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NEW QUESTION: 1
On the following graphic, you will find layers of policies.
What is a precedence of traffic inspection for the defined polices?
A. A packet arrives at the gateway, it is checked against the rules in the networks policy layer and then if implicit Drop Rule drops the packet, it comes next to IPS layer and then after accepting the packet it passes to Threat Prevention layer.
B. A packet arrives at the gateway, it is checked against the rules in IPS policy layer and then it comes next to the Network policy layer and then after accepting the packet it passes to Threat Prevention layer.
C. A packet arrives at the gateway, it is checked against the rules in the networks policy layer and then if there is any rule which accepts the packet, it comes next to IPS layer and then after accepting the packet it passes to Threat Preventionlayer
D. A packet arrives at the gateway, it is checked against the rules in the networks policy layer and then if there is any rule which accepts the packet, it comes next to Threat
Prevention layer and then after accepting the packet it passes to IPS layer.
Answer: C
Explanation:
To simplify Policy management, R80 organizes the policy into Policy Layers. A layer is a set of rules, or a Rule Base.
For example, when you upgrade to R80 from earlier versions:
When the gateway matches a rule in a layer, it starts to evaluate the rules in the next layer.
All layers are evaluated in parallel
NEW QUESTION: 2
IBM Tivoli Endpoint Manager (TEM) Relay Selections must be checked to ensure all clients are automatically selecting the relay. What is the quickest way to determine this information?
A. Open Web Reports and execute the report BES Relay Selection Method
B. All Content > Computers and click the BES Relay Selection column to count the systems that are reporting automatic and manual
C. BigFix Management > Deployment Overview and review the BES Relay Selection Method
D. All Content > Deployment Overview and review the BES Relay Selection Method
Answer: C
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. 208.149.23.64[110/3] via 190.172.23.10, 00:00:07, Serial1/0 208.149.23.96[110/3] via 190.173.23.10,
00:00:16, Serial1/0
B. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, FastEthemet0/0 208.149.23.96[110/13] via
190.173.23.10, 00:00:16, FastEthemet0/0
C. 208.149.23.64[110/13] via 190.173.23.10, 00:00:07, Serial1/0 208.149.23.96[110/13] via
190.173.23.10,
00:00:16, Serial1/0 208.149.23.96[110/13] via 190.173.23.10, 00:00:16, FastEthemet0/0
D. 208.149.23.64[110/1] via 190.172.23.10, 00:00:07, Serial1/0 208.149.23.96[110/3] via 190.173.23.10,
00:00:16, FastEthemet0/0
Answer: B
Explanation:
Explanation/Reference:
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 -> 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. 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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