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NEW QUESTION # 131
Which statement is correct when comparing 5 GHz and 6 GHz channels with identical channel widths?
Answer: A
Explanation:
Explanation
The correct statement when comparing 5 GHz and 6 GHz channels with identical channel widths is that 5 GHz channels travel different distances and provide different throughputs to clients compared to 6 GHz channels.
This statement reflects the fact that higher frequency signals tend to have higher attenuation Attenuation is a general term that refers to any reduction in signal strength during transmission over distance or through an object or medium . Higher attenuation means that higher frequency signals have shorter range and lower throughput than lower frequency signals. Some facts about this statement are:
5 GHz channels have lower frequency than 6 GHz channels, which means they have lower attenuation than 6 GHz channels.
Lower attenuation means that 5 GHz channels can travel longer distances and provide higher throughputs to clients than 6 GHz channels with identical channel widths.
However, the difference in distance and throughput between 5 GHz and 6 GHz channels may not be significant in indoor environments where there are many obstacles and reflections that affect signal propagation.
The advantage of using 6 GHz channels over 5 GHz channels is that they offer more spectrum availability, less interference, and more non-overlapping channels than 5 GHz channels.
The other options are not correct because:
5 GHz channels travel the same distances and provide different throughputs to clients compared to 6 GHz channels: This option is false because 5 GHz channels do not travel the same distances as 6 GHz channels due to higher attenuation of higher frequency signals.
5 GHz channels travel the same distances and provide the same throughputs to clients compared to 6 GHz channels: This option is false because 5 GHz channels do not travel the same distances or provide the same throughputs as 6 GHz channels due to higher attenuation of higher frequency signals.
5 GHz channels travel different distances and provide the same throughputs to clients compared to 6 GHz channels: This option is false because 5 GHz channels do not provide the same throughputs as
6 GHz channels due to higher attenuation of higher frequency signals.
References: https://www.wi-fi.org/discover-wi-fi/wi-fi-certified-6e
https://www.wi-fi.org/file/wi-fi-alliance-spectrum-needs-study
https://www.cisco.com/c/en/us/support/docs/wireless-mobility/wireless-lan-wlan/82068-power-levels.html
https://www.cisco.com/c/en/us/products/collateral/wireless/spectrum-expert-wi-fi/prod_white_paper0900aecd80
NEW QUESTION # 132
You are working with a pair of 6300M switches in a VSF stack. The switch has 48 SmartRate 5G ports, 2 SFP28 ports, and 2 SFP56 ports. Both SFP56 ports are used for stacking.
You need to provide an LACP connection to another identical stack with the maximum available bandwidth possible. What should you configure?
Answer: C
Explanation:
To provide an LACP connection with the maximum available bandwidth, one should configure a link aggregation group (LAG) using all available ports that can be used for data transfer. Since the SFP56 ports are used for stacking, the next best option is to use the 2 SFP28 ports and as many SmartRate 5G (SR5) ports as possible on each switch, which would allow for a 16-member LAG, with 2 SFP28 and 6 SR5 ports on each switch contributing to the LAG.
NEW QUESTION # 133
A network technician is troubleshooting one new AP at a branch office that will not receive Its configuration from Aruba Central The other APs at the branch are working as expected The output of the 'show ap debug cloud-server command' shows that the "cloud conflg received" Is FALSE.
After confirming the new AP has internet access, what would you check next?
Answer: D
Explanation:
Explanation
If the AP has internet access but does not receive its configuration from Aruba Central, one possible reason is that the AP does not have a license assigned in Aruba Central. A license is required for each AP to be managed by Aruba Central.
References:https://www.arubanetworks.com/techdocs/Central/2.5.2-GA/HTML_frameset.htm#GUID-8F0E7E8B
NEW QUESTION # 134
Which part of the WPA Key Hierarchy is used to encrypt and/or decrypt data?
Answer: B
Explanation:
The part of WPA Key Hierarchy that is used to encrypt and/or decrypt data is Pairwise Temporal Key (PTK). PTK is a key that is derived from PMK Pairwise Master Key (PMK) is a key that is derived from PSK Pre-shared Key (PSK) is a key that is shared between two parties before communication begins, ANonce Authenticator Nonce (ANonce) is a random number generated by an authenticator (a device that controls access to network resources, such as an AP), SNonce Supplicant Nonce (SNonce) is a randomnumber generated by supplicant (a device that wants to access network resources, such as an STA), AA Authenticator Address (AA) is MAC address of authenticator, SA Supplicant Address (SA) is MAC address of supplicant using Pseudo-Random Function (PRF).
PTK consists of four subkeys:
- KCK Key Confirmation Key (KCK) is used for message integrity check
- KEK Key Encryption Key (KEK) is used for encryption key distribution
- TK Temporal Key (TK) is used for data encryption
- MIC Message Integrity Code (MIC) key
The subkey that is specifically used for data encryption is TK Temporal Key (TK). TK is also known as Pairwise Transient Key (PTK). TK changes periodically during communication based on time or number of packets transmitted.
The other options are not part of WPA Key Hierarchy because:
- PMK: PMK is not part of WPA Key Hierarchy, but rather an input for deriving PTK.
- KCK: KCK is part of WPA Key Hierarchy, but it is not used for data encryption, but rather for message integrity check.
- Nonce: Nonce is not part of WPA Key Hierarchy, but rather an input for deriving PTK.
NEW QUESTION # 135
What command is used to add a static route to a network 192.168.10.0/24 via gateway 192.168.1.1 on an Aruba router?
Answer: D
NEW QUESTION # 136
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