Here's the question bank on all the computer networks topics.
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?
Concept:Using concept of pipelining:For packet 1 we have to take \(3 \times {T_t}\) (because 3 links are there ) for all subsequent packets there will be one \({{\rm{T}}_{\rm{t}}}\) time.Data:Packet size = 103 bytesHead size = 100 bytesFormula:\({\rm{Transmisson\;time\;}}\left( {{\rm{\;}}{{\rm{T}}_{{\rm{t\;}}}}} \right) = \frac{{{\rm{Data}} + {\rm{Header}}}}{{{\rm{Bandwidth}}}}\)Explanation:Case 1:Number of packets = 1\({{\rm{T}}_{\rm{t}}} = \frac{{1000 + 100}}{{1000000}} = \frac{{1100}}{{1000}} = 1.1{\rm{ms}}\) \({\rm{T}}1 = 3 \times {{\rm{T}}_{\rm{t}}} = 3.3{\rm{ms}}\) Case 2:Number of packets = 10Size of one packet = \(\frac{{1000}}{{10}} = 100\;bytes\)\({{\rm{T}}_{\rm{t}}} = \frac{{100 + 100}}{{1000000}} = \frac{{200}}{{1000}} = 0.2{\rm{ms}}\) \({\rm{T}}2 = 3 \times 0.2{\rm{\;}} + 0.2 \times 9 \times 1 = 2.4{\rm{\;ms}}\) Case 3:Number of packets = 20Size of one packet = \(\frac{{1000}}{{20}} = 50\;bytes\)\({{\rm{T}}_{\rm{t}}} = \frac{{50 + 100}}{{1000000}} = \frac{{150}}{{1000}} = 0.15{\rm{ms}}\)\({\rm{T}}3 = 3 \times 0.15{\rm{\;}} + 0.15 \times 19 \times 1 = 3.3{\rm{\;ms}}\)So T1 = T3, T3 > T2 is the correct answer.
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