InClassActivity24-sol

InClassActivity24-sol - Activity#24(Solution 8:10 AM...

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ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2003 Page 1 of 3 Activity #24 (Solution) 4/23/2003 8:10 AM Activity 24.1 Circuit and packet switching Consider a pair of nodes connected by a network which has link bandwidth of 80 Mbps. a. (8 points) Find the total time needed to transfer a 4,000,000 byte message if the transfer is based on circuit switching with link setup time of 80ms and de-allocation time of 40ms. (Mbps = 10 6 bits/sec, Gbps = 10 9 bits/sec, KB= 2 10 bytes, MB = 2 20 bytes.) It takes 80ms to setup the link, plus transfer time for 32 million bits (i.e. 32,000,000 / 80x10 6 = 400ms), plus 40ms to de- allocate the link. Total = 520ms = 0.52 sec. b. (8 points) Calculate the throughput in bits per second. Throughput is the actual number of “payload” bits that make it, after the overheads. In this case, 32,000,000 bits made it in 0.4 sec, so the throughput 32,000,000 / 0.4 = 80 Mbps. One can also consider the setup and de-allocation times, which is also acceptable. Then, the throughput would be 32,000,000 / 0.52 = 61.5 Mbps. c. (8 points) Find the number of packets that this message is broken into. Assume that the packet size is 512 bytes, with header size of 12 bytes. Payload = 512 – 12 = 500 bytes So, the number of packets = 4,000,000 bytes / 500 bytes = 8,000 packets d. (8 points) Find the total number of bits being sent for case (c) above. 8,000 packets with 512 bytes each sums up to 8,000x512x8 = 32,768,000 bits e. (8 points) Find the total time needed to transfer the 4,000,000 byte message if the transfer is based on packet switching . Assume that the data can be sent continuously. The initial setup time is 20ms. From (d), it takes 32,768,000 bits / 80,000,000 bps = 0.4096s (without setup time)

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InClassActivity24-sol - Activity#24(Solution 8:10 AM...

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