Download Local Area Networks

Network Architecture and Client-Server Computing
Assignment
Total Points: 45
Due: 3/3/11
1. A 4480-byte datagram is to be transmitted and needs to be fragmented because it
will pass through an Ethernet with a maximum payload of 1500 bytes. Show the
Total Length, More Flag, and Fragment Offset values in each of the resulting
fragments.
8 points
2. Consider the effect of using slow start on a line with a 10-msec round-trip time
and no congestion. The receive window is 32 KB and the maximum segment size
is 2 KB. How long does it take before the first full window can be sent?
5 points
3. If the TCP round-trip time, RTT, is currently 30 msec and the following
acknowledgements come in after 26, 32, and 24 msec, respectively, what is the
new RTT estimate after each acknowledgement?
5 points
4. A transport layer message consisting of 1500 bits of data and 160 bits of header is
sent to an internet layer, which appends another 160 bits of header. This is then
transmitted through two networks, each of which uses a 24-bit packet header. The
destination network has a maximum packet size of 800 bits. How many bits,
including headers, are delivered to the network layer protocol at the destination?
5 points
5.
a. What is the fastest line speed at which a host can blast out 1500-byte TCP
payloads with a 120-sec maximum packet lifetime without having the sequence
numbers wrap around? Take TCP, IP, and Ethernet overhead into consideration.
Assume that Ethernet frames may be sent continuously and that the Ethernet
overhead is 26 bytes.
5 points
b. To get around the problem of sequence numbers wrapping around while old
packets still exist, one could use 64-bit sequence numbers. However,
theoretically, an optical fiber can run at 75 Tbps. What maximum packet lifetime
is required to make sure that future 75 Tbps networks do not have wraparound
problems even with 64-bit sequence numbers? Assume that each byte has its own
sequence number, as TCP does.
5 points
6. One key design decision for ATM was whether to use fixed or variable length
cells. Let us consider this decision from the point of view of efficiency. We can
define transmission efficiency as
U = number of information bytes / (number of information bytes + number of
overhead bytes)
a) Consider the use of fixed-length cells. In this case the overhead consists of the
header bytes. Define:
L = data field size of the cell in bytes
H = header size of the cell in bytes
X = number of information bytes to be transmitted as a single message
Derive an expression for U. (Hint: The number of cells needed to transmit B
bytes is  X/ L.
b) If cells have variable length, overhead is determined by the header, plus the
flags to delimit the cells or an additional length field in the header. Let Hv =
additional overhead bytes required to enable the use of variable-length cells.
Derive an expression for U in terms of X, H, and Hv.
c)
Let L = 48, H = 5, and Hv = 2. Plot U versus message size for fixed and
variable-length cells (use message size, X = 24, 48, 72, 96, 120, 144, 168, and
192). Comment on the results for U for fixed length cells when X is small and
when X becomes large. Do the same for U for variable-length cells. Does the
efficiency of variable-length cells provide significant gains over the
efficiency of fixed-length cells form most values of X?
12 points