Balanced Sequences and Optimal Routing

  • Thread starter Thread starter rhj23
  • Start date Start date
  • Tags Tags
    Sequences
rhj23
Messages
20
Reaction score
0
I've been reading the paper on Balanced Sequences and Optimal Routing (Altman, Gaujal, Hordijk; 2000). However, there are a couple of proofs given that I don't quite follow. There are statements made that are assumed to trivially follow, but I can't see how and am hoping someone will be able to help me.

The first is in the proof of Proposition 2.16. The fact that l_i >= (n-1)l_1 + n is easily shown, but 'on the other hand' l_i <= n(l_1) - 3 does not seem to follow from any similar method.

The second is in the proof of Theorem 2.21, Step (2). Where does the fact that |s_1| >= max{4, 2(n+m)+1} come from. (Obviously the 4 is trivial, but I do not understand the 2(n+m)+1)

The paper is attached; I hope that someone with a better understanding than me will be able to follow the proofs and let me know where I'm missing something obvious!

Thanks
 

Attachments

Physics news on Phys.org
A question

Sorry not to provide any answer to your questions? But if you don't mind, I hav a question about that paper on Balanced sequences and optimal routing. I hav been looking for some particle applications of the results presented in that paper, but unfornately I havn't been abled to find a good one, so far. May be it is due to my lack of knowledge about queuing networks.
Therefore, I will be more than please, if you can provide me with an application, or any link or article where I will be able to find some.
Thanks.

ml7
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top