Ethernet CSMA/CD behavior after collisions: Which statements correctly describe how hosts determine when to transmit on a CSMA/CD Ethernet LAN, especially after a collision occurs?

Difficulty: Easy

Correct Answer: 2 and 5

Explanation:


Introduction / Context:
Classic shared-media Ethernet uses Carrier Sense Multiple Access with Collision Detection (CSMA/CD). Understanding the rules for sensing, deferring, and retransmitting after collisions is fundamental to legacy Ethernet operations and exam objectives.



Given Data / Assumptions:

  • Multiple hosts share one collision domain (hub or half-duplex link).
  • CSMA/CD dictates when a host can transmit.
  • After collisions, a defined backoff algorithm prevents repeated simultaneous retransmissions.


Concept / Approach:

Correct statements are: (2) stations must wait until the medium is idle before transmitting, and (5) after a collision, all stations perform a random backoff; when their timers expire, they all have equal priority to transmit. Statement (1) is false because simultaneous successful transmissions cannot occur on the same collision domain. Statements (3) and (4) mischaracterize hubs and priority rules.



Step-by-Step Solution:

Sense carrier; if busy, defer (rule 2).If a collision occurs, jam signal is sent.All stations invoke exponential random backoff (rule 5).When backoff expires and medium is idle, a station attempts retransmission with no innate priority.


Verification / Alternative check:

Ethernet specifications define binary exponential backoff; trace captures show jam sequences and randomized retransmission intervals after collisions.



Why Other Options Are Wrong:

1: Simultaneous success is impossible on a shared collision domain.

3: Adding hubs increases collisions; it does not improve CSMA/CD efficiency.

4: No host gains first priority after a collision; all defer and randomize.



Common Pitfalls:

Applying CSMA/CD rules to full-duplex switched Ethernet (no collisions); assuming priority mechanisms exist where they do not.



Final Answer:

2 and 5

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