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NEW QUESTION: 1
A customer is considering an implementation of SAP HANA in their current environment either on x86 or IBM Power Systems.
Which IBM Power Systems advantage over x86 should be discussed with the customer?
A. Higher memory to core ratios.
B. IBM Power servers support VMware and PowerVM hypervisors.
C. Lower hardware cost.
D. A Scale-out IBM Power clustered configuration can have more nodes than x86.
Answer: A
NEW QUESTION: 2
Which is true about caller security principal propagation for asynchronous EJB method Invocations?
A. Caller security principal propagates along with an asynchronous EJB method invocation only if the calling bean has at least one protected method.
B. Caller security principal propagates along with an asynchronous EJB method invocation only if the target bean has at least one protected method.
C. Caller security principal propagates along with an asynchronous EJB method invocation.
D. Caller security principal does not propagate along with an asynchronous EJB method invocation.
Answer: C
NEW QUESTION: 3
Which command will update an rpm package? (Choose two.)
A. rpm -F package_name.rpm
B. rpm -v package_name.rpm
C. update package_name.rpm
D. rpm -U package_name.rpm
E. rpm -i package_name.rpm
Answer: A,D
Explanation:
Explanation/Reference:
References:
http://www.rpm.org/max-rpm/rpm.8.html
NEW QUESTION: 4
Refer to the exhibit.
Link1 is a copper connection and Link2 is a fiber connection. The fiber port must be the primary port for all forwarding. The output of the show spanning-tree command on SW2 shows that the fiber port is blocked by spanning tree. An engineer enters the spanning- tree port-priority 32 command on GO/1 on SW2, but the port remains blocked. Which command should be entered on the ports that are connected to Link2 to resolve the issue?
A. Enter spanning-tree port-priority 224 on SW1.
B. Enter spanning-tree port-priority 32 on SW1.
C. Enter spanning-tree port-priority 4 on SW2.
D. Enter spanning-tree port-priority 64 on SW2.
Answer: B
Explanation:
Explanation
SW1 needs to block one of its ports to SW2 to avoid a bridging loop between the two switches.
Unfortunately, it blocked the fiber port Link2. But how does SW2 select its blocked port? Well, the answer is based on the BPDUs it receives from SW1. answer 'Enter spanning-tree port-priority 32 on SW1' BPDU is superior than another if it has:
1. answer 'Enter spanning-tree port-priority 32 on SW1' lower Root Bridge ID
2. answer 'Enter spanning-tree port-priority 32 on SW1' lower path cost to the Root
3. answer 'Enter spanning-tree port-priority 32 on SW1' lower Sending Bridge ID
4. answer 'Enter spanning-tree port-priority 32 on SW1' lower Sending Port ID These four parameters are examined in order. In this specific case, all the BPDUs sent by SW1 have the same Root Bridge ID, the same path cost to the Root and the same Sending Bridge ID.
The only parameter left to select the best one is the Sending Port ID (Port ID = port priority + port index). And the port index of Gi0/0 is lower than the port index of Gi0/1 so Link 1 has been chosen as the primary link.
Therefore we must change the port priority to change the primary link. The lower numerical value of port priority, the higher priority that port has. In other words, we must change the port-priority on Gi0/1 of SW1 (not on Gi0/1 of SW2) to a lower value than that of Gi0/0.