Four people must evacuate from an island to the mainland. The only link is a narrow bridge which allows passage of two people at a time. Moreover, the bridge must be illuminated, and the four people have only one lantern among them. After each passage to the mainland, if there are still people on the island, someone must bring the lantern back. Crossing the bridge individually, the four people take 2, 4, 8 and 16 minutes respectively. Crossing the bridge in pairs, the slower speed is used. What is the minimum time for the whole evacuation?
I took various combinations of 2,4,8 and 16 and added the various combinations. This is a very long method of calculation. First 2m,4m -> 4m ... Then 2m will return so 4+2=6m ... Then 2m,8m -> 8m ... So 6+8=14m ... Then 2m will return so 14+2=16m ... Then 2m,16m -> 16m ... So 16+16=32m
So, I get the answer as 32minutes. Is this correct?
I presented a shorter solution with 30 minutes length in the comment above:
2 and 4 go -> takes 4 mins, 2 goes back, then 8 and 16 go -> takes 16 mins, then 4 goes back, then 2 and 4 go again, all in all 4+2+16+4+4=30 minutes.
And I thought a bit if you can prove that it's the shortest one. I think you can as follows:
One obviously needs (at least) 3 rounds to bring all four to the other end of the bridge. Every passage of the bridge away from the island lasts at least 4 minutes, since the best you can do is send the 2 minute and the 4 minute guy. One passage will have to feature the 16 minute guy, so the times for the three crossings away from the island will be at least $4+4+16=24$ minutes. The fastest combination for the two crossings back is $2 \cdot 2=4$, but this is only possible if the 2-minute-guy is always at the end of the bridge, which means that he has to move all the time, leading to the solution proposed by the OP above which lasts 32 minutes. The second best option leads to $2+4$ minutes and gives my solution. This shows, admittedly with a lot of words, that my suggested solution is indeed the shortest possible one.