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NEW QUESTION: 1
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 /30 subnet
B. a /25 subnet
C. a /27 subnet
D. a /29 subnet
E. a /28 subnet
F. a /26 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: 2
다음 중 컴퓨터 프로그램에 의한 수학적 정리의 증명에 관한 것은 어느 것입니까?
A. 분석 정리 증명
B. 자동 정리 처리
C. 자동 정리 증명
D. 선택 사항이 없습니다.
E. 자동화 된 기술 증명
Answer: C
Explanation:
설명:
ATP (Automated theorem proving)는 컴퓨터 프로그램에 의한 수학 정리의 증명입니다. 기본 논리에 따라 정리의 타당성을 결정하는 문제는 사소한 것에서 불가능한 것으로 다양합니다. 자동화 된 정리 증명의 상업적 사용은 주로 집적 회로 설계 및 검증에 집중되어있다.

NEW QUESTION: 3
Which of the following statements best describes what a Public Certificate Authority (CA) is?
A. The FortiGate process in charge of generating digital certificates on the fly for SSL inspection purposes
B. A service that validates digital certificates for certificate-based authentication purposes
C. A service that provides a digital certificate each time a user is authenticating
D. An entity that certifies that the information contained in a digital certificate is valid and true.
Answer: B

NEW QUESTION: 4
ITパフォーマンスに関するエグゼクティブレベルへのレポートは、
A. 運用改善に関連するITパフォーマンス
B. 組織慣行に対するサードパーティのコンプライアンス
C. 組織戦略に対するIT成果物
D. 組織内のキャパシティプランニングの有効性
Answer: C