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

Proof for Region 2
Triangular region 2 is de ned by the conditions: y s and x s and x + y
U(1 + ). This region is completely enclosed in the fairness region. The new point
is given by:
Observe that:
x 0 =
x(1 ; )
z
and y 0 =
y(1 ; )
z
y0 y
=
0 x x and x0 + y 0 (x + y)(1 ; )
=
z
= U(1 ; )
That is, the new point isattheintersection of the line joining the origin and the old
point and the lower boundary of the TUB. This intersection is in the fairness region.
This proves Claim C2 for all points in region 2.
Proof for Region 4
Triangular region 4 is de ned by the conditions: y < s and x < s and x + y
U(1 ; ). The new point isgiven by:
Observe that:
x 0 =
x(1+ )
z
y0 y
=
0 x x and x0 + y 0 =
and y 0 =
(x + y)(1 + )
z
y(1 + )
z
= U(1+ )
That is, the new point isattheintersection of the line joining the origin and the old
point and the upper boundary of the TUB.
As shown in Figure 5.25(b), region 4 consists of 3 parts: 4a, 4b, and 4c. All
points in region 4a are inside the fairness region and remain so after the application
of the TUB algorithm. All points in region 4b move to region 1a where subsequent
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