Applied Test Series

Question

Consider a selective repeat sliding window protocol, the window size of the sender is 128. What will be the sequence number for the 800th frame? [ Assume the available sequence number are the minimum required for window size of 128]

 

Answer 1

Window size of selective repeat  $2^{m-1} = 128$

 

$Min_{m} = m-1 = 7\\ \Rightarrow m = 7+1 = 8$

 

No of available seq number = $2^{8} = 256$

First 256 frames = 0 to 255

For next  256 frames = 0 to 255

For next 256 frames = 0 to 255 

For last 32 frames = 0 to 31

 

The 800th frame 31 is the sequence  number