The round trip delay between A and B is 80 msec and the bottleneck bandwidth on the path between A and B is 128 Kbps. Station A uses 32 byte packets to transmit messages to station B using a sliding window protocol. = Sender window size / Optimal window sizeĬalculating Maximum Achievable Throughput- Tserver public static void main (String args)throws IOException ServerSocket serverSocketnull. Transmission delay = Packet size / Bandwidth Multiple frames sent by source are acknowledged by receiver using a single ACK frame. Sender window size = Receiver window size = 5 In sliding window method, multiple frames are sent by sender at a time before needing an acknowledgment.What is the maximum achievable throughput in this communication? The propagation delay over the link is 200 μs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 μs. The send and receive window sizes are 5 packets each. A and B are using the sliding window protocol for flow control. Host A is sending data to host B over a full duplex link. We use only (1+2a) sequence numbers and rest remains unused.With 9 bits, number of sequence numbers possible = 512.Minimum number of bits required in sequence number field = 9.Assuming that each frame carries 1 KB of data, what is the minimum number of bits needed for the sequence number? Thus, approximate optimal window size = 21938 frames.Ī sliding window protocol is designed for a 1 Mbps point to point link to the moon which has a one way latency (delay) of 1.25 sec. We use only (1+2a) = 109 sequence numbers and rest remains unused.Ĭompute approximate optimal window size when packet size is 53 bytes, RTT is 60 msec and bottleneck bandwidth is 155 Mbps.With 7 bits, number of sequence numbers possible = 128.Minimum number of bits required in sequence number field = 7.5.If your frames reach the server it will send ACK signal to client otherwise it will send NACK signal to client. 4.To send frames to server from the client side. 2.Get the frame size from the user 3.To create the frame based on the user request. For 3000 km, propagation delay = 3000 x 6 μsec = 18000 μsecĬalculating Bits Required in Sequence Number Field- EX.NO:1(a) Implementation of Sliding Window Protocol AIM: To write a java program to perform sliding window protocol ALGORITHM: 1.Start the program.

Sliding window protocol java program