computer network

Question

If window receiver size is 64 KB and The round trip time of a connection is 50 msec and maximum segment size used is 4 KB.In slow start phase the current congestion window size is 16 MSS and sender gets 4 ACK then then what should be value of congestion window (In KB)

Answer 1

In the slow start phase, the congestion window size (cwnd) is increased exponentially. The formula for the slow start phase is given by:

cwnd=cwnd+MSScwnd=cwnd+MSS

where MSS is the Maximum Segment Size. Given that the current cwnd is 16 MSS and the sender receives 4 ACKs, the cwnd after the 4 ACKs would be:

cwnd=16+(4×MSS)cwnd=16+(4×MSS)

You mentioned that the maximum segment size (MSS) used is 4 KB. So,

cwnd=16+(4×4)cwnd=16+(4×4)

cwnd=16+16cwnd=16+16

cwnd=32cwnd=32

Now, let's convert the cwnd from segments to kilobytes. If MSS is 4 KB, then 32 segments would be 32×432×4 KB:

cwnd in KB=32×4cwnd in KB=32×4

cwnd in KB=128cwnd in KB=128

Therefore, the value of the congestion window after receiving 4 ACKs in the slow start phase would be 128 KB.