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HPE7-A01 Study Materials and Aruba Certified Campus Access Professional Exam Test Dumps - HPE7-A01 PDF Guide - PassExamDumps
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HP Aruba Certified Campus Access Professional Exam Sample Questions (Q14-Q19):
NEW QUESTION # 14
A customer wants to provide wired security as close to the source as possible The wired security must meet the following requirements:
-allow ping from the IT management VLAN to the user VLAN
-deny ping sourcing from the user VLAN to the IT management VLAN
The customer is using Aruba CX 6300s
What is the correct way to implement these requirements?
- A. Apply an outbound ACL on the user VLAN allowing temp echo-reply traffic toward the IT management VLAN
- B. Apply an inbound ACL on the user VLAN denying icmp echo traffic toward the IT management VLAN
- C. Apply an outbound ACL on the user VLAN denying icmp echo traffic toward the IT management VLAN
- D. Apply an inbound ACL on the user VLAN allowing icmp echo-reply traffic toward the IT management VLAN
Answer: B
Explanation:
Explanation
An inbound ACL is applied to traffic entering a port or VLAN. An outbound ACL is applied to traffic leaving a port or VLAN4. To deny ping sourcing from the user VLAN to the IT management VLAN, an inbound ACL on the user VLAN should be used to filter icmp echo traffic toward the IT management VLAN. Icmp echo-reply traffic is not needed to be allowed because it is already permitted by default5. References: 4
https://techhub.hpe.com/eginfolib/Aruba/OS-CX_10.04/5200-6692/GUID-9B8F6E8F-9C7A-4F0D-AE7B-9D8E
5
https://techhub.hpe.com/eginfolib/Aruba/OS-CX_10.04/5200-6692/GUID-0C3A9D0F-6E5B-4E1A-AF3C-8D8
NEW QUESTION # 15
Match each PoE power class to Its corresponding 802.3 standard. (Options may he used more than once or not at all)
Answer:
Explanation:
* Class 3 (15.4W): 802.3af
* Class 4 (30W): 802.3at
* Class 6 (60W): 802.3bt
* Class 8 (90W): 802.3bt
NEW QUESTION # 16
What are two advantages of splitting a larger OSPF area into a number of smaller areas? (Select two)
- A. It extends the LSDB
- B. It reduces processing overhead.
- C. It increases stability
- D. It reduces the total number of LSAs
- E. it simplifies the configuration.
Answer: B,C
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 # 17
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. 24 dBm
- B. 8 dBm
- C. 2dBm
- D. 22 dBm
Answer: B
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)
