Ethernet (IEEE 802.3) frame format:
1, preamble: 7 bytes 0x55, a series of 1 0 intervals, used for signal synchronization.
2. Frame start delimiter: 1 byte 0xd5 (101010/), indicating the start of the frame.
3.DA (destination MAC): 6 bytes.
4.SA (source MAC): 6 bytes
5. Type/length: 2 bytes, with 0 ~ 1500 reserved as the length field value and 1536 ~ 65535 reserved as the type field value (0x0600~0xFFFF).
6. Data: 46 ~ 1500 bytes
7. Frame Check Sequence (FCS): 4 bytes, which is a checksum obtained by calculating the contents from the destination MAC to the data domain using CRC.
The LAN with CSMA/CD as the MAC algorithm is called Ethernet. The methods to avoid CSMA/CD conflict are: listen before sending, send while listening, and resend after random delay. Once a conflict occurs, every host must be able to detect it. Minimum transmission gap and minimum frame length are also specified to avoid collision.
Consider the following extreme cases: the frame sent by the host is very small, and the two conflicting hosts are far apart. Just before the frame sent by host A is transmitted to B, B starts to send frames. In this way, when the frame of A arrives at B, B detects the collision and sends a collision signal. If A's frame has been sent before B's conflict signal is sent to A, A will not detect the conflict and mistakenly think that it has been sent successfully. Because there is a time delay in signal propagation, it takes some time to detect the collision. This is why there must be a minimum frame length limit.
According to the standard, when 10Mbps Ethernet uses repeaters, the maximum connection length is 2500m, and it can pass through four repeaters at most. Therefore, the minimum transmission time of 10Mbps Ethernet frame is 5 1.2 microseconds. The data that can be transmitted during this period is 5 12 bits, so it is also called 5 12 bits. This time is defined as an Ethernet slot or collision slot. 5 12 bits = 64 bytes, which is why the smallest Ethernet frame is 64 bytes.
Bit 5 12 is the time when the host captures the channel. If the 64 bytes of a frame sent by the host still have no conflict, there will be no conflict in the future, and it is said that the host has captured the channel. Because the channel is shared by all hosts, in order to avoid a single host occupying the channel for too long, the maximum frame length of Ethernet frames is specified as 1500.
When the time slot of 100Mbps Ethernet is still 5 12 bits, Ethernet stipulates that the minimum transmission time of a frame must be 5.12 μ s.
The time slot of 1000Mbps Ethernet is increased to 5 12 bytes, which is 4.096μs when 4096 bits are used.
Article 2:
2. Conflict slot time
Assume that the medium length of the bus is s, the transmission speed of the frame on the medium is 0.7C (light speed), and the transmission speed of the network is R(bps).
The frame length is L(bps), and tPHY is the physical layer delay of a station; ?
Have:?
Collision slot time =2S/0.7C+2tPHY?
Because Lmin/R= collision slot time?
So: Lmin =(2S/0.7C+2tPHY )×R?
Lmin is called the minimum frame length. ?
Collision slot time is an extremely important parameter in Ethernet, which has the following characteristics:?
(1) This is the longest time required to detect conflicts. ?
(2) The frame length is required to have a lower limit. (i.e. the shortest frame length)
(3) In case of collision, frame fragments will appear. ?
(4) In case of collision, wait for a certain time. T = rt (t is the collision slot time)?
2. Let's estimate the time required to detect the conflict in the worst case.
(1)A and B are the two farthest hosts on the Internet. Let the signal propagation delay between A and B be τ, and let A be at t?
When a frame starts to be sent, it will arrive at B at t+τ. If B sends a frame at t+τ-ε, it will arrive at t+τ.
A collision is detected and a blocking signal is issued. ?
(2) The blocking signal will reach a at t+2τ. Therefore, A must still be transmitting at t+2τ to detect the collision, so the frame's?
The sending time must be greater than 2τ. ?
(3) According to the standard, when 10Mbps Ethernet uses repeaters, the maximum connection length is 2500m, and it can pass through four repeaters at most.
This regulation stipulates that for 10Mbps Ethernet, the minimum transmission time of a frame must be 51.2μ s. ?
(3)5 1.2μs is the propagation time of 5 12 bit data at 10Mbps Ethernet rate, which is usually called 5 12 bit time. Is this time fixed?
It means Ethernet time slot. 5 12 bits =64 bytes, so the minimum length of an Ethernet frame is 5 12 bits =64 bytes. ?
2, Ethernet 18 bytes, I think it should be "destination MAC(6)+ source MAC(6)+ type (2)+CRC (4).
3. As for the IP maximum transmission unit 1500, it is only stipulated.
Article 3:
Lmin/R = 2*S/(0.7*C) + 2*Tphy + n*Tr？
Minimum frame length?
R- network rate (10M, etc. )?
S- conflict domain (*2 means once and for all, just draw a picture)?
C- standard speed of light (*0.7 means that in twisted pair, optical fiber is used)?
Tphy- Physical layer delay (*2 is due to going through the physical layers of two hosts)?
Tr-repeater delay (generally equivalent to the delay of two physical layers?
Principle:?
As shown in the figure:?
Host a repeater b host b?
- ?
|? Physical layer |? |? |?
- ?
|? |? |? |?
+ - +? + - +?
| & lt-S->； |?
According to the Ethernet media access control mechanism, the length of a transmission is La?
If the collision indication returned by B is received, stop sending frames and send 32bit?
