Optimal strategy for cutting a sausage? You are a student, assigned to work in the cafeteria today, and it is your duty to divide the available food between all students. The food today is a sausage of 1m length, and you need to cut it into as many pieces as students come for lunch, including yourself.
The problem is, the knife is operated by the rotating door through which the students enter, so every time a student comes in, the knife comes down and you place the cut. There is no way for you to know if more students will come or not, so after each cut, the sausage should be cut into pieces of approximately equal length. 
So here the question - is it possible to place the cuts in a manner to ensure the ratio of the largest and the smallest piece is always below 2?
And if so, what is the smallest possible ratio?
Example 1 (unit is cm):


*

*1st cut: 50 : 50     ratio: 1  

*2nd cut: 50 : 25 : 25   ratio: 2 - bad


Example 2


*

*1st cut: 40 : 60              ratio: 1.5

*2nd cut: 40 : 30 : 30            ratio: 1.33

*3rd cut: 20 : 20 : 30 : 30    ratio: 1.5

*4th cut: 20 : 20 : 30 : 15 : 15  ratio: 2 - bad


Sorry for the awful analogy, I think this is a math problem but I have no real idea how to formulate this in a proper mathematical way.
 A: YES, it is possible!
You mustn't cut a piece in half, because eventually you have to cut one of them, and then you violate the requirement. 
So in fact, you must never have two equal parts. 
Make the first cut so that the condition is not violated, say $60:40$. 
From now on, assume that the ratio of biggest over smallest is strictly less than $2$ in a given round, and no two pieces are equal. (This holds for the $60:40$ cut.) 
We construct a good cut that maintains this property.
So at the next turn, pick the biggest piece, and cut it in two non-equal pieces in an $a:b$ ratio, but very close to equal (so $a/b\approx 1$).  All you have to make sure is that 


*

*$a/b$ is so close to $1$ that the two new pieces are both smaller that the smallest piece in the last round. 

*$a/b$ is so close to $1$ that the smaller piece is bigger than half of the second biggest in the last round (which is going to become the biggest piece in this round).  


Then the condition is preserved.
For example, from $60:40$ you can move to $25:35:40$, then cut the fourty to obtain $19:21:25:35$, etc.
A: You can't do better than a factor of $2$.
Assume to the contrary that you have a strategy such that the ratio between the largest and smallest remaining piece is always $<R$ for some $R<2$.
Then, first we can see that for every $\varepsilon>0$ there will eventually be two pieces whose ratio is at most $1+\varepsilon$. Otherwise the ratio between largest and smallest piece would be at least $(1+\varepsilon)^n$ which eventually exceeds $R$.
Once you have two pieces of length $a$ and $b$, with $a < b < (1+\varepsilon)a$, eventually you will have to cut one of them.
If you cut $a$ first, one of the pieces will be at most $\frac12 a < \frac12b$, so your goal has immediately failed.
However, if you cut $b$ first, one of the pieces will be at most $\frac12b < \frac{1+\varepsilon}2 a$, which means you've lost if we choose $\varepsilon$ to be small enough that $\frac{1+\varepsilon}2 < \frac1R$. And that is always possible if $R<2$.
