| 31. | Consider a 128×10 3 bits/second satellite communication link with one way propagation delay of 150 milliseconds. Selective retransmission (repeat) protocol is used on this link to send data with a frame size of 1 kilobyte. Neglect the transmission time of acknowledgement. The minimum number of bits required for the sequence number field to achieve 100% utilization is ___________ |
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Answer: (b).4
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| 32. | How many bytes of data can be sent in 15 seconds over a serial link with baud rate of 9600 in asynchronous mode with odd parity and two stop bits in the frame? |
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Answer: (b).12,000 bytes
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| 33. | A 1Mbps satellite link connects two ground stations. The altitude of the satellite is 36,504 km and speed of the signal is 3 × 10^8 m/s. What should be the packet size for a channel utilization of 25% for a satellite link using go-back-127 sliding window protocol? Assume that the acknowledgment packets are negligible in size and that there are no errors during communication. |
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Answer: (a).120 bytes
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| 34. | The minimum frame size required for a CSMA/CD based computer network running at 1 Gbps on a 200m cable with a link speed of 2 × 10^8m/s is |
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Answer: (b).250 bytes
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| 35. | In an IPv4 datagram, the M bit is 0, the value of HLEN is 10, the value of total length is 400 and the fragment offset value is 300. The position of the datagram, the sequence numbers of the first and the last bytes of the payload, respectively are |
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Answer: (c).Last fragment, 2400 and 2759
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| 36. | Consider a source computer(S) transmitting a file of size 106 bits to a destination computer(D)over a network of two routers (R1 and R2) and three links(L1, L2, and L3). L1connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D.Let each link be of length 100 km. Assume signals travel over each link at a speed of 108 meters per second.Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D? |
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Answer: (a).1005 ms
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| 37. | One of the header fields in an IP datagram is the Time to Live(TTL)field.Which of the following statements best explains the need for this field? |
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Answer: (d).It can be used to prevent packet looping
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| 38. | For which one of the following reasons does Internet Protocol (IP) use the time-to- live (TTL) field in the IP datagram header |
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Answer: (c).Prevent packets from looping indefinitely
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| 39. | Consider the following three statements about link state and distance vector routing protocols, for a large network with 500 network nodes and 4000 links. [S1] The computational overhead in link state protocols is higher than in distance vector protocols. [S2] A distance vector protocol (with split horizon) avoids persistent routing loops, but not a link state protocol. [S3] After a topology change, a link state protocol will converge faster than a distance vector protocol. Which one of the following is correct about S1, S2, and S3 ? |
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Answer: (d).S1 and S3 are true, but S2 is false
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| 40. | Which one of the following is TRUE about interior Gateway routing protocols - Routing Information Protocol (RIP) and Open Shortest Path First (OSPF)? |
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Answer: (a).RIP uses distance vector routing and OSPF uses link state routing
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