CS 635 Fall 1998
Homework 1 (W. Stallings Lan&Wan)
1.3 What are the key factors that determine the response time
and throughput of a local area network such as that of Figure 1.2 (bus)?
Of a centralized system such as that of Problem 1.2 (point-to-point links
to the central server)?
A. For a LAN, performance depends primarily on how efficiently
the communications medium is used. For a centralized system, performance
depends primarily on the ability of the central system to keep up with
the load imposed on it.
2.4 For the bit stream 0101110, scetch the waveforms for NRZ,
NRZI, Manchester, and Differential Manchester as well as for Miller coding
and E-NRZ.
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Nonreturn-to-Zero-Level (NRZ-L)
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0 = high level
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1 = low level
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Nonreturn-to-Zero Inversted (NRZI)
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0 = no transition at beginning of interval (one bit time)
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1 = transition at beginning of interval
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Manchester
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0 = transition from high to low in middle of interval
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1 = transition from low to high in middle of interval
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Differential Manchester
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Always a transition in middle of interval
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0 = transition at beginning of interval
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1 = no transition at beginning of interval
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Enhanced Nonreturn-to-Zero (E-NRZ)
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Separate NRZ-L into 7-bit words
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Invert bits 2,3,6,7
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Add one partiy bit to each word to make the total number of 1's in the
8-bit word and odd count
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Miller coding
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0 = transition at the end of the bit period if the next bit is 0, no transition
if the next bit is 1
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1 = transition in the middle of the interval
Homework 2
3.1 An asynchronous device such as a teletype transmits characters
one at a time with unpredictable delays between characters.What problem,
if any, do you forsee if such a device is connected to a local network
and allowed to transmit at will (subject to gaining access to the medium)?
How might such problems be resolved? Answer for ring, bus and star.
A. For bus and ring networks, each individual character could
be sent out as a separate packet, resulting in tremendous overhead. This
problem could be overcome by buffering characters and only sending out
blocks of characters. With a circtui -switched star network, a certain
capacity is deficated to the attached station, and therefore no problem
arises.
3.3 Consider the tranfser of a file containing one million
characters from one section to another. What is the total elasped time
and effective throughput for the following cases?
a. A circuit-switched, star topology network. Call setup time
is negligible, and the data rate of the medium is 64 kbps
b. A bus topology local network with two stations a distance
D apart, a data rate of B bps, and a packet size P with 80 bits of overhead.
Each packet is acknowledged with an 88-bit packet before the text is sent.
The propagation speed on the bus is 200m/microseconds. Solve for
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D=1km, B=1 Mbps, P=256 bits
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D=1km, B=10 Mbps, P=256 bits
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D=10km, B=1 Mbps, P=256 bits
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D=1km, B=50 Mbps, P=10,000 bits
c. A ring topology with a total circular length of 2D, with the
two stations a distance D apart. Aknowledgement is achieved by allowing
a packet to circualte past the destination station, back to the source
station . There are N repeaters on the ring, each of which introduces a
delay of 1 bit time. Repeat the calculation for each of b1 through b4 for
N=10, 100, 1000.
A.
a. T=(8x10**6 bits)/(64x10**3 bps) = 125 seconds
b. Transfer consists of a sequence of cycles. On cycle consists
of:
Data Packet = Data Packet Transmission Time + Propagation Time
ACK Packet = ACK Packet Transmission Time + Propagation Time
Define:
c = Cycle time
Q = data biits per packet
T = total time required
Td = Data Packet Transmit Time
Ta = ACK Packet Tramit Time
Tp = Propagation Time
Then
Tp = D/(200x10**6m/sec)
Ta = (88 bits)/(B bps)
Td= (P bits)/(B bps)
T = (8x10**6 bits x C sec/cyle)/(Q bits/cycle)
C= Ta + Td + 2 Tp
Q = P-80
b1
C= 354x10**(-6)
Q=176
T = 16 sec
b2
C=44.4x10**(-6)
Q=176
T=2sec
b3
C=444*10**(-6)
Q=176
T=20sec
b4
C=211.6x10**(-6)
Q=9,920
T=0.17 seconds
c. Define
Tr = Total repeater Delay = N/B
Then
C = Td+2Ttp+Tr
Q=P-80
T=(8x10**6 x C)/Q
We show results for N=100
c1.
C=366x10**(-6)
T=16.6 sec
c2.
C=45.6x10**(-6)
T=2.1 sec
c3.
C=456x10**(-6)
T=20.7 sec
c4.
C=212x10**(-6)
T=0.17 sec
