HOW TO FIND WINDOW SIZE IN SLIDING WINDOW PROBLEM 1


Sliding Window:

In sliding window method, Sender can sent multiple frames at a time before needing an acknowledgment from receiver’s side.

The maximum window size for data transmission using the selective reject protocol with n-bit frame sequence numbers is  2^(n-1)

The maximum window size for data transmission using the SR protocol with n-bit frame sequence numbers is (N+1)/2

The maximum window size for both sender and receiver with n-bit frame sequence numbers is N/2.

Numerical question based on sliding window protocol

This question asked in JAN 2017 Paper 3 in CBSE NET

Question no 26:
Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milliseconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is
(A) 20
(B) 10
(C) 30
(D) 40
Explanation:
Always write given data in question
 Round Trip propagation delay = 40ms
Frame size = 32 bytes=32*8 bits ( because 1 byte = 8 bits)
Bandwidth = 64 kbps
Transmission Time = Frame size/Bandwidth = 32*8/(64) ms = 4 ms
Let n be the window size.
Utilization = n/(1+2a)
where a = Propagation time / transmission time = n/(1+40/4)
For maximum utilization: n = 11  
Ans: (B)

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