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NEW QUESTION: 1
How does the system handle data with consistent planning in DP?
Note: There are 3 correct answers to this question.
A. It keeps the data for one key figure consistent at all levels.
B. It stores data consistently in aggregates and at the lowest level.
C. It keeps data for different characteristic value combinations consistent.
D. It keeps data in the internal BW and liveCache consistent.
E. It keeps data for different key figures consistent at all levels.
Answer: A,B,C

NEW QUESTION: 2
Ihr Netzwerk enthält eine Active Directory-Domäne mit dem Namen contoso.com.
Die Domäne enthält einen Server mit dem Namen Server1, auf dem Windows Server 2012 R2 ausgeführt wird und auf dem die DHCP-Serverrolle installiert ist.
Server1 hat einen Bereich mit dem Namen Scope1. Eine Richtlinie mit dem Namen Policy1 ist für Scope1 konfiguriert.
Policy1 ist so konfiguriert, dass virtuelle Hyper-V-Maschinen für einen Tag geleast werden.
Alle anderen Computer erhalten eine Mietdauer von acht Tagen.
Sie implementieren einen zusätzlichen DHCP-Server mit dem Namen Server2, auf dem Windows Server 2012 R2 ausgeführt wird.
Auf Server1 konfigurieren Sie Scopel für DHCP-Failover.
Sie stellen fest, dass virtuelle Maschinen, die IP-Adressen von Server2 erhalten, eine Lease-Dauer von acht Tagen haben.
Sie müssen sicherstellen, dass die Lease-Dauer einen Tag beträgt, wenn Server2 den virtuellen Hyper-V-Maschinen IP-Adressen zuweist.
Die Lösung muss sicherstellen, dass andere Computer, die IP-Adressen von Server2 erhalten, eine Lease-Dauer von acht Tagen haben.
Was tun?
A. Klicken Sie auf Server1 mit der rechten Maustaste auf Bereich1, und klicken Sie dann auf Bereich replizieren.
B. Klicken Sie auf Server2 mit der rechten Maustaste auf Bereich1, und klicken Sie dann auf Abgleichen.
C. Erstellen Sie auf Server2 eine neue DHCP-Richtlinie.
D. Löschen Sie auf Server1 Richtlinie1 und erstellen Sie die Richtlinie neu.
Answer: A
Explanation:
Erläuterung
Bereich 1 wurde für DHCP-Failover eingerichtet. Jetzt müssen wir es von Server1 auf Server2 replizieren.
http://technet.microsoft.com/en-us/library/dd183579(v=ws.10).aspx
http://technet.microsoft.com/en-us/library/cc772101.aspx

NEW QUESTION: 3
Refer to the exhibit. The network is converged.
After link-state advertisements are received from Router_A, what information will Router_E contain in its routing table for the subnets 208.149.23.64 and
208.149.23.96?

A. O 208.149.23.64 [110/1] via 190.172.23.10, 00:00:07, Serial 1/0 O 208.149.23.96 [110/3] via 190.173.23.10, 00:00:16, FastEthernet 0/0
B. O 208.149.23.64 [110/13] via 190.172.23.10, 00:00:07, Serial 1/0 O 208.149.23.96 [110/13] via 190.172.23.10, 00:00:16, Serial 1/0 O 208.149.23.96 [110/13] via
190.173.23.10, 00:00:16, FastEthernet 0/0
C. O 208.149.23.64 [110/3] via 190.172.23.10, 00:00:07, Serial 1/0 O 208.149.23.96 [110/3] via 190.172.23.10, 00:00:16, Serial 1/0
D. O 208.149.23.64 [110/13] via 190.173.23.10, 00:00:07, FastEthernet 0/0 O 208.149.23.96 [110/13] via 190.173.23.10, 00:00:16, FastEthernet 0/0
Answer: D
Explanation:
Router_E learns two subnets subnets 208.149.23.64 and 208.149.23.96 via Router_A through FastEthernet interface. The interface cost is calculated with the formula 108 / Bandwidth. For FastEthernet it is 108 / 100 Mbps = 108 / 100,000,000 = 1. Therefore, the cost is 12 (learned from Router_A) + 1 = 13 for both subnets
-B is not correct. The cost through T1 link is much higher than through T3 link (T1 cost = 108 / 1.544 Mbps = 64; T3 cost = 108 / 45 Mbps = 2) so surely OSPF will choose the path through T3 link -> Router_E will choose the path from Router_A through FastEthernet0/0, not Serial1/0 -C & D are not correct.
In fact, we can quickly eliminate answers B, C and D because they contain at least one subnet learned from Serial1/0 -they are surely incorrect.

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