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NEW QUESTION: 1
If R1 and R2 ran OTP to each other, R1 would learn about the network 10.0.2.0/24 from R2 through EIGRP, treat the prefix 10.0.2.0/24 as an EID prefix, and take the advertising next hop
198.51.100.62 as the RLOC for this EID prefix. Similarly, R2 would learn from R1 about the network 10.0.1.0/24 through EIGRP, treat the prefix 10.0.1.0/24 as an EID prefix, and take the advertising next hop 192.0.2.31 as the RLOC for this EID prefix. On both routers, this information would be used to populate the LISP mapping tables. Whenever a packet from 10.0.1.0/24 to
10.0.2.0/24 would arrive at R1, it would use its LISP mapping tables just like in ordinary LISP to discover that the packet has to be LISP encapsulated and tunneled toward 198.51.100.62, and vice versa. The LISP data plane is reused in OTP and does not change; however, the native LISP mapping and resolving mechanisms are replaced by EIGRP.
Reference: CCIE Routing and Switching V5.0 Official Cert Guide, Volume 1, Fifth Edition Which feature is supported by EIGRP but is not supported by OSPF?
A. unequal-cost load balancing
B. route filtering
C. route summarization
D. equal-cost load balancing
Answer: A
Explanation:
EIGRP support unequal-cost load balancing via the "variance ..." while OSPF only supports equalcost load balancing.

NEW QUESTION: 2
あなたは、Microsoft SQL Server 2014サーバーでホストされているアプリケーションのデータベース開発者です。データベースには、次の定義を持つ2つのテーブルがあります。

世界中の顧客がいくつかの国から発注します。各顧客が最も注文を出した国を表示する必要があります。
どのTransact-SQLクエリを使用しますか？
A. c.CustomerID、c.CustomerName、o。顧客cINNER JOIN（SELECT customerID、ShippingCountry、RANK（）の順に並べ替え、RnkFROM OrdersGROUP BY customerID、ShippingCountryの順に選択） CustomerID = o.CustomerIDWHERE o.Rnk = 1
B. SELECT c.CustomerID、c.CustomerName、o.ShippingCountry FROM（SELECT c.CustomerID、c.CustomerName、o.ShippingCountry、OVER（PARTITION BY BY顧客ID BY BY COUNT（o.OrderAmount）ASC）AS RnkFROM顧客cINNER JOIN注文c.CustomerID = o.CustomerIDGROUP BY c.CustomerID、c.CustomerName、o.ShippingCountry）cs WHERE Rnk = 1
C. SELECT c.CustomerID、c.CustomerName、o.ShippingCountry FROM Customer cINNER JOIN（SELECT CustomerID、ShippingCountry、COUNT（OrderAmount）DESC）AS OrderAmountからFROM OrdersGROUP BY CustomerID、ShippingCountry）ASとしてc.CustomerID = o.CustomerIDORDER BY OrderAmount DESC
D. SELECT c.CustomerID、c.CustomerName、o.ShippingCountry FROM Customer cINNER JOIN（SELECT CustomerID、ShippingCountry、RANK（）OVER（PARTITION、BY CustomerIDORDER BY COUNT（受注額）DESC）AS RnkFROM OrdersGROUP BY CustomerID、ShippingCountry）AS oON c.CustomerID = o.CustomerIDWHERE o.Rnk = 1
Answer: D
Explanation:
Use descending (DESC) ordering.
To order by the number of orders we use ORDER BY COUNT(OrderAmount).
Finally a WHERE close is needed: WHERE o.Rnk = 1

NEW QUESTION: 3
Welches der folgenden Ziele ist ein gültiges Testziel? [K1]
A. Sicherstellen, dass keine Mängel vorliegen
B. Fehler vermeiden
C. Fehler beheben
D. Lokalisieren von Fehlern im Code
Answer: B

NEW QUESTION: 4
An LTM Specialist is troubleshooting an HTTP monitor. The pool member is accessible directly through a browser, but the HTTP monitor is marking the pool member as down.
GET / HTTP/1.1
HTTP/1.1 400 Bad Request
DatE. Tue, 23 Oct 2012 21:39:07 GTM
Server: Apache/2.2.22 (FreeBSD) PHP/5.4.4
mod_ssl/2.2.22 OpenSSL/0.9.8q DAV/2
Content-LengtH. 226
Connection: close
Content-TypE. text/html; charset=iso-8859-1
Which issue is the pool member having?
A. The pool member is rejecting the request because it is invalid.
B. The pool member has too many concurrent connections.
C. The pool member lacks the object requested by the monitor.
D. The pool member is NOT accepting requests from the LTM device IP address.
Answer: A