# Which forum, if any, is appropriate for this problem?

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• Dec 25th 2010, 11:47 PM
JohnnyG
Which forum, if any, is appropriate for this problem?
The solution is math based surely but the problem is regarding rotating schedules but I just cannot figure out how to break this problem down into manageable pieces to solve. Could someone suggest which forum to post this problem, or don't post it not appropriate for this site?

Issue:
How to equally disperse hours at 9 positions throughout a 24 hour day on a 8 day schedule. Some positions are not in play during part of the day. There are 25 people to rotate who work 10 hour shifts, the quantity and hour of shifts are 3/05-1500, 4/07-1700, 2/09-1900, 2/11-2100, 1/13-2300, 4/15-0100, 3/17-0300, 3/19-0500, 3/21-0700.

No one can be in a position for more then 6 hours daily and want to limit the number of position moves per day to no more then 3, two is preferable.

It's just boggling my mind on the complexity of how to break this down into a repeating pattern that even comes close to equal dispersal amongst all shifts.

Anyways some idea of where, if any, forum I could post this too for someone to suggest a route to follow on this? I'm not seeking a full solution its too much work but some logical process on how to proceed. I have a excel sheet that makes it easier to understand.
• Dec 26th 2010, 02:51 AM
tom@ballooncalculus
I may be missing something, but I'm finding it easy enough to assign workers to new positions as they either clock in or clock up 6 hours in a position, so that all nine positions are close enough to having been equally long occupied after 24 hours, and also nobody has had to move more than once. But perhaps that's the point of it - you just need to use the spreadsheet (with a column for each of the nine positions) to apply a common-sense algorithm assigning workers to positions every two hours. As I say, it seems to be safe to let everyone stay in position for up to 3 rows (2-hour sessions). Then (a) assign new positions for those due their 6-hour change (because their choice of positions is limited). Then (b) assign the new arrivals. Before (a) and before (b) update the number of hours occupied, for each position - so you can always assign to lowest-rated positions.

E.g.

http://www.ballooncalculus.org/draw/sched.png
• Dec 26th 2010, 03:19 PM
JohnnyG
Well there's just more variables then I revealed since I really just wanted to know where to post this. Because as you can see from the number of shifts there isn't enough people to fill every space throughout the day because some of the positions don't exist for the whole 24 hours.

And at times there are more people then I have actual positions but they are still working so I put them into a place saver position. The problem arises when someone is not present I of course have to just use my place saver people to fill holes which is fine at first but the more I use ad hoc changes the more out of whack the system gets till it's just a mess.

I think I will actually just withdraw the issue I don't think it's even possible to find a pattern or logical progression that fits with this many variables, doing it ad hoc each day based on whatever was done the day before seems to still be the best plan, even though its often too much work daily to put the pieces back together equally.

I guess I was hoping for a master plan that although deviations occur I could get everyone back into "the" progression plan eventually and only have to allow for the odd absence. Once I start using excel and cell shading whatever plan I come up with starts to break down around day 5 but it appears to have a pattern that I cannot figure out yet and was hoping to nail down what my brain is seeing but my mind is not.
• Dec 27th 2010, 01:10 AM
tom@ballooncalculus
I don't see the problem... although you keep saying there is a wealth of detail to the problem that you don't want to mention. This is generally the wrong approach at this forum. Give us the whole problem!

Anyway, I can see that maybe I shouldn't have assumed that positions can be occupied by more than one at a time. Fine - just have a tenth (idling) column, which will be the exception to this rule. Even easier, surely? I don't see how things have to get 'out of whack'. Have you tried continuing my chart down to 5 a.m.? Easy enough to get a repeatable pattern. Or why not? (Tell us the whole problem.)