UDP has only 8 octets of "overhead" whereas TCP has 20 octets of
"overhead". Why does UDP require so little "overhead"?
The following figure illustrates typical density functions of
arrival times in the transport layer.
What is the implication of this density functions for the
timeout interval that should be used? Explain!
In general, what are the drawbacks of using a small timeout interval?
In general, what are the drawbacks of using a large timeout interval?
The process used to release a TCP connection is fairly complicated.
Describe the process.
Is there any reason to acknowledge aknowledgments?
Recall that TCP sequence numbers are 4 octets long and that they
"wrap" (i.e., they reset to 0 after the last one is used).
How many unique sequence numbers are there?
Why is it less than ideal to exhaust sequence numbers?
Suppose a host elects to use the default TCP maximum
segment size (MSS) of 536 octets (to avoid IP
fragmentation). What is the largest file that can be
transferred without exhausting the sequence numbers.
The SCTP server does not keep state information after it responds
with an INIT ACK, it waits until after it receives a COOKIE ACK.
Why is this a good idea (from a resources perspective)?
Given this, and the fact that the server authenticates the cookie in the
COOKIE ECHO, what must be true of the authentication process?
Multipath TCP (MPTCP) is of particular interest to smartphone users.
Why?
What can happen on a smartphone (that does not support MPTCP) if you
initiate a TCP connection using WiFi, start to use that connection,
and then the WiFi signal is lost?
Suppose we have two devices that are communicating with each other
using a scheme in which every message that is transmitted
by the sender must be acknowledged by the receiver.
(If the acknowledgment is not received within a prescribed amount of time,
the sender will try again.)
Suppose further
that the probability that the sender's message does not reach the
receiver is \(p\) and that the probability that the
receiver's acknowledgment does not reach the sender is \(q\).
What is the probability, \(P\{\mbox{Success}\}\),
that a message will be sent and acknowledged
correctly the first time it is transmitted?
What is the probability that the sender will "time out" on the first
attempt (and need to
try again)? That is, what is \(P\{\mbox{Failure}\}\)?
What is the probability that the sender will "time out" on the first
attempt and succeed on the second attemp? That is, what is
\(P\{\mbox{Success in 2}\}\)?
What is the probability that the sender will "time out" on the first
\(n-1\) attempts and succeed on attemp \(n\)?
That is, what is \(P\{\mbox{Success in }n\}\)?
What is the expected number of attempts?
(Hint: Letting \(P\{\mbox{Success}\} = 1 - q - p + pq\), the
\(P\{\mbox{Success in }n\}\) is given by the geometric
distribution
\((1 - P\{\mbox{Success}\})^{n-1} \cdot P\{\mbox{Success}\}\).
So, this question is asking you to use a reliable source
and find the expected value of the geometric distribution.)
The textbook claims that the utilization rate for the stop-and-wait
protocol is:
\[
U = \frac{L/R}{RTT + L/R}
\]
In lecture, we said that the utilization rate for the stop-and-wait
protocol is:
\[
U = \frac{1}{1 + \left( \frac{RTT}{d_{\mbox{trans}}} \right)}
\]
Show that they are the same.
Suppose \(L = 1000\) (in octets) and \(R = 10^9\)
(in bits per second), what must the round trip travel time be
when using the stop-and-wait protocol for the
utilization to be 0.90?
Using the same assumptions as in the previous question and also
assuming that \(w = 10\), what must the round trip travel
time be when using the fixed window flow control for the
utilization to be 0.90?
