I currently prepared my ONT certification Exam.
In the preparation of my exam certification I but ONT HowtoPass pratice Exam to make sure I'm ready to go at test center.
One question in the practice Exam is :
A RED profile is configured with a mark probability denominator of 1. What will be the result about this configuration on the packet lost as the average queue length reaches the maximum threshold ?
Base on my understand is since the average queue size is lower than minimum threshold, no drop occur, when minimum treshold is reach the mark probablylity denominator say drop 1 paquet on 1 paquet which is mean 100%. How can maximum threshold be be reach ?
Because the answer was : When the average queue lenght is at the maximum threshold 100% are drop.
sorry If I've confused you.
However, the p of the mark denominator tells what is the arrival point of drop probability when average queue size is reaching max threshold.
x= average queue size
D= probability of be dropped
thr1= min threshold
thr2= max threshold
if x < thr1 D= 0 no packets dropped
if x > thr2 D=1 packets are dropped
if thr1 < x < thr2
D= (x - thr1) * 1 / [(thr2-thr1)*p]
D=1 as x -> thr2-
if p>1 the arrival point of D function is not 1 when x -> thr2 but 1/p
For this reason I say that using a p>1 when permitted provides a better treatment for packets when x average queue size is between the two thresolds
For mark-prob-denominator, enter the denominator for the fraction of packets dropped when the average queue size is at the >>maximum threshold<<. The range is 1 to 65535. The default is 10. For example, if the denominator is 512, one out of every 512 packets is dropped when the queue is at the maximum threshold.
there were some nice graphs about this.
From a maths point of view the D function has a discontinuity point in x=thr2 when p>1
1/p for x= thr2-
1 for x= thr2+
Hope to help