11. 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).40

12. Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 ms. The minimum frame size is
a. 94
b. 416
c. 464
d. 512
Answer: (c).464

13. How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit, eight data bits, two stop bits, and one parity bit ?
a. 600
b. 800
c. 876
d. 1200
Answer: (b).800

14. A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is:
a. 0.5
b. 0.625
c. 0.75
d. 1.0
Answer: (b).0.625

15. A 2 km long broadcast LAN has 10^7 bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2 × 108 m/s. What is the minimum packet size that can be used on this network?
a. 50 bytes
b. 100 bytes
c. 200 bytes
d. None of these
Answer: (d).None of these

16. Host A is sending data to host B over a full duplex link. A and B are using the sliding window protocol for flow control. The send and receive window sizes are 5 packets each. Data packets (sent only from A to B) are all 1000 bytes long and the transmission time for such a packet is 50 µs. Acknowledgement packets (sent only from B to A) are very small and require negligible transmission time. The propagation delay over the link is 200 us. What is the maximum achievable throughput in this communication?
a. 7.69 × 10^6 bytes per second
b. 11.11 × 10^6 bytes per second
c. 12.33 × 10^6 bytes per second
d. 15.00 × 10^6 bytes per second
Answer: (b).11.11 × 10^6 bytes per second

17. In serial data transmission, every byte of data is padded with a ‘0’ in the beginning and one or two ‘I’ s at the end of byte because
a. Receiver is to be synchronized for byte reception
b. Receiver recovers lost ‘0’ and ‘1’s from these padded bits
c. Padded bits are useful in parity computation
d. None of these
Answer: (a).Receiver is to be synchronized for byte reception

18. Consider a LAN with four nodes S1, S2, S3 and S4. Time is divided into fixed-size slots, and a node can begin its transmission only at the beginning of a slot. A collision is said to have occurred if more than one node transmit in the same slot. The probabilities of generation of a frame in a time slot by S1, S2, S3 and S4 are 0.1, 0.2, 0.3 and 0.4, respectively. The probability of sending a frame in the first slot without any collision by any of these four stations is _________.
a. 0.462
b. 0.711
c. 0.5
d. 0.652
Answer: (a).0.462

19. Suppose that the stop-and-wait protocol is used on a link with a bit rate of 64 kilobits per second and 20 milliseconds propagation delay. Assume that the transmission time for the acknowledgment and the processing time at nodes are negligible. Then the minimum frame size in bytes to achieve a link utilization of at least 50% is _________.
a. 160
b. 320
c. 640
d. 220
Answer: (b).320

20. Consider a CSMA/CD network that transmits data at a rate of 100 Mbps (108 bits per second) over a 1 km (kilometer) cable with no repeaters. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable?
a. 8000
b. 10000
c. 16000
d. 20000
Answer: (d).20000

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