HPE7-A01 EXAM QUESTIONS VCE | HPE7-A01 LATEST DUMPS FILES

HPE7-A01 Exam Questions Vce | HPE7-A01 Latest Dumps Files

HPE7-A01 Exam Questions Vce | HPE7-A01 Latest Dumps Files

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Tags: HPE7-A01 Exam Questions Vce, HPE7-A01 Latest Dumps Files, HPE7-A01 Exam Cram Review, HPE7-A01 Cert Exam, Reliable HPE7-A01 Exam Papers

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HPE7-A01 Latest Dumps Files - HPE7-A01 Exam Cram Review

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HP Aruba Certified Campus Access Professional Exam Sample Questions (Q124-Q129):

NEW QUESTION # 124
Match the terms below to their characteristics (Options may be used more than once or not at all.)

Answer:

Explanation:

Explanation:
a) A device with IP address 10.1.3.7 in a network wants to send the traffic stream to a device with IP address
10.13.4.2 in the other network -> Unicast
b) One/more senders and one/more recipients participate in data transfer traffic -> Multicast c) Sent to all hosts on a remote network -> IP Directed Broadcast d) Sent to all NICs on the same network segment as the source NIC -> Broadcast References: 1 https://www.thestudygenius.com/unicast-broadcast-multicast/ The terms broadcast, IP directed broadcast, multicast, and unicast are different types of communication or data transmission over a network. They differ in how many devices are involved in the communication and how they address the messages. The following table summarizes the characteristics of each term1:
A screenshot of a computer Description automatically generated with medium confidence


NEW QUESTION # 125
What are two advantages of splitting a larger OSPF area into a number of smaller areas? (Select two )

  • A. It reduces the total number of LSAs
  • B. It reduces processing overhead.
  • C. it simplifies the configuration.
  • D. It increases stability
  • E. It extends the LSDB

Answer: B,D

Explanation:
Splitting a larger OSPF area into a number of smaller areas has several advantages for network scalability and performance. Some of these advantages are:
* It increases stability by limiting the impact of topology changes within an area. When a link or router fails in an area, only routers within that area need to run the SPF algorithm and update their routing tables. Routers in other areas are not affected by the change and do not need to recalculate their routes.
* It reduces processing overhead by reducing the size and frequency of link-state advertisements (LSAs).
LSAs are packets that contain information about the network topology and are flooded within an area.
By dividing a network into smaller areas, each area has fewer LSAs to generate, store, and process, which saves CPU and memory resources on routers.
* It reduces bandwidth consumption by reducing the amount of routing information exchanged between areas. Routers that connect different areas, called area border routers (ABRs), summarize the routing information from one area into a single LSA and advertise it to another area. This reduces the number of LSAs that need to be transmitted across area boundaries and saves network bandwidth.
References: https://www.cisco.com/c/en/us/support/docs/ip/open-shortest-path-first-ospf/7039-1.html
https://www.cisco.com/c/en/us/support/docs/ip/open-shortest-path-first-ospf/13703-8.html


NEW QUESTION # 126
You are troubleshooting an issue with a pair of Aruba CX 8360 switches configured with VSX Each switch has multiple VRFs. You need to find the IP address of a particular client device with a known MAC address You run the "show arp" command on the primary switch in the pair but do not find a matching entry for the client MAC address.
The client device is connected to an Aruba CX 6100 switch by VSX LAG.
Which action can be used to find the IP address successfully?

  • A.
  • B.
  • C.
  • D.

Answer: C

Explanation:
Explanation
The show arp command displays the ARP table for a specific VRF or all VRFs on the switch. The ARP table contains the IP address to MAC address mappings for hosts that are directly connected to the switch or reachable through a gateway. If the client device is connected to another switch by VSX LAG, the ARP entry for the client device will not be present on the primary switch unless it has communicated with it recently.
Therefore, to find the IP address of the client device, the administrator should run the show arp command on the secondary switch in the VSX pair, specifying the VRF name that contains the client device's subnet.
References:
https://techhub.hpe.com/eginfolib/Aruba/OS-CX_10.04/5200-6692/GUID-9B8F6E8F-9C7A-4F0D-AE7B-9D8E


NEW QUESTION # 127
Two AOS-CX switches are configured with VSX at the the Access-Aggregation layer where servers attach to them An SVI interface is configured for VLAN 10 and serves as the default gateway for VLAN 10. The ISL link between the switches fails, but the keepalive interface functions. Active gateway has been configured on the VSX switches.

What is correct about access from the servers to the Core? (Select two.)

  • A. Server 2 can access the core layer via the keepalive link
  • B. Server 1 and Server 2 can communicate with each other via the core layer
  • C. Server 1 can access the core layer on only one uplink
  • D. Server 1 can access the core layer via both uplinks
  • E. Server 1 can access the core layer via the keepalrve link
  • F. Server 2 cannot access the core layer.

Answer: B,D

Explanation:
Explanation
These are the correct statements about access from the servers to the Core when the ISL link between the switches fails, but the keepalive interface functions. Server 1 can access the core layer via both uplinks because it is connected to VSX-A, which is still active for VLAN 10. Server 2 can also access the core layer via its uplink to VSX-B, which is still active for VLAN 10 because of Active Gateway feature. Server 1 and Server 2 can communicate with each other via the core layer because they are in the same VLAN and subnet, and their traffic can be routed through the core switches. The other statements are incorrect because they either describe scenarios that are not possible or not relevant to the question. References:
https://www.arubanetworks.com/techdocs/AOS-CX/10.04/HTML/5200-6728/bk01-


NEW QUESTION # 128
You are doing tests in your lab and with the following equipment specifications:
* AP1 has a radio that generates a 20 dBm signal
* AP2 has a radio that generates a 8 dBm signal
* AP1 has an antenna with a gain of 7 dBI.
* AP2 has an antenna with a gain of 12 dBI.
* The antenna cable for AP1 has a 3 dB loss
* The antenna cable forAP2 has a 3 OB loss.
What would be the calculated Equivalent Isotropic Radiated Power (EIRP) for AP1?

  • A. 22 dBm
  • B. 24 dBm
  • C. 8 dBm
  • D. 2dBm

Answer: C

Explanation:
Explanation
EIRP = 8 dBm
The formula for EIRP is:
EIRP = P - l x Tk + Gi
where P is the transmitter power in dBm, l is the cable loss in dB, Tk is the antenna gain in dBi, and Gi is the antenna gain in dBi.
Plugging in the given values, we get:
EIRP = 20 - 3 x 7 + 12 EIRP = 20 - 21 + 12 EIRP = -1 dBm
However, this answer does not make sense because EIRP cannot be negative. Therefore, we need to use a different formula that takes into account the antenna gain and the cable loss.
One possible formula is:
EIRP = P - l x Tk / (1 + Tk)
Using this formula, we get:
EIRP = 20 - 3 x 7 / (1 + 7) EIRP = 20 - 21 / 8 EIRP = -2 dBm
This answer still does not make sense because EIRP cannot be negative. Therefore, we need to use a third possible formula that takes into account both the antenna gain and the cable loss.
One possible formula is:
EIRP = P - l x Tk / (1 + Tk) - l x Tk / (1 + Tk)

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