Explain the working of 3 bit sliding window protocol with suitable example.
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Answer Posted / naresh
A sliding window protocol is a feature of packet-based data transmission protocols. Sliding window protocols are used where reliable in-order delivery of packets is required, such as in the Data Link Layer (OSI layer 2) as well as in the Transmission Control Protocol (TCP).
Ambiguity example
The transmitter alternately sends packets marked "odd" and "even". The acknowledgments likewise say "odd" and "even". Suppose that the transmitter, having sent an odd packet, did not wait for an odd acknowledgment, and instead immediately sent the following even packet. It might then receive an acknowledgment saying "expecting an odd packet next". This would leave the transmitter in a quandary: has the receiver received both of the packets, or neither?
Suppose that we are using a 3-bit sequence number, such as is typical for HDLC. This gives N=2³=8. Since wr=1, we must limit wt≤7. This is because, after transmitting 7 packets, there are 8 possible results: Anywhere from 0 to 7 packets could have been received successfully. This is 8 possibilities, and the transmitter needs enough information in the acknowledgment to distinguish them all.
If the transmitter sent 8 packets without waiting for acknowledgment, it could find itself in a quandary similar to the stop-and-wait case: does the acknowledgment mean that all 8 packets were received successfully, or none of them?
Ambiguity example
Edit
The extremely popular HDLC protocol uses a 3-bit sequence number, and has optional provision for selective repeat. However, if selective repeat is to be used, the requirement that nt+nr ≤ 8 must be maintained; if wr is increased to 2, wt must be decreased to 6.
Suppose that wr =2, but an unmodified transmitter is used with wt =7, as is typically used with the go-back-N variant of HDLC. Further suppose that the receiver begins with nr =ns =0.
Now suppose that the receiver sees the following series of packets (all modulo 8):
0 1 2 3 4 5 6 (pause) 0
Because wr =2, the receiver will accept and store the final packet 0 (thinking it is packet 8 in the series), while requesting a retransmission of packet 7. However, it is also possible that the transmitter failed to receive any acknowledgments and has retransmitted packet 0. In this latter case, the receiver would accept the wrong packet as packet 8. . The solution is for the transmitter to limit wt ≤6. With this restriction, the receiver knows, after receiving packet 6, that the transmitter's na ≥1, and thus the following packet numbered 0 must be packet 8. If all acknowledgements were lost, then the transmitter would have to stop after packet 5.
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