NTA UGC NET Aug 2024 Paper-2

Question 13

Consider a CSMA/CD network that transmits data at the rate of 100 Mbps over a 1 Kilometre cable with no repeater. If the minimum frame size required for this network is 1250 bytes, what is the signal speed (km/sec) in the cable.

A	8000
B	16000
C	12000
D	20000

computer-networksCSMA/CD

Question 13 Explanation:

To calculate the signal speed (in km/sec) in the cable, we need to use the relationship between the transmission speed, propagation delay, and frame size in a CSMA/CD network.

### Given:
– Transmission rate = 100 Mbps
– Frame size = 1250 bytes
– Distance = 1 kilometer

### Step 1: Convert the frame size to bits
1 byte = 8 bits, so:
Frame size = 1250 bytes = 1250 × 8 = 10000 bits

### Step 2: Calculate the transmission time for the frame
Transmission time = Frame size in bits / Transmission rate
Transmission time = 10000 bits / (100 × 10^6 bits/sec) = 0.0001 sec = 100 microseconds

### Step 3: Use the propagation delay formula
In a CSMA/CD network, the minimum frame size must be large enough to ensure the sender can detect a collision. The round-trip propagation delay should be at least the time to send the frame. The round-trip propagation delay time is given by:

Propagation delay = Transmission time = 100 microseconds
One-way propagation delay = 100 microseconds / 2 = 50 microseconds

### Step 4: Calculate the signal speed
Propagation delay = Distance / Signal speed
Rearranging for signal speed:
Signal speed = Distance / Propagation delay
Signal speed = 1 km / (50 × 10^(-6) sec) = 20000 km/sec

### Final Answer:
The signal speed in the cable is 20000 km/sec.

Correct Answer: D