */
   set TASKS    = 1..&NumTasks;
   set MACHINES = 1..&NumMachines;

   /* declare parameters */
   num profit   {MACHINES, TASKS};
   num weight   {MACHINES, TASKS};
   num capacity {MACHINES};

   /* read data sets to populate data */
   read data profit_data   into [i=_n_] {j in TASKS} <profit[i,j]=col('p'||j)>;
   read data weight_data   into [i=_n_] {j in TASKS} <weight[i,j]=col('w'||j)>;
   read data capacity_data into [_n_] capacity=b;

The following statements declare the optimization model:

   /* declare decision variables */
   var Assign {MACHINES, TASKS} binary;

   /* declare objective */
   max TotalProfit =
      sum {i in MACHINES, j in TASKS} profit[i,j] * Assign[i,j];

   /* declare constraints */
   con Assignment {j in TASKS}:
      sum {i in MACHINES} Assign[i,j] = 1;

   con Knapsack {i in MACHINES}:
      sum {j in TASKS} weight[i,j] * Assign[i,j] <= capacity[i];

The following statements use two different decompositions to solve the problem. The first decomposition defines each Assignment constraint as a block and uses the pure network simplex solver for the subproblem. The second decomposition defines each Knapsack constraint as a block and uses the MILP solver for the subproblem.

   /* each Assignment constraint defines a block */
   for{j in TASKS}
      Assignment[j].block = j;

   solve with milp / logfreq=1000
      decomp        =()
      decomp_subprob=(algorithm=nspure);

   /* each Knapsack constraint defines a block */
   for{j in TASKS}
      Assignment[j].block = .;
   for{i in MACHINES}
      Knapsack[i].block = i;

   solve with milp / decomp;
quit;

The solution summaries are displayed in Output 15.2.1.

Output 15.2.1: Solution Summaries

The OPTMODEL Procedure

Solution Summary
Solver MILP
Algorithm Decomposition
Objective Function TotalProfit
Solution Status Optimal within Relative Gap
Objective Value 563
   
Relative Gap 0.0000925018
Absolute Gap 0.0520833333
Primal Infeasibility 6.661338E-16
Bound Infeasibility 2.220446E-16
Integer Infeasibility 6.661338E-16
   
Best Bound 563.05208333
Nodes 1763
Iterations 1802
Presolve Time 0.00
Solution Time 4.46

Solution Summary
Solver MILP
Algorithm Decomposition
Objective Function TotalProfit
Solution Status Optimal
Objective Value 563
   
Relative Gap 0
Absolute Gap 0
Primal Infeasibility 0
Bound Infeasibility 0
Integer Infeasibility 0
   
Best Bound 563
Nodes 3
Iterations 33
Presolve Time 0.01
Solution Time 0.24



The iteration log for both decompositions is shown in Output 15.2.2. This example is interesting because it shows the tradeoff between the strength of the relaxation and the difficulty of its resolution. In the first decomposition, the subproblems are totally unimodular and can be solved trivially. Consequently, each iteration of the decomposition algorithm is very fast. However, the bound obtained is as weak as the bound found in direct methods (the LP bound). The weaker bound leads to the need to enumerate more nodes overall. Alternatively, in the second decomposition, the subproblem is the knapsack problem, which is solved using MILP. In this case, the bound is much tighter and the problem solves in very few nodes. The tradeoff, of course, is that each iteration takes longer because solving the knapsack problem is not trivial. Another interesting aspect of this problem is that the subproblem coverage in the second decomposition is much smaller than that of the first decomposition. However, when dealing with MILP, it is not always the size of the coverage that determines the overall effectiveness of a particular choice of decomposition.

