By default when we say about MSS for TCP ethernet packet 1460 and MTU is 1500.
MSS = MTU - 20(IP header) - 20(TCP Header) = 1460
from the above the TCP header is calculated without any options in TCP header.
In case if any packet consists option value in TCP header it will reduce the MSS size or not?
Then what will be the MSS size presence of option in TCP header
Your correct in thinking the MSS shrinks when IP/TCP options are added.
MSS = MTU - (20(IP header) + len(IP Options)) - (20(TCP Header) + len(TCP Options))
The other main reason it would be lowered is if the packet is being encapsulated in some way (IPsec/GTP) since that adds overhead to the packet.
The MSS won't decrease at all, however if IP, TCP options were there the MSS value will be 1460 only for ethernet. The options will be the part of sender data part, so sender need to reduce the data part while adding IP or TCP options on the header.
Refer the following page
http://tools.ietf.org/html/rfc6691
Related
I have been learning wireshark lately. While inspecting TCP segments, I saw a strange situation, at least for me. There was a mismatching SEQ,ACK numbers. Then I realized that difference between two ACK's same as 1 and half package size. However, as far as I know, ACKs are only growing by the full packet size. So what happened here?
SEQ ACK
1 1
2897 8689
5793 13033 <--
8689 14481
11585
14481
Well, there's not much to go by here, but I'm going to guess that you are capturing packets on the machine sending the large packets, i.e., the packets with 2896 bytes of TCP payload. As such, you are seeing the packets before they are IP-fragmented and actually transmitted out on the wire.
But you can't just send 2896 bytes of data onto the wire; Ethernet links typically impose a 1500 byte MTU (Maximum Transmission Unit), and when you account for IP and TCP header overhead, you usually end up with 1460 bytes of available payload or MSS (Maximum Segment Size). In your case it looks like you're only getting 1448 bytes of available MSS, most likely due to the addition of one or more IP and/or TCP header options.
In any case, the 2896 bytes of payload are going to be fragmented over 2 IP fragments, each containing 1448 bytes of TCP payload. I'm pretty sure that what you're seeing is an ACK from the receiver after having received 1 full segment plus 1 IP fragment from the next segment.
The previous ACK number was 8689, and 8689 + 2896 = 11585. Now add 1/2 of the data segment (2896 / 2 = 1448) and you get 11585 + 1448 = 13033. That's the ACK number you're seeing. Now add the other 1/2 and you get 13033 + 1448 = 14481, which is the ACK number of the next packet.
I hope that makes sense?
For an in depth look at the drawbacks of local packet captures, I direct you to a well-written blog by Jasper Bongertz titled, "The drawbacks of local packet captures".
Since I do not have reputation to comment the answer of Christopher Maynard, I comment it here.
As Christopher stated correctly the MTU (Maximum Transmission Unit) is standard Ethernet networks is 1500Bytes. However, crafting 1460Bytes (1500Bytes - 20Bytes IP Header - 20Bytes TCP Header, ignoring possible TCP options) TCP segments imposes a high load on the network stack since every segment must be handled separately.
As an example, if I want to transmit 1Mb (equals to 1024*1024Bytes = 1048576Bytes) of data, it would result in 1Mb/1460Bytes = 719 segments. In contrast to 1Mb/65525Bytes (theoretical max TCP segment size) = 16 segments. Since the overhead in the Kernel is mostly independent of the segment/packet size, small segments require much more processing.
To counteract this fact, TCP Segmentation Offloading (TSO) was developed. TSO allows the Kernel to craft TCP segments with maximum size and the NIC (Network Interface Card) uses hardware acceleration to split these segments into the separate TCP segments of 1460Bytes. Therefore, no IP fragmentation is required.
I have data 1245 MB for trasmission via UDP over IPv4.
For calculation of expected number of packet transmission from A to B then B relay to C, if the data transmitted in blocks of size 320bytes (i.e; payload = 320bytes), and header is 20 bytes, do we minus 20 from 320 or add in?
