Consider the store and forward packet switched network given below. Assume…
2014
Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 106 bytes / sec. A user on host A sends a file of size 103 bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?
Answer: D. T1 = T3, T3 > T2 — Key idea: store-and-forward across three links (A → R1 → R2 → B) allows pipelining of packets. Transmission time per packet on a link = packet size /…
- A.
T1 < T2 < T3
- B.
T1 > T2 > T3
- C.
T2 = T3, T3 < T1
- D.
T1 = T3, T3 > T2
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Correct answer: D
Key idea: store-and-forward across three links (A → R1 → R2 → B) allows pipelining of packets. Transmission time per packet on a link = packet size / bandwidth. Total time for N equal packets through L links = (L + N - 1) * (packet size / bandwidth).
Given values: file size F = 1000 bytes, header H = 100 bytes, bandwidth R = 10^6 bytes/s, number of links L = 3.
Case 1 (single packet): packet size S1 = F + H = 1100 bytes. Transmission time per link = 1100/10^6 = 0.0011 s. Total time T1 = L * (1100/10^6) = 3 * 0.0011 = 0.0033 s.
Case 2 (10 parts): user data per packet = F/10 = 100 bytes, packet size S2 = 100 + 100 = 200 bytes, per-link time = 200/10^6 = 0.0002 s. Total time T2 = (L + 10 - 1) * 0.0002 = 12 * 0.0002 = 0.0024 s.
Case 3 (20 parts): user data per packet = F/20 = 50 bytes, packet size S3 = 50 + 100 = 150 bytes, per-link time = 150/10^6 = 0.00015 s. Total time T3 = (L + 20 - 1) * 0.00015 = 22 * 0.00015 = 0.0033 s.
Conclusion: T1 = 0.0033 s, T2 = 0.0024 s, T3 = 0.0033 s. Therefore the correct relationship is T1 = T3 and both are greater than T2.