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 6Mbps 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.
 Fall '13
 Schlumprecht
 Networking

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