For instance,
1245MB = 1305477120 bytes
Total UDP Payload = 320 - 20 or 320 + 20?
The packet consists of:
the IP header (20 bytes)
the UDP header (8 bytes)
your payload (320 bytes).
Total: 348 bytes.
For calculating the number of packets you don't need to take into account the size of the transport or network layer headers. You specified a payload size of 320 bytes, which is well within the maximum size of a UDP payload without fragmentation.
Each time you call send() or sendto(), this will create a datagram (packet), so the math is simply dividing the total size by your 320 byte chunks:
1305477120 / 320 = 4079616 packets
As a side point, if you were to make your UDP payload larger, that would reduce the total number of packets. On a lot of networks, the MTU is 1500 bytes, so you can send:
1500 bytes - IP header (20 bytes) - UDP header (8) bytes = 1472 bytes for payload
As a second side point, if your UDP payload is too big, i.e. payload + IP/UDP headers exceeds the MTU, then your single call to send() would result in multiple IP fragment packets.
what is the difference between a link MTU max transmission unit and a link bandwidth? MTU refers to the max size of a packet that a link can send. link bandwidth refers to the max number of bits that a link can send. arent they the same thing
MTU = maximum size of 1 packet. For example, ethernet is 1500 bytes.
Bandwidth = bits you can send trough a link, for example, 1 gigabit per second.
So to send 1 megabyte of data over a line, you first need to cut it into small packets that fit the protocol. So you will create about 700 packets of 1500 bytes to fit one megabyte into. Those will go over your line at the bandwidth specified.
But there is overhead: in TCP/IP every packet needs its IP and TCP headers. So the smaller the MTU, the more packets you will need, so the header overhead will become more significant, meaning you can actually use less of the total bandwidth
Suppose I have two hosts, host A and host B, and host A sends a 1500 byte packet to B and there are two routers between them with a MTU of 800 bytes for the first link, and MTU of 500 bytes for the second link between them.
As I understand it the packet will have to be fragmented both times. The packet will have to be fragmented into 3 packets (2 equally sized packets and one smaller one).
Then when we hit this second router we will have each of the first two fragmented packets be fragmented once more. The first two original fragments will produce 1 larger fragment (close to 500 bytes), and one smaller one.
This is where I am confused.
I know that the 'More Fragment' bit should be set to 1 on all of the first 3 fragments except the last indicating that the packets are part of a fragment up to and including the next packet with 'More Fragment' set to 0. However, for the second set of fragments I am not so sure. If this was the first time the packet had been fragmented, the last fragment would have 'More Fragment' of 0, but since this is actually just a fragment somewhere in the middle of the original message, I have a feeling it should be 1.
I am hoping someone with more experience can shed some light on this for me. Is the 'More Fragment' bit used in reassembly? If so I would imagine that when the fragments are fragmented then we do not set the last fragment to 0.
So the algorithm would go:
If fragmenting a packet with 'More Fragment' set to 0,
then set 'More Fragment' to 1 in all fragmented packets except the last.
Else If fragmenting a packet with 'More Fragment' set to 1,
then set 'More Fragment' to 1 in all of the fragmented packets.
Is my assumption here correct?
You are more or less right. Routers besides adding MF (more fragments) flag are doing some more processing:
To fragment/segment a long internet packet, a router (R1 in the figure
below) creates a new IP packet and copies the contents of the IP
header fields from the long packet into the new IP header. The data of
the long packet is then divided into two portions on a 8 byte (64 bit)
boundary, so that the first packet is less than the MTU of the
out-going interface. The more-fragments flag (MF) in the first packet
is set to one (to indicate that more fragments of this packet follow).
The More Flag may already be set in this packet if it has already been
fragmented by another system. This packet is forwarded.
The second created new packet is then processed. The packet header
field is identical to that of the original packet (including the same
value of the packet ID, the total length field, the more-fragments
flag (MF) and the fragment offset field in the original packet). The
packet header field is updated with a new offset field, by adding the
number of payload bytes sent in the first fragment. If this new packet
is larger than the allowed link MTU, the packet is again fragmented.
