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NEW QUESTION: 1
Which of the following is NOT true of the Kerberos protocol?
A. It performs mutual authentication
B. Only a single login is required per session.
C. The KDC is aware of all systems in the network and is trusted by all of them
D. The initial authentication steps are done using public key algorithm.
Answer: D
Explanation:
Kerberos is a network authentication protocol. It is designed to provide strong authentication for client/server applications by using secret-key cryptography. It has the following characteristics:
* It is secure: it never sends a password unless it is encrypted.
* Only a single login is required per session. Credentials defined at login are then passed between resources without the need for additional logins.
* The concept depends on a trusted third party - a Key Distribution Center (KDC). The
KDC is aware of all systems in the network and is trusted by all of them.
* It performs mutual authentication, where a client proves its identity to a server and a server proves its identity to the client.
Kerberos introduces the concept of a Ticket-Granting Server/Service (TGS). A client that wishes to use a service has to receive a ticket from the TGS - a ticket is a time-limited cryptographic message - giving it access to the server. Kerberos also requires an
Authentication Server (AS) to verify clients. The two servers combined make up a KDC.
Within the Windows environment, Active Directory performs the functions of the KDC. The following figure shows the sequence of events required for a client to gain access to a service using Kerberos authentication. Each step is shown with the Kerberos message associated with it, as defined in RFC 4120 "The Kerberos Network Authorization Service
(V5)".
Kerberos Authentication Step by Step
* Step 1: The user logs on to the workstation and requests service on the host. The workstation sends a message to the Authorization Server requesting a ticket granting ticket
(TGT).
* Step 2: The Authorization Server verifies the user's access rights in the user database and creates a TGT and session key. The Authorization Sever encrypts the results using a key derived from the user's password and sends a message back to the user workstation.
The workstation prompts the user for a password and uses the password to decrypt the incoming message. When decryption succeeds, the user will be able to use the TGT to request a service ticket.
* Step 3: When the user wants access to a service, the workstation client application sends a request to the Ticket Granting Service containing the client name, realm name and a timestamp. The user proves his identity by sending an authenticator encrypted with the session key received in Step 2
* Step 4: The TGS decrypts the ticket and authenticator, verifies the request, and creates a ticket for the requested server. The ticket contains the client name and optionally the client
IP address. It also contains the realm name and ticket lifespan. The TGS returns the ticket to the user workstation. The returned message contains two copies of a server session key
- one encrypted with the client password, and one encrypted by the service password.
* Step 5: The client application now sends a service request to the server containing the ticket received in Step 4 and an authenticator. The service authenticates the request by decrypting the session key. The server verifies that the ticket and authenticator match, and then grants access to the service. This step as described does not include the authorization performed by the Intel AMT device, as described later.
* Step 6: If mutual authentication is required, then the server will reply with a server authentication message.
The Kerberos server knows "secrets" (encrypted passwords) for all clients and servers under its control, or it is in contact with other secure servers that have this information.
These "secrets" are used to encrypt all of the messages shown in the figure above.
To prevent "replay attacks," Kerberos uses timestamps as part of its protocol definition. For timestamps to work properly, the clocks of the client and the server need to be in synch as much as possible. In other words, both computers need to be set to the same time and date. Since the clocks of two computers are often out of synch, administrators can establish a policy to establish the maximum acceptable difference to Kerberos between a client's clock and server's clock. If the difference between a client's clock and the server's clock is less than the maximum time difference specified in this policy, any timestamp used in a session between the two computers will be considered authentic. The maximum difference is usually set to five minutes.
Note that if a client application wishes to use a service that is "Kerberized" (the service is configured to perform Kerberos authentication), the client must also be Kerberized so that it expects to support the necessary message responses.
For more information about Kerberos, see http://web.mit.edu/kerberos/www/.
References:
Introduction to Kerberos Authentication from Intel
and
http://www.zeroshell.net/eng/kerberos/Kerberos-definitions/#1353
and
http://www.ietf.org/rfc/rfc4120txt

NEW QUESTION: 2
You have an Exchange organization. All servers in the organization have Exchange Server 2010 SP1 installed.
The organization contains a database availability group (DAG) named DAG1. DAG1 contains two Mailbox server named Server1 and Server2. Five databases are replicated in DAG1.
You need to install Exchange rollup updates on Server1. The solution must not prevent users from accessing their mailboxes.
What should you do first?
A. Run the Suspend-MailboxDatabaseCopy cmdlet, and then switch over all mailbox databases to Server2.
B. run the Set-MailboxDatabase cmdlet, and then configure all mailbox database copies as lagged copies.
C. run the Set-DatabaseAvailabilityGroup cmdlet, and then enable Datacenter Activation Coordination for DAG1.
D. run the Update-MailboxDatabaseCopy cmdlet, and then modify the mailbox database Activiation preference.
Answer: A
Explanation:
Explanation/Reference: In situations where you have a planned outage/maintenance window or if you perhaps need to seed a database, the first step is to suspend replication for the involved database(s). This can be done both via the Exchange Management Console (EMC) and the Exchange Management Shell (EMS). To do so via the EMC, you simply right-click on the respective database copy/copies and select suspend in the context menu.
To do the same via the EMS, you can use the following command:
Suspend-MailboxDatabaseCopy -Identity MDB02\E14EX02
http://www.msexchange.org/articles_tutorials/exchange-server-2010/high-availability-recover y/uncovering-exchange-2010-database-availability-groups-dags-part4.html

NEW QUESTION: 3
귀사는 클라우드 인프라를 확장하고 많은 플랫 파일과 정적 자산을 S3로 옮깁니다. 현재 VPN을 사용하여 컴퓨팅 인프라에 액세스하지만 모든 중요한 데이터를 AWS로 오프로드 할 때 정적 파일에 대한 안정성이 더 필요합니다. 비용을 낮추면서 가장 좋은 행동 방법은 무엇입니까? 정답을 선택하십시오.
A. 프라이빗 VIF를 사용하여 컴퓨팅 및 S3 리소스에 모두 액세스하기 위한 직접 연결 연결을 만듭니다.
B. 두 개의 직접 연결 연결을 만듭니다. 각각은 프라이빗 VIF에 연결되어 최대 복원력을 보장합니다.
C. 퍼블릭 VIF를 사용하여 직접 연결 연결을 생성하고 DX 연결을 통해 VPN을 VPN 엔드 포인트로 라우팅합니다.
D. S3 끝점을 만들고 VPN의 끝점 접두사 목록으로 연결하여 S3 리소스에 액세스 할 수 있도록합니다.
Answer: C
Explanation:
An S3 endpoint cannot be used with a VPN. A Private VIF cannot access S3 resources. A Public VIF with a VPN will ensure security for your compute resources and access to your S3 resources.
Two DX connections are very expensive and a Private VIF still won't allow access to your S3 resources.

NEW QUESTION: 4
A PL/SQL procedure queries only those columns of a redefined table that were unchanged by the online table redefinition.
What happens to the PL/SQL procedure after the online table redefinition?
A. It becomes invalid for all options of online table redefinition but automatically gets revalidated the next time it is used.
B. It becomes invalid for all options of online table redefinition and is automatically recompiled during online redefinition of the table.
C. It remains valid.
D. It becomes invalid only if the storage parameters have been modified and it automatically gets revalidated the next time it is used.
Answer: C