ComputerNetworks
September 9, 2024ComputerNetworks
September 9, 2024SlidingWindowProtocol
Question 14

The distance between two stations M and N is L kilometers. All frames are K bits long. The propagation delay per kilometer is t seconds. Let R bits/second be the channel capacity. Assuming that processing delay is negligible, the minimum number of bits for the sequence number field in a frame for maximum utilization, when the sliding window protocol is used, is:
⌈log_{2}(2LtR+2K/K)⌉


⌈log_{2}(2LtR/K)⌉


⌈log_{2}(2LtR+K/K)⌉


⌈log_{2}(2LtR+K/2K)⌉

Question 14 Explanation:
Maximum window size at sender side,
N = (T_{k} + 2T_{p})/T_{k}
T_{p} = l × l sec
T_{k} = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver’s window size.
So, no. of bits required,
⌈log_{2} K+2LtR/K⌉
N = (T_{k} + 2T_{p})/T_{k}
T_{p} = l × l sec
T_{k} = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver’s window size.
So, no. of bits required,
⌈log_{2} K+2LtR/K⌉
Correct Answer: C
Question 14 Explanation:
Maximum window size at sender side,
N = (T_{k} + 2T_{p})/T_{k}
T_{p} = l × l sec
T_{k} = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver’s window size.
So, no. of bits required,
⌈log_{2} K+2LtR/K⌉
N = (T_{k} + 2T_{p})/T_{k}
T_{p} = l × l sec
T_{k} = K/R sec
So, N = (K/R+2Lt)/(K/R) = K+2LtR/K
Note that when asked for general sliding window protocol (not GBN nor SR) then we do not care about receiver’s window size.
So, no. of bits required,
⌈log_{2} K+2LtR/K⌉
Subscribe
Login
0 Comments