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NEW QUESTION: 1
The implementations group has been using the test bed to do a 'proof-of-concept' that requires both Client
1 and Client 2 to access the WEB Server at 18.104.22.168. After several changes to the network addressing, routing scheme, DHCP services, NTP services, layer 2 connectivity, FHRP services, and device security, a trouble ticket has been opened indicating that Client 1 cannot ping the 22.214.171.124 address.
Use the supported commands to isolated the cause of this fault and answer the following questions.
What is the solution to the fault condition?
A. Under the interface Serial0/0/0 configuration enter the ip nat outside command.
B. Under the ip access-list standard nat_trafic configuration enter the permit 10.2.0.0 0.0.255.255 command.
C. Under the interface Serial0/0/0 configuration enter the ip nat inside command.
D. Under the ip access-list standard nat_trafic configuration enter the permit 126.96.36.199 0.0.0.255 command.
On R1 we need to add the client IP address for reachability to server to the access list that is used to specify which hosts get NATed.
The main screen consists of two parts; the Main scenario and the Topology tabs. The main scenario describes TSHOOT.com test bed. The Topology tabs allow you to display the appropriate and select the trouble ticket.
To complete the item, you will first need to familiarize yourself with the TSHOOT.com test bed by clicking on the master scenario first and then the topologies tabs. Once you are familiar with the test bed and the topologies, you should start evaluating the trouble ticket. You will be presented with a Trouble Ticket scenario that will describe the fault condition. You will need to determine on which device the fault condition is located, to which technology the fault condition is related, and the solution to each trouble ticket. This will be done by answering three questions.
To begin, click on the Ticket on the Topology tabs.
Please note. Some of the questions will require you to use the scroll bar to see all options.
Read the ticket scenario to understand the fault condition.
Open the appropriate topology, based upon the ticket scenario.
Open the console of the desired device by clicking on that device in the topology, based upon your
Use the supported show, ping and trace commands to begin your fault isolation process.
Move to other devices as need by clicking on those devices within the topology.
The trouble ticket will include three questions that you will need to answer:
1. Which device contains the fault
2. Which technology the fault condition is related to
3. What is the solution to the issue
To advance to the next question within the ticket click on "Next Question".
When you click "DONE", the trouble ticket will turn RED and will no longer be accessible.
You may also use the "Previous Question" button to review questions within that specific ticket.
To complete a trouble ticket, answer all three questions and click "DONE". This will store your response
to the questions. Do not click on "DONE" unless you have answered all questions within the ticket.
Click the NEXT button on the bottom of the screen once a ticket is RED. This action moves you to the
Topology Overview (Actual Troubleshooting lab design is for below network design) Client Should have IP 10.2.1.3
EIGRP 100 is running between switch DSW1 & DSW2
OSPF (Process ID 1) is running between R1, R2, R3, R4
Network of OSPF is redistributed in EIGRP
BGP 65001 is configured on R1 with Webserver cloud AS 65002
HSRP is running between DSW1 & DSW2 Switches
The company has created the test bed shown in the layer 2 and layer 3 topology exhibits.
This network consists of four routers, two layer 3 switches and two layer 2 switches.
In the IPv4 layer 3 topology, R1, R2, R3, and R4 are running OSPF with an OSPF process number 1.
DSW1, DSW2 and R4 are running EIGRP with an AS of 10. Redistribution is enabled where necessary.
R1 is running a BGP AS with a number of 65001. This AS has an eBGP connection to AS 65002 in the ISP's network. Because the company's address space is in the private range.
R1 is also providing NAT translations between the inside (10.1.0.0/16 & 10.2.0.0/16) networks and outside (188.8.131.52/24) network.
ASW1 and ASW2 are layer 2 switches.
NTP is enabled on all devices with 184.108.40.206 serving as the master clock source.
The client workstations receive their IP address and default gateway via R4's DHCP server.
The default gateway address of 10.2.1.254 is the IP address of HSRP group 10 which is running on DSW1 and DSW2.
