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Data Link Control

Explore the functions of data link layer including framing, flow and error control, and protocols for reliable communication between nodes. Learn about character-oriented and bit-oriented framing strategies.

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Data Link Control

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  1. Data Link Control B. R. Chandavarkar CSE Dept., NITK Surathkal

  2. The two main functions of the data link layer are data link control and media access control. • The first, data link control, deals with the design and procedures for communication between two adjacent nodes: node-to-node communication. • Data link control functions include framing, flow and error control, and software implemented protocols that provide smooth and reliable transmission of frames between nodes. • To implement data link control, we need protocols. • Each protocol is a set of rules that need to be implemented in software and run by the two nodes involved in data exchange at the data link layer. • We discuss five protocols: two for noiseless (ideal) channels and three for noisy (real) channels.

  3. Framing • The data link layer, on the other hand, needs to pack bits into frames, so that each frame is distinguishable from another. • Framing in the data link layer separates a message from one source to a destination, or from other messages to other destinations, by adding a sender address and a destination address. • The destination address defines where the packet is to go; the sender address helps the recipient acknowledge the receipt. • Although the whole message could be packed in one frame, that is not normally done. One reason is that a frame can be very large, making flow and error control very inefficient. When a message is carried in one very large frame, even a single-bit error would require the retransmission of the whole message. When a message is divided into smaller frames, a single-bit error affects only that small frame.

  4. Frames can be of fixed or variable size. • In fixed-size framing, there is no need for defining the boundaries of the frames; the size itself can be used as a delimiter. • An example of this type of framing is the ATM (Asynchronous Transfer Mode) wide-area network, which uses frames of fixed size called cells. • In variable-size framing, we need a way to define the end of the frame and the beginning of the next. • Historically, two approaches were used for this purpose: a character-oriented approach and a bit-oriented approach.

  5. Character-Oriented Protocols • In a character-oriented protocol, data to be carried are 8-bit characters from a coding system such as ASCII. • The header, which normally carries the source and destination addresses and other control information, and the trailer, which carries error detection or error correction redundant bits, are also multiples of 8 bits. • To separate one frame from the next, an 8-bit (1-byte) flag is added at the beginning and the end of a frame. • The flag, composed of protocol-dependent special characters, signals the start or end of a frame.

  6. Character-oriented framing was popular when only text was exchanged by the data link layers. • The flag could be selected to be any character not used for text communication. • Now, however, we send other types of information such as graphs, audio, and video. Any pattern used for the flag could also be part of the information. • If this happens, the receiver, when it encounters this pattern in the middle of the data, thinks it has reached the end of the frame. • To fix this problem, a byte-stuffing strategy was added to character-oriented framing. • In byte stuffing (or character stuffing), a special byte is added to the data section of the frame when there is a character with the same pattern as the flag. • The data section is stuffed with an extra byte. This byte is usually called the escape character (ESC), which has a predefined bit pattern. • Whenever the receiver encounters the ESC character, it removes it from the data section and treats the next character as data, not a delimiting flag.

  7. Bit-Oriented Protocols • In a bit-oriented protocol, the data section of a frame is a sequence of bits to be interpreted by the upper layer as text, graphic, audio, video, and so on. • However, in addition to headers (and possible trailers), we still need a delimiter to separate one frame from the other. • Most protocols use a special 8-bit pattern flag 01111110 as the delimiter to define the beginning and the end of the frame.

  8. That is, if the flag pattern appears in the data, we need to somehow inform the receiver that this is not the end of the frame. • We do this by stuffing 1 single bit (instead of 1 byte) to prevent the pattern from looking like a flag. The strategy is called bit stuffing. • In bit stuffing, if a 0 and five consecutive 1 bits are encountered, an extra 0 is added. This extra stuffed bit is eventually removed from the data by the receiver. • Note that the extra bit is added after one 0 followed by five 1s regardless of the value of the next bit. • This guarantees that the flag field sequence does not inadvertently appear in the frame.

  9. Flow and Error Control • The most important responsibilities of the data link layer are flow control and error control. Collectively, these functions are known as data link control. Flow Control • Flow control coordinates the amount of data that can be sent before receiving an acknowledgment and is one of the most important duties of the data link layer. • In most protocols, flow control is a set of procedures that tells the sender how much data it can transmit before it must wait for an acknowledgment from the receiver. • The flow of data must not be allowed to overwhelm the receiver. • Any receiving device has a limited speed at which it can process incoming data and a limited amount of memory in which to store incoming data.

  10. Error control in the data link layer is often implemented simply: Any time an error is detected in an exchange, specified frames are retransmitted. This process is called automatic repeat request (ARQ).

