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This question already has an answer here:

You are living in the 22nd century. The fuel prices are unimaginably high and it is very rare to get, so the Government has introduced some new laws to control fuel smuggling.

Rule 1: One cannot carry more fuel than their vehicle's tank capacity (fuel cannot be carried by hand or any other method. The only way to transfer fuel from one place to another place is by filling it in the vehicles tank and driving it to the destination. Then the fuel can be drained from the tank).

Rule 2: All vehicles have to be run on their own fuel (one cannot fill the tank and then push the vehicle, or pull with another vehicle to the destination or use alternate energy).

Rule 3: Fuel can only be sold at the Fuel Market run by the Government.


Now, you have 300 liters of petrol at your home (saved by your ancestors). You have decided to sell it at the Fuel Market. But the problem is that the Fuel Market is 1000 km away from your home. Your car has a tank capacity of 100 liters and a mileage of 10 km/liter. You know that there are storage stations run by old mafia members every 1 km and they will charge 1 million dollars for storage (you can drain as much amount of fuel from your car and store there if you want). Once you reach the Fuel Market, you will get a price of 1 million dollars per liter of petrol. What is the maximum amount of money you can make by selling the fuel at the Fuel Market and how you can achieve it?


IMPORTANT: You have to use your car. You cannot borrow or buy another car with more tank capacity and/or mileage. Also you can abandon your car after selling the fuel if you want since your car costs only a few dollars


Note: I don't know the answer. So you have to prove your answer with details to get accepted.

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marked as duplicate by Joe Z., Psychemaster, skv, Kevin Nov 27 '14 at 6:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Doesn't it depend how much money you have at the beginning?? $\endgroup$ – Rand al'Thor Nov 26 '14 at 12:31
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    $\begingroup$ With such fuel prices, why are cars consuming so much more gas/kilometer than they do today??? $\endgroup$ – oerkelens Nov 26 '14 at 12:53
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    $\begingroup$ @AeJey - if you have practically unlimited funds, why do you want to sell the fuel anyway? $\endgroup$ – Rand al'Thor Nov 26 '14 at 13:01
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    $\begingroup$ If fuel-guzzling cars are so rare, why is there such a demand of fuel to justify such prices? Higher fuel prices lead to more efficient cars because high fuel prices justify the extra R&D. I'm suspecting government involvement in creating an artificially high demand and thus maintaining the high fuel prices. $\endgroup$ – oerkelens Nov 26 '14 at 13:33
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    $\begingroup$ I don't think this is a complete duplicate. In the camel question, storing bananas is free; but in this question, storing gas reduces your profits. This may change what the optimal strategy is. $\endgroup$ – Kevin Nov 26 '14 at 13:39
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1 million dollars is such a small storage charge that the solution is identical to the camel and bananas puzzle. See here. Use storage areas at the 200 km and 534 km points. We arrive with 53.3 liters of fuel. Since the problem uses integer kilometer distances only, hence multiples of .1 liters only, this is the closest we can get to the camel and bananas puzzle's 53 1/3 bananas.

Can we do better using fewer storage steps? Well, we only used 3 storage steps in this answer. Good luck using 0 storage steps. If you use 1 storage step, then it is straightforward to prove that you can't get anywhere near 53 liters. (Casework on the storage's position.) 2 storage steps works out the same way. It's intuitively true, but it's a messy argument, so I'll avoid putting it here for both of our sakes. In conclusion, the camel and bananas answer is the best we can do. Hey, it gets us a cool 50.3 million dollars.

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  • $\begingroup$ They do charge per storage, though, not only once. $\endgroup$ – No. 7892142 Nov 26 '14 at 15:27
  • $\begingroup$ In that case, this solution uses 3 entire storages instead of 2. $\endgroup$ – Lopsy Nov 26 '14 at 15:30
  • $\begingroup$ I was just saying! $\endgroup$ – No. 7892142 Nov 26 '14 at 15:31
  • $\begingroup$ Alright :) Answer edited to reflect that. $\endgroup$ – Lopsy Nov 26 '14 at 15:36
  • $\begingroup$ If you use 533 km as the drop-off point, you can get 50.4 million. $\endgroup$ – Joe Z. Nov 28 '14 at 16:12
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I did 4 sets of analysis. On the way, i realised that whichever way he takes, he has to have approximately 100 litres, no matter how many intervals he plan. If he will have lesser than 100 litres on the next interval trip, he is better off driving to the end and give up the extra fuels on the previous interval trip.

FIRST SET

The first set is in intervals of 100km. He brings first 100 litre to first checkpoint, puts down 80 and drive back. Bring 2nd 100 litre, puts 80 and drive back. Take last set of 100 litre and drive to 1st checkpoint.

Mafia Storage: 2 times
Remaining Fuel: 250 litres
Distance Covered: 100 km

He repeats again, bringing first 100 litres to 2nd checkpoint, puts 80 down and drive back. Takes 2nd set, puts 80 down and drive back. Takes last 50 litre and drive to 2nd checkpoint.

