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0 votes
Four bits are used for packet sequence numbering in a sliding window protocol used in a computer network. What is the maximum window size?
(a) 4

(b) 15
(c) 8

(d) 16.
asked in Computer Networks by Boss (34.4k points) | 2.3k views

3 Answers

+2 votes
Best answer
15 is the maximum windows size for 4 bit sequences and it will happen in Go back N.
answered by Active (2.6k points)
selected by
How solved it?

Explain with logic/steps!!, please
See in Stop and Wait we have receiver and sender each of size one.

In Go Back N we can have Sender size as N-1 but receiver as only 1.

In Selective Repeat we have Sender and Receiver each of equal size.(N/2,N/2)

Now in question it is given 4 bit sequence , so we can have 16 sequence number in total  and the maximum size is only possible with GBN.

so (2^n - 1)
Consider sender window size as 2  which is also the sequence number space and receiver window size is 1 always.

Now, sender sends 2 packets. Both are acknowledged by receiver but the acknowledgements are lost.


Say, timeout occurs at sender. Now, sender will send initial 0 but receiver is expecting next 0. This will create confusion.


So, we are sure that qindows size cant be equal to sequence number space.


Now, make the sender window size = 1 and see if it works.


So, sender window size <= sequnce number space - 1 for sliding window protocol
0 votes
Incase of GoBackN protocol

Sender window size is n-1

and Receiver window size is 1

the size is (2^n)-1

the answer is 15
answered by (73 points)
0 votes

In the stop and wait sender windows size and receiver windows size is only=1

In the Go back N ARG sender windows size and size is only=2^n-1 and receiver windows =1

In selective Repeat ARG windows size and Reciever Size =2^n-1 

So that maximum windows size is possible in only Go back ARG only =2^4-1= 15 

answered ago by (243 points)

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