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How many packets I can send per second?

George-Sl
Level 1
Level 1

Hello Net Architects,

 

1)

Let's say I have a connection of 300KB(kbytes) DL and 150KB(kbytes) UL P/S,

How many HTTPS packets of 674Bytes I can send?considering that that HTTPs packet is a request to a data that is 406Bytes

Bytes Sent: 674
Bytes Received: 406

 

so

(150*1024)/674=227 Packets P/S which means it should take 4 milliseconds to send a packet, so the question is, is that the send speed to my ISP(from my router to my ISP router) or to the Server(My router to Server's Router)?

 

2) I sniffed my https packet to that particular server

ClientConnected: 18:08:51.598
ClientBeginRequest: 18:09:18.980
GotRequestHeaders: 18:09:18.980
ClientDoneRequest: 18:09:18.980
Determine Gateway: 0ms
DNS Lookup: 0ms
TCP/IP Connect: 0ms
HTTPS Handshake: 0ms
ServerConnected: 18:09:16.195
ServerGotRequest: 18:09:18.980
ServerBeginResponse: 18:09:19.047
GotResponseHeaders: 18:09:19.047
ServerDoneResponse: 18:09:19.047
ClientBeginResponse: 18:09:19.047
ClientDoneResponse: 18:09:19.047

Overall Elapsed: 0:00:00.067

 

why the Overall Elapsed does not count the time from ClientConnected to ClientDoneResponse?

what is serverconnected and clientconnected?, their timestamp is off the chart?!

 

3)

I also found this

ESTIMATED WORLDWIDE PERFORMANCE
--------------
The following are VERY rough estimates of download times when hitting servers based in Seattle.

US West Coast (Modem - 6KB/sec)
RTT: 0.10s
Elapsed: 0.10s

 

what type of packet we refer to when we say RTT? ICMP or any packet, because it's different

 

thx a lot

2 Accepted Solutions

Accepted Solutions

Giuseppe Larosa
Hall of Fame
Hall of Fame

Hello George,

as already noted the link speeds in telecommunications are expressed in bit/s and not byte/s.

Also in telecommunications we dot not use a 1024 multiplier but 1000 like international metric system

 

1 Kbps = 1000 bps

1 Mbps = 1000000 bps

And so on.

Of course packet size is in bytes and 1 byte  = 8 bit.

 

However, in telecommunications you need to take in account the so called protocol overhead, that depends on the OSI layer 2 technology in use.

example for ethernet:

 

if the WAN link is ethernet, the IP packet is carried within an ethernet frame.

An ethernet frame has a 14 bytes header (6 bytes for destination MAC, 6 bytes for source MAC, 2 bytes ethertype = protocol type) and at the end there is a 4 byte FCS (a checksum of the whole frame).

However, ethernet is a LAN protocol that misses the use of a clock.

So actually on the physical layer the equivalent of 8 bytes are sent as a pre-amble just to allow the receiver side to synchronize before the frame is sent.

In addition to this a silence is needed between two frames to allow the receiver to understand and detect the end of the previous frame.

The total of pre-amble and inter-frame gap is 20,1 bytes equivalent.

 

So given a packet size in IP like 674 bytes, on wire they count like 674+18+20,1 = 712,1 bytes.

 

RTT = Round Trip Time  = two way delay and it is a measure performed with ICMP or UDP with timestamps inside.

 

What is important to understand is that the total one way delay includes different components:

-serialization component  (the one that I have explained above for the ethernet case)

-time to wait in queue variable on the router interface itself.

-propagation delay : time for the signal to arrive at destination can be important on long distance WAN links.

 

TCP connection setup requires a three way handshake that requires exchange of at least three packets

TCP SYN ---->

                    <----- SYN, ACK

TCP ACK

 

So the time to setup a TCP connection is not equivalent to RTT, your understanding is correct.

 

Hope to help

Giuseppe

 

View solution in original post

Just to add a bit to Giuseppe's information, keep in mind packets do not have to conform to the frame size. We normally assume that the packet will just be contained by the frame, but that may not be the case (although it usually is). If it doesn't, the packet per second rate will be impacted.

Further when you're measuring PPS, distance latency will also impact how fast TCP ramps up and also how fast it recovers from any packet loss. Additionally if the RWIN doesn't support the bandwidth delay product, TCP will never get to wire-rate.

View solution in original post

5 Replies 5

Martin L
VIP
VIP

 

they gave you 300KB(kbytes) in bytes not bits? shouldn't speed be in bits per second

 

https://www.speedtest.net/result/8321330655

 

Giuseppe Larosa
Hall of Fame
Hall of Fame

Hello George,

as already noted the link speeds in telecommunications are expressed in bit/s and not byte/s.

Also in telecommunications we dot not use a 1024 multiplier but 1000 like international metric system

 

1 Kbps = 1000 bps

1 Mbps = 1000000 bps

And so on.

Of course packet size is in bytes and 1 byte  = 8 bit.

 

However, in telecommunications you need to take in account the so called protocol overhead, that depends on the OSI layer 2 technology in use.

example for ethernet:

 

if the WAN link is ethernet, the IP packet is carried within an ethernet frame.

An ethernet frame has a 14 bytes header (6 bytes for destination MAC, 6 bytes for source MAC, 2 bytes ethertype = protocol type) and at the end there is a 4 byte FCS (a checksum of the whole frame).

However, ethernet is a LAN protocol that misses the use of a clock.

So actually on the physical layer the equivalent of 8 bytes are sent as a pre-amble just to allow the receiver side to synchronize before the frame is sent.

In addition to this a silence is needed between two frames to allow the receiver to understand and detect the end of the previous frame.

The total of pre-amble and inter-frame gap is 20,1 bytes equivalent.

 

So given a packet size in IP like 674 bytes, on wire they count like 674+18+20,1 = 712,1 bytes.

 

RTT = Round Trip Time  = two way delay and it is a measure performed with ICMP or UDP with timestamps inside.

 

What is important to understand is that the total one way delay includes different components:

-serialization component  (the one that I have explained above for the ethernet case)

-time to wait in queue variable on the router interface itself.

-propagation delay : time for the signal to arrive at destination can be important on long distance WAN links.

 

TCP connection setup requires a three way handshake that requires exchange of at least three packets

TCP SYN ---->

                    <----- SYN, ACK

TCP ACK

 

So the time to setup a TCP connection is not equivalent to RTT, your understanding is correct.

 

Hope to help

Giuseppe

 

Just to add a bit to Giuseppe's information, keep in mind packets do not have to conform to the frame size. We normally assume that the packet will just be contained by the frame, but that may not be the case (although it usually is). If it doesn't, the packet per second rate will be impacted.

Further when you're measuring PPS, distance latency will also impact how fast TCP ramps up and also how fast it recovers from any packet loss. Additionally if the RWIN doesn't support the bandwidth delay product, TCP will never get to wire-rate.

I was thinking the same thing, about the Ethernet frame.

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