PW0-100 Wireless LAN Administration, Certified Wireless Network Administrator (CWNA) Exam Questions PDF download

QUESTION: 105

Wireless LAN throughput is partially based on which of the following variables?

Use of acknowledgements on the network.
Whether Infrared or Spread Spectrum is used.
Packet size is use.
Manufacturer's hardware limitations.

Answer(s): B,C

Explanation:

Infrared (IR) is a light-based transmission technology and is not spread spectrum-spread
spectrum technologies all use RF radiation. IR devices can achieve a maximum data rate of 4 Mbps at close range, but as a light-based technology, other sources of IR light can interfere with IR transmissions. The typical data rate of an IR device is about 115 kbps, which is good for exchanging data between handheld devices. An important advantage of IR networks is that they do not interfere with spread spectrum RF networks. For this reason, the two are complementary and can easily be used together.

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QUESTION: 106

What is the minimum hop rate for FHSS as specified by the FCC?

1 hop every 1.0 seconds.
1 hop every 0.4 seconds.
1 hop every 0.8 seconds.
None of the above.

Answer(s): B

Explanation:

As mentioned in the study guide, according to the rules specified by the FCC, the maximum dwell time that can be set for FHSS systems is 400 ms. Thus after, the frequency must be changed. Thus, the answer should be B (0.4).

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QUESTION: 107

Wireless LAN throughput can change based on: (Choose two)

Use of polling on the access point.
Use of Token Ring or Ethernet on the network backbone.
Type of antenna(s) in use.
Type of spread spectrum technology in use.

Answer(s): A,D

Explanation:

Wireless LAN throughput can change based on use of PCF mode for the access point and also on the type of spread spectrum technology used (using explanation for question 98). Antenna type only effects system coverage/range. Thus the answer should be A, D.

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QUESTION: 108

Using DSSS technology with two access points in the same physical area, the first access point is configured to use channel 6.
Which channel on the second access point should be selected to result in the greatest network throughput?

channel 1
channel 4
channel 8
channel 9

Answer(s): A

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QUESTION: 109

Because of the limited number of non-overlapping channels available in the 2.4 GHz 802.11b standard, what is the maximum number of access points that can be co-located without causing contention in the wireless network?

1
2
3
15

Answer(s): C

Explanation:

This question is slightly ambiguous. Since it does not use the word "theoretical max." or "practical max.", there really can be 2 answers to this. However, since the study guide mentions that for most purposes (including channel re-use), 3 co-locatable systems can be setup, thus the answer should be C (3) Note: It is clearly stated in the study guide that FHSS systems are more co-locatable than DSSS systems. Specifically you can co-locate upto 26 FHSS systems (non-synchronised), where as only 3 are allowed with DSSS systems.

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QUESTION: 110

1mW of power is the equivalent to which of the following?

0dB
+10dB
+10dBi
0 dBm

Answer(s): D

Explanation:

For absolute measurements, the reference power P2 is defined at 1 milliwatt, hence dBm, and 0 dBm = 1 mW:
dBmPower unit mWpower unit dBmPower unit mWpower unt
+20 100 mW -30 1 µW
+10 10 mW -40 100 nW
0 1 mW -50 10 nW
-10 100 µW -60 1 nW
-20 10 µW -70 100 pW

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