...the
There is a common misunderstanding about modem speed. a modem has 2400 baud means it can send 2400 such waveforms per second. But if there are 16 distinct waveforms (4bits per waveform), the modem speed is 9600bps. See also Fig 2-12 on textbook.
...constraint.
without such a constraint, we can of course make the signal infinitely strong so that it completely overwhelm the noise in the channel, making the transmission completely error free.
...noise
35.450, 35.651 and other ECSE courses deal specifically with these analytical issues
...found.
Although much progress has been made since Shannon's first paper, developing effective error protection schemes remains to be one of the most challenging research problems today.
...TDM
see page 101 on textbook.
...switched,
see 2.5.2-2.5.4 on text.
...transmitting.
Mathematically, this can be made much more precise by using probability theory.
...process.
More precisely, the number of transmission requests is a random variable, with probability:
...listen
only during, not before the transmission
...occur.
why use random delay, not fixed delay?
...channel.
How so?
....
Fig.3-12 on text
...throughput.
Why?
...ring
Actually it's more like a star topology for easy maintenance, see text for details
...token.
This is called release after transmission (RAT). There is also a release after reception (RAR), i.e., the sender releases the token only after the frame comes around the ring back to the sender.
....
clearly, , , , ... are the times the token arrives at node 1, etc.
...diagram:
Different people may have different understanding of each of the components, what's more important are the functions they have, not their names.
...diagram:
From G.E. Keiser's book
...reduced.
Suppose a channel has a data rate of 100bps, with bit error rate of 0.01, we can add so much redundancy to the data bit stream, that the actual transmission rate is 100, at a error rate of 0.49, you can check that this is still a better channel than the one without error codes. Also think why I didn't use 0.99 as the error rate?
...example
From Bertsekas and Gallagher
...Let
Because we are dealing with binary operations, the above can also be written as
...and
From Walrand's book.
...bandwidth.
Here by bandwidth we mean the maximum digital transmission rate, which is different from the physical channel bandwidth defined in Fourier sense.
...public.
The standard book can be purchased in bookstore.
...header.
We will see that an IP address is 4bytes long, if we want to include both source and destination, that'll be 8bytes just for the addresses.
...applied
Internet RFC1591, J. Postel, March 1994
...bytes
The first 8 bytes of the IP datagram contains further information about the service carried in the IP datagram, such as TCP, UDP, etc. In fact it is the header for higher layer packets encapsulated in the IP datagram.
...source
Remember IP header contains source IP address
...time.
This assumption is not necessary for general Poisson processes, but we make it here to simplify our analysis later.
...seconds
To be precise, three packets served and one arrival, or four packets served and two arrivals, ... should also be included in the calculation. But it can be shown that these extra terms are all negligible with respect to .
...chain
On the other hands, the solution for an ergodic Markov chain, or ergodic, but before reaching the steady state, is called transient solution.
...,
Other well knowns ports: 21 for , 69 for , etc.
...128.113.42.66.1056.
Whether to use : or . in the port address depends on the OS implementations.
...variable
actually this definition is not precise, more precisely it should be that the cumulative distribution function is continuous.
...Tanenbaum.
To make the solution trackable, the problem made the implicit assumption that the Binary Exponential Backoff algorithm does not terminate when more than consecutive 10 collisions have occured.
...the
the machine rcs-ibm3.rpi.edu has the program.

zhangj@
Mon May 23 15:29:10 EDT 1994