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In the above figure, assume that the network uses 16 bits addresses and that the prefix `01001010` has been assigned to the entire network. Since the network contains four routers, the network operator could assign one block of sixty-four addresses to each router. `R1` would use address `0100101000000000` while `A` could use address `0100101000000001`. `R2` could be assigned all addresses from `0100101001000000` to `0100101001111111`. `R4` could then use `0100101011000000` and assign `0100101011000001` to `B`. Other allocation schemes are possible. For example, `R3` could be allocated a larger block of addresses than `R2` and `R4` could use a sub-block from `R3` 's address block.
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the `data packets`
 
the `control packets`
 
For example, consider again the network topology above. `A` sends a control packet towards `B`. The initial `record route` is empty. When `R1` receives the packet, it adds its own address to the `record route` and forwards a copy to `R2` and another to `R3`. `R2` receives the packet, adds itself to the `record route` and forwards it to `R3`. `R3` receives two copies of the packet. The first contains the `[R1,R2]` `record route` and the second `[R1]`. In the end, `B` will receive two control packets containing `[R1,R2,R3,R4]` and `[R1,R3,R4]` as `record routes`. `B` can keep these two paths or select the best one and discard the second. A popular heuristic is to select the `record route` of the first received packet as being the best one since this likely corresponds to the shortest delay path.
 
With the received `record route`, `B` can send a `data packet` to `A`. For this, it simply reverses the chosen `record route`. However, we still need to communicate the chosen path to `A`. This can be done by putting the `record route` inside a control packet which is sent back to `A` over the reverse path. An alternative is to simply send a `data packet` back to `A`. This packet will travel back to `A`. To allow `A` to inspect the entire path followed by the `data packet`, its `source route` must contain all intermediate routers when it is received by `A`. This can be achieved by encoding the `source route` using a data structure that contains an index and the ordered list of node addresses. The index always points to the next address in the `source route`. It is initialized at `0` when a packet is created and incremented by each intermediate node.
 
Flat or hierarchical addresses
 
The last, but important, point to discuss about the `data plane` of the networks that rely on the datagram mode is their addressing scheme. In the examples above, we have used letters to represent the addresses of the hosts and network nodes. In practice, all addresses are encoded as a bit string. Most network technologies use a fixed size bit string to represent source and destination address. These addresses can be organized in two different ways.
 
The first organization, which is the one that we have implicitly assumed until now, is the `flat addressing` scheme. Under this scheme, each host and network node has a unique address. The unicity of the addresses is important for the operation of the network. If two hosts have the same address, it can become difficult for the network to forward packets towards this destination. `Flat addresses` are typically used in situations where network nodes and hosts need to be able to communicate immediately with unique addresses. These `flat addresses` are often embedded inside the network interface cards. The network card manufacturer creates one unique address for each interface and this address is stored in the read-only memory of the interface. An advantage of this addressing scheme is that it easily supports unstructured and mobile networks. When a host moves, it can attach to another network and remain confident that its address is unique and enables it to communicate inside the new network.
 
This hierarchical allocation of addresses can be applied in any type of network. In practice, the allocation of the addresses must follow the network topology. Usually, this is achieved by dividing the addressing space in consecutive blocks and then allocating these blocks to different parts of the network. In a small network, the simplest solution is to allocate one block of addresses to each network node and assign the host addresses from the attached node.
 
In the above figure, assume that the network uses 16 bits addresses and that the prefix `01001010` has been assigned to the entire network. Since the network contains four routers, the network operator could assign one block of sixty-four addresses to each router. `R1` would use address `0100101000000000` while `A` could use address `0100101000000001`. `R2` could be assigned all addresses from `0100101001000000` to `0100101001111111`. `R4` could then use `0100101011000000` and assign `0100101011000001` to `B`. Other allocation schemes are possible. For example, `R3` could be allocated a larger block of addresses than `R2` and `R4` could use a sub-block from `R3` 's address block.
 
