Policing 5Mbit/s

[moitani]

If I have a 5M circuit, how will the SP police?

police 5000000 625000

OR:

police 5000000 156250

Because if it was the second option, I wouldn't be really getting full bandwidth right?

Thanks,

- Mohammad

[Mahesh Gohil]

Hi Mohammad,

This vary from provider to provider.

like i as an provider is using police 5120000 3200000 (Formula bc=cir*1.5/8 )and some people are using

like  5120000 640000 (bc=cir/8).

No matter what formula used you will get 5mb in both cases..reason being bc decides your tc interval (tc=bc/cir in ms)...so bc is amount of

burst of data allowed in that tc interval.

so if your cir=5meg you surely get that one but bc decided what amount of data passed in tc interval

hope this is helpful to you

Regards

Mahesh

[moitani]

But what about the second option I stated:

police 5000000 156250

This makes the bc as (cir/8)/4 (which btw is the default if you don't specify).

If I was sending at a constant rate I wouldn't be getting full bandwidth if this bc was used right?

Thanks,

- Mohammad

[Mahesh Gohil]

Hi mohammad,

In that case also you will get 5meg bw..reason is for this setting tc will be 31.25 ms. so for every 31.25 ms

your 156250 bits of data will be allowed.

and for 1000ms it will be 156250*1000/31.25 = 5000000.

I hpe this is clear now

Regards

Mahesh

[moitani]

Thank you Mahesh.

In the statement:

police 5000000 156250

156250 is in bytes not in bits.

It equals to 1250000 bits and so tc will be equal to 250 ms.

So as per your explanation I will be able to send 156250 bytes every 250 ms.

Having said that why do I have 50 percent packet drops with a high packet size?

ping 172.16.133.57 size 5000 repeat 10

Type escape sequence to abort.

Sending 10, 5000-byte ICMP Echos to 172.16.133.57, timeout is 2 seconds:

!.!.!.!.!.

Success rate is 50 percent (5/10), round-trip min/avg/max = 1/1/4 ms

What difference does the bc make?

Thanks,

- Mohammad

[Mahesh Gohil]

Hi mohammad,

Policing work little differently.It allow bytes of data based on packet arrival time. for ex.

you allowed to pass 5000000 bits of data in one second. so if 1 second elapsed since last packet it allows

5000000  bits of data and suppose 0.1 second elapsed then it allows 5000000/10 bits of data and so on...

so the formula is [current packet arrival time - previous packet arrival time * police rate /8]

so in your case if 1250000 bits of data allowed per 250 ms and

suppose if you send packet of 1250000 bits for first time period of 250ms and say there is silent period of 250 ms

so allowed bits of data is 1250000*2....but what if you send 1250000*3 bits? ofcourse the last 1250000 bits will be dropped

(if there is single tocken bucket used)

But the thing is how do you mesure this..i don't think you will see anything on output for every 250ms....all you can see if drop counter increased

in output of "sh policy-map interface xx" command.

Regards

Mahesh