I am doing something like this with glpk-4.23:
lp = lpx_read_cpxlp("../prob.cplex");
glp_create_index(lp);
glp_simplex(lp, simplex_control_params);
for(i = 0; i < 40; i++ ) {
printf("Iteration %d\n", i);
add_rows(lp, buffer, index, value, col_status, subprob);
printf("Added rows...\n");
simplex_control_params->meth = GLP_DUALP;
simplex_control_params->presolve = GLP_ON;
glp_simplex(lp, simplex_control_params);
}
Where my "add_rows" function does something like this:
for( i = 1; i <= glp_get_num_cols(lp); i++ )
col_status[i] = glp_get_col_stat(lp, i);
count = 0;
do {
row = glp_add_rows(lp, 1);
read a row from another file and parse it to set ind and val arrays...
glp_set_row_bnds...
glp_set_mat_row(lp, row, 2, ind, val);
} while(++count < n );
for( i = 1; i <= glp_get_num_cols(lp); i++ )
glp_set_col_stat(lp, i, col_status[i]);
If I keep the presolver on, the basis information from the previous solve is
lost and the 2-phase primal algorithm kicks off from scratch (instead of the
dual simplex using the previous basis). If I turn the presolver off, I know my
problem is bigger than it needs to be, but the dual simplex does kick in
starting with the previously optimal basis. So I guess my question is: is
there a way to use the presolver AND supply an initial basic solution? If not,
is there a mathematical/algorithmic reason this isn't possible? If so, I'd be
interested in hearing it because it will probably change my approach (and
understanding) of my problem. My understanding from the glpk code is that the
presolver uses a copy of the original problem to perform all of it
transformations and then the solution to the presolved problem is translated
back to the original problem when the simplex completes. This transformation
must interfere with any basis that the original problem had, thus eliminating
it I suppose? Hope all of these questions make sense.
I'll gladly supply more info if it's helpful. Thanks in advance for any
insights.
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