Mafia Storage: 4 times
Remaining Fuel: 200 Litres
Distance Covered: 200 km

He repeats the process over again and he will arrive at 3rd checkpoint with:

Mafia Storage: 5 times
Remaining Fuel: 170 Litres
Distance Covered: 300 km

He repeats the 4th time:

Mafia Storage: 6 times
Remaining Fuel: 140 Litres
Distance Covered: 400 km

5th time:

Mafia Storage: 7 times
Remaining Fuel: 110 Litres
Distance Covered: 500 km

From here, he doesn't have to go back and collect the 10 litres as he can drive to the end and would have made:

100 - 50 - 7 = 43 Mil

Instead of:

80 - 40 - 8 = 32 Mil

SECOND SET

The 2nd set is in intervals of 200 km. He brings the first 100 litre to the first checkpoint and he will be left with 80. Puts down 60, drive back with remaining 20. Takes next 100 litre and repeats. The data will be:

Mafia Storage: 2 times
Remaining Fuel: 200 litres
Distance Covered: 200km

He repeats again for the 2nd time to 2nd checkpoint, bringing 100 litres. Puts 60 down, drive back. Takes last set of 100 and drive back to 2nd checkpoint.

Mafia Storage: 3 times
Remaining Fuel: 140 litres
Distance Covered: 400km

From here on, he can take the 100 litres and continue driving to the end. He would have made:

100 - 60 - 3 = 37 Mil

THIRD SET

The 3rd analysis is in intervals of 300 km. He brings first 100 litre to first checkpoint, puts down 40 and drive back. He brings 100 again, puts down 40 and drive back. He takes last set of 100 and drive straight to 1st checkpoint.

Mafia Storage: 2 times
Remaining Fuel: 150 litres
Distance Covered: 300 km

He can now drive straight to the end and he would have made:

100 - 70 - 2 = 28 Mil

Instead of:

60 - 40 - 3 = 17 Mil

FOURTH SET

The last analysis will be intervals of 400km.

Starts off taking 100 to 1st checkpoint, puts 20 down and drive back. Takes 100 and drive to 1st checkpoint, drop 20 and drive back. Take last set of 100 and drive to 1st checkpoint.

Mafia Storage: 2 times
Remaining Fuel: 100 km
Distance Covered: 400 km

From here on, he can drive directly to the end because he can contain all his fuel, therefore he would have made:

100 - 60 - 2 = 38 Mil

Through all these data, we can say that the first set is at the top with 43 mil followed by 4th set with 38 mil. I have tested a few more sets privately to see if lower interval trips are better, but it seems like the storage is killing the bank (intervals of 50 and 1). Therefore, i am confident to say that the first set will earn him the most.

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Edited:
52.6 million, if the mafia will take 1 liter of fuel instead of 1 million

Go out to 196km, pay 1 liter, drop off 59.8, drive back. Repeat. Go back to 196km. At this point you have 200 liters.

Go out 526km (+330km), pay 1 liter, drop off 33, drive back. Go out 526km, fill up, drive to depot and sell the 52.6 liters.

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I thought this was an exact duplicate of the camel transporting bananas problem, but some people have pointed out the difference that in this case, it costs money to deposit fuel.

The way it's laid out, however, means that the problem is numerically identical to the camel problem except with the added condition that it costs 10 bananas to set down bananas at any location. So I will give the solution framed in those terms for easier analogy.

This answer by durron597 gives us a formula for the cost of moving bananas assuming the camel can carry $c$ bananas:

For bananas $1$ through $c$, it costs $1$ banana per km to move a banana.
For bananas $c+1$ through $2c$, it costs $3$ banana per km to move a banana.
For bananas $2c+1$ through $3c$, it costs $5$ banana per km to move a banana.

Since we want a solution that uses as few "banana-setdowns" as possible, we can't just shunt the pile of bananas along one kilometre at a time. We need a solution that moves $200$, then $333$, then $1,000$ bananas all at once.

Consider the following list of steps:

  • Load up 1,000 bananas, carry them 200 km forward.
  • Set down 600 (-10) bananas, carry the rest back.
  • Load up 1,000 bananas, carry them 200 km forward.
  • Set down 600 (-10) bananas, carry the rest back.
  • Load up 1,000 bananas, carry them 200 km forward.

You now have 1,980 bananas 200 km ahead, 800 of which are on the camel.

  • Load up 200 more bananas, carry them 333 km forward.
  • Set down 334 (-10) bananas, carry the rest back.
  • Load up 980 bananas, carry them 333 km forward.

You now have 324 + 647 = 971 bananas 533 km ahead, 647 of which are currently on the camel.

  • Load up the remaining 324 bananas, and carry them 467 km ahead.

You're left with 504 bananas. In terms of the car's fuel price, this means you can sell the fuel in your car for 50.4 million dollars.

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