Consider A Shared Ethernet Network With Only Two Computers
Consider A Shared Ethernet Network With Only Two Computers Each Compu
Consider a shared Ethernet network with only two computers. Each computer transmits at 10 Mbps. At time t=0, each computer attempts to transmit a packet for the first time, but these packets collide. Computer #1 chooses a back-off time of 0 mini-slots, whereas computer #2 chooses a back-off time of 1 mini-slot. Here is information that you might need: The length of a mini-slot is equal to the time required to transmit 512 bits. Each packet contains 10,000 bits. A jam signal consists of 32 bits. The propagation delay from each computer to the Ethernet hub is 0.20 milliseconds. (A millisecond is 10^{-3} seconds.) Ignore the time required to sense that the channel is idle.
Paper For Above instruction
This paper analyzes the timing and collision resolution process in a simplified shared Ethernet environment comprising only two computers. Both computers start transmitting simultaneously at time t=0, leading to a collision due to simultaneous transmission. The goal is to determine the precise times at which each computer detects the collision, when each will attempt retransmission, and whether their retransmissions will collide again.
a) Times when each computer detects the collision
The detection of a collision in an Ethernet network occurs when a transmitting device detects a signal conflict, which is usually rapid due to the physical property that collision signals propagate quickly through the medium. The key parameters are:
- Transmission rate (R): 10 Mbps
- Packet size: 10,000 bits
- Mini-slot size: 512 bits (time to transmit 512 bits) = 512 bits / 10 Mbps = 512 / 10^7 seconds = 51.2 microseconds
- Propagation delay (τ): 0.20 milliseconds = 200 microseconds
The collision occurs at time t=0 as both computers attempt to transmit simultaneously. The collision signal propagates back and forth along the medium. Each computer detects the collision as soon as the collision signal reaches back to them, which occurs after the propagation delay.
Since the propagation delay from each computer to the hub is τ=200 microseconds, the collision signal reaches each computer after τ=200 microseconds. Therefore, both computers detect the collision at:
Time of collision detection: t_collision = 200 microseconds
b) When computer #1 retransmits its packet
Computer #1 selected a back-off of 0 mini-slots. In Ethernet, the back-off time is selected randomly from a range of slots to reduce repeated collisions. Here, with only two devices and small contention window, the back-off delay is calculated as:
Back-off delay = (chosen mini-slots) × (mini-slot time)
Since mini-slot time = time to transmit 512 bits = 51.2 microseconds, and computer #1 chose 0 mini-slots, the back-off delay for computer #1 is zero.
However, transmission cannot occur until the carrier sense detects the medium is idle. Given an initial collision, each computer must wait at least for the propagation delay plus the jam signal transmission to ensure the channel is clear.
The transmission process for retransmission will occur after the collision detection delay plus the back-off.
Thus, computer #1 will start retransmission immediately after its back-off, which is zero, but only after the collision has been resolved and the medium considered idle. Assuming the collision was detected at 200 microseconds, computer #1 will transmit again after a short contention window.
In practice, Ethernet's exponential back-off algorithm doubles the contention window after each collision, but for simplicity, we take the immediate retransmission after the detected collision time. Therefore, computer #1 will retransmit at approximately:
t_retransmit, #1 ≈ t_collision + small delay ≈ 200 microseconds
c) When computer #2 retransmits its packet
Computer #2 chose a back-off time of 1 mini-slot, which is 51.2 microseconds. Its retransmission time will be:
t_retransmit, #2 = t_collision + back-off time (51.2 μs)
Additionally, in Ethernet, after the collision detection at 200 microseconds, the transmitting devices wait for their back-off period before transmitting again. So, computer #2 starts its retransmission approximately at:
t_retransmit, #2 ≈ 200 μs + 51.2 μs ≈ 251.2 μs
d) Will the retransmissions from computer #1 and computer #2 collide?
To evaluate if the retransmissions will collide, consider their transmission start times:
- Computer #1: approximately at 200 μs (back-off=0)
- Computer #2: approximately at 251.2 μs (back-off=1 mini-slot)
The retransmission durations are determined by packet size and transmission rate:
Transmission time = Packet size / Rate = 10,000 bits / 10^7 bits/sec = 0.001 sec = 1 mswhich is much longer than the difference in their retransmission start times. Specifically, computer #1 begins transmitting at approximately 200 μs, and computer #2 begins transmitting at approximately 251.2 μs, which is 51.2 μs later.
Given that the duration of each transmission is 1 ms (1000 μs), and the second retransmission starts only 51.2 μs after the first, the two transmissions will overlap and thus collide again.
In conclusion, the initial collision is detected after 200 microseconds. The first computer retransmits immediately after detection, and the second computer waits 51.2 microseconds before retransmitting. Because their transmission durations overlap, these subsequent transmissions will collide, necessitating further back-off and retransmission attempts.
Summary of Key Timing Events
- Collision detection time: 200 μs
- Computer #1 retransmission: approximately 200 μs
- Computer #2 retransmission: approximately 251.2 μs
- Collision of retransmissions: probable due to overlapping transmission durations
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