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Computer Network Questions and Answers

Answer: A

Correct answer: source routing

In IBM's Token Ring the sending station discovers and carries the whole route in the frame, so the bridges simply follow the directions given rather than deciding for themselves. That is source routing, and it is what distinguishes Token Ring bridging from Ethernet's transparent bridging.

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52.

In which circuit switching, delivery of data is delayed because data must be stored and retrieved from RAM 

Answer: B

Correct answer: time-division

A time-division switch works by writing incoming samples into memory and reading them out in a different order, so each sample must be stored and fetched again. That store-and-fetch is the source of the delay; a space-division crossbar makes a direct path and adds none.

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53.

If a network designer wants to connect 5 routers as point-to-point simplex line, then total number of lines required would be

Answer: B

Correct answer: 20

A simplex line carries traffic in one direction only, so each pair of routers needs two of them - one each way. With 5 routers there are \( 5\times 4 = 20 \) ordered pairs, and so 20 lines. Full-duplex lines would need only \( \binom{5}{2}=10 \).

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54.

To interface a computer terminal with a modem, required physical layer standard is

Answer: D

Correct answer: Either (b) or (c)

Both RS-232-C and RS-449 are physical-layer standards for the interface between a terminal and a modem; RS-449 was introduced to carry higher rates over longer distances, but either will serve.

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55.

A 100 km long cable runs at the T1 data rate. The propagation speed in the cable is two-thirds the speed of light. How many bits are in the cable?

Answer: C

Correct answer: 772

First the propagation delay:

\[ t = \frac{100\times 10^{3}}{2\times 10^{8}} = 500\ \mu\text{s} \]

Then the bits in flight at the T1 rate of 1.544 Mbps:

\[ 1.544\times 10^{6}\times 500\times 10^{-6} \approx 772\ \text{bits} \]

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56.

How much bandwith is there in 0.1 micron of spectrum at a wavelength of 1 micron ?

Answer: C

Correct answer: 30,000 GHz

Differentiating \( f = c/\lambda \) gives \( \Delta f = c\,\Delta\lambda/\lambda^{2} \):

\[ \Delta f = \frac{3\times 10^{8}\times 0.1\times 10^{-6}}{(1\times 10^{-6})^{2}} = 3\times 10^{13}\ \text{Hz} = 30{,}000\ \text{GHz} \]

That enormous figure is why a single fibre has so much more capacity than any copper medium.

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57.

It is desired to send a sequence of computer screen images over an optical fiber. The screen is 480 x 640 pixels, each pixel being 24 bits. There are 60 screen images per second. How much bandwidth is needed, and how many microns of wavelenth are needed for this band at 1.30 microns?

Answer: C

Correct answer: 1.5 × 10−6 microns

The screen needs

\[ 480\times 640\times 24\times 60 \approx 4.42\times 10^{8}\ \text{bps} \]

and the spectral width for that bandwidth at \( \lambda = 1\ \mu\text{m} \) is

\[ \Delta\lambda = \frac{\lambda^{2}\Delta f}{c} = \frac{(10^{-6})^{2}\times 4.42\times 10^{8}}{3\times 10^{8}} \approx 1.5\times 10^{-12}\ \text{m} \]

which is \( 1.5\times 10^{-6} \) microns - a minute slice of the fibre's available spectrum.

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Answer: C

Correct answer: placing a telephone call prior to a conversion

A virtual circuit is set up before any data flows and torn down afterwards, so the route is fixed for the duration - exactly as dialling establishes a path before the conversation begins. Nothing is physically switched into place, which is what makes the circuit virtual.

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59.

How many digits of the network user Address are known as the DNIC (Data Network Identiication Code) ?

Answer: B

Correct answer: First four

The DNIC is the first four digits of an X.121 network user address: three for the country and one to identify the network within it.

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