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COEN233

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COEN 233 Computer Network

Mid-term review

  1. Besides bandwidth and latency, what other parameter is needed to give a good characterization
    of the quality of service offered by a network used for (i) digitized voice traffic?
    (ii) video traffic? (iii) financial transaction traffic?

A uniform delivery time is needed for voice as well as video, so the amount
of jitter in the network is important. This could be expressed as the standard
deviation of the delivery time. Having short delay but large variability is actually
worse than a somewhat longer delay and low variability.

For financial
transaction traffic, reliability and security are very important.

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  1. A client-server system uses a satellite network, with the satellite at a height of 40,000
    km. What is the best-case delay in response to a request?

The request has to go up and down, and the response has to go up and down.
The total path length traversed is thus 160,000 km. The speed of light in air
and vacuum is 300,000 km/sec, so the propagation delay alone is
160,000/300,000 sec or about 533 msec.

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  1. Match the layers—Link, Network, and Transport—with the guarantees that each layer
    could provide to higher layers.

Guarantee Layer

Best effort delivery Network
Reliable Delivery Transport
In-order Delivery Transport
Byte-stream abstraction Transport
Point-to-point link abstraction Data link

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  1. Two networks each provide reliable connection-oriented service. One of them offers
    a reliable byte stream and the other offers a reliable message stream. Are these identical?
    If so, why is the distinction made? If not, give an example of how they differ.

Message and byte streams are different. In a message stream, the network
keeps track of message boundaries. In a byte stream, it does not. For example,
suppose a process writes 1024 bytes to a connection and then a little
later writes another 1024 bytes. The receiver then does a read for 2048 bytes.
With a message stream, the receiver will get two messages, of 1024 bytes
each.

With a byte stream, the message boundaries do not count and the receiver
will get the full 2048 bytes as a single unit. The fact that there were
originally two distinct messages is lost.

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  1. What does ‘‘negotiation’’ mean when discussing network protocols? Give an example.

Negotiation has to do with getting both sides to agree on some parameters or
values to be used during the communication. Maximum packet size is one
example, but there are many others.

__

  1. What is the main difference between TCP and UDP?

TCP is connection oriented, whereas UDP is a connectionless service.

__

  1. Suppose the algorithms used to implement the operations at layer k is changed. How
    does this impact operations at layers k − 1 and k + 1?

This has no impact on the operations at layers k-1 or k+1.

  1. Suppose there is a change in the service (set of operations) provided by layer k. How
    does this impact services at layers k-1 and k+1?

There is no impact at layer k-1, but operations in k+1 have to be reimplemented.

  1. Provide a list of reasons for why the response time of a client may be larger than the
    best-case delay.

One reason is request or response messages may get corrupted or lost during
transmission. Another reason is the processing unit in the satellite may get
overloaded processing several requests from different clients.

  1. What are the disadvantages of using small, fixed-length cells in ATM?

Small-sized cells result in large header-to-payload overhead.

Fixed-size cells result in wastage of unused bytes in the payload.

ATM cell : 5 byte header + 48 byte data

  1. Nyquist’s theorem
    Maximum data rate = 2Blog2 V bits/sec, where B is the bandwidth and V is the number of discrete state

    •Shannon’s formula for capacity of a noisy channel

    *Blog2 (1 + S/N)bits/sec*

  1. Is the Nyquist theorem true for high-quality single-mode optical fiber or only for
    copper wire?

The Nyquist theorem is a property of mathematics and has nothing to do with
technology. It says that if you have a function whose Fourier spectrum does
not contain any sines or cosines above f, by sampling the function at a frequency
of 2f you capture all the information there is. Thus, the Nyquist
theorem is true for all media.

  1. What is the difference, if any, between the demodulator part of a modem and the coder
    part of a codec? (After all, both convert analog signals to digital ones.)

A coder accepts an arbitrary analog signal and generates a digital signal from
it. A demodulator accepts a modulated sine wave only and generates a digital
signal.

  1. Three packet-switching networks each contain n nodes. The first network has a star
    topology with a central switch, the second is a (bidirectional) ring, and the third is
    fully interconnected, with a wire from every node to every other node. What are the
    best-, average-, and worst-case transmission paths in hops?

The three networks have the following properties:
Star: best case = 2, average case = 2, worst case = 2.
Ring: best case = 1, average case = n/4, worst case = n/2.
Full interconnect: best case = 1, average case = 1, worst case = 1.

  1. A bit stream 10011101 is transmitted using the standard CRC method described in the
    text. The generator polynomial is x 3 + 1. Show the actual bit string transmitted. Suppose
    that the third bit from the left is inverted during transmission. Show that this
    error is detected at the receiver’s end. Give an example of bit errors in the bit string
    transmitted that will not be detected by the receiver.

