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

have:
and
y 0 =
y(1 ; )
z
x + y 0 =
xz + y(1 ;
z
)
= U(1 ;
xfz ; (1 ;
)+
z
)g
= U(1+
xf(1 +
);
) ; zg +2y
z
(5.11)
(5.12)
(5.13)
(5.14)
Since, the last terms of equations 5.12 and 5.13 are both positive, the new point is
still in the TUB. This proves Claim C1.
Further, we have:
Therefore,
y 0
x
y 0
x
< y
x
y
= (1 ; )
x
and y0
x
(1 ; )
That is, the slope of the line joining the operating point to the origin decreases but
does not overshoot the fairness region.
Note that when z =1; , y 0 = y. That is, the operating point doesnot change.
Thus, the points on the lower boundary of the TUB ( x + y = U(1 ; ) ) do not
move, and hence the fairness for these points does not improve in this step. It will
change only in the next step when the operating point moves from (x� y 0 ) to (x 0 �y 0 ).
The proof for the case (x 0 �y) is similar. This completes the proof of C1 and C2
for region 1. The proof for region 3 is similar.
146