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posted by martyb on Friday December 01 2017, @01:36PM   Printer-friendly
from the time-to-reschedule dept.

A glitch in American Airlines Group's scheduling system has given too many pilots time off during the busy holiday season in mid-late December. The airline may be forced to pay pilots more to show up:

American Airlines Group Inc. is rushing to resolve a scheduling fault that gave time off to too many pilots in December -- a flaw that has left more than 15,000 flights without sufficient crew during the holiday rush, according to a union for the carrier's pilots.

The Allied Pilots Association estimated the number of affected flights, from Dec. 17 to Dec. 31, based on information provided by the carrier, said Dennis Tajer, a spokesman for the union. American spokesman Matt Miller declined to quantify the potential number of flights involved, saying the airline expects to correct the problem in time to prevent service disruptions.

"We are working diligently to address the issue and expect to avoid cancellations this holiday season," Miller said. The number of flights involved will decline each day as the carrier reassigns them, he said.

The computer-system problem will force American to rebuild its staffing schedule, similar to what airlines must do after major weather disruptions, said John Cox, chief executive officer of consultant Safety Operating Systems and a former commercial airline pilot. Revenue will take a hit if American has to scrub many flights. At a minimum, the carrier is likely to face higher labor costs just as investors are stepping up scrutiny of airline expenses.

Also at Wired and Reuters.


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  • (Score: 0) by Anonymous Coward on Friday December 01 2017, @03:32PM (1 child)

    by Anonymous Coward on Friday December 01 2017, @03:32PM (#603933)

    the beauty of having a computer solve an NP hard problem for you ... whatever route it comes up with it should be trivial to check ...

    This property that solutions have efficient verifiers is a property of problems in NP. (Small nit: NP specifically contains only decision problems -- which are problems with yes/no solutions. Often the problem you care about is not exactly a decision problem, but usually this is not a huge obstacle for analyzing it).

    NP-hard is a much larger set, that includes many problems outside of NP and therefore such problems will not in general have efficient verifiers. (NP-hard roughly means "the set of problems that are at least as hard as the hardest problems in NP"). When the problem is both in NP-hard and in NP, we call it an NP-complete problem.

  • (Score: 0) by Anonymous Coward on Friday December 01 2017, @03:51PM

    by Anonymous Coward on Friday December 01 2017, @03:51PM (#603939)

    This property that solutions have efficient verifiers is a property of problems in NP. (Small nit: NP specifically contains only decision problems -- which are problems with yes/no solutions. Often the problem you care about is not exactly a decision problem, but usually this is not a huge obstacle for analyzing it). NP-hard is a much larger set, that includes many problems outside of NP and therefore such problems will not in general have efficient verifiers. (NP-hard roughly means "the set of problems that are at least as hard as the hardest problems in NP"). When the problem is both in NP-hard and in NP, we call it an NP-complete problem.

    According to wikipedia [wikipedia.org], "Although any given solution to an NP-complete problem can be verified quickly (in polynomial time), there is no known efficient way to locate a solution in the first place; indeed, the most notable characteristic of NP-complete problems is that no fast solution to them is known." This seems to contradict your assertion about efficient verifiers.