Overhead need to be conside

Overhead need to be considerwhen calculate traffic performance

Issue 0: Feb/18/2003 by [email protected]

Describe:

When we say the traffic performance, almost people did not consider about the protocol’s overhead, because that is less the the traffic, but the overhead actually need to be consider because protocol to protocol transfer and transmitter media change. Every different media require different transmit function.Here is a example:

Host A send a file via FTP to Host B through the Ethernet/ADSL/IMA/Ethernet networks.the scenario like below.

Some protocol format and header describe below.

A. IP Packet to IMA E1 Interface Over Head Formula

1. 64 byte IP packet, via 1483 Bridge mode.Known conditions:

IP packet size 64 byteLLC SNAP: LLC frame bridge mode 18 byteAAL5 :CPCS PDU tail 8 bytesIMA frame size = per 127 cell plus 1 IMA ICP cellPacket in ATM = 64+8+18=90/48=2 cellATM Size = 2 x 53 = 106 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps

overhead rate = 64/106 = 0.60378Rate = 8192 Kbps x 0.60378 = 4946.166 KbpsIMA Cell =((53/(53*127))byte but too small

2. 64 byte IP packet via 1483 Route ModeKnown conditions:

IP packet size 64 byteLLC SNAP: LLC frame route mode 8 byteAAL5 : CPCS PDU tail 8 byteIMA frame size = per 128 cell plus 1 IMA ICP cellPacket in ATM = 64+8+8=80/48=2 cellATM Size = 2 x 53 = 106 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps

overhead rate = 64/106 = 0.60378Rate = 8192 Kbps x 0.60378= 4946.166 KbpsIMA Cell =((53/(53*127))byte but too small

3. 1500 byte IP packet with 1483 Bridge ModeKnown conditions:

IP packet size 1500 byteIMA frame size = per 128 cell plus 1 IMA ICP cellAAL5 : CPCS PDU tail 8 byteLLC SNAP: LLC frame bridge mode 18 bytePacket in ATM = 1500+8+18=1526/48=32 cellATM size 32x53 = 1696 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps

Overhead rate = 1500/1696 = 0.8844IMA Cell =((53*32/(53*127))byte but too smallActually rate = 8192 Kbps x 0.8844 = 7245.005 Kbps #

4. 1500 byte IP packet with 1483 Route ModeKnown conditions:

IP packet size 1500 byteIMA frame size = per 128 cell plus 1 IMA ICP cellAAL5 : CPCS PDU tail 8 byteLLC SNAP: LLC frame route mode 8 bytePacket in ATM = 1500+8+8=1516/48=32 cellATM size = 32x53 = 1696 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


B. TCP Packet to IMA E1 Interface Over Head Formula

1. 24 byte data via TCP, through 1483 Bridge mode.Known conditions:

TCP packet size 24 byteTCP header 20 bytesIP header 20 bytesLLC SNAP: LLC frame bridge mode 18 byteAAL5 :CPCS PDU tail 8 bytesIMA frame size = per 127 cell plus 1 IMA ICP cellPacket in ATM = 24+20+20+8+18=90/48=2 cellATM Size = 2 x 53 = 106 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


2. 24 byte data via TCP through 1483 Route ModeKnown conditions:

Data size 24 bytesTCP header 20 bytesIP header 20 bytesLLC SNAP: LLC frame route mode 8 byteAAL5 : CPCS PDU tail 8 byteIMA frame size = per 128 cell plus 1 IMA ICP cellPacket in ATM = 24+20+20+8+8=80/48=2 cellATM Size = 2 x 53 = 106 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


3. 1460 byte data via TCP through 1483 Bridge ModeKnown conditions:

Data size 1460 bytesTCP header 20 bytesIP header 20 bytesIMA frame size = per 128 cell plus 1 IMA ICP cellAAL5 : CPCS PDU tail 8 byteLLC SNAP: LLC frame bridge mode 18 bytePacket in ATM = 1460+20+20+8+18=1526/48=32 cellATM size 32x53 = 1696 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


