2 votes 2 votes Station $A$ uses $32-byte$ packets to transmit messages to Station $B$ using a sliding window protocol. The round trip time delay between $A$ and $B$ is $40\,ms$ and the bottleneck bandwidth on the path $A$ and $B$ is $64\, kbps.$What is the optimal window size that $A$ should use? $5$ $10$ $40$ $80$ Computer Networks isrodec2017 + – gatecse asked Dec 17, 2017 recategorized Feb 11, 2018 by srestha gatecse 1.5k views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply joshi_nitish commented Dec 20, 2017 reply Follow Share it should be 11, but i will go with nearest option i.e n=10 2 votes 2 votes Akshay Koli 4 commented Dec 20, 2017 reply Follow Share yes it should be 11. 0 votes 0 votes Anu007 commented Dec 20, 2017 reply Follow Share maximum bits transmited in Rtt = 64 Kbps * 40 msec = 2560 frame size = 32* 8 bits = 256 so size of sender window = = 2560 /256= 10 0 votes 0 votes srivivek95 commented Dec 20, 2017 reply Follow Share Anu007 Sir, so size of sender window = = 2560 /256= 10 Here, units of denominator & numerator are not matching. It should be so size of sender window = = 2621.44 /256= 10.24 0 votes 0 votes Please log in or register to add a comment.
7 votes 7 votes In one second link can send 64 kb In RTT link can send 64kb*40ms = 64* 1000* 40/1000 = 2560 bits = 320 bytes Size of one packet =32 bytes Size of optimal window =320/32 =10 Answer is B PS: kbps means 1000 bits per second, not 1024. Please refer https://en.wikipedia.org/wiki/Data_rate_units#Decimal_multiples_of_bits sh!va answered Jan 11, 2018 sh!va comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Transmission delay $T_t=\frac{32\times8}{64\times10^{3}}=$ 4 msec. Assuming 100% bandwidth utilization, we get $1=\frac{N\times T_t}{RTT}$ $\rightarrow1=\frac{N\times 4}{40}\rightarrow N=10.$ habedo007 answered Apr 20, 2018 habedo007 comment Share Follow See all 0 reply Please log in or register to add a comment.