In the IPv6 layer 3 topology R1, R2, and R3 are running OSPFv3 with an OSPF process number 6.
DSW1, DSW2 and R4 are running RIPng process name RIP_ZONE.
The two IPv6 routing domains, OSPF 6 and RIPng are connected via GRE tunnel running over the underlying IPv4 OSPF domain. Redistribution is enabled where necessary.
Recently the implementation group has been using the test bed to do a 'proof-of-concept' on several implementations. This involved changing the configuration on one or more of the devices. You will be presented with a series of trouble tickets related to issues introduced during these configurations.
Note: Although trouble tickets have many similar fault indications, each ticket has its own issue and solution.
Each ticket has 3 sub questions that need to be answered & topology remains same.
Question-1 Fault is found on which device,
Question-2 Fault condition is related to,
Question-3 What exact problem is seen & what needs to be done for solution
Client is unable to ping IP 220.127.116.11...
Steps need to follow as below:-
1. When we check on client 1 & Client 2 desktop we are not receiving DHCP address from R4
2. Ipconfig ----- Client will be receiving IP address 10.2.1.3
3. IP 10.2.1.3 will be able to ping from R4 , R3, R2, R1
4. Look for BGP Neighbourship
5. Sh ip bgp summary ----- State of BGP will be in active state. This means connectivity issue between serial
6. Check for running config. i.e sh run --- over here check for access-list configured on interface as BGP is down (No need to check for NAT configuration as its configuration should be right as first need to bring BGP up)
7. In above snapshot we can see that access-list of edge_security on R1 is not allowing wan IP network
8. Change required: On R1, we need to permit IP 18.104.22.168/30 under the access list.
NEW QUESTION: 2
John works as a Database Administrator for Bluewell Inc. The company has a SQL Server database. A table in the database has a candidate key and an attribute that is not a constituent of the candidate key. The non-key attribute depends upon the whole of the candidate key rather than just a part of it. Which of the following normal forms is represented in the scenario?
A. 1 NF
B. 4 NF
C. 2 NF
D. 3 NF
Second normal form (2NF) is used in database normalization. A table that is in first normal form must meet additional criteria if it is to qualify for second normal form. Specifically, a 1NF table is in 2NF if and only if, given any candidate key and any attribute that is not a constituent of a candidate key, the non-key attribute depends upon the whole of the candidate key rather than just a part of it.
Answer C is incorrect. The 1NF is a normalization form in which each column in a row contains a single value, i.e., each attribute of the entity is single valued. Single valued attributes are also known as atomic attributes, as they cannot be decomposed into smaller units. There are mainly three kinds of attributes that prevent a table from being in the first normal form. They are as follows: Composite attributes Plural attributes (attributes that have more than one value) Attributes with complex data types The table below is in 1 NF, as all the columns in each row contain a single value.
Answer D is incorrect. Third normal form (3NF) is used in database normalization. A table is in 3NF if and only if the relation S (table) is in second normal form (2NF) and every non-prime attribute of S is non-transitively dependent on every key of S.
Answer A is incorrect. Fourth normal form (4NF) is a normal form used in database normalization. Introduced by Ronald Fagin in 1977, 4NF is the next level of normalization after Boyce-Codd normal form (BCNF). Whereas the second, third, and Boyce-Codd normal forms are concerned with functional dependencies, 4NF is concerned with a more general type of dependency known as a multivalued dependency. Symbolically, it can be represented as follows: If A -> > B|C, then A -> B and A -> C Here, A, B, and C are attributes.
NEW QUESTION: 3
Which two are features of GETVPN but not DMVPN and FlexVPN? (Choose two.)
A. enabled use of ESP or AH
B. preservation of IP protocol in outer header
C. sequence numbers that enable scalable replay checking
D. one IPsec SA for all encrypted traffic
E. design for use over public or private WAN
F. no requirement for an overlay routing protocol