Data link layer 2

1
Data link layer (2)
2
Review

• How Protocol 3 deals with
• Lost packet
• Lost ACK
• Premature timeout

3
Correctness Part is Done!

• Protocol 3 will work without causing errors,
  meaning that the data link layer will deliver the
  packets from the network layer to the other side
  in order.
• ARQ. (automatic repeat request)

4
Optimization

• So the next step is the optimization.
• At the end of last lecture, considered an example
  of
• 500ms delay. 50kbps. Each frame is 1000 bits.
• Very low link efficiency.

5
Solution

• So, increase the window size from 1 to n.
• Would want to have a large window size whenever
  the bandwidth delay product is large.
• If bandwidth is large, sending is fast
• If delay is large, even a moderate bandwidth will
  cause a poor efficiency

6
Question

• Given link speed and link delay, how large should
  the window be to achieve 100 efficiency?

7
Sender window

• The sender has a window from m1 to mn means
  that
• the sender has received ack from the receiver for
  frame up to m,
• the sender is allowed to send m1, m2, , mn.
  These packets can be outstanding without ack.
• Sender has to maintain the sender window
  (buffering all outstanding frames), maintain
  timeout for all outstanding frames.
• The size of the sender window is determined by
  the bandwidth and link delay

8
Go-Back-N

• Lets first consider the receivers window size
  is only 1.
• This is called Go-Back-N.

9
Go-Back-N (Protocol 4)

• Sender.
• while (1)
• If network layer got data, if current
  window not full yet, read data, send to physical
  layer with the frame ID. Start timer. Increment
  frame ID by one.
• If got ACKm, if m is outside the
  window, dont do anything. If m is inside the
  window, consider all frames in the window with
  frame ID less than the m acked. Forward window to
  m.
• If timeout for frame m, resend all
  frames in the window with frame ID greater than
  m.
• Receiver
• while (1)
• Get data from the physical layer. If
  it is with the expected frame ID, give it to the
  network layer, increment the frame ID by 0ne.
  Send ACK.
• Note assuming the sequence number is the frame
  ID.

10
Go-Back-N

• Two new things.
• Cumulative ACK
• Resend every frame after the timeout frame

11
Go-Back-N
12
Go-Back-N

• What if ACK is lost?

13
Go-Back-N

• Works fine is error is rare. Achieves full
  capacity.
• Else wastes bandwidth.

14
Sequence Number

• If the sequence number field is large enough,
  just use frame ID as sequence number.
• Else, how many bits do we need?

15
Sequence number

• Using the sequence number is to make sure that
  the receiver can be sure that a received frame is
  not a duplicate of some other frames.
• Using finite number of bits means that some
  frames will have the same sequence number.
• So, should make sure that the receiver is sure
  that when a frame with seqm is received, the
  sender has got ACK for all previous frames with
  the same seq.

16
Sequence Number

• Say the sequence number field is 2 bits.
• Will this be true if the window size is 4?
• No, because the sender may send 0,1,2,3. Then the
  receiver sends ACK1,2,3,4. If one of the ACK got
  through, the sender will send frame 4 with seq0.
  If none of the ACK got through, the sender will
  send frame 0 with seq0.

17
Sequence number

• Solution?
• Use window size less than 4. Say, 3.
• Now, if the receiver gets frame 0,1,2, it sends
  ACK1,2,3. Then, if it receives a frame with seq0
  again, will it be able to tell whether it is a
  new frame or a retransmission?
• Yes, because the receiver will accept frame with
  seq3 first. It knows that if the sender sends
  seq3, it must have got ack for frame 0.