Q98.Marks: +2.0UGC NET Paper 2: Computer Science 8th Oct 2022
📄 Passage
Based on the following passage, answer the Questions:
A 3000 km long trunk operates at 1.536 mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6 μ sec/km.
The transmission and propagation delays are respectively
1.Tr = 333.33 μ sec, Tp = 18000 μ sec✓ Correct
2.Tr = 300 μ sec, Tp = 15360 μ sec
3.Tr = 33.33 μ sec, Tp = 1800 μ sec
4.Tr = 1800 μ sec, Tp = 33.33 μ sec
Solution
Transmission Delay
Tr = Packet size / Bandwidth
= 64 bytes / 1.536 Mbps
= (64 × 8 bits) / (1.536 × 106 bits per sec)
= 333.33 μsec
Propagation Delay
For 1 km, propagation delay = 6 μsec
For 3000 km, propagation delay Tp= 3000 × 6 μsec = 18000 μsec