Computer Networks | Set 11
Following questions have been asked in GATE CS 2006 exam. 1) Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps. What is the optimal window size that A should use? (A) 20 (B) 40 (C) 160 (D) 320 Answer (B)
Round Trip propagation delay = 80ms Round Trip Time = Tt + 2*Tp Frame size = 32*8 bits Bandwidth = 128kbps Transmission Time = 32*8/(128) ms = 2 ms RTT = Tt + 2*Tp 80 = 2 + 2*Tp Tp = 39 Let n be the window size. UtiliZation = n/(1+2a) where a = Propagation time / transmission time Utilization = 1 (Max) n = 1+2a = 1 + 2 * Tp/Tt = 1 + 2 * 39/2 n = 40 Sliding Window Size = 40.
2) Two computers C1 and C2 are configured as follows. C1 has IP address 188.8.131.52 and netmask 255.255.128.0. C2 has IP address 184.108.40.206 and netmask 255.255.192.0. which one of the following statements is true? (A) C1 and C2 both assume they are on the same network (B) C2 assumes C1 is on same network, but C1 assumes C2 is on a different network (C) C1 assumes C2 is on same network, but C2 assumes C1 is on a different network (D) C1 and C2 both assume they are on different networks. Answer (C)
Network Id of C1 = bitwise '&' of IP of C1 and subnet mask of C1 = (220.127.116.11) & (255.255.128.0) = 18.104.22.168 C1 sees network ID of C2 as bitwise '&' of IP of C2 and subnet mask of C1 = (22.214.171.124) & (255.255.128.0) = 126.96.36.199 which is same as Network Id of C1. Network Id of C2 = bitwise '&' of IP of C2 and subnet mask of C2 = (188.8.131.52) & (255.255.192.0) = 184.108.40.206 C2 sees network ID of C1 as bitwise '&' of IP of C1 and subnet mask of C2 = (220.127.116.11) & (255.255.192.0) = 18.104.22.168 which is different from Network Id of C2.
Therefore, C1 assumes C2 is on same network, but C2 assumes C1 is on a different network.
3) Station A needs to send a message consisting of 9 packets to Station B using a sliding window (window size 3) and go-back-n error control strategy. All packets are ready and immediately available for transmission. If every 5th packet that A transmits gets lost (but no acks from B ever get lost), then what is the number of packets that A will transmit for sending the message to B? (A) 12 (B) 14 (C) 16 (D) 18 Answer (C) Total 16 packets are sent. See the following table for a sequence of events. Since go-back-n error control strategy is used, all packets after a lost packet are sent again.
Sender Receiver 1 2 1 3 2 4 3 5 4 6 7 6 7 [Timeout for 5] 5 6 5 7 6 8 9 8 9 [Timeout for 7] 7 8 7 9 8 [Timeout for 9] 9 9
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