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NEW QUESTION: 1
An engineer has performed a post-deployment site survey and noticed that access points were installed in suboptimal locations due to physical restrictions and obstacles. Which action must the engineer take to resolve this deployment issue?
A. Set DCA Channel Assignment Method to freeze.
B. Set TPC maximum and minimum power levels.
C. Set all Auto RE parameters to factory default
D. Set static channels on all AP radios.
Answer: B
Explanation:
Explanation
Suspect poor coverage/not enough AP density -- If AP transmit power is already at the maximum, you can enable the lower data rates to allow clients to connect from farther away. This has a negative impact on performance, but it can help the customer understand that additional APs might be necessary.
https://www.cisco.com/c/en/us/support/docs/wireless/5500-series-wireless-controllers/116057-site-survey-guidelineswlan-00.html

NEW QUESTION: 2
DHCPの操作に関して正しい記述はどれですか？
A. アドレスの競合が検出された場合、アドレスはプールから削除され、サーバーが再起動されるまで再利用されません。
B. DHCPクライアントは、pingを使用してアドレスの競合を検出します。
C. アドレスの競合が検出された場合、管理者が設定可能な期間、アドレスはプールから削除されます。
D. アドレスの競合が検出された場合、アドレスはプールから削除され、管理者は競合を解決する必要があります。
E. DHCPサーバーはgratuitous ARPを使用してDHCPクライアントを検出します。
F. DHCPクライアントは、Gratuitous ARPを使用してDHCPサーバーを検出します。
Answer: D
Explanation:
2つのホストが同じIPアドレスを使用すると、アドレスの競合が発生します。アドレスの割り当て中に、DHCPはpingとgratuitous ARPを使用して競合をチェックします。競合が検出された場合、アドレスはプールから削除されます。管理者が競合を解決するまで、アドレスは割り当てられません。
http://www.cisco.com/en/US/docs/ios/12_1/iproute/configuration/guide/1cddhcp.html

NEW QUESTION: 3
実行時間の長いジョブを一度だけ処理する必要があります。どうすればいいですか？
A. SQSキューを使用して、可視性タイムアウトをジョブが処理するのに十分な長さに設定します。
B. SQSキューを使用して、再処理タイムアウトをジョブが処理するのに十分な長さに設定します。
C. SNSキューを使用して、可視性タイムアウトをジョブが処理するのに十分な長さに設定します。
D. SNSキューを使用して、再処理タイムアウトをジョブが処理するのに十分な長さに設定します。
Answer: A
Explanation:
The message timeout defines how long after a successful receive request SQS waits before allowing jobs to be seen by other components, and proper configuration prevents duplicate processing.
http://docs.aws.amazon.com/AWSSimpleQueueService/latest/SQSDeveloperGuide/MessageLife cycle.ht ml

NEW QUESTION: 4



A. Option D
B. Option B
C. Option E
D. Option A
E. Option C
Answer: A
Explanation:
The IP address 192.168.1.17 255.255.255.0 specifies that the address is part of the 192.168.1.0/24 subnet
24 mask bits = 255.255.255.0 28 mask bits = 255.255.255.240 192.168.1.0/24 subnet has a host range of 192.168.1.1 to 192.168.1.254 (0 being network and 255 being broadcoast) 192.168.1.17/28 subnet has a host range of 192.168.1.17 to 192.168.1.30 (16 being network and 31 being broadcast) 192.168.1.65/28 subnet has a host range of 192.168.1.65 - 192.168.1.78 (64 being network and 79 being broadcast)
if fa0/0 was left as /24, you can see that the host range includes the host range of 192.168.1.64/28 which conflicts.
Simply speaking, you can't overlap the subnets.
By changing the subnet mask of fa0/0 to 255.255.255.240, these networks would no longer overlap.