part 1: Suppose value of K is 1 (Ethernet CSMA/CD), compute backoff time when link capacity (data rate) is: 10 kbps 100 kbps part 2: Suppose a frame (F) experiences collision forth time on a shared Ethernet link of I Mbps: What is the maximum possible value of K in this case Suppose maximum value of K is taken, compute backoff bit time for frame F Convert bit time into absolute time (second or fraction of a second) part 3: Suppose two hosts (Host-A and Host-B) are connected through a shared Ethernet bus of 1 Mbps. The distance between two nodes is 5 meters. The signal propagation speed over physical media is 2.5 x 10^5 meters/second. Compute the propagation delay in this case. What will be the propagation delay for a shared bus of 10 Mbps instead Suppose both hosts sense idle channel at the same time and starts transmission. Will it result in collision (yes or no) If yes: How much time is needed to detect this collision at Host-A? If yes: How much time is needed to detect this collision at Host-B?
part 1:
Suppose value of K is 1 (Ethernet CSMA/CD), compute backoff time when link capacity (data rate) is:
- 10 kbps
- 100 kbps
part 2:
Suppose a frame (F) experiences collision forth time on a shared Ethernet link of I Mbps:
- What is the maximum possible value of K in this case
- Suppose maximum value of K is taken, compute backoff bit time for frame F
- Convert bit time into absolute time (second or fraction of a second)
part 3:
Suppose two hosts (Host-A and Host-B) are connected through a shared Ethernet bus of 1 Mbps. The distance between two nodes is 5 meters. The signal propagation speed over physical media is 2.5 x 10^5 meters/second.
- Compute the propagation delay in this case.
- What will be the propagation delay for a shared bus of 10 Mbps instead
- Suppose both hosts sense idle channel at the same time and starts transmission. Will it result in collision (yes or no)
- If yes: How much time is needed to detect this collision at Host-A?
- If yes: How much time is needed to detect this collision at Host-B?
Step by step
Solved in 2 steps