Chem_402_Key_Homework_2_Fall_2010 - mentioned above and the...

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Chem 402: Chemical Information Systems Dr. George A. Papadantonakis Fall 2010 Key to Homework No. 2 Due on Friday, September 17, 2010 There are in all 132 bytes in the message below, or 1056 bits, This is the first assignment from the Chem 402 class. It is about packets and their transmission over links with different bandwidths. Each letter of the message above represents 1 byte (8 bits). That includes blanks and periods. (a) (6 points) Find the time it will take to transmit this message on end to end links with bandwidths of 56kbps (phone modem), 1.5Mbps and 4Mbps(DSL), assuming circuit switching. Link Speed time 56kbps 18.86 1.5Mbps 0.704 4 Mbps 0.264 (b)(9 points) If you were to break up the message into packets of size 20 bytes each, with 4 bytes for header, how long will it take to reach the destination if it must go through all three links
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Unformatted text preview: mentioned above, and the routers are operating in store and forward mode. The message will get divided into 7 packets of size 24 bytes, or 192 bits each (for simplicity, we'll assume that the last one is padded out). The delays introduced by the three links are 3.43, 0.128 and 0.048 msecs each. The time it takes the message to reach the destination is really the time at which the seventh packet starts plus the delays introduced at the three links. The time at which 7th packet starts, assuming that 56kbps is the first link is 6x3.43. Add all the delays and you get 7x3.43+0.128+0.048 msec as the answer. Since the order of the links is not specified, assuming any one of them to be the starting link would be OK....
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