Here's the question bank on all the computer networks topics.
Consider a source computer (S) transmitting a file of size 106 bits to a destination computer (D) over a network of two routers (R1 and R2) and three links (L1, L2, and L3). L1 connects S to R1; L2 connects R1 to R2; and L3 connects R2 to D. Let each link be of length 100 km. Assume signals travel over each link at a speed of 108 meters per second. Assume that the link bandwidth on each link is 1Mbps. Let the file be broken down into 1000 packets each of size 1000 bits. Find the total sum of transmission and propagation delays in transmitting the file from S to D?
The correct answer is option 1Data:Number of packets = n =1000Size of each Packet = L = 1000 bitsThe link bandwidth on each link is = BW = 1Mbps =106 bits/sEach link be of length = d= 100 kmSignals travel over each link at a speed = v= 108m/sTransmission delay = TPropagation delay = PFormula:\(T = \frac{L}{{BW}}\)\(P = \frac{d}{v}\)Calculation:Propagation delay for the first packet to transmit S to R1\(P = \frac{{100×10^3}}{{ {{10}^8}}} = {1\times10^{ - 3}}\;s = 1\;ms\)So, the propagation delay for the first packet to transmit S to DP =1 × 3=3msTransmission delay for the first packet to transmit S to R1\(T = \frac{{1000}}{{{{1\times10}^6}}} = 1 × {10^{ - 3}}\;s = 1\;ms\;\)So, the transmission delay for the first packet to transmit S to DT=1 × 3=3msSo, total time is taken to transmit the first packet from S to D =P +T=6msUsing pipelining, the remaining packet follows the first packetthen remaining every packet take 1ms to transmit from S to DSo, the total time is taken to transmit all 1000 packets from S to D is? time for the first packet + time for the remaining 999 packets? 6ms + 999ms? 1005msSo, the total time is taken to transmit all 1000 packets from S to D is 1005ms.
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