2 2 votes The round trip delay between X and Y is given as 60 ms and bottle neck bandwidth of link between X and Y is 512 KBps. What is the optimal window size (in packets) if the packet size is 64 bytes and channel is full duplex Computer Networks + – thor 2.6k views answer comment Share Follow Print See all 6 Comments 6 6 Comments reply Show 3 previous comments thor commented Dec 15, 2016 reply Follow flag You sure about it. Does using full duplex affect it anyway? 0 0 replyShare Rajesh Raj commented Dec 15, 2016 reply Follow flag window size =(1+2a) where a =Tp/Tt so it comes 481 bits 0 0 replyShare santhoshdevulapally commented Dec 16, 2016 reply Follow flag @Rajesh,dont use formula 0 0 replyShare Please log in or register to add a comment.
2 2 votes it is 240 packets only. if the link is half duplex then in 1 rtt you will be able to send 1 window. but if link is full duplex you can send 2 windows in 1 rtt. so 2 windows size will become 60*512B. so number of packets per window will be 240 balagangadhar12 answered Dec 26, 2016 balagangadhar12 comment Share Follow See all 2 Comments 2 2 Comments reply soumyagupta commented May 6, 2021 reply Follow flag Why in full duplex we can send 2 Windows in RTT 0 0 replyShare divine_paragon commented Nov 14, 2025 reply Follow flag Here, full duplex only suggests that while Ack of receiver end is on the way to reach sender end, we can keep going on sending packets from sender side thus achieving full duplex i.e both device can send data simultaneously. 0 0 replyShare Please log in or register to add a comment.
2 2 votes first find the bandwidth delay product = bandwidth * RTT = 512KBps * 60ms = 30720B. If the size of the each packet is 64B. then total no of packets = 30720B/64B = 480 packets. IT is given wrong in made easy workbook ankitron answered May 10, 2019 ankitron comment Share Follow 0 reply Please log in or register to add a comment.
