Solution1

Solution1 - Problem 6 Coding delay is (48*8) / (64*103) = 6...

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Problem 6 Coding delay is (48*8) / (64*10 3 ) = 6 ms, transmission delay is 48*8/10 6 = 0.384 ms, propagation delay is 2 ms. Total delay = 6 ms + 0.384 ms + 2 ms = 8.384 ms Problem 7 a) 10 users can be supported because each user requires one tenth of the bandwidth. b) c) . d) Problem 12 The queuing delay is 0 for the first transmitted packet, L/R for the second transmitted packet, and generally, (n-1)L/R for the n th transmitted packet. Thus, the average delay for the N packets is (L/R + 2L/R + . ...... + (N-1)L/R)/N = L/RN(1 + 2 + . .... + (N-1)) = LN(N-1)/(2RN) = (N-1)L/(2R) Note that here we used the well-known fact that 1 + 2 + . ...... + N = N(N+1)/2 Problem 18 a) 1,000,000*(10,000,000/250,000,000) = 40,000 bits b) 40,000 bits c) The bandwidth-delay product of a link is just the maximum number of bits that can be in the link d) 1 bit is 250 meters long, which is longer than a football field e) s/R Problem 24 a) It takes (7.5*10 6 ) / (1.5* 10 6 ) = 5 seconds to move the message from source to the
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This note was uploaded on 11/09/2010 for the course ECE 441 taught by Professor Davidstarobinski during the Fall '09 term at BU.

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Solution1 - Problem 6 Coding delay is (48*8) / (64*103) = 6...

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