What is the minimum frame size for a CSMA/CD network running at 1 Gbps over a 1-km cable with no repeaters?
The minimum frame size for this CSMA/CD network is 0.00125 bytes.
Understanding Minimum Frame Size Calculation
To determine the minimum frame size for a CSMA/CD network running at 1 Gbps over a 1-km cable with no repeaters, we need to consider the signal speed in the cable, which is given as 200,000 km/sec.
Round-Trip Time (RTT) Calculation:
The round-trip time (RTT) is the time it takes for a signal to travel from the sender to the receiver and back. In this case, the cable length is 1 km, so the total distance the signal needs to travel is 2 km (1 km to the receiver and 1 km back to the sender).
Using the signal speed of 200,000 km/sec, we can calculate the round-trip time (RTT) as follows:
RTT = total distance / signal speed
RTT = 2 km / 200,000 km/sec
RTT = 0.00001 sec
Minimum Frame Size Calculation:
Since the sender needs to detect collisions before the entire frame is transmitted, the minimum frame size should be long enough to ensure that at least one bit is transmitted within the round-trip time (RTT).
To calculate the minimum frame size, we convert the transmission speed of 1 Gbps to bytes per second:
1 Gbps = 1,000 Mbps = 1,000,000 kbps = 1,000,000,000 bps.
Thus, 1 Gbps = 1,000,000,000 bps.
Minimum frame size = RTT / transmission speed
Minimum frame size = 0.00001 sec / 1,000,000,000 bps * 8 bits/byte
Minimum frame size = 0.00125 bytes
Therefore, the minimum frame size for a CSMA/CD network running at 1 Gbps over a 1-km cable with no repeaters is 0.00125 bytes. A frame of this size ensures collision detection within the specified round-trip time.