Here is a genuine logistical scenario from my other life as an Erector of Fences. This happened a few days ago
Me and Ade were at a livery yard.
We had 100m of post and rail fence to do, all the supplies were tipped at one end of the fence line, and because of the poor access we had to carry them out by hand.
The rails are 3.6m long and so required a post every 1.8m. It was to be a 3 rail fence, so the rails were to be laid out in piles of three, each pile touching end to end with the next pile. Along side each pile there were to be two posts, as the posts are 1.8 long this means the posts are also laid end to end .
Me and Ade are both farmers with dodgy knees and can only manage to carry either three rails or two posts at a time, we can also only manage to travel at 1m per second ,including picking up and putting down.
Now in the cut-throat world of Fence Erecting time is money, so Ade went to get help and I began to carry the stuff out. I'd done 10 trips when he arrived back with Richard and they began to help . Richard, the yard owner, is an ex-rugby league player with dodgy knees and could only travel at our pace, BUT he decided he could carry 3 posts each trip. He did this for 6 trips then sat down on the grass and watched me and Ade do the rest.
How long did it take us? Please show workings out, as I haven't done it yet either.
Assume the first trip was 3.6 metres. Remember this is a real life situation, so the numbers may not be exact, but there needs to be enough rails and posts to complete 100 metres of fence.
Hmmm. If you bill for time, and we're working out the time for you, it seems to me a large consultancy fee is in order.
Oh law, I've been rumbled.
Before I find myself in the embarrassing situation of answering my own question wrongly (billl,where are you?) I now realize there is not enough infomation above to answer it, as you need to know exactly which posts Richard took out.
So assume that while Richard was helping, me and Ade carried out rails only.
I'm working on it now, under the assumption that Richard laid out the posts for the sections of fence that would be nearest to the pile of supplies. Let me know if we want him to do the most distant piles, though.
ALSO (obviously, just nit-picking), it is assumed that this is a straight fence (no short-cuts to distant sections).
ALSO, I am assuming that the first pile of three rails (and the first three posts) are a distance of 0m from the supply pile, and therefore take "no time", even though picking up/putting down/and arranging end-to-end are all part of the process. However long this would take (something less than 3.6 seconds...) can of course be added to my final answer to the puzzle. This assumption allows me to (ideally, though not realistically) consider the second section as being simply 3.6m from the supplies, with no need for information about the distance of the supplies from the fence, or how that distance increases as the supplies dwindle.
EDIT: I decided it would be simpler (mathematically) to have the first THREE posts be at a distance of zero from the pile of supplies.
Last edited by billl; 05-02-2011 at 04:25 PM. Reason: the first three posts are at a distance of zero meters, not the first two.
I've got 41.94 minutes (41 minutes, 56 seconds).
I'll type up the work pretty soon here...
Yeah. For sure. 41.94 minutes. That's what I got too. 41.94 minutes.
Um, some points regarding what it takes to lay the first section of rails out has altered my result. (I'll be done soon, just keeping you posted.)
NUMBER OF "SECTIONS" OF FENCE = 28
TOTAL LENGTH OF FENCE= 100.8m (3.6m x 28)
As explained in the puzzle, each section will have 3 rails of fence, and two posts. Note, the first section will actually have a third post (the 'end' post), and I will assume that this post can simply be left where it is in the pile of supplies at the beginning of the fence, and doesn't have to be carried anywhere.
Note that I'm not worrying about the unmentioned (and apparently minimal, but *realistically* increasing as supplies are used up) distance of the supplies from the beginning of the fencing.
First, let's see how much distance must be covered to lay out all of the rails from end to end. Each section needs one "round trip" by a worker--even the first section, which just needs to be "arranged" at a distance of "zero" from the pile of supplies will need the worker to carry the far end of the rails the full 3.6 meters. So the first section is a round trip of 7.2 meters.
The Second Section would thus be 7.2m from the supplies (round trip of 14.4m), the Third Section would be 10.8m away (21.6m round trip), etc.
The furthest section needing a pile of rails would be the 28th Section, at a distance of 100.8m (round trip = 201.6m).
7.2m = Shortest Trip
201.6m = Longest Trip
Average Trip = (Shortest + Longest) divided by 2 = 104.4m
Total Ground Covered By Workers Carrying Rails = 104.4m x 28 round trips = 2923.2m
Well, while Mick and Ade are working on the fencing, Richard gives them a head start on the posts by doing the first 9 sections (that's his 18 posts, plus the first one that's just lying at the beginning, with the supplies). When we realize that he gets work done faster than Mick or Ade, AND we assume that, what with his knee problems and everything, he only bothers helping with the nearest sections, it is safe to assume that Mick and Ade are still quite busy on the rails by the time Richard's contribution is complete. So we don't need to figure out how long it takes Richard to finish his part of the job (Although it is pretty interesting, not at all straight-forward, and *might *make a good puzzle). All we need to realize is that he has given the other two guys a head start on the posts, once they finish with the rails.
