1. Frames are transmitted in both directions.

2. It requires a full duplex communication channel.

   i. It uses Piggy Backing which means that the outgoing acknowledgement is delayed so that they can be hooked to the next outgoing data frame.

   ii. Piggy backing is used so that there is no need to send a separate acknowledgement frame; thus saving bandwidth.

   iii. In piggy backing we have to only set/reset the ack field which consists of a few bits. If a separate acknowledgement frame had to be sent it would need a header and a checksum, thus using more bits.

3. Sending Window: Represents the frame numbers that have been sent but are yet not acknowledged.

4. Receiving Window: Represents the frame numbers that the receiver can accept.

GO BACK N Protocol

1. In this case when a damaged frame arrives the receiver simply discards all the subsequent frames.

2. It can transfer more than one frame at a time thus it is faster than the 1-bit sliding window protocol.

3. Working: Sender sends N frames and waits for an acknowledgement, if Rth frame is in error, packets received after Rth frame will be discarded and it starts resending from Rth to Nth frame. E.g.: Assume window size= 4 frames.

   Suppose sender sends frames 1 to 3.

   It has to now wait for the acknowledgement before it can proceed.

   As each successive acknowledgement is received the window slides forward and the sender can send the next frames.

   Suppose an acknowledgement of the 0 frame is lost then the sender discards all the frames after the lost frame (i.e. 1, 2, 3) and retransmits from the 0th frame.

4. Max Senders Window Size = 2k – 1

   Reason:

   a. Suppose that we keep window size = 2k

   b. Assume k=3, therefore 2k =8

   c. At time t1 sends frames 0 to 7 to /b.

   d. B receives each of them in order and at time t2 sends acknowledgement for the most recent frame i.e. frame 7.

   e. Suppose that this acknowledgement gets lost.

   f. Now B does not know that the acknowledgement it send is lost.

   g. A does not receive the acknowledgement and times out. On timing out it resends the frames 0 to 7 to B.

   h. But it is a duplicate of the frames that were already received by B.

   i. This problem occurs because 2 consecutive windows contain the same frame numbers. Reducing the frame size by 1 solves this problem.

   j. Hence maximum sender window size 2k – 1.

5. Max Receivers Window size =1.

   Reason: The frames are always received in order.

6. The frames must be received in order.