Directions (Q36–40): A 3000 km long trunk operates at 1.536 Mbps and transmits 64-byte frames using a sliding window protocol. The propagation speed is 6 μs/km.
The maximum achievable throughput is
Pick an option to check yourself, or open the answer below.
Show answer & explanation
Correct answer: C. 0.901
Explanation
Throughput = efficiency × bandwidth = 0.587 × 1.536 Mbps ≈ 0.901 Mbps.
More Data Communication and Computer Networks questions
- A classless address is given as 167.199.170.82/27. The number of addresses in the network is
- Match List I with List II. List I: (A) DES, (B) AES, (C) 3DES, (D) RSA. List II: (I) Key size 256, (II) Key size 1024, (III) Key size 56, (…
- Match List I with List II. List I: (A) BIND, (B) LISTEN, (C) ACCEPT, (D) SOCKET. List II: (I) Block the caller until a connection attempt a…
- Match List I with List II. List I: (A) Physical layer, (B) Data link layer, (C) Network layer, (D) Presentation layer. List II: (I) Routing…
- Which of the statements given below is/are correct? It is always important and useful to include an 'alt' attribute on 'img' tag in HTML be…
- Match List I with List II. List I: A. Data link layer, B. Network layer, C. Transport layer, D. Session layer. List II: I. True end-to-end…