Based on the following passage, answer the Questions : A 3000 km long trunk…
2022
Based on the following passage, answer the Questions :
A 3000 km long trunk operates at 1.536mbps and is used to transmit 64 bytes frames and uses sliding window protocol. The propagation speed is 6𝜇 sec/km.
The transmission and propagation delays are respectively
- A.
Tt=333.33 𝜇 sec,Tp=18000 𝜇 sec
- B.
𝑇𝑡=300 𝜇 sec,𝑇𝑝=15360 𝜇 sec
- C.
Tt=33.33 𝜇 sec,Tp=1800 𝜇 sec
- D.
𝑇t=1800 𝜇 sec,𝑇p=33.33 𝜇 sec
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Correct answer: A
Solution: Compute transmission and propagation delays.
Transmission delay (Tt): Frame size = 64 bytes = 64 × 8 = 512 bits. Link rate = 1.536 Mbps = 1.536 × 10^6 bits/s. Tt = 512 / 1.536×10^6 s = 0.000333333... s = 333.33 μs.
Propagation delay (Tp): Propagation time per km = 6 μs/km. Distance = 3000 km. Tp = 3000 × 6 μs = 18000 μs = 0.018 s.
Therefore, transmission delay = 333.33 μs and propagation delay = 18000 μs.
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