In the network diagram shown below, Host A and Host B are connected via four swi
ID: 3747980 • Letter: I
Question
In the network diagram shown below, Host A and Host B are connected via four switching devices S1, S2, S3 and S4 and five links L1, L2, L3, L4 and L5. Each ink has a transmission rate R-2 Mbps, a physical length of d250 Km (1Km-1000 m) and propagation speed S-3x10 m/s Host A Host B Si S2 S3 S4 4 LI 2 L3 t Assume that the network is a packet network with negligible processing and queuing delays. Host A is to send Host B a file of size L=73000 bytes. The file is divided into packets. Each packet is appended with a 40 byte header. The packet size, including the header is to be P=1500 bytes. 1- How long does it take to transfer the file? [5 points 2- What is the percentage of the overhead data transferred in the headers? [5 points]Explanation / Answer
Size of file, L = 73000 bytes
Packet size including header = 1500 bytes
Header size = 40 bytes
Therfore, Maximum data that can be transferred in 1 packet = 1500 - 40 = 1460 bytes
No.of packets needed = 73000/1460 = 50 packets
Transmission rate = 2 Mbps = 2 Mega bits per second = (2 * 10^6) / 8 bytes = 0.5 * 10^6 bytes
(Note that Mbps = Mega bits per second and MBps = Mega Bytes per second)
Since every packet has to reach the host B, every packet has to travel through the 5 links
Propagation time for each link = 250 km / (3 * 108) = 25/(3*104) = 8.33/104 = 0.833 milli sec
Propogation time for 5 links = 5 * (Propagation time for each link) = 5 * 0.833 = 4.165 milli sec
Transmission time for 1 link = 1500 / (0.5 * 106) = 3 milli seconds
Transmission time for 5 links = 5*3 = 15 milli seconds
Now, the above calculated transmission time is for 1 packet
For the first packet, transmission delay will be 0
Hence transmission delay for 49 packets = 49 * 15 msec = 735 msec
Total delay = 735 msec + 4.165 msec = 79.165 msec
Hence it takes approximately 79.165 msec to transfer the file.
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