Originally posted by NemesioI don't think there is a "best"; unless maybe it's one camel pace. It seems like a limiting problem.
If I take it in a total of 25 forty-mile increments, my amounts are:
40 - 9240 kg (920 kg each time, plus 40 kg on the final trip).
80 - 8480 kg
120 - 7800 kg
160 - 7200 kg
200 - 6600 kg
240 - 6080 kg
280 - 5560 kg
320 - 5120 kg
360 - 4680 kg
400 - 4320 kg
440 - 3960 kg
480 - 3680 kg
520 - 3400 kg
560 - 3120 kg
600 - 2840 kg
640 - 2640 kg ...[text shortened]... despite the problems
with my summary above).
Davegage, am I on the right track?
Nemesio
Originally posted by NemesioNemesio:
If I take it in a total of 25 forty-mile increments, my amounts are:
40 - 9240 kg (920 kg each time, plus 40 kg on the final trip).
80 - 8480 kg
120 - 7800 kg
160 - 7200 kg
200 - 6600 kg
240 - 6080 kg
280 - 5560 kg
320 - 5120 kg
360 - 4680 kg
400 - 4320 kg
440 - 3960 kg
480 - 3680 kg
520 - 3400 kg
560 - 3120 kg
600 - 2840 kg
640 - 2640 kg ...[text shortened]... despite the problems
with my summary above).
Davegage, am I on the right track?
Nemesio
You and AThousandYoung are both are on the right track in as much as you should arrive at the right answer if you consider the limit as you go to smaller and smaller intervals. If you want to formulate the problem in that way, be my guest, but it may be tricky to set-up the problem and figure out the solution (I'm speculating here because I didn't solve the problem this way).
Fortunately there is an easier (more elegant?) way to get the right answer.
I would like to give you some hints, but I don't want to give away too much yet. For now, consider the following hints:
1) Consider two different cases -- case A, where the increment is 10 miles, and case B where each increment is 1 mile. At the ten mile marker, both methods will get 9810 KG there. After twenty miles, both methods will get 9620 KG there. After thirty miles, both methods result in 9430 KG. Now you might be saying how can this be when I thought that case B should be doing better than case A. Well, if you carry the analysis through, case B will get more grass to Damascus than case A, but the question is at what points along the way does case B do the better job?
2) There is nothing that says the increments you analyze must be constant, or uniform, throughout the trip.
Admittedly, I am not sure how clear or helpful these hints are because I am trying to not be too obvious. But if they don't help, let me know and I will provide a hint that is much more laser-guided...
Originally posted by davegageAh. I think I got it.
1) Consider two different cases -- case A, where the increment is 10 miles, and case B where each increment is 1 mile. At the ten mile marker, both methods will get 9810 KG there. After twenty miles, both methods will get 9620 KG there. After thirty miles, both methods result in 9430 KG. Now you might be saying how can this be when I thought that case B ...[text shortened]... cus than case A, but the question is at what points along the way does case B do the better job?
Bring your load through as few multiple trips as possible, cutting it as close as possible to the
nearest thousand kg mark (your first step will be 53 miles forward, with 8993 kg). Then creep
forward one mile (so, mile marker 54 with 8976 kg). Repeat. The next step will be somewhat
larger (because you have fewer trips to make). I worked it out kinda wrong and its too late to
figure out the precise figure, but this seems to optimize the kg.
Did I get the principle davegage?
Nemesio
1 edit
Originally posted by NemesioYes, you are onto the right idea.
Ah. I think I got it.
Bring your load through as few multiple trips as possible, cutting it as close as possible to the
nearest thousand kg mark (your first step will be 53 miles forward, with 8993 kg). Then creep
forward one mile (s ...[text shortened]... o optimize the kg.
Did I get the principle davegage?
Nemesio
If you want to get the answer you would get if you considered infinitesimally small increments, don't limit yourself to whole mile increments, ie, on the first leg go 1000/19 = 52.63157..... miles, etc.
