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Exam 1 Review (ISQS 3349)
Chapter 2
Review Focuses
1.
Three network architectures: host-based, client-based, and client-server
2.
Four general functions of network application programs:
Data storage, data access logic, application logic, and presentation logic.
3.
Network terminology
Intranet: an organization's internal network that uses Internet technology
Extranet: The intranet that some of its functions are accessible to the organization's business partners
Client: the input/output hardware device on a network.
Server: the computer provides services to the clients on the network.
Circuit: the pathway through which the messages travel.
Mainframe: a large, general-purpose computer.
Media: is the matter or substance that carries the voice or data transmission.
Front-end processors: a computing device as an extension of mainframe for communication activities.
4.
Internet applications
Web, email, ftp
5.
ISO’s OSI network model and Internet 5-layer model
Short-answer Questions:
1) What is middleware and what does it do?
Middleware is software that sits between the application software on the client and the application
software on the server.
Middleware does two things. First, it provides a standard way of communicating that can translate
between software from different vendors. The second function of middleware is to manage the
message transfer from clients to servers (and vice versa) so that clients need not know the specific
server that contains the application’s data.
2) Which is less expensive: host-based networks or client-server networks? Explain.
Network expenses depend on many factors. In general, the hardware for a client-server network is less
expensive than that required for a host-based network. It is also less expensive to expand a client-server
network to handle increased loads. However, the application software and management of a client-server
network can be more expensive than a host-based network.
3) How does a thin client differ from a fat client?
A “thin client” approach places little or no application logic on the client, while a “fat client” approach
places all or almost all of the application logic on the client. Thin clients are much easier to manage. If an
application changes, only the server with the application logic needs to be updated. With a fat client, the
software on all of the clients would need to be updated.
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True/False questions:
1.
In a host-based network, the client computer performs the majority of the work.
2.
The work in a network is shared between the clients and servers in a client-server network.
3.
The host-based network is predicted to become the dominant network architecture of the future.
4.
Application logic refers to the processing required to access data.
5. Currently, it is much less expensive to develop and maintain application software for client-server
networks.
Chapter 3
Review focuses:
1.
Three types of data flows: Simplex, half-duplex, and full-duplex.
2.
Transmission media
Guided Media: Coaxial cable, twisted-pair cable, fiber optic cable
Unguided media: radio, infrared light, microwave, satellite
3.
Digital transmission of digital data
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4.
Analog transmission of digital data
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5.
Modulation: the technique that modifies the form of an electrical signal so the signal can carry
information on a communication medium.
Modem: MOdulator/DEModulator
Amplitude, frequency, and phase.
Types of modulation: AM, FM, PM, QAM (quadrature amplitude modulation, 3-bit by PM and 1
bit by AM), TCM (Trellis-coded modulation, up to 10-bit per symbol).
Bit rate vs. symbol (baud) rate (information rate and signaling speed)
Bandwidth: refers to a range of frequencies.
Data compression has been used by the modem (Huffman encoding, Lempel-Ziv encoding)
Digital transmission of analog data
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6.
ASCII and EBCDIC coding
Serial transmission and Parallel transmission
Digital encoding: Bipolar signaling and unipolar signaling, Manchester encoding
CODEC: COder/DECoder are a pair of devices for analog data conversion.
Pulse code modulation (PCM)
V.92 modem
ADSL and cable modem
Multiplexing and multiplexers
A multiplexer is a hardware device that puts several simultaneous transmissions on a single
communication circuit. There are FDM, TDM, STDM and IMUX.
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Questions
Short-answer questions:
1.
Describe how data could be transmitted using amplitude modulation.
With amplitude modulation (AM), the amplitude or height of the wave is changed. One amplitude is
defined to be zero, and another amplitude is defined to be a one.
2.
What is a modem?
Modem is an acronym for MOdulator/DEModulator. A modem takes the digital electrical pulses received
from a computer, terminal, or microcomputer and converts them into a continuous analog signal that is
needed for transmission over an analog voice grade circuit. Modems are either internal (i.e., inside the
computer) or external (i.e., connected to the computer by a cable).
