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NEW QUESTION: 1
Which shape should be used to call a connector in a flow? (Choose One)
A. Shape 4
B. Shape 2
C. Shape 1
D. Shape 3
Answer: D
NEW QUESTION: 2
You are a database administrator of a Microsoft SQL Server 2012 environment. The environment contains two servers named SQLServer01 and SQLServer02. The database Contoso exists on SQLServer01. You plan to mirror the Contoso database between SQLServer01 and SQLServer02 by using database mirroring. You need to prepare the Contoso database for database mirroring. Which three actions should you perform in sequence? (To answer, move the appropriate actions from the list of actions to the answer area and arrange them in the correct order.)
Build List and Reorder:
Answer:
Explanation:
NEW QUESTION: 3
View the Exhibit.
You are designing an IP addressing scheme for the network in the exhibit above.
Each switch represents hosts that reside in separate VLANs. The subnets should be allocated to match the following host capacities:
Router subnet: two hosts
SwitchA subnet: four hosts
SwitchB subnet: 10 hosts
SwitchC subnet: 20 hosts
SwitchD subnet: 50 hosts
You have chosen to subnet the 192.168.51.0/24 network.
Which of the following are you least likely to allocate?
A. a /26 subnet
B. a /25 subnet
C. a /29 subnet
D. a /30 subnet
E. a /27 subnet
F. a /28 subnet
Answer: B
Explanation:
Explanation/Reference:
Section: Addressing and Routing Protocols in an Existing Network Explanation Explanation:
Of the available choices, you are least likely to allocate a /25 subnet. The largest broadcast domain in this scenario contains 50 hosts. A /25 subnet can contain up to 126 assignable hosts. In this scenario, allocating a /25 subnet would reserve half the 192.168.51.0/24 network for a single virtual LAN (VLAN).
The total number of hosts for which you need addresses in this scenario is 86. Therefore, you would only need to use half the /24 subnet if all 86 hosts were residing in the same VLAN.
You should begin allocating address ranges starting with the largest group of hosts to ensure that the entire group has a large, contiguous address range available. Subnetting a contiguous address range in structured, hierarchical fashion enables routers to maintain smaller routing tables and eases administrative burden when troubleshooting.
You are likely to use a /26 subnet. In this scenario, the largest VLAN contains 50 hosts. If you were to divide the 192.168.51.0/25 subnet into two /26 subnets, the result would be two new subnets capable of supporting up to 62 assignable hosts: the 192.168.51.0/26 subnet and the 192.168.51.64/26 subnet.
Therefore, you should start subnetting with a /26 network. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.64/26 subnet to SwitchD's VLAN.
You are likely to use a /27 subnet. The nextlargest broadcast domain in this scenario is the SwitchC subnet, which contains 20 hosts. If you were to divide the 192.168.51.0/26 subnet into two /27 subnets, the result would be two new subnets capable of supporting up to 30 assignable hosts: the 192.168.51.0/27 subnet and the 192.168.51.32/27 subnet. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.32/27 subnet to SwitchC's VLAN.
You are likely to use a /28 subnet. The nextlargest broadcast domain in this scenario is the SwitchB subnet, which contains 10 hosts. If you were to divide the 192.168.51.0/27 subnet into two /28 subnets, the result would be two new subnets capable of supporting up to 14 assignable hosts: the 192.168.51.0/28 subnet and the 192.168.51.16/28 subnet. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.16/28 subnet to SwitchB's VLAN.
You are likely to use a /29 subnet. The nextlargest broadcast domain in this scenario is the SwitchA subnet, which contains four hosts. If you were to divide the 192.168.51.0/28 subnet into two /29 subnets, the result would be two new subnets capable of supporting up to six assignable hosts: the 192.168.51.0/29 subnet and the 192.168.51.8 subnet. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.8/29 subnet to SwitchA's VLAN.
You are likely to use a /30 subnet. The final subnet in this scenario is the link between RouterA and RouterB, which contains two hosts. If you were to divide the 192.168.51.0/29 subnet into two /30 subnets, the result would be two new subnets capable of supporting two assignable hosts each: the
192.168.51.0/30 subnet and the 192.168.51.4/30 subnet. To maintain a logical, hierarchical IP structure, you could then allocate the 192.168.51.4/30 subnet to the link between RouterA and RouterB. This would leave the 192.168.51.0/30 subnet unallocated. However, you could further divide the 192.168.51.0/30 subnet into single /32 host addresses that could then be used for loopback IP addressing on the routers.
Reference:
CCDA 200-310 Official Cert Guide, Chapter 8, IPv4 Address Subnets, pp. 302-310 CCDA 200-310 Official Cert Guide, Chapter 8, Plan for a Hierarchical IP Address Network, pp. 311-312 Cisco: IP Addressing and Subnetting for New Users
NEW QUESTION: 4
A. Protect any device on or off the network
B. Uncover current & emergent threats
C. Real-time sandboxing
D. Identify the internet infrastructure used for attacks
E. Data Loss Prevention
F. Retrospective Analysis
Answer: A,B,D
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Aalk - 2014-05-05 16:45:18
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