4. 1460 byte IP packet with 1483 Route ModeKnown conditions:

Data size 1460 bytesTCP header 20 bytesIP header 20 bytesIMA frame size = per 128 cell plus 1 IMA ICP cellAAL5 : CPCS PDU tail 8 byteLLC SNAP: LLC frame route mode 8 bytePacket in ATM = 1460+20+20+8+8=1516/48=32 cellATM size = 32x53 = 1696 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


C. UDP Packet to IMA E1 Interface Over Head Formula

1. 36 byte data via UDP, through 1483 Bridge mode.Known conditions:

Data size 36 bytesUDP header 8 bytesIP header 20 byteLLC SNAP: LLC frame bridge mode 18 byteAAL5 :CPCS PDU tail 8 bytesIMA frame size = per 127 cell plus 1 IMA ICP cellPacket in ATM = 36+8+20+8+18=90/48=2 cellATM Size = 2 x 53 = 106 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


2. 36 byte IP packet via 1483 Route ModeKnown conditions:

Data size 36 bytesUDP header 8 bytesIP header 20 bytesLLC SNAP: LLC frame route mode 8 byteAAL5 : CPCS PDU tail 8 byteIMA frame size = per 128 cell plus 1 IMA ICP cellPacket in ATM = 36+8+20+8+8=80/48=2 cellATM Size = 2 x 53 = 106 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


3. 1472 byte UDP packet with 1483 Bridge ModeKnown conditions:

Data size 1472 bytesUDP header 8 bytesIP header 20 bytesIMA frame size = per 128 cell plus 1 IMA ICP cellAAL5 : CPCS PDU tail 8 byteLLC SNAP: LLC frame bridge mode 18 bytePacket in ATM = 1472+8+20+8+18=1526/48=32 cellATM size 32x53 = 1696 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


4. 1472 byte UDP packet with 1483 Route ModeKnown conditions:

Data size 1472UDP header 8 bytesIP header 20 bytesIMA frame size = per 128 cell plus 1 IMA ICP cellAAL5 : CPCS PDU tail 8 byteLLC SNAP: LLC frame route mode 8 bytePacket in ATM = 1472+8+20+8+8=1516/48=32 cellATM size = 32x53 = 1696 byte4 E1 bandwidth = 2048 x 4 =8192 Kbps


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71092782

70521854

70521854

72454946

72454946

RouteBridgeRouteBridgeRouteBridge

Large pktSmall pkt

UDP packetTCP packetIP packetSize

These 3 kinds of packet test result but still not consider about any detail, such like the TCP traffic will has handshake algorithm so, after a window size packet transmitted, the transmit will stop and wait until the receiver side send a ACK packet back for confirm the data is correct, and continue send next window, so the current TCP traffic will below the rate for this table.

So if you are test the traffic performance via IP packet ( send ICMP echo and reply), then you need to consider that the host send out ICMP packet’s rate, in normal, the MS Windows is send a ICMP packet 1 second by default, but if you are use Linux, the you can use this command for fast send “ping -i0.01 IP_address”, that it could send ICMP packet every 0.01 second.

So if your ICMP packet is 32 bytes, in MS Windows will send out :32 byte* 1 Sec = 32 * 8 * 1 = 256 bps

but with Linux with “-i” condition :32 bytes * (1/0.01) = 32 * 8 * 100 = 25600 bps

That is very different test condition.

So when you are test traffic performance with FTP application, that you will see the traffic rate from FTP application, that need to be consider about the server round trip time for that FTP service. FTP is work with TCP packet, so the overhead is more then others packet, and need to consider the TCP windows size and handshake. That is very complex to calculate in real life, but one true is that FTP protocol can make sure your file is correctly send and received but it need more overhead to make sure that.

So when you are calculate the TCP/IP packet performance rate, please consider what media you have and what kind of traffic you send and receive when the packet size different, the performance will change.

Technology

Overhead need to be conside