Katherine Dawicki
CSC 8560
Dr. Schragger
April 1, 2003
Homework 5
Pg. 513 problems section
1. Suppose the information content of a packet is the bit pattern 1010101010101011 and
an even parity scheme is being used. What would the value of the checksum field be
for the case of a two-dimensional parity scheme? (one’s compliment) Your answer
should be such that a minimum-length checksum field is used.
11101
8. p should equal .01 to maximize efficiency. Np(1-p)^(N-1)= .37*p(1-p)^(.37-1)
18. Suppose two nodes, A and B, are attached to opposite ends of a 900 m cable, and that
they each have one frame of 1,000 bits (including all headers and preambles) to send
to each other. Both nodes attempt to transmit at time t=0. Suppose there are four
repeaters between A and B, each inserting a 20-bit delay. Assume the transmission
rate is 10 Mbps, and CSMA/CD with backoff intervals of multiples of 512 bits is
used. After the first collision, A draws K=0 and B draws K=1 in the exponential
backoff protocol. Ignore the jam signal.
a) What is the one-way propagation delay (including repeater delays ) between A
and B in seconds? Assume that the signal propagation speed is 2 * 10^8 m/sec.
900m/(2 * 10^8)m/s + (4*20+1,000)/10000000) = 0.0001125 seconds
b) At what time (in seconds) is A’s packet completely delivered at B?
1000/10000000 + 900m/(2 * 10^8)m/s + (4*20+512)/10000000)=0.0000637
seconds
+ 0.0001125 seconds=0.0002672 seconds
c) Now suppose that only A has a packet to send and that the repeaters are replaced
with bridges. Suppose that each bridge has a 20-bit processing delay in addition
to a store-and-forward delay. At what time, in seconds, is A’s packet delivered at
B? 900m/(2 * 10^8)m/s + (4*20+1,000)/10000000) = 0.0002125 seconds
22. Recall that ATM uses 53-byte packets consisting of five header bytes and 48 payload
bytes. Fifty- three bytes is unusually small for fixed-length packets; most networking
protocols (IP, Ethernet, Frame Relay, and so forth) use packets that are, on average,
significantly larger. One of the drawbacks of a small packet size is that a large
fraction of link bandwidth is consumed by overhead bytes; in the case of ATM,
almost 10 percent of the bandwidth is “wasted” by the ATM header. In this problem
we investigate why such a small packet size was chosen. To this end, suppose that
the ATM cell consists of P bytes (possibly different from 48) and 5 bytes of header.
a) Consider sending a digitally encoded voice source directly over ATM. Suppose
the source is encoded at a constant rate of 64 Kbps. Assume each cell is entirely
filled before the source sends the cell into the network. The time required to fill a
cell is the packetization delay. In terms of L, determine the packetization delay
in milliseconds.
Voice is low bandwidth, and thus has a large packetization.
packetization delay in seconds= 8L / 64kbps = L/8kb
b) Packetization delays greater than 20 msec can cause a noticeable and unpleasant
echo. Determine the packetization delay for L = 1,500 bytes (roughly
corresponding to a maximum-sized Ethernet Packet) and for L = 48
(corresponding to an ATM cell).
L= 1,500bytes packetization delay in seconds = .1875
L= 48 bytes packetization delay in seconds =0.006
c) Calculate the store-and-forward delay at a single ATM switch for a link rate of
R=155Mbps (a popular link speed for ATM) for L= 1,500 bytes, and for L = 48
bytes.
8L/R
L=1500= 7.74*10^-5
L=48=2.47*10^-6
d) Comment on the advantages of using a small cell size.
The packetization delay increases with cell size, therefore small cell size is more
advantageous since it has short cell creation time.