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added 14 commits
July 27, 2026 00:25
…every 20 iterations in lu_factorize_bump
Codecov Report❌ Patch coverage is
Additional details and impacted files@@ Coverage Diff @@
## latest #3185 +/- ##
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- Coverage 73.19% 72.95% -0.24%
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Files 432 436 +4
Lines 105499 105985 +486
Branches 16991 17066 +75
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+ Hits 77216 77323 +107
- Misses 28007 28386 +379
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Closes #3120 |
filikat
reviewed
Jul 29, 2026
| if (info_.status_ipm != IPX_STATUS_not_run) | ||
| return; | ||
| } | ||
| printf("LpSolver::RunIPM before BuildStartingBasis(); timer.Elapsed() = %s\n", time(this->control_.Elapsed()).c_str()); |
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These 2 printf should be removed
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Apart from that, it looks good |
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Timeout when forming the initial basis is achievable with
control_.InterruptCheck(), andbool interruptused to communicate interruption down to levels whereIPX_ERROR_time_interruptcan be returned or assigned toinfo->errflag.Spotted that
Int Load(const int* basic_status);is redundant.Timeout within
basiclu_factorizeneeds an independent clock, time start and time limit.When IPX takes "forever" there was no logging, so I've added that at the end of each pass through the loop in
BasicLu::_Factorize, with an interval of at least 5s. It's interesting to see howBasicLu::_Factorizekeeps returning to pick up more memory. There can still be long gaps, but it's not feasible to put HiGHS logging intoBasicLu::_Factorize.As with all time-out code, unit tests are troublesome, but I've developed it with the examples that made me realise that there's no time-out when forming the initial basis and factorization in IPX