Timeline for Modelling a complex work schedule
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Nov 22, 2016 at 20:33 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 23, 2016 at 20:24 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 23, 2016 at 19:09 | answer | added | Michael | timeline score: 1 | |
| Sep 3, 2016 at 22:29 | history | edited | Tulains Córdova | CC BY-SA 3.0 |
grammar
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| Apr 22, 2016 at 5:51 | comment | added | bigstones | Sounds like a job for Operations Research! | |
| Apr 22, 2016 at 3:02 | history | tweeted | twitter.com/StackProgrammer/status/723346106445402117 | ||
| Apr 21, 2016 at 16:28 | comment | added | Snoop | Maybe come up with some sort of a system where you load all the work that needs to be done into a queue. Then, do the work at scheduled intervals until the queue is exhausted. | |
| Apr 21, 2016 at 15:00 | comment | added | Gareth Lloyd | Yes, more trips to the exchange is a worse outcome. | |
| Apr 21, 2016 at 14:53 | comment | added | Dan Pichelman | Is there a cost (money or time) to go to the exchange? | |
| Apr 21, 2016 at 13:52 | comment | added | radarbob | Since workers are obviously trusted with keys and may keep keys over night and for as long as necessary - as long as an exchange is not required - why not make a set of keys for each worker that they keep permanently? Alternatively, create a set of keys for each worker for all places they go for a given time period, say, a week. Keys are duplicated as needed to make a week-set for every worker. All workers exchange keys once a week. | |
| Apr 21, 2016 at 12:47 | history | edited | Gareth Lloyd | CC BY-SA 3.0 |
added 1 character in body
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| Apr 21, 2016 at 12:43 | comment | added | Gareth Lloyd | Yes, workers can carry any number of keys. Regarding "average" I think I expressed myself poorly. I was thinking more about fairness, that no worker should have to complete a disproportionate number of trips, so a low variance. (edited question to match) | |
| Apr 21, 2016 at 12:14 | comment | added | Doc Brown | Average number of trips to E per worker = "total number of trips" / m. So if m is constant, your two goals are the same goal. More interesting: I guess each worker can carry more than one key at the same time? | |
| Apr 21, 2016 at 11:53 | review | First posts | |||
| Apr 22, 2016 at 2:26 | |||||
| Apr 21, 2016 at 11:50 | history | asked | Gareth Lloyd | CC BY-SA 3.0 |