MICK AND ADE CARRYING POSTS
After Richard's head start, Mick and Ade still need to do the bulk of the post-carrying. Sections 10-28 remain to be done.
(3.6 x 10) x 2 = 72 = Shortest Trip (remember, these are "Round Trips", so we multiply distance by 2)
(3.6 x 28) x 2 = 201.6 = Longest Trip
Average Trip = 273.6 (Shortest + Longest) divided by 2 = 136.8m
Total Ground Covered By Mick and Ade Carrying Posts = 136.8m x 19 (sections) = 2599.2m
At this point, we could simply add the distances covered by Mick and Ade carrying the posts and rails, divide by two (because they are working simultaneously), and the result would be the number of seconds they spent doing the work. HOWEVER, Mick did the first 10 sections of rails on his own, and so we need to find the amount of time spent on that (and subtract THAT distance from the time that they were working on the rails *together*). I'm going to assume that Mick did the nearest sections first--but this assumption has a BIG affect on the final results... If he had done the 10 most distant sections first, that would've meant that Ade was off looking for Richard for a much, much longer time.
JUST MICK ON THE RAILS
Section 1 (round trip = 7.2m) to Section 10 (round trip = 72m).
Average Trip = 79.2m divided by 2 = 39.6m
Total Ground Covered By Just Mick Carrying The First 10 Sections Of Rails = 39.6m x 10 trips = 396m
The total time spent while Ade was off looking for help = 396 seconds.
At the 396-seconds mark, Richard joins in (chipping away at the scope of Mick and Ade's task) and Ade and Mick begin to work together.
TOTAL DISTANCE COVERED BY MICK ALONE *AND* MICK AND ADE = Rail trips ground covered + "post Richard" Posts ground covered
TOTAL DISTANCE COVERED BY MICK ALONE *AND* MICK AND ADE = 2923.2m + 2599.2m = 5522.4m
5522.4m is the distance that needs to be covered, after Richard's head-start on the posts is taken into account.
TIME = (MICK ALONE) divided by 1m/sec + (MICK AND ADE WORKING TOGETHER) divided by 2m/sec
note that Mick and Ade working together cut the time in half...
TIME = (396)/1 + (5522.4-396)/2
note that Mick's early work is subtracted from the total work done, in order to find the amount done by them working together.
TIME = 396 + 2563.2 = 2959.2 seconds
Total time = 49.32 minutes = 49 minutes, 19.2 seconds.
Again, I assumed that Richard and Mick began the rails and posts by doing the nearest sections first.
Finally, I want to point out that, in all likelihood (unless I've been very lucky here) we would end up with either Mick or Ade standing around, arms folded, watching the other guy walking back from his job of carrying the final pair of posts to the most distant section (and my procedure here considers the walk back to be part of the job, but that's beside the point). Since that walk would take 108 seconds at the most, I have to say that my answer therefore has a 108 second margin of error--(because I'm not going to figure out the exact amount of time that one of them is left to work on the last bit by himself, I haven't even begun to think about how to do that...). It's worth noting that this margin of error would be cut down to a mere 3.6 seconds if Mick and Richard had begun the whole operation by starting on the most distant section first, but, as I mentioned earlier, we would've then ended up with a much longer time-frame for this job in such a case, and it frankly wouldn't bring me any sort of joy to run though the motions again and figure out the math on it. I mean, this was pretty fun (seriously!), but that's my best effort. Feel free to look it over--really, I think I have taken a reasonable stab at it, but I wouldn't be surprised at all if there were errors of even the most basic kind in this "solution"! I used a calculator app on my computer that required me to point and click on the numbers, for example.
billl wins at life
Well, thanks. Upon reflection, I think I'm possibly way off, in regards to the margin of error--one guy might begin the last section just as the guy doing the next to last is returning (I think), and that sort of situation could push the final guy's work to a 216 second trip. However, another thing I forgot to take into consideration is the fact that time spent with just the one guy working at the end is still *half* as efficient as when they are working together, which brings the margin for error back down to 108 seconds. Hmph... Oh, and the bit about how long it takes Richard to do his part isn't as tricky as I initially imagined, either.
Let's just say I've gotten it right enough, until someone else finds some other problem with it. In the meantime, that means I should be providing the next puzzle.
Last edited by billl; 05-02-2011 at 10:43 PM.
A sheep herder has seven sheep: four white, two brown, and one black. How many of them can each say that they are the same color as another of the herder's sheep?