Dealing with heterogeneous datalink layers
 
Sometimes, the network layer needs to deal with heterogeneous datalink layers. For example, two hosts connected to different datalink layers exchange packets via routers that are using other types of datalink layers. Thanks to the network layer, this exchange of packets is possible provided that each packet can be placed inside a datalink layer frame before being transmitted. If all datalink layers support the same frame size, this is simple. When a node receives a frame, it decapsulates the packet that it contains, checks the header and forwards it, encapsulated inside another frame, to the outgoing interface. Unfortunately, the encapsulation operation is not always possible. Each datalink layer is characterized by the maximum frame size that it supports. Datalink layers typically support frames containing up to a few hundreds or a few thousands of bytes. The maximum frame size that a given datalink layer supports depends on its underlying technology. Unfortunately, most datalink layers support a different maximum frame size. This implies that when a host sends a large packet inside a frame to its nexthop router, there is a risk that this packet will have to traverse a link that is not capable of forwarding the packet inside a single frame. In principle, there are three possibilities to solve this problem. To discuss them, we consider a simple scenario with two hosts connected to a router as shown in the figure below.
 
Consider in the network above that host `A` wants to send a 900 bytes packet (870 bytes of payload and 30 bytes of header) to host `B` via router `R1`. Host `A` encapsulates this packet inside a single frame. The frame is received by router `R1` which extracts the packet. Router `R1` has three possible options to process this packet.
 
The packet is too large and router `R1` cannot forward it to router `R2`. It rejects the packet and sends a control packet back to the source (host `A`) to indicate that it cannot forward packets longer than 500 bytes (minus the packet header). The source could react to this control packet by retransmitting the information in smaller packets.
 
The network layer is able to fragment a packet. In our example, the router could fragment the packet in two parts. The first part contains the beginning of the payload and the second the end. There are two possible ways to perform this fragmentation.
 
Each of the packet fragments is a valid packet that contains a header with the source (host `A`) and destination (host `B`) addresses. When router `R2` receives a packet fragment, it treats this packet as a regular packet and forwards it to its final destination (host `B`). Host `B` reassembles the received fragments.
 
These three solutions have advantages and drawbacks. With the first solution, routers remain simple and do not need to perform any fragmentation operation. This is important when routers are implemented mainly in hardware. However, hosts must be complex since they need to store the packets that they produce if they need to pass through a link that does not support large packets. This increases the buffering required on the hosts.
 
Furthermore, a single large packet may potentially need to be retransmitted several times. Consider for example a network similar to the one shown above but with four routers. Assume that the link `R1->R2` supports 1000 bytes packets, link `R2->R3` 800 bytes packets and link `R3->R4` 600 bytes packets. A host attached to `R1` that sends large packet will have to first try 1000 bytes, then 800 bytes and finally 600 bytes. Fortunately, this scenario does not occur very often in practice and this is the reason why this solution is used in real networks.
 
Fragmenting packets on a per-link basis, as presented for the second solution, can minimize the transmission overhead since a packet is only fragmented on the links where fragmentation is required. Large packets can continue to be used downstream of a link that only accepts small packets. However, this reduction of the overhead comes with two drawbacks. First, fragmenting packets, potentially on all links, increases the processing time and the buffer requirements on the routers. Second, this solution leads to a longer end-to-end delay since the downstream router has to reassemble all the packet fragments before forwarding the packet.
 
The last solution is a compromise between the two others. Routers need to perform fragmentation but they do not need to reassemble packet fragments. Only the hosts need to have buffers to reassemble the received fragments. This solution has a lower end-to-end delay and requires fewer processing time and memory on the routers.
 
The first solution to the fragmentation problem presented above suggests the utilization of control packets to inform the source about the reception of a too long packet. This is only one of the functions that are performed by the control protocol in the network layer. Other functions include :
 
sending a control packet back to the source if a packet is received by a router that does not have a valid entry in its forwarding table
 

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In the above figure, assume that the network uses 16 bits addresses and that the prefix `01001010` has been assigned to the entire network. Since the network contains four routers, the network operator could assign one block of sixty-four addresses to each router. `R1` would use address `0100101000000000` while `A` could use address `0100101000000001`. `R2` could be assigned all addresses from `0100101001000000` to `0100101001111111`. `R4` could then use `0100101011000000` and assign `0100101011000001` to `B`. Other allocation schemes are possible. For example, `R3` could be allocated a larger block of addresses than `R2` and `R4` could use a sub-block from `R3` 's address block.
6 years ago
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Source string location
../../principles/network.rst:295
String age
6 years ago
Last updated
a month ago
Source string age
6 years ago
Translation file
locale/cs/LC_MESSAGES/principles/network.po, string 64