By IEEE 802.3, an Ethernet frame has to carry a payload of at least 46 bytes. This is for collision detection-- collisions of smaller frames (may) go undetected.
The Q is: what if the payload to be carried is shorter? what kind of padding is used to scale the frame up to the slot size-- 64 bytes?
TIA.
To quote from Data and Computer Network Communication (emphasis mine);
If the network layer wishes to send less than 46 bytes of data the MAC protocol adds sufficient number of zero bytes (0x00, is also known as null padding characters) to satisfy the requirement.
Some buggy drivers fail to do this though as noted by Adaptec.
Related
In Intel 8086, we have 20 bit address bus so we have 2^(20) locations.
I am confused on how this 16 bit data is stored in 20 bit address.
Example:
If I want to store 2 in first memory location, then my address will be 00000000000000000001(20 bits) and data in it will be: 00000000000000010(16 bits).
Is it this way or am I missing something?
Thank you.
Each 20-bit physical address slot holds one 8-bit byte. A 16-bit word takes 2 consecutive address slots.
Physical addresses (in original 8086) are 20 bits wide, data is (up to) 16 bits wide.
8086 does in fact have a 16-bit data bus, while 8088 had an 8-bit data bus and took 2 cycles to transfer a word but most of the rest of the chip is identical to 8086 right down to the transistor layout.
Address width and data width don't have to be related in any particular way.
You don't store 20-bit linear addresses as data in memory. A pointer stored in memory is typically a 32-bit segment:offset pair.
You could compact a pointer down to 3 bytes, by normalizing the segment value so only a 4-bit offset is needed, or whatever other technique to remove some of the redundancy caused by the seg<<4 + offset real-mode address calculation.
Let's say we have two machines on a network MA and MB,
MA considers little endian the order of the bits in a byte,
on the contrary MB considers big endian the order of the bits in a byte.
How do MA and MB agree on what "endianess" to use for the bits in a byte during the
communication over the network ?
Is there a standard "network endianess" or what ?
Do socket programmers have to take any actions in ensuring a correct communication ?
For example HTTP is a text protocol, that means that machines send and receive bytes which represent characters,what if the encoding of the characters is different in the endianess of the bits ?
Yes, the hardware protocols specify the bit order of bytes on all network links. This is generally handled automatically by the NIC hardware.
See, for example, this description of Ethernet frame format.
Ethernet transmission is strange, in that the byte order is big-endian (leftmost byte is sent first), but bit order little-endian (rigthmost, or LSB (Least Significant Bit) of the byte is sent first).
Check this page: http://www.comptechdoc.org/independent/networking/protocol/protlayers.html
It suggests that byte ordering is done at the Presentation Layer which is quite high up. This however will relate specifically to the application that you are using. I suspect data at lower levels (wrapping the higher levels) has a predetermined byte and bit order.
I am working on a project in which i have to send data in bits.
Is it possible to send 1 bit from one computer to another through internet.Most of the people
said to me that minimum internet packet size is 64bytes.If i send 1bit from one computer to another then packet bandwidth is 64bytes.
A TCP or UDP packet consists of a header and data. Maybe you could have one bit of data in the data section, but you would need the header as well. Without the header it would be impossible to send the packet. The header contains all the information required for sending the packet where it is supposed to go and making sure it arrives safely.
I will take "internet packet" to mean Ethernet frame based on the value you give.
An Ethernet frame has a minimum total size of 64 bytes including both header and payload, this is to ensure that the time to transmit a single frame is greater than the round trip time between nodes.
This requirement is a feature of any network that uses CSMA/CD (specifically the CD (collision detection) part), it allows a sensing node to detect a collision whilst still transmitting a frame.
Whilst Ethernet can be used to send a frame smaller than 64 bytes "padding" will be added to ensure the frame is at least 64 bytes.
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.
What is the overhead for PPP and Ethernet sending 5000 bytes?
Frame size for Point-to-Point Protocol: 8 bytes
MTU: 500 bytes
Frame size for Ethernet: 18 bytes
MTU: 1500 bytes
Both sending 5000 bytes..
I know this is just a calculation, but I am not sure how to do it. I can't find it anywhere. I would think that since a PPP frame takes 8 bytes and maximum transmission unit is 500 then it can send (500 - 8)bytes of information in one go. It sends 10 frames, resulting in 4920 bytes sent. Then sends the final (80+8)bytes with the last frame.
Similar for Ethernet. (1500 - 18)bytes with each frame. 3 frames sent means 4446 bytes sent. Sending (554+18)bytes in the last frame.
This obviously doesn't answer the "overhead" question. Anyone have any ideas?
It really depends on how you define overhead. This answer will assume overhead is the number of bytes that you need to transmit in addition to the data itself.
For Ethernet, assuming the 5000 byte payload is not encapsulated in an IP + TCP/UDP frame, you will have 18 bytes of overhead for every packet sent. That means each transmission with an MTU of 1500 will be able to hold 1482 bytes. To transmit 5000 bytes, this means 4 packets must be transmitted, which means an overhead of 72 bytes (18 * 4). Note that the overhead becomes bigger when you include things like the IP frame which contains a TCP frame.
For PPP, as you've already shown, you can send 492 bytes per frame. Eleven frames means 88 bytes of overhead (11 * 8) - again, not including any additional protocol frames within the payload.
In both these examples any protocols that build on top of these link layer protocols will contribute to overhead. For example, an Ethernet packet sent with an IPv4 frame which contains a UDP datagram will have an additional 28 bytes consumed by headers and not data (20 bytes in an IPv4 header and 8 in a UDP header, assuming no IP options). Considering the original Ethernet example, this means the amount of data per packet becomes 1454 bytes, which luckily still comes to 4 packets (the extra spills over into the smaller 4th packet), with 144 bytes of overhead.
You can read more here (I find that page a little hard to read though).