1
GATE CSE 2007
MCQ (Single Correct Answer)
+2
-0.6
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:
A
$$\left\lceil {{{\log }_2}{{2LtR + 2K} \over K}} \right\rceil $$
B
$$\left\lceil {{{\log }_2}{{2LtR} \over K}} \right\rceil $$
C
$$\left\lceil {{{\log }_2}{{2LtR + K} \over K}} \right\rceil $$
D
$$\left\lceil {{{\log }_2}{{2LtR + 2K} \over {2K}}} \right\rceil $$
2
GATE CSE 2006
MCQ (Single Correct Answer)
+2
-0.6
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
3
GATE CSE 2006
MCQ (Single Correct Answer)
+2
-0.6
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
4
GATE CSE 2005
MCQ (Single Correct Answer)
+2
-0.6
In a packet switching network, packets are routed from source to destination along a single path having two intermediate nodes. If the message size is 24 bytes and each packet contains a header of 3 bytes, then the optimum packet size is:
A
4
B
6
C
7
D
9
GATE CSE Subjects
Software Engineering
Web Technologies
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Medical
NEET
Graduate Aptitude Test in Engineering
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CBSE
Class 12