What is MTU? Why is the MTU generally 1500?

In fact, a standard Ethernet data frame size is: 1518, the header information is 14 bytes, the tail checksum FCS is 4 bytes, so the real size for the upper layer protocol transmission is: 1518-14-4 = 1500, so where does the value 1518 come from?

Let’s say I take a larger value

If the MTU value is the same as the IP packet size, and the size of an IP packet is 65535, then the size of an Ethernet frame is 65535 + 14 + 4 = 65553, which seems perfect. The sender does not need to unpack and the receiver does not need to reassemble.

So let’s say our current bandwidth is: 100Mbps, because Ethernet frames are the smallest identifiable unit in transmission, and then the optical signal corresponding to 0101, we can only send one Ethernet frame at a time. If multiple frames are sent at the same time, the peer end will not be able to reassemble into a single Ethernet frame. At 100Mbps bandwidth (assuming no intermediate loss), let’s calculate the time required to send this frame:

(65553 * 8)/(100 * 1024 * 1024) ≈ 0.005(s)Copy the code

It takes 5ms to transmit a frame on a 100M network, which means no other process can send any data. If you dial an earlier phone and the network speed is only 2M:

(65553 * 8)/(2 * 1024 * 1024) ≈ 0.100(s)Copy the code

100ms, this is a nightmare. In fact, this is like a traffic light, the time to set a reasonable, alternate traffic, otherwise the same direction if the green light has been, then the other direction will be blocked into xiang.

Since it’s not ok to make it bigger, is it ok to make it smaller?

Assuming the MTU value is set to 100, the time for a single frame to be transmitted at 2Mbps bandwidth requires:

(100 * 8)/(2 * 1024 * 1024) * 1000 ≈ 5(ms)Copy the code

No matter what the MTU is set to, the size of the Ethernet header frame and the end frame is fixed, 14 + 4. Therefore, when the MTU is 100, the transmission efficiency of an Ethernet frame is:

(100-14-4) / 100 = 82%Copy the code

The formula is: (t-14-4)/T, when T tends to infinity, the transmission efficiency is close to 100%, that is, the larger the MTU value is, the highest transmission efficiency, but based on the transmission time of the previous point, let’s compromise, since the head plus the tail is 18, let’s make a whole 1500, the total size is 1518, transmission efficiency:

1500/1518 = 98.8%Copy the code

100Mbps Transmission time:

(1518 * 8)/(100 * 1024 * 1024) * 1000 = 0.11(ms)Copy the code

2Mbps Transmission time:

(1518 * 8)/(2 * 1024 * 1024) * 1000 = 5.79(ms)Copy the code

Overall the time is acceptable

The default value of most routers is 1500, theoretically there is no problem, so why can I change it to 1480 when playing the game? The reason is that I was using ADSL Internet access, ADSL using PPPoE protocol.

PPPoE

PPPoE is between Ethernet and IP. The PPPoE Protocol is divided into two parts: Point to Point Protocol (PPP) and Over Ethernet (oE). The PPPoE header is as follows:

| VER(4bit) | TYPE(4bit) | CODE(8bit) | SESSION-ID(16bit) | LENGTH(16bit) |
Copy the code

So we have 48 bits, which is 6 bytes, so what are the other 2 bytes? The ANSWER is the PPP ID, which occupies two bytes. Therefore, in the PPPoE environment, the optimal MTU value should be: 1500-6-2 = 1492.

The way I surf the Internet

My Internet access was as follows:

PC -> Router -> TelecomCopy the code

I dial up from the router, and then the PC connects to the router for Internet access.

Root cause

For router dial-up, the PC encapsulates Ethernet frames according to MTU:1492. Even if the packets pass the router, the router only forwards the packets without unpacking them.

When dial-up with the router, the PC does not know the communication mode of the router and encapsulates Ethernet frames with the MTU of 1500 by default set by the network card. When it arrives at the router, the router needs to encapsulate PPPoE and add 8-byte header information. In this way, the packet must be unpacked. The router split this frame into two frames, one frame is 1492 and the other frame is 8, and then added the PPPoE header to each frame.

Usually there is no card when playing games, because the router can still process the amount of data, and when carrying out AOE for group monsters, because the amount of data in a short time is too large, the router can not process it, there will be packet loss and delay, and it will be disconnected.

The 1480 mentioned in the post may be set as small as possible to avoid another PPPoE encapsulation caused by the second dial-up, because the time is long, there is no way to go back to the scene at that time to capture packets.