Homework 5 Solution

# The we get t is 25 msec number we get s t b 65540 s

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Unformatted text preview: terval, 50\$ tains 50 However, TEquating sequence because during that + 10\$t = more65,540 is received instead of 4 (a bucket. \$t Mbytes.this answer these two quantities, is ever corrupted andt. ecorrect packets can be obtained by will be the formula obsolete, since the current seque rror), answer 5 through 65,540 using rejected as Solving this equation, tokens arrive. The we get \$t is 25 msec. number we get S = t/ b ! 65,540. S = C/ (M ! #). Substituting, is thought 8 o(6 e 1) or 1.6 sec. 28. Call the length of the e solution burstllinterval problems is to include the age of each packet after the seque Th maximum to a these \$t. In the extreme case, the bucket is full at thenumber the d decrement it once per second. tWhen the age hits zero, the information f start of an interval (1 Mbyte) and another 10\$ Mbytes come in during the router is discarded. during the transmission burst mes in, say, every 10 sec, so router in interval. The output Normally, a new packet co contains 50\$t Mbytes. only times out when a router wedown (o10\$t = onsecutive packets have been lost, an u Equating these two quantities, is get 1 + r six c 50\$t. the buffer are exhausted. 24. The router has to do approximately the same amount of work queueing a packet, no matter how big it is. There is little doubt that processing 10 packets of 100 bytes each is much more work than processing 1 packet of 1000 bytes. 25. 5It is not possible toasend any network greater than by a token bucket. The token bucket is filled at . A computer on 6-Mbps packets is regulated 1024 bytes, ever. a a token Mbps. sec, 200,000 cells/sec can with 8 Each cell holds 48 26. Withrate of 1 every 5Itµis initially filled to capacitybe sent. megabits. How long can the computer t bytes or the full Mbps? dataransmit at384 bits.6The net data rate is then 76.8 Mbps. 27. The naive answer says that at 6 Mbps it takes 4/3 sec to drain an 8 megabit bucket. However, this answer is wrong, because during that interval, more tokens arrive. The correct answer can be obtained by using the formula S = C/ (M ! #). Substituting, we get S = 8 / (6 ! 1) or 1.6 sec. 28. Call the length of the maximum burst interval \$t. In the extreme case, the bucket is full at the start of the interval (1 Mbyte) and another 10\$t Mbytes 6come in during the interval. The outputPoisson Processes of rates λ1 and λ2 produces a new . Show that merging two independent during the transmission burst contains 50\$t Mbytes. Equating these two quantities, we get 1 + 10\$t = 50\$t. Poisson Process with rate λ1+λ . Solving this equation, we get \$t is2 25 msec. Show that the probability of receiving k arrivals is given by a Poisson distribution with parameter λ1+λ2. You should also show the independent increment property....
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## This note was uploaded on 01/16/2014 for the course ECEN 424 taught by Professor Schlumprecht during the Fall '13 term at Texas A&M.

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