It shouldn't take too long before you arrive at the optimal answer...
EDIT: The idea behind this approach is that the camel never starts any leg of the trip in the direction of Damascus without carrying his maximum load of 1000 KG. Hence it makes it optimal.
Originally posted by davegageOk. Well the best I could do was this (I don't have the patience to use anything
Yes, you are onto the right idea.
If you want to get the answer you would get if you considered infinitesimally small increments, don't limit yourself to whole mile increments, ie, on the first leg go 1000/19 = 52.63157..... miles, etc.
It shouldn't take too long before you arrive at the optimal answer...
EDIT: The idea behind this approach i ...[text shortened]... direction of Damascus without carrying his maximum load of 1000 KG. Hence it makes it optimal.
smaller than whole numbers):
Make 10 trips of 53 miles, which leaves you with 8993 kg at mile 53.
Make 9 trips of 59 miles, which leaves you with 7990 kg at mile 112.
Make 8 trips of 67 miles, which leaves you with 6985 kg at mile 179.
Make 7 trips of 76 miles, which leaves you with 5997 kg at mile 255.
Make 6 trips of 91 miles, which leaves you with 4996 kg at mile 346.
Make 5 trips of 111 miles, which leaves you with 3997 kg at mile 456.
Make 4 trips of 143 miles, which leaves you with 2996 kg at mile 599.
Make 3 trips of 200 miles, which leaves you with 1996 kg at mile 799.
Make 2 trips of the remaining 201 miles, which leave you with 1393 kg at Damascus.
I'm going to take a stab and guess that doing fractional milage increments saves
you an additional 50 kg at most (the difference if I landed on even 1000-kg increments
above).
Nemesio
Originally posted by NemesioWhen I do the analysis with non-whole number increments, I get the optimal number to be 1400 KG (actually, just slightly less than 1400, but it's 1400 if you round to the nearest KG), so your analysis above is quite close to this.
Ok. Well the best I could do was this (I don't have the patience to use anything
smaller than whole numbers):
Make 10 trips of 53 miles, which leaves you with 8993 kg at mile 53.
Make 9 trips of 59 miles, which leaves you with 7990 kg at mile 112.
Make 8 trips of 67 miles, which leaves you with 6985 kg at mile 179.
Make 7 trips of 76 miles, which le ...[text shortened]... itional 50 kg at most (the difference if I landed on even 1000-kg increments
above).
Nemesio
Originally posted by davegageSince the problem is cracked open now, do you care to give your analysis?
When I do the analysis with non-whole number increments, I get the optimal number to be 1400 KG (actually, just slightly less than 1400, but it's 1400 if you round to the nearest KG), so your analysis above is quite close to this.
Nemesio
Originally posted by NemesioMy analysis is similar to yours, except that I consider non-whole number increments. It does slightly better than yours only because it is more optimal right near the 1000 KG increment points.
Since the problem is cracked open now, do you care to give your analysis?
Nemesio
So there will be 9000 KG after 1000/19 miles
8000 KG after 1000(1/19 + 1/17) miles
7000 KG after 1000(1/19 + 1/17 + 1/15) miles
6000 KG after 1000(1/19 + 1/17 + 1/15 + 1/13) miles
5000 KG after 1000(1/19 + 1/17 + 1/15 + 1/13 + 1/11) miles
4000 KG after 1000(1/19 + 1/17 + 1/15 + 1/13 + 1/11 + 1/9) miles
3000 KG after 1000(1/19 + 1/17 + 1/15 + 1/13 + 1/11 + 1/9 + 1/7) miles
2000 KG after 1000(1/19 + 1/17 + 1/15 + 1/13 + 1/11 + 1/9 + 1/7 + 1/5) miles
So at this point, we have 2000 KG and only about 200.0778032.... miles left. So in Damascus, we are left with 2000 - 3(200.0778032) = 1399.76659... KG
So in terms of whole number increments of KG, 1399 would be the most we can get there. If we round to the nearest KG, then we get 1400 KG.