1 votes 1 votes What should be the throughput in sliding window protocol for a window size of $100$ bits in terms of mbps is?. If RTT is equal to $100$ $\mu$sec and bandwidth is $10$mbps. Computer Networks computer-networks sliding-window + – sourav. asked Aug 31, 2015 • edited Aug 3, 2017 by sourav. sourav. 3.8k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 8 votes 8 votes Efficiency = Amount of data sent/Amount of data that could be sent = 100 / (RTT * Bandwidth) = 100/(100 * 10) = 0.1 Throughput = Efficiency * Bandwidth = 0.1 * 10Mbps = 1 Mbps Arjun answered Aug 31, 2015 Arjun comment Share Follow See all 4 Comments See all 4 4 Comments reply swagatika sahoo commented Jan 27, 2016 reply Follow Share THROUGHT=PACKET LENGTH/RTT=100bits/100microsec=100/100*10^-6=1*10^6=1Mbps 0 votes 0 votes Pranav Kapur commented Apr 10, 2016 reply Follow Share According to me the answer should be 0.9. Plz correct me if i am wrong Efficiency= Tt / (Tt+2*Tp) ...................................(i) RTT= 2 * Tp Tp= 50 ms ................................................................(ii) Tt = L/B Tt= 100 * 10^6/ 10 * 10^6 Tt = 10 ms .................................................................(iii) Put (ii) and (iii) in (i) efficiency = 1 / 11 Also, Throughput = efficiency * Bandwidth Throughput= 1/11 * 10 = 0.9 2 votes 2 votes Archies09 commented Jun 13, 2016 reply Follow Share @Arjun Sir Why did u multiply bandwidth with RTT? 0 votes 0 votes Nirmal Gaur commented Jan 31, 2020 reply Follow Share @PranavKapur Given 100 bits is the size of the window not the size of a packet, so L cannot be equal to 100. 0 votes 0 votes Please log in or register to add a comment.
