Traffic Flow - Junction, Delay, Webster, Greatest Upper Bound

MidnightR
Messages
39
Reaction score
0
[PLAIN]http://img96.imageshack.us/img96/7816/12530747.jpg

Hopefully this will post successfully...

Erm its the first part I'm not sure on, after that it's easy. I'm just not understanding the wording.

I need to work out the effective green time during the cycle
 
Last edited by a moderator:
Physics news on Phys.org
I believe since q = 0.7 then g > or = 0.35c what would this make the greatest lower bound? g = 0.35c?
 
Last edited:
Think I got it,

c=? q=0.7 s=2 therefore

g/c <=1
0.35c/g <= 1

therefore

g <= c <= g/0.35

we know c = g +10 so g= 10-c & sub in

c-10 <= c <= (c-10)/0.35

so assuming c is positive then

c >= 200/13 so take greatest lower bound = 200/13

however I'm still not sure that c = g+10 rather than c=2g+10
 
Last edited:
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...

Similar threads

Replies
0
Views
5K
Replies
48
Views
66K
Replies
2
Views
9K
Replies
7
Views
3K
Replies
7
Views
4K
Replies
2
Views
4K
Replies
29
Views
6K
Back
Top