Homework 5 Solution

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

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.

Ask a homework question - tutors are online