3.
What is QAM?
One popular technique is quadrature amplitude modulation (QAM). QAM involves splitting the symbol
into eight different phases (three bits) and two different amplitudes (one bit), for a total of 16 different
possible values. Thus, one symbol in QAM can represent four bits.
4.
How do ADSL and cable modem use multiplexing techniques?
ADSL uses FDM and cable modem uses STDM (see slides).
True/False questions:
4. In an analog signal, bandwidth is the difference between the highest and lowest frequencies in a band.
5. Some of the bandwidth of traditional telephone circuits is taken up by guardbands.
Multi-choice questions:
6.
Which of the following is a byte, under the normal definition?
a. 0
b. 1
c. 8 consecutive bits, such as 00011100, that represent one character
d. 325
e. a special character that is used for Apple computers only
7.
Which of the following is not an important characteristic of sound waves?
a. phase
b. frequency
c. amplitude
d. bandwidth
e. all of the above are important characteristics of sound waves
Chapter 4
Review focuses:
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1.
Three functions of the data link layer:
Media access control, error control, and message delineation
2.
Media access control
X-ON/X-OFF, polling, Contention
3.
Error control in networks
1) Error prevention
2) Error detection
a. parity checking
b. LRC
c. Checksum
d. CRC
3) Error correction
a. Stop-and-wait ARQ
b. Continuous ARQ
c. Forward error correction
4.
Data link protocols
1)
2)
3)
4)
Start-stop
Asynchronous file transfer: XMODEM
Synchronous transmission: SDLC, Ethernet, Token ring, SLIP, PPP
Transparency problem
Key terms:
Attenuation, message delineation, bit-oriented protocol, byte-oriented protocol, byte-count protocol,
burst-error, Hamming code, white noise
Questions:
1.
Briefly describe how even parity and odd parity work.
Even parity is when the seven bits of an ASCII character have an even (2, 4, or 6) number of 1s, and
therefore a 0 is placed in the eighth parity position. Odd parity is when the seven bits of an ASCII character
have an odd (1, 3, 5, or 7) number of 1s, and therefore a 0 is placed in the eighth parity position.
2.
Compare and contrast stop-and-wait ARQ and continuous ARQ.
With stop-and-wait ARQ, the sender stops and waits for a response from the receiver after each message or
data packet. After receiving a packet, the receiver sends either an acknowledgment (ACK) if the message
was received without error, or a negative acknowledgment (NAK) if the message contained an error. If it is
an NAK, the sender resends the previous message. If it is an ACK, the sender continues with the next
message. Stop-and-wait ARQ is by definition, a half duplex transmission technique.
With continuous ARQ, the sender does not wait for an acknowledgment after sending a message; it
immediately sends the next one. While the messages are being transmitted, the sender examines the stream
of returning acknowledgments. If it receives an NAK, the sender retransmits the needed messages.
Continuous ARQ is by definition a full duplex transmission technique, because both the sender and the
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receiver are transmitting simultaneously (the sender is sending messages, and the receiver is sending ACKs
and NAKs).
3.
What is transparency, and why is this a problem with SDLC?
SDLC and other bit-oriented protocols suffer from a transparency problem; that is, the protocol is not
“transparent” because it cannot automatically send all types of data with any bit patterns. It is possible that
the user’s data to be transmitted contains the same bit pattern as the flag (e.g., 01111110). If this is not
prevented, the receiver will mistakenly believe that this data marks the end of the frame and ignore all the
data that follows it.
4.
How does SDLC overcome their transparency problems?
The solution is called bit stuffing. Anytime the sender encounters five 1’s in a row in the user’s data to be
transmitted, the sender “stuffs” one extra bit, a 0, into the message and continues to transmit. Anytime the
receiver encounters five 1’s followed by a 0 (i.e., 111110), the receiver automatically deletes the 0 and
continues to process the data stream. Conversely, if the received encounters five 1’s followed by a 1 (i.e.,
111111) it knows to expect another zero as part of the flag. This technique works, but increases the
complexity of the protocol.
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