The frame is 10011101. The generator is 1001. The message after appending
three zeros is 10011101000. The remainder on dividing 10011101000 by
1001 is 100. So, the actual bit string transmitted is 10011101100. The received
bit stream with an error in the third bit from the left is 10111101100.
Dividing this by 1001 produces a remainder of 100, which is different from 0.
Thus, the receiver detects the error and can ask for a retransmission. If the
14 PROBLEM SOLUTIONS FOR CHAPTER 3
transmitted bit stream is converted to any multiple of 1001, the error will not
be detected. A trivial example is if all ones in the bit stream are inverted to
zeros.

https://www.gatevidyalay.com/cyclic-redundancy-check-crc-error-detection/

  1. Give at least one reason why PPP uses byte stuffing instead of bit stuffing to prevent
    accidental flag bytes within the payload from causing confusion.

PPP was clearly designed to be implemented in software, not in hardware as
bit-stuffing protocols such as HDLC nearly always are. With a software implementation,
working entirely with bytes is much simpler than working with
individual bits.

In addition, PPP was designed to be used with modems, and
modems accept and transmit data in units of 1 byte, not 1 bit.

  1. Consider the delay of pure ALOHA versus slotted ALOHA at low load. Which one is
    less? Explain your answer.

With pure ALOHA, transmission can start instantly. At low load, no collisions
are expected so the transmission is likely to be successful. With slotted
ALOHA, it has to wait for the next slot. This introduces half a slot time of
delay.

  1. What is the baud rate of classic 10-Mbps Ethernet?

Classic Ethernet uses Manchester encoding, which means it has two signal
periods per bit sent. The data rate is 10 Mbps, so the baud rate is twice that,
or 20 megabaud.

Baud rate is the measure of the number of changes to the signal (per second) that propagate through a transmission medium.

IPG: 9.6 microsecond

Maximum throughput: 10,000,000 / (96 + (8+64)*8) = 10, 000, 000 / 672

  1. Sketch the Manchester encoding on a classic Ethernet for the bit stream 0001110101.

The signal is a square wave with two values, high (H) and low (L). The pattern
is LHLHLHHLHLHLLHHLLHHL.

0: LH

1: HL

  1. An IP packet to be transmitted by Ethernet is 60 bytes long, including all its headers.
    If LLC is not in use, is padding needed in the Ethernet frame, and if so, how many
    bytes?

The minimum Ethernet frame is 64 bytes, including both addresses in the
Ethernet frame header, the type/length field, and the checksum. Since the
header fields occupy 18 bytes and the packet is 60 bytes, the total frame size
is 78 bytes, which exceeds the 64-byte minimum. Therefore, no padding is
used.

  1. Some books quote the maximum size of an Ethernet frame as 1522 bytes instead of
    1500 bytes. Are they wrong? Explain your answer.

The payload is 1500 bytes, but when the destination address, source address,
type/length, and checksum fields are counted, plus the VLAN header, the
total is indeed 1522. Prior to VLANs, the total was 1518.

Maximum VLAN node: 4096-2 = 4094

  1. Name two networks that allow frames to be packed back-to-back. Why is this feature
    worth having?

Gigabit Ethernet has it and so does 802.16 (IEEE 802.16 is a series of wireless broadband standards).
It is useful for bandwidth efficiency (one preamble, etc.) but also when there is a lower limit on frame size.

  1. Give two reasons why networks might use an error-correcting code instead of error
    detection and retransmission.

One reason is the need for real-time quality of service. If an error is discovered,
there is no time for a retransmission. The show must go on. Forward
error correction can be used here. Another reason is that on very low-quality
lines (e.g., wireless channels), the error rate can be so high that practically all
frames would have to be retransmitted, and the retransmissions would probably
damaged as well. To avoid this, forward error correction is used to increase
the fraction of frames that arrive correctly.

  1. List two ways in which WiMAX is similar to 802.11, and two ways in which it is different
    from 802.11.

Like 802.11, WiMAX wirelessly connects devices, including mobile devices
to the Internet at Mbps speeds. Also, like 802.11, WiMAX is based on
OFDM and MIMO technologies.

However, unlike 802.11, WiMAX base stations
are much more powerful than 802.11 access points. Also, transmissions
in WiMAX are carefully scheduled by the base station for each subscriber
without any possibility of collisions unlike CSMA/CA used in 802.11.

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  1. List some of the security concerns of an RFID system.

One key security concern is unauthorized tracking of RFID tags. An adversary
with an appropriate RFID reader can track the locations of the items
tagged using RFID tags. This becomes quite serious if the item is sensitive in
nature, for example, a passport, and the tag can be used to retrieve further
information, for example, the nationality and other personal information of
the person holding the passport.

Another security concern is the ability of a
reader to change tag information. This can be used by an adversary to, for example,
change the price of a tagged item he plans to buy.

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  1. Store-and-forward switches have an advantage over cut-through switches with respect
    to damaged frames. Explain what it is.

Store-and-forward switches store entire frames before forwarding them.
After a frame comes in, the checksum can be verified. If the frame is damaged,
it is discarded immediately. With cut-through, damaged frames cannot
be discarded by the switch because by the time the error is detected, the frame
is already gone. Trying to deal with the problem is like locking the barn door
after the horse has escaped.

forward corrupted frames