Q8.Marks: +2.0UGC NET Paper 2: Computer Sc 6th Jan 2025 Shift 1
The network using CSMA/CD has a bandwidth of 20 mbps. If the maximum propagation time is 25μ sec (microsecond), what is the minimum size of the frame?
1.500 bits
2.1000 bits✓ Correct
3.1500 bits
4.2000 bits
Solution
The correct answer is 2) 1000 bits.
Key Points
The network uses CSMA/CD, which stands for Carrier Sense Multiple Access with Collision Detection.
To determine the minimum frame size in a CSMA/CD network, we need to ensure that the transmission time is at least twice the maximum propagation delay. This is to detect collisions before the entire frame is transmitted.
Given:
Bandwidth (B) = 20 Mbps
Maximum Propagation Time (T) = 25 μs
The minimum frame size (S) can be calculated using the formula: S = 2 * B * T
Calculating the minimum frame size:
S = 2 * 20 Mbps * 25 μs
Converting units: 1 Mbps = 10^6 bits per second and 1 μs = 10^-6 seconds
However, the actual minimum frame size to avoid collision detection issues is often standardized, and in this case, it is 1000 bits.
Additional Information
CSMA/CD is used in Ethernet networks to manage the transmission of data and avoid collisions. The calculation ensures that the round-trip time for a signal to propagate across the network and back is taken into account.
Understanding frame size calculations is crucial for network design and ensuring efficient data transmission.