The following code uses PROC OPTMODEL in SAS/OR or SAS Optimization to maximize the number of zero-sum groups: data have; input amount; datalines; -56 76 10 -76 56 20 -45 -3 15 -35 40 4 4 ; proc optmodel; /* declare parameters and read input data set */ set OBS; num amount {OBS}; read data have into OBS=[_N_] amount; num numGroups = card(OBS)/2; set GROUPS = 1..numGroups; /* declare decision variables */ var UseGroup {GROUPS} binary; var Assign {OBS, GROUPS} binary; /* declare objective */ max NumGroupsUsed = sum {g in GROUPS} UseGroup[g]; /* declare constraints */ con OneGroupPerObs {i in OBS}: sum {g in GROUPS} Assign[i,g] <= 1; con GroupSumZero {g in GROUPS}: sum {i in OBS} amount[i] * Assign[i,g] = 0; con AssignImpliesUse {i in OBS, g in GROUPS}: Assign[i,g] <= UseGroup[g]; con UseImpliesAssign {g in GROUPS}: 2*UseGroup[g] <= sum {i in OBS} Assign[i,g]; con Symmetry {g in GROUPS diff {1}}: UseGroup[g] <= UseGroup[g-1]; /* call MILP solver */ solve; /* create output data set */ num assignedGroup {OBS}; for {i in OBS} do; for {g in GROUPS: Assign[i,g].sol > 0.5} do; assignedGroup[i] = g; leave; end; end; create data want from [i] amount assignedGroup; quit; The resulting optimal solution has 4 groups: Obs i amount assignedGroup 1 1 -56 4 2 2 76 2 3 3 10 . 4 4 -76 2 5 5 56 4 6 6 20 3 7 7 -45 1 8 8 -3 1 9 9 15 3 10 10 -35 3 11 11 40 1 12 12 4 1 13 13 4 1   See also Solved: Net Groups of Lines in a Table to $0.00 - SAS Support Communities for an alternative greedy approach. ... View more

The following reads from your three data sets, solves the problem, and creates two output data sets: one in sparse form like your costs data set, and one in dense matrix form, like you requested.  Note the use of the COMPRESS function because New York is not a valid data set variable name.   proc optmodel; /* specify parameters */ set <str> O; set <str> D; number c{O,D}; number a{O}; number b{D}; /* read data */ read data supply into O=[origination] a=supply; read data demand into D=[destination] b=demand; read data costs into [origination destination] c=cost; /* model description */ var x{O,D} >= 0; min total_cost = sum{i in O, j in D}c[i,j]*x[i,j]; constraint supply{i in O}: sum{j in D}x[i,j]=a[i]; constraint demand{j in D}: sum{i in O}x[i,j]=b[j]; /* solve and output */ solve; print x; /* create data */ create data sparse from [origination destination] x; create data dense from [origination]=O {j in D} <col(compress(j))=x[origination,j]>; quit; ... View more

Not quite sure I understand, but does the following do what you want? %let numEntities = 5; data indata; input dataset state $ e1-e&numEntities; datalines; 1 S1 5.2 5.8 8.2 3.1 7.2 1 S2 8.1 3.6 5.0 4.9 1.8 2 S1 2.3 3.1 9.1 7.3 4.0 2 S2 4.2 5.3 7.1 4.7 7.3 ; proc optmodel; set DATASETS; read data indata into DATASETS=[dataset]; set <str> STATES; read data indata into STATES=[state]; set ENTITIES = 1..&numEntities; num performance {DATASETS, STATES, ENTITIES}; read data indata into [dataset state] {entity in ENTITIES} <performance[dataset,state,entity]=col('e'||entity)>; var IsState {ENTITIES, STATES} binary; max Objective = sum {e in ENTITIES, s in STATES} (prod {d in DATASETS} performance[d,s,e]) * IsState[e,s]; con OneStatePerEntity {e in ENTITIES}: sum {s in STATES} IsState[e,s] = 1; for {e in ENTITIES, s in STATES: min {d in DATASETS} performance[d,s,e] < 2} fix IsState[e,s] = 0; solve; print IsState; print {e in ENTITIES, s in STATES: IsState[e,s].sol > 0.5} (prod {d in DATASETS} performance[d,s,e]); quit; ... View more