Output 15.2.2: Log

NOTE: There were 8 observations read from the data set WORK.PROFIT_DATA.                        
NOTE: There were 8 observations read from the data set WORK.WEIGHT_DATA.                        
NOTE: There were 8 observations read from the data set WORK.CAPACITY_DATA.                      
NOTE: Problem generation will use 4 threads.                                                    
NOTE: The problem has 192 variables (0 free, 0 fixed).                                          
NOTE: The problem has 192 binary and 0 integer variables.                                       
NOTE: The problem has 32 linear constraints (8 LE, 24 EQ, 0 GE, 0 range).                       
NOTE: The problem has 384 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 192 variables, 32 constraints, and 384 constraint coefficients. 
NOTE: The MILP solver is called.                                                                
NOTE: The Decomposition algorithm is used.                                                      
NOTE: The Decomposition algorithm is executing in single-machine mode.                          
NOTE: The DECOMP method value USER is applied.                                                  
NOTE: The subproblem solver chosen is an LP solver but at least one block has integer variables.
NOTE: The problem has a decomposable structure with 24 blocks. The largest block covers 3.13%   
      of the constraints in the problem.                                                        
NOTE: The decomposition subproblems cover 192 (100.00%) variables and 24 (75.00%) constraints.  
NOTE: The deterministic parallel mode is enabled.                                               
NOTE: The Decomposition algorithm is using up to 4 threads.                                     
      Iter         Best       Master         Best       LP       IP  CPU Real                   
                  Bound    Objective      Integer      Gap      Gap Time Time                   
NOTE: Starting phase 1.                                                                         
         1       0.0000       8.9248            . 8.92e+00        .    0    0                   
         4       0.0000       0.0000            .    0.00%        .    0    0                   
NOTE: Starting phase 2.                                                                         
         5     574.0000     561.1587            .    2.24%        .    0    0                   
         6     568.8833     568.5610            .    0.06%        .    0    0                   
         8     568.6464     568.6464     560.0000    0.00%    1.52%    0    0                   
NOTE: Starting branch and bound.                                                                
         Node  Active   Sols         Best         Best      Gap    CPU   Real                   
                                  Integer        Bound            Time   Time                   
            0       1      1     560.0000     568.6464    1.52%      0      0                   
            5       7      2     563.0000     568.4782    0.96%      0      0                   
         1000     432      2     563.0000     564.6212    0.29%      2      2                   
         1762       0      2     563.0000     563.0521    0.01%      4      4                   
NOTE: The Decomposition algorithm used 4 threads.                                               
NOTE: The Decomposition algorithm time is 4.46 seconds.                                         
NOTE: Optimal within relative gap.                                                              
NOTE: Objective = 563.                                                                          
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 192 variables, 32 constraints, and 384 constraint coefficients. 
NOTE: The MILP solver is called.                                                                
NOTE: The Decomposition algorithm is used.                                                      
NOTE: The Decomposition algorithm is executing in single-machine mode.                          
NOTE: The DECOMP method value USER is applied.                                                  
NOTE: The problem has a decomposable structure with 8 blocks. The largest block covers 3.13% of 
      the constraints in the problem.                                                           
NOTE: The decomposition subproblems cover 192 (100.00%) variables and 8 (25.00%) constraints.   
NOTE: The deterministic parallel mode is enabled.                                               
NOTE: The Decomposition algorithm is using up to 4 threads.                                     
      Iter         Best       Master         Best       LP       IP  CPU Real                   
                  Bound    Objective      Integer      Gap      Gap Time Time                   
NOTE: Starting phase 1.                                                                         
         1       0.0000      10.0000            . 1.00e+01        .    0    0                   
         8       0.0000       0.0000            .    0.00%        .    0    0                   
NOTE: Starting phase 2.                                                                         
        11     717.5556     540.0000     540.0000   24.74%   24.74%    0    0                   
        13     670.3333     548.0000     548.0000   18.25%   18.25%    0    0                   
        14     627.9557     548.0000     548.0000   12.73%   12.73%    0    0                   
        16     592.2500     549.8750     548.0000    7.15%    7.47%    0    0                   
        19     592.2500     558.0000     558.0000    5.78%    5.78%    0    0                   
         .     592.2500     558.0000     558.0000    5.78%    5.78%    0    0                   
        20     577.6667     558.0000     558.0000    3.40%    3.40%    0    0                   
        23     574.6667     560.6667     560.0000    2.44%    2.55%    0    0                   
        24     574.6667     563.0000     563.0000    2.03%    2.03%    0    0                   
        25     569.5000     563.5000     563.0000    1.05%    1.14%    0    0                   
        26     566.1905     563.7143     563.0000    0.44%    0.56%    0    0                   
        28     564.5000     564.0000     563.0000    0.09%    0.27%    0    0                   
        29     564.0000     564.0000     563.0000    0.00%    0.18%    0    0                   
NOTE: Starting branch and bound.                                                                
         Node  Active   Sols         Best         Best      Gap    CPU   Real                   
                                  Integer        Bound            Time   Time                   
            0       1      8     563.0000     564.0000    0.18%      0      0                   
            4       0      8     563.0000     563.0000    0.00%      0      0                   
NOTE: The Decomposition algorithm used 4 threads.                                               
NOTE: The Decomposition algorithm time is 0.15 seconds.                                         
NOTE: Optimal.                                                                                  
NOTE: Objective = 563.