0 10 10 1 ... strengthen the collision; And delay t = r*T for such a long time; ?
Where T = 1/C, r is a random number within (0,2 k), and k = min(n, 10)?
N is the number of consecutive collisions. If n is too large, it will be handled by the superior. ?
If we specify the conflict domain s first. The situation is as follows:
Host A sends at the speed of R, arrives at B at the speed of 0.7*C, and then collides?
Back to a, the distance of a * * * is 2*S, and the corresponding frame length is Lmin. Because?
If Los Angeles
The conflict indication lost its meaning, and the media access control failed. ?
If Lmin is defined first, S can also be determined. ?
Such as IEEE 802.3 Lmin = 64B；; Let r = 65438+100 Mbps tphy =15 us?
Tr = 0, then S = 2.3km can be calculated.
In practical application, the effective distance of the signal on the medium should also be considered.
For twisted pair, the effective distance should be less than s, so the repeater will be introduced to extend it?
Transmission distance, but it can be seen that this will reduce S, so this is a contradiction.
This should be considered when designing Ethernet.
Why is the minimum Ethernet frame 64 bytes?
Article source:? /thread- 12833- 1- 1 . html？ Thank you!
Hello, everyone. I remember reading about the origin of the minimum 64-byte Ethernet frame on the forum before. It is a formula to calculate that this is the minimum transmission time for CSMA/CD algorithm to detect conflicts, but I have just turned it over on the forum for a long time and can't see it. Which prawn can explain it to my younger brother again? thank you
In addition, I saw in some materials that the overhead of Ethernet frame header is 18 bytes, but how did this come about? Because Dst MAC+Src MAC+Type (length) is only 12 bytes (note: it should be 14 bytes here), where do the other six bytes come from?
Finally, why does Ethernet stipulate that the maximum transmission unit is 15 18 bytes?
-
The problem is terrible. Do not laugh at me.
1, about the origin of the smallest byte.
2. Conflict slot time
Assume that the medium length of the bus is s, the transmission speed of the frame on the medium is 0.7C (light speed), and the transmission speed of the network is R(bps).
The frame length is L(bps), and tPHY is the physical layer delay of a station; ?
Have:?
Collision slot time =2S/0.7C+2tPHY?
Because Lmin/R= collision slot time?
So: Lmin =(2S/0.7C+2tPHY )×R?
Lmin is called the minimum frame length. ?
Collision slot time is an extremely important parameter in Ethernet, which has the following characteristics:?
(1) This is the longest time required to detect conflicts. ?
(2) The frame length is required to have a lower limit. (i.e. the shortest frame length)
(3) In case of collision, frame fragments will appear. ?
(4) In case of collision, wait for a certain time. T = rt (t is the collision slot time)?
2. Let's estimate the time required to detect the conflict in the worst case.
(1)A and B are the two farthest hosts on the Internet. Let the signal propagation delay between A and B be τ, and let A be at t?
When a frame starts to be sent, it will arrive at B at t+τ. If B sends a frame at t+τ-ε, it will arrive at t+τ.
A collision is detected and a blocking signal is issued. ?
(2) The blocking signal will reach a at t+2τ. Therefore, A must still be transmitting at t+2τ to detect the collision, so the frame's?
The sending time must be greater than 2τ. ?
(3) According to the standard, when using repeaters in 10Mbps Ethernet, the maximum connection length is 2500m, and it can pass through four repeaters at most.
This regulation stipulates that for 10Mbps Ethernet, the minimum transmission time of a frame must be 51.2μ s. ?
(3)5 1.2μs is the propagation time of 5 12 bit data at 10Mbps Ethernet rate, which is usually called 5 12 bit time. Is this time fixed?
It means Ethernet time slot. 5 12 bits =64 bytes, so the minimum length of an Ethernet frame is 5 12 bits =64 bytes. ?
2, Ethernet 18 bytes, I think it should be "destination MAC(6)+ source MAC(6)+ type (2)+CRC (4).
3. As for the IP maximum transmission unit 1500, it is only stipulated.
Some will include the leading characters of Ethernet frames in the frame header.
The first edition of IEEE 802.3 standard was released on June 24th, 1983. Because Xerox handed over four patents on CSMA/CD to IEEE and authorized manufacturers to use the corresponding patents at a very low price, there is no problem of high patent cost in using IEEE 802.3 standard to produce products. Subsequently, the 802.3 standard was recognized by ANSI and ISO, making the IEEE 802.3 standard an open and authoritative standard.
Although the release time is similar to that of Ethernet II standard, and the standard level of IEEE 802.3 is much higher than the former, more than ten years after the release of the standard, the 802.3 standard can only be equally applied with Ethernet II. The essential reason is that the frame structure in the 802.3 standard must be used in conjunction with the 802.2 standard, which greatly reduces the flexibility of application and significantly increases the overhead of the protocol. Complexity and inefficiency make it difficult for 802.3 to prevail.
1998, the 802.3 working group made major modifications to the 802.3 standard, and finally completed the new version of the standard IEEE 802.3-2002. The 2002 edition of the 802.3 standard has greatly improved the MAC frame structure, making it flexible with Ethernet II encapsulation, thus completely replacing Ethernet II in application, making IEEE 802.3-2002 the only standard of Ethernet.
Since 2005, a new version of the 802.3 standard has been formulated.
The 802.3-2002 standard defines the header structure of Ethernet as da (6)+sa (6)+len/type (2) =14 bytes.
If my answer is helpful to you, I hope you can adopt it, thank you!