  11. Simplest Protocol

  12. Stop-and-Wait Protocol

  13. Stop-and-Wait ARQ Protocol

  14. Assume that, in a Stop-and-Wait ARQ system, the bandwidth of the line is 1 Mbps, and 1 bit takes 20 ms to make a round trip. What is the bandwidth-delay product? If the system data frames are 1000 bits in length, what is the utilization percentage of the link? • The bandwidth-delay product is 20,000 bits during the time it takes for the data to go from the sender to the receiver and then back again. However, the system sends only 1000 bits. We can say that the link utilization is only 1000/20,000, or 5 percent. For this reason, for a link with a high bandwidth or long delay, the use of Stop-and-Wait ARQ wastes the capacity of the link.

  15. Sliding Window on Sender Sliding Window on Receiver LAR – Last Acknowledgment Received LFS – Last Frame Sent SWS – Send Window Size LFR – Last Frame Received LAF – Largest Acceptable Frame RWS – Receive Window Size

  16. When a frame with sequence number SeqNum arrives, the receiver takes the following action. • If SeqNum ≤ LFR or SeqNum > LAF, then the frame is outside the receiver’s window and it is discarded. • If LFR < SeqNum ≤ LAF, then the frame is within the receiver’s window and it is accepted. • Now the receiver needs to decide whether or not to send an ACK. • Let SeqNumToAck denote the largest sequence number not yet acknowledged, such that all frames with sequence numbers less than or equal to SeqNumToAck have been received. The receiver acknowledges the receipt of SeqNumToAck, even if higher-numbered packets have been received. • This acknowledgment is said to be cumulative. • It then sets LFR = SeqNumToAck and adjusts LAF = LFR + RWS.

  17. Example: • Suppose LFR = 5 (i.e., the last ACK the receiver sent was for sequence number 5), and RWS = 4. This implies that LAF = 9. • Should frames 7 and 8 arrive, they will be buffered because they are within the receiver’s window. However, no ACK needs to be sent since frame 6 is yet to arrive. Frames 7 and 8 are said to have arrived out of order. (Technically, the receiver could resend an ACK for frame 5 when frames 7 and 8 arrive.) • Should frame 6 then arrive—perhaps it is late because it was lost the first time and had to be retransmitted, or perhaps it was simply delayed—the receiver acknowledges frame 8, bumps LFR to 8, and sets LAF to 12. • If frame 6 was in fact lost, then a timeout will have occurred at the sender, causing it to retransmit frame 6.

  18. Issue: • We observe that when a timeout occurs, the amount of data in transit decreases, since the sender is unable to advance its window until frame 6 is acknowledged. This means that when packet losses occur, this scheme is no longer keeping the pipe full. The longer it takes to notice that a packet loss has occurred, the more severe this problem becomes. • Solution: • The receiver could have sent a negative acknowledgment (NAK) for frame 6 as soon as frame 7 arrived. • It would have been legitimate to send additional acknowledgments of frame 5 when frames 7 and 8 arrived; • In some cases, a sender can use duplicate ACKs as a clue that a frame was lost. • Selective Acknowledgments-The receiver could acknowledge the receipt of frames 7 and 8. Giving more information to the sender makes it potentially easier for the sender to keep the pipe full, but adds complexity to the implementation.

  19. Design (Sliding Window on Sender and Receiver): • The sending window size is selected according to how many frames we want to have outstanding on the link at a given time; SWS is easy to compute for a given delay× bandwidth product. • On the other hand, the receiver can set RWS to whatever it wants. • Two common settings are RWS = 1, which implies that the receiver will not buffer any frames that arrive out of order, and RWS = SWS, which implies that the receiver can buffer any of the frames the sender transmits. • It makes no sense to set RWS > SWS since it’s impossible for more than SWS frames to arrive out of order.

  20. Go-Back –N ARQ Protocol • Sliding Window

  21. Selective Repeat ARQ

  22. Station A needs to send a message consisting of 9 frames to Station B using a sliding window (window size 3) and go-back-n error control strategy. All packets are ready and immediately available for transmission. If every 5th frame that A transmits gets lost (but no acks from B ever get lost), then what is the number of frames that A will transmit for sending the message to B? Justify your answer with the timing diagram. • Window size is 3, so maximum 3 frames can be remained unacknowledged. In go back ‘n’ if acknowledge for a frame is not received then frames after that frame is also retransmitted. Frame sequence for 9 frame is shown below. 1 2 3 4 5 6 7 5 6 7 8 9 7 8 9 9

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