So in nutshell for already fragmented packets whole IP header is being copied to new packets (smaller fragments of fragment). It means that if IP packet has MF set it will be also copied to last fragmented packet. Thanks to changing offset and packets ID host receiving fragmented packets will be able to reassemble those.
What is the size of an empty UDP datagram? And that of an empty TCP packet?
I can only find info about the MTU, but I want to know what is the "base" size of these, in order to estimate bandwidth consumption for protocols on top of them.
TCP:
Size of Ethernet frame - 24 Bytes
Size of IPv4 Header (without any options) - 20 bytes
Size of TCP Header (without any options) - 20 Bytes
Total size of an Ethernet Frame carrying an IP Packet with an empty TCP Segment - 24 + 20 + 20 = 64 bytes
UDP:
Size of Ethernet frame - 24 Bytes
Size of IPv4 Header (without any options) - 20 bytes
Size of UDP header - 8 bytes
Total size of an Ethernet Frame carrying an IP Packet with an empty UDP Datagram - 24 + 20 + 8 = 52 bytes
Himanshus answer is perfectly correct.
What might be misleading when looking at the structure of an Ethernet frame [see further reading], is that without payload the minimum size of an Ethernet frame would be 18 bytes: Dst Mac(6) + Src Mac(6) + Length (2) + Fcs(4), adding minimum size of IPv4 (20) and TCP (20) gives us a total of 58 bytes.
What has not been mentioned yet is that the minimum payload of an ethernet frame is 46 byte, so the 20+20 byte from the IPv4 an TCP are not enough payload! This means that 6 bytes have to be padded, thats where the total of 64 bytes is coming from.
18(min. Ethernet "header" fields) + 6(padding) + 20(IPv4) + 20(TCP) = 64 bytes
Hope this clears things up a little.
Further Reading:
Ethernet_frame
Ethernet
See User Datagram Protocol. The UDP Header is 8 Bytes (64 bits) long.
The mimimum size of the bare TCP header is 5 words (32bit word), while the maximum size of a TCP header is 15 words.
Best wishes,
Fabian
If you intend to calculate the bandwidth consumption and relate them to the maximum rate of your network (like 1Gb/s or 10Gb/s), it is necessary, as pointed out by Useless, to add the Ethernet framing overhead at layer 1 to the numbers calculated by Felix and others, namely
7 bytes preamble
1 byte start-of-frame delimiter
12 bytes interpacket gap
i.e. a total of 20 more bytes consumed per packet.
If you're looking for a software perspective (after all, Stack Overflow is for software questions) then the frame does not include the FCS, padding, and framing symbol overhead, and the answer is 54:
14 bytes L2 header
20 bytes L3 header
20 bytes L4 header
This occurs in the case of a TCP ack packet, as ack packets have no L4 options.
As for FCS, padding, framing symbol, tunneling, etc. that hardware and intermediate routers generally hide from host software... Software really only cares about the additional overheads because of their effect on throughput. As the other answers note, FCS adds 4 bytes to the frame, making it frame 58 bytes. Therefore 6 bytes of padding are required to reach the minimum frame size of 64 bytes. The Ethernet controller adds an additional 20 bytes of framing symbols, meaning that a packet takes a minimum of 84 byte times (or 672 bit times) on the wire. So, for example, a 1Gbps link can send one packet every 672ns, corresponding to a max packet rate of roughly 1.5MHz. Also, intermediate routers can add various tags and tunnel headers that further increase the minimum TCP packet size at points inside the network (especially in public network backbones).
However, given that software is probably sharing bandwidth with other software, it is reasonable to assume that this question is not asking about total wire bandwidth, but rather how many bytes does the host software need to generate. The host relies on the Ethernet controller (for example this 10Mbps one or this 100Gbps one) to add FCS, padding, and framing symbols, and relies on routers to add tags and tunnels (A virtualization-offload controller, such as in the second link, has an integrated tunnel engine. Older controllers rely on a separate router box). Therefore the minimum TCP packet generated by the host software is 54 bytes.