One Stop GATE Forum: networking queestions

Message

   1) Suppose you are designing a sliding window
protocol for a 1 Mbps point-to-point link to a stationary satellite
revolving around the earth at 3 * 10000 Km attitude. Assuming that each
frame carries 1 Kb of data, what is the minimum number of bits you need
for the sequence number when RWS = 1 (RWS = receiver window size) . 

   a)  20 bits             b)  32 bits               c)  5 bits                     d)  16 bits 

2) Compute approximate optimal window size when packet size is 53
bytes, propagation delay is 30 msec and bandwidth is 155 Mbps. 

    a)  32,000 packets                     b)  12,000 packets                c)  22,000 packets               d)  42,000 packets

Author	Message
Adhiti Newbie Joined: 02Feb2007 Online Status: Offline Posts: 32	Topic: networking queestions Posted: 02Feb2007 at 11:38am
	1) Suppose you are designing a sliding window protocol for a 1 Mbps point-to-point link to a stationary satellite revolving around the earth at 3 * 10000 Km attitude. Assuming that each frame carries 1 Kb of data, what is the minimum number of bits you need for the sequence number when RWS = 1 (RWS = receiver window size) . a) 20 bits b) 32 bits c) 5 bits d) 16 bits 2) Compute approximate optimal window size when packet size is 53 bytes, propagation delay is 30 msec and bandwidth is 155 Mbps. a) 32,000 packets b) 12,000 packets c) 22,000 packets d) 42,000 packets Post Resume: Click here to Upload your Resume & Apply for Jobs

	IP Logged

Alok Newbie Joined: 02Feb2007 Online Status: Offline Posts: 19	Posted: 02Feb2007 at 3:59pm
	1) Transmission time : 1Kb / 1Mbps = 2^10 / 10^6 Now propagation delay is : distance / signal speed here since sattelite communication is used signal speed is same as speed of light = 3 * 10^8 m/s Thus the RTT(round trip time) = 2 * propagation delay Thus total time it takes to complete transmission of one packet is TotalTime = Transmission time + RTT Since Sliding window protocol is used number of packets that can be transmitted in TotalTime= TotalTime / transmission time from the number of packets you can easily make out the number of bits required in sequence number 2^n >= Number of packets The same method can be used for second Question.
	IP Logged


One Stop GATE Forum : GATE Previous Years Test Papers - Discuss Here : CS Papers