The Decomposition Algorithm

The generalized assignment problem (GAP) is that of finding a maximum profit assignment from tasks to 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 , which, if set to , indicates that machine is assigned to task . For ease of notation, define two index sets and . A GAP can be formulated as a MILP as follows:

In this formulation, constraints (assignment) ensure that each task is assigned to exactly one machine. Inequalities (knapsack) ensure that for each machine, the capacity restrictions are met.

Consider the following example taken from Koch et al. (2011) with tasks to be assigned to 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 AssignmentCon {j in TASKS}: 
      sum {i in MACHINES} Assign[i,j] = 1;

   con KnapsackCon {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} 
      AssignmentCon[j].block = j;
      
   solve with milp / logfreq=1000 
      decomp        =() 
      decomp_subprob=(algorithm=nspure);

   /* each knapsack constraint defines a block */
   for{j in TASKS} 
      AssignmentCon[j].block = .;
   for{i in MACHINES} 
      KnapsackCon[i].block = i; 
      
   solve with milp / decomp=();
quit;

The solution summaries are displayed in Output 13.2.1.

The iteration log for both decompositions is shown in Output 13.2.2. 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.