清華大學 計算機網路 第2週小考

(+10)	(10%) 1. Which of the following components is NOT used in 10Base5 Ethernet ? (單選)
		(a)Transceivers
		(b)Terminators
		(c)Optical Fiber
		(d)Repeaters
		(e)Coaxial cable

(+10)	(10% )2. Consider 3 stations are connected by “Hub1” of 10baseT Ethernet, another 3 stations are connected by “Hub2” of 10baseT Ethernet, and Hub1 and Hub2 are connected to each other directly. At most how many stations can transmit simultaneously without a collision ?
	Ans:	ANS:1

(+0)	(10%)3. Consider 3 stations are connected by “Hub1” of 10baseT Ethernet, another 3 stations are connected by “Switch1” of 10baseT Ethernet, and Hub1 and Switch1 are connected to each other directly. At most how many stations can transmit simultaneously without a collision ?
	Ans:	ANS:4

(+10)	(10%)4. Consider 3 stations are connected by “Switch1” of 10baseT Ethernet, another 4 stations are connected by “Switch2” of 10baseT Ethernet, and Switch1 and Switch2 are connected to each other directly. At most how many stations can transmit simultaneously without a collision ?
	Ans:	ANS:7

(+10)	(10%)5. Which of the following are correct for CSMA/CD (Carrier Sense Multiple Access with Collision Detection) protocol ?(多選)
		(a)Listen (carrier sense) before transmit
		(b)If channel is sensed idle, then transmit entire frame even a collision is detected
		(c)If channel is sensed idle, then transmit a frame and stop the transmission when a collision is detected
		(d)If channel is sensed busy, then defer transmission for a random time.
		(e)If channel is sensed busy, then defer transmission until the channel is idle.

(+10)	(10%) 6. Which of the following are correct when a “collision” happens for CSMA/CD protocol ?(多選)
		(a)The entire frame will still be completely transmitted
		(b)A jamming sequence will be sent by the sending host to enforce the collision.
		(c)The transmission is stopped and after a random delay, the frame is transmitted again from the “carrier sense” step.
		(d)The transmission is stopped and after a random delay, the frame is transmitted again immediately.
		(e)The transmission is stopped. But the frame is transmitted again as long as the channel becomes idle.

(+10)	(10%) 7. The “collision window” of the CSMA/CD protocol is the time required to detect a collision. Which of the following are correct for collision window ? Assume the one way propagation time of the network is “a”. (多選)
		(a)The collision window is equal to a
		(b)The collision window is equal to 2a
		(c)The worst case scenario to detect a collision happens when the two hosts are close to each other.
		(d)The best case scenario to detect a collision happens when the two hosts are close to each other.
		(e)The worst case scenario to detect a collision happens when the two hosts are at opposite ends of the Ethernet

(+10)	(10%) 8. The minimum length of a frame for the Ethernet is 64 bytes. Which of the following is correct when a node receives a frame shorter than 64 bytes ?(單選)
		(a)It is impossible for a node to receive a frame shorter than 64 bytes.
		(b)Frame shorter than 64 bytes can still be accepted as some of the fields such as MAC addresses may be correct.
		(c)Frame shorter than 64 bytes will be treated as “collided” frame and will be dropped directly.
		(d)Frame shorter than 64 bytes can still be accepted as long as it is a “broadcast” frame to all nodes.
		(e)A jamming sequence will be sent to enforce the collision.

(+10)	(10%) 9. The “exponential backoff algorithm” used by CSMA/CD protocol works as each time a network adaptor (card) tries to transmit but fails (collision), it “doubles” the maximum amount of time it waits before trying again. Assume the “collision window” (or slot time) of the network is “T”. Then For the 3rd collision, the waiting time for the node to randomly select is among 0 and __(Ans)__ T times
	Ans	ANS:7

(+10)	(10%) 10. Which of the following are correct for traditional Ethernet CSMA/CD performance ?(多選)
		(a) Ethernets work best under lightly loaded conditions.
		(b) Ethernets work very well under heavy loads as all collisions will be detected.
		(c) It’s a fair network that all nodes may share the bandwidth very equally.
		(d) Under heavy loads, too much of the network’s capacity is wasted by collisions.
		(e) The throughput of an Ethernet depends on the traffic load, heavy load will have better throughput.