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When a router wants to signal congestion to the sender, it adds a bit in the header of packets sent. When a packet arrives at the router, the router calculates the average queue length for the last (busy + idle) period plus the current busy period. (The router is busy when it is transmitting packets, and idle otherwise). When the average queue ...
With the normal untagged Ethernet frame overhead of 18 bytes and the 1500-byte payload, the Ethernet maximum frame size is 1518 bytes. If a 1500-byte IP packet is to be carried over a tagged Ethernet connection, the Ethernet frame maximum size needs to be 1522 bytes due to the larger size of an 802.1Q tagged frame.
TCP window scale option is needed for efficient transfer of data when the bandwidth-delay product (BDP) is greater than 64 KB [1].For instance, if a T1 transmission line of 1.5 Mbit/s was used over a satellite link with a 513 millisecond round-trip time (RTT), the bandwidth-delay product is ,, =, bits or about 96,187 bytes.
Under IPv4, a router that receives a network packet larger than the next hop's MTU has two options: drop the packet if the Don't Fragment (DF) flag bit is set in the packet's header and send an Internet Control Message Protocol (ICMP) message which indicates the condition Fragmentation Needed (Type 3, Code 4), or fragment the packet and send it ...
The default TCP Maximum Segment Size is for IPv4 is 536. For IPv6 it is 1220. [1]: §3.7.1 Where a host wishes to set the maximum segment size to a value other than the default, the maximum segment size is specified as a TCP option, initially in the TCP SYN packet during the TCP handshake. The value cannot be changed after the connection is ...
The payload length field of IPv6 (and IPv4) has a size of 16 bits, capable of specifying a maximum length of 65 535 octets for the payload. In practice, hosts determine the maximum usable payload length using Path MTU Discovery (yielding the minimum MTU along the path from sender to receiver), to avoid having to fragment packets.
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The packet transmission time in seconds can be obtained from the packet size in bit and the bit rate in bit/s as: Packet transmission time = Packet size / Bit rate. Example: Assuming 100 Mbit/s Ethernet, and the maximum packet size of 1526 bytes, results in Maximum packet transmission time = 1526×8 bit / (100 × 10 6 bit/s) ≈ 122 μs
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