Computer Networking

Test 1, 28 February 2013

This test has six questions; answer at least five of them.

Up to the first 75 words appearing after a question are taken to be the answer to the question. Words following the 75th word are not part of the answer and will be ignored.

  1. The Application Layer is the top of the standard OSI ISO stack. What would be the consequences in terms of protocol-stack operation of swapping the Application and Presentation Layers? Justify your answer.

    The presentation layer manipulates the data to make it suitable for transport, and the application layer knows the nature of the data and the suitable transport manipulations. Swapping the order of the two layers prevents each from doing its job. The presentation layer has no guidance about the best manipulations to apply, and the application layer can't influence the data manipulations, and may have trouble dealing with manipulated data (as with encrypted data).

  2. A particular antenna works best as a receiver when its length is 1/2i of the wavelength it receives for non-negative integer i, and larger values of 1/2i are better than smaller values. Assuming 2.8 GHz transmissions, what antenna length would you recommend for a cell phone? Show your work.

    Taking the formula λf = c and plugging in the known values gives

    λ · 2.8 · 109 cyc/sec = 186,000 mi/sec
    which gives
    λ=(186,000 mi/sec)/(2.8 · 109 cyc/sec)
    =(0.000186/2.8) mi/cyc
    =(0.000186/2.8) mi/cyc · 5,280 ft/mi
    =0.00007 mi/cyc · 5,280 ft/mi
    =0.35 ft/cyc
    0.65 feet is about 4 inches, which is about the size of a cell phone, in which case i should be 0 or perhaps 1 for smaller cell phones.
  3. Some e-mail systems support a Content Return: header field. It specifies whether the body of a message is to be returned in the event of nondelivery. Does this field belong to the envelope or to the header? Justify your answer.

    The Content Return header should most likely be in the envelope, because it helps message delivery (or nondelivery). Content Return in the message header is of little use because either the recipient never sees it (because the message isn't delivered) or the recipient sees it, in which case it has no purpose, the message having been delivered.

  4. You are deciding whether or not to change a communication line from twisted pair to fiber optics. Given the choice between Shannon’s theorem and Nyquist’s theorem, which would be more helpful for you to consider in your decision making? Justify your answer.

    Nyquist's theorem is a sampling theorem; it determines the sampling rate needed to accurately reconstruct the signal being sampled. Shannon's theorem is an information carrying theorem, determining the maximum information flow over a communication channel with various characteristics (signal to noise ratio, primarily). Of the two theorems, it seems most likely that Shannon's theorem would be the most helpful in deciding between two transmission media.

  5. Addressing, the naming of systems connected to a network, is a service provided by the protocol stack. How appropriate is it to put addressing services in the Application Layer, as opposed to elsewhere further down in the stack? Justify your answer.

    Addressing gives each system in the network a unique identifier, which should be (relatively) permanent and independent of how the system uses and is used by the network. Making the application layer responsible for addressing is impractical for many reasons: addressing is big job, which each application should not have to take on; applications should coordinate to provide consistent addressing; most systems don't use most applications, leading to wasted addressing effort in each application.

  6. Justify the claim that the maximum frequency produced by unipolar non-return to zero line encoding is half the bit rate.

    UNRZ uses a non-zero value and a zero value (1 and 0 volts, say) to represent bits. A sequence of zero and one bits forces the signal to alternate between the non-zero and zero values with a pair of bits driving the signal through one cycle. If bits are sent at B bits/sec, then each pair of bits produces a cycle at half that rate, or B/2 cyc/sec.

This page last modified on 2013 March 4.