question about RED

Answered Question

Hi !


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.

Correct Answer by Giuseppe Larosa about 8 years 6 months ago

Hello Christian,

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]


if p=1

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


see


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.


http://www.cisco.com/en/US/docs/switches/metro/catalyst3750m/software/release/12.2_35_se/configuration/guide/swqos.html#wp1281938


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

Giuseppe


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Giuseppe Larosa Thu, 01/08/2009 - 12:15
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Hello Christian,

the WRED allows for


0% loss for average queue size < thershold1

0 < x < = 1/p for queue size > thr1 <= thr2


where p the is mark probability denominator

for queue size > thr2 the packet drop probability is 1


to be noted no all hardware allows to configure and use a p >1


if p>1 the packets receive a better treatment between the two thresholds.


Then you need to think that you can have different thresholds for different traffic classes (up to 8 for TOS based WRED up to 64 for DSCP based QoS)


So I think that the answer provided by the book is correct.


Hope to help

Giuseppe


Hello Giuseppe,

You has lost me here : 0 < x < = 1/p for queue size > thr1 <= thr2 (from your text)


What is the "x" in your equation ? I think it's avreage queue size. But the equation it self let me confused...

also you wrote :

"to be noted no all hardware allows to configure and use a p >1 "

is it possible you had made a mistake and you want wrote : p >= 1 (equal or grather than) ?

also you wrote :

if p>1 the packets receive a better treatment between the two thresholds. (I don't understand what you mean exactly here.... I know as the queue lengh grow from minimum threshold to maximum threshold, random dropping rates also grow to finally reach 100% dropping at maximum threshold)

When I read on RED and WRED, I has understand these 3 dropping mode : no drop mode, random drop mode and finally full drop mode. Still the average queue size is lower than minimum threshold we are in no drop mode (no packet is drop). If minimum threshold is pass, we pass in random mode where 1 packet of each (mark probability denominator) value is drop and this is grow to ratio (mark probability denominator)/(mark probability denominator) at maximum threshold value. Finally when the maximum threshold is reach, as I saied all packet are drop.

My problem is here : if I choose value 1 to set as (mark probability denominator) we obtain ratio 1/1 which mean 100%, I it should be the queue is fill still is the average queue is below then the minimum threshold but when minimum threshold is reach we have 1/(mark probability denominator) to be drop or 1/1 in the present case. How I can reach maximum threshold ? Because it was the answer of the question on the pratice test ?

Correct Answer
Giuseppe Larosa Sat, 01/10/2009 - 03:14
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Hello Christian,

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]


if p=1

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


see


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.


http://www.cisco.com/en/US/docs/switches/metro/catalyst3750m/software/release/12.2_35_se/configuration/guide/swqos.html#wp1281938


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

Giuseppe


Hi Giuseppe,


Now I had understand you, I was understand previously when I read the book, the drop probability is establish for the minimum threshold not the maximum. After explication, and I put some fictif number in your formula every become clear enough and now make sence...


thanks a lot for your help....


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