Traffic Management for the Available Bit Rate (ABR) Service in ...

Proof for Region 2
Points in the triangular region 2 satisfy the conditions: y s, x s, and x + y
U(1 + )
In this region, both x and y are greater than or equal to the fair share s = U=2.
Therefore, the new point is given by : (x0�y0 x(1; )
)=( � z
x 0 + y 0 =
x(1 ; )+y(1 ; )
z
= (x + y)(1 ; )
z
= Uz(1 ; )
y(1; )
).Hence,
z
z
= U(1 ; )
This indicates that the new point isonthelower line of the TUB (which is a part of
the TUB) This proves claim C1 for all points in region 2.
The proof of claim C1 for regions 3 and 4 is similar to that of regions 1 and 2,
respectively.
5.8.2 Proof of Claim C2
We show convergence to the fairness region (claim C2) as follows. Any point
in the fairness region remains in the fairness region. Further, any point (x� y) in
the TUB but not in the fairness region moves towards the fairness region at every
step. Consider the line L joining the point (x� y) to the origin (0� 0) as shown in
Figure 5.25(a). As the angle between this line and the fairness line (x = y) decreases,
the operation becomes fairer. We show that in regions outside the fairness zone, the
angle between the line L and the fairness line either decreases or remains the same. If
the angle remains the same, the point moves to a region where the angle will decrease
in the subsequent step.
We introduce four more lines to Figure 5.25(a). These lines correspond to y =
(1 + )x� y = (1 ; )x� y =
(1; )
(1+ )
x and y = x. This results in the TUB
(1+ ) (1; )
141