Answer:
approx 306 frames need to be in transit to keep this pipe full
maximum reasonable size is 306 frames
sequence number = 10 bit
Explanation:
given data
transmission rate = 1 Mbps
size of frame Tt = 1 KB
one way latency Tp = 1.25 s
to find out
How many frames need and What is the maximum reasonable size for SWS and RWS and what is the number of bits needed for the sequence number
solution
we know useful time is =
so total time is
total time = Tt + 2Tp
total time = + 2 × 1.25
so
possible frames are
≤ 1
n ≤ 1 +
n ≤ 306.17
so approx 306 frames need to be in transit to keep this pipe full
and
we know here
size of sending window is equal to size of receiving window
so maximum reasonable size is 306 frames
and
bit need for sequence number is
sequence number =
sequence number = 10 bit
