CCIE Level question

Calculating flows for WFQ.

Flow 1 - 500 byte packets with IP precedence 3.
Flow 2 - 250 byte packets with IP precedence 3.
Flow 3 - 250 byte packets with IP precedence 1.

For every 1 packet of flow 1 that is serviced, how many packets of flow 2 and flow 3 will be serviced?

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Ricka182

I have no clue, but my guess is:

Uhhh, WFQ=WTF!!!

1/2 packet for flow 2,and 1/2 for flow 2...????? Sorry, I have never touched a Cisco book, I just like trivia...

viper75

mama!!!

Drakonblayde

tunerX wrote:

Calculating flows for WFQ.

Flow 1 - 500 byte packets with IP precedence 3.
Flow 2 - 250 byte packets with IP precedence 3.
Flow 3 - 250 byte packets with IP precedence 1.

For every 1 packet of flow 1 that is serviced, how many packets of flow 2 and flow 3 will be serviced?

I'm going to assume the weight being used is post IOS 12.0(5)T.

Let's see, the weight for flow 1 would be....

32384 / 2000 = 16.192

Flow 2 would be

32384 / 1000 = 32.384

Flow 3 would be

32384 / 500 = 64.768

So for every one packet that flow 1 gets through, flow 2 will get two through and flow 3 will get 4 through.

At least, I think

QoS isn't something I'm delving into in depth at present.

Orion82698

I'm going to assume the weight being used is post IOS 12.0(5)T.

Let's see, the weight for flow 1 would be....

32384 / 2000 = 16.192

Flow 2 would be

32384 / 1000 = 32.384

Flow 3 would be

32384 / 500 = 64.768

So for every one packet that flow 1 gets through, flow 2 will get two through and flow 3 will get 4 through.

At least, I think icon_smile.gif QoS isn't something I'm delving into in depth at present.

Wow! Man...... I thought addition was hard

Drakonblayde

hehe, I'm not sure if I'm right, I remember reading over Weighted Fair Queueing in one of my books, and I had to look up the weight for the post 12.0(5)T IOS (pre it's 4096), but after that, it's a fairly simple formula...

Weight = 32384 ( or 4096) / (IP Precedence + 1) * PacketSize

After that, just do the math and whatever comes out with the higher weight has more priority

tunerX

You got the formula right but the weights backwards. With WFQ lower weights are better.

flow 1: 32384/(3+1)=8096
flow 2: 32384/(3+1)=8096
flow 3: 32384/(1+1)=16192

If we assume that all flows started at the same time. The initial value of the SN (sequence numbers) for each flow will be zero. To make things easier, I will drop all of the numbers on the right.

flow 1: SN1=(8096*500)=4048000
flow 2: SN1=(8096*250)=2024000
flow 3: SN1=(16192*250)=4048000

all following serial numbers are calculated by adding the previous serial number to the product of weight and length.

SN = (previous SN + (weight*length))

On average for every 1 packet that flow 1 sends, flow 2 will have sent 2 and flow 3 will send 1.

Queue

F1: SN3~12, SN2~8, SN1~4
F2: SN3~6, SN2~4, SN1~2
F3: SN3~12, SN2~8, SN1~4
Queue

Drakonblayde

tunerX wrote:

You got the formula right but the weights backwards. With WFQ lower weights are better.

DOH! Well, I'll never forget that little tidbit again

Hrm, my math was also wrong. I guess the proper way for the formula then would be Weight = (32384 / (IP Precedence + 1)) * PacketSize.