EMC E10-001 Online Practice Questions and Exam Preparation

EMC E10-001 Online Questions & Answers

• Question 41:

  What is an activity in the security risk assessment process?

  A. Determining the extent of potential threats in the IT infrastructure
  B. Defining security measures to direct the safe execution of operations
  C. Ensuring that security measures comply with processes and policies
  D. Deploying security measures to mitigate the impact of risk occurrences
  A. Determining the extent of potential threats in the IT infrastructure

• Question 42:

  What is a characteristic of a VSAN?

  A. Management traffic in a VSAN is restricted within its boundary
  B. Multiple zone sets can be active at the same time in a VSAN
  C. Consolidates multiple physical SANs to create a logical SAN
  D. Uses 16 bits of the 24-bit FC address and supports up to 127 nodes
  A. Management traffic in a VSAN is restricted within its boundary

• Question 43:

  Which iSCSI name requires an organization to own a registered domain name?

  A. IQN
  B. EUI
  C. WWNN
  D. WWPN
  A. IQN

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  iSCSI Name

  A unique worldwide iSCSI identifier, known as an iSCSI name, is used to identify the initiators and targets within an iSCSI network to facilitate communication. The unique identifier can be a combination of the names of the department,

  application, or manufacturer, serial number, asset number, or any tag that can be used to recognize and manage the devices. Following are two types of iSCSI names commonly used:

  -iSCSI Qualified Name (IQN): An organization must own a registered domain name to generate iSCSI Qualified Names. This domain name does not need to be active or resolve to an address. It just needs to be reserved to prevent other

  organizations from using the same domain name to generate iSCSI names. A date is included in the name to avoid potential conflicts caused by the transfