Apples

[evaporade]

find a way to minimise the forward km, also least number of forward km at below full capacity.

[kamil9876]

Yes that sounds similair to my strategy:

E.g: say you have 2000 apples instead of 3000. We can split this into two lots of 1000. Say we take one of those lots to x. That means you drop off 1000-x apples at x km. Now take the other 1000 and go the whole way. Along the way pick up the apples at x. Hence:

(you have 1000-x apples and want to pick up 1000-x)


Which implies that if you drop off 500 apples at x=500 and then pick them up then that is the minimum x value. Ie: if you drop off 400 at x=600 you have less since you have travelled a total of 1000+600 hence eaten that many apples. However if you drop off apples before x=500 then they will not fit into the truck.

My solution for 3000 does this as well but for two drop of points since we have to make at least 3 different journeys. 750 is most with two drop off points. (imagine that instead of dropping of 250 at x=750 u drop off 600 at x=400. Then those won't fit once your at 600 since at 600 you would have 900 apples in the truck by filling up at x=500), which brings me to my lemma that the best strategy involves you filling up the capacity each time you pick up.
Questions still remain, among others, regarding whether more than the minimum journeys gives more apples.
I think I can prove that maxmimum delivery(assuming you can only deliver once(which is probably the case for apples at A below or equal to 3000)) occurs with only 2 drop off points if 2 drop of points is the minimum  number of drop off points.

[shinny]

in the first table, From step 9 onwards all wrong

Ah sigh, thanks for picking that up. Was pretty sure I had a mistake somewhere (since it didn't make sense that I was only losing 2 apples for 3 distinct trips) but I couldn't find it.

[kamil9876]

I did some research and apparently 533 is the best the source can come up with http://www.keymath.com/documents/daa1/CondensedLessonPlans/DAA_CLP_00.pdf

I did some research and apparently 533 is the best the source can come up with http://www.keymath.com/documents/daa1/CondensedLessonPlans/DAA_CLP_00.pdf

Ahh that problem assumes that the camel needs to eat on the way back, but we havn't assumed so hence our greater number. I googled it to check what it said on the sauce i first got the question from and it said "The car driver has developed an addiction to apples: when he has apples aboard he eats 1 apple with each mile made" So our efforts are not in vain

[evaporade]

First dropoff point at 1000/3 km from A, next dropoff point at 500 km from the first.

[evaporade]

I did some research and apparently 533 is the best the source can come up with http://www.keymath.com/documents/daa1/CondensedLessonPlans/DAA_CLP_00.pdf

534 because the truckie did not eat an apple the last 2/3 km (mi).

[evaporade]

2.) What's the minimum value of apples that he should start of with (rather than 3000) at A so that he can deliver 1000 apples to B?

3666 apples, first dropoff at 1000/6 km, next dropoff at 1000/3 km from the first.

[dcc]

Using an algorithm which is similar to kamil's (possibly the same one), I have shown that if you use drop-off points then at the end you will have:

at the end.

Therefore, I am going to say that if you allow me to have drop-off points, then I will have an INFINITE number of apples at the end.

For sensible answers however, notice that the function is decreasing, so the best solution over the natural numbers is when , that is, a single drop-off point (750 apples).  As to whether there is a better method, that is another question entirely

p.s. I have discovered a truly marvellous proof of the above relation, however this post is too small to contain it.

[dcc]

Actually thats not the best method, obviously.  Its a bit greedy, if you catch my drift.

[evaporade]

Another question: If the capacity of the truck is 999, what is the max delivery to B?

Answer: 833 apples

[dcc]

evaporade, could you post the steps that one must go through to arrive at the destination with 834? I can't seem to replicate your answer :|

[evaporade]

First dropoff point at 1000/3 km from A, next dropoff point at 500 km from the first.

Have to make 3 forward trips (1000 apples each trip) from A to the first dropoff. Total distance = 1000/3 x 3 = 1000 km. .: 1000 apples eaten and 2000 apples left at the first dropoff point.

Have to make 2 forward trips (1000 apples each trip) from the first dropoff to the next dropoff. Total distance = 500 x 2 = 1000 km. .: 1000 apples eaten and 1000 apples left at the second dropoff point.

The remaining distance = 1000 - 1000/3 - 500 = 166 2/3 km.

One final forward trip (1000 apples) of 166 2/3 km to B. .: 166 apples eaten, 834 delivered.

I let you think about why those two dropoff points were chosen.

[kamil9876]

This is similair to shinny's solution. He takes 1000 apples and carries them 1km, then goes back for the next 1000 and carries them 1km, then the last lot carries 1km. So by travelling 1km he loses 3 apples. He loses 1000 apples by travelling 1000/3 km. At this point, he only requires two back trips. And so he just takes one lot, goes back for the other and carries that 1km as well, so he is losing 2 apples per km. until 1000/2=500 km later where he has 1000 apples, and now he only does not have to return, but just go straight there, hence the total apples he comes back with is 1000-(1000 - (1000/3 + 500)).

Note how this is equivalent to evaporade's method, just in a different order ie: instead of going back for each load after 1km, why not 0.5km? 0.1, or even the whole 1000/3 since it's all quite arbitrary. This is because you will lose those apples in the second load by one 333.3 km of trip as much as you will by 333.3 lots of those smaller 1km trips.

[evaporade]

To see one really understand, try to explain

2.) What's the minimum value of apples that he should start of with (rather than 3000) at A so that he can deliver 1000 apples to B?

and

Another question: If the capacity of the truck is 999, what is the max delivery to B?

[kamil9876]

Oh and as a proof:

imagine if he only had to travel 1km. The obvious solution would be 3 lots of 1000, rather than 6 lots of 500 of anything like that. of course 1 lot of 3000 would be best but we cannot have this, hence minimum number of possible trips ())should always be taken. This minimum is occurs until 1000/3.

In comparison to my attempt, or any other one for that matter, is that we went over the same domain too many times. e.g: i went over the (0,500) domain 3 times, whereas evaporade only went over the (333.3,500) domain twice and so he saved 166.7 apples apples there. Going minimum lots by 1km is a safe way to guarantee unnecesary trips, and by looking ahead we can see it has the same outcome as 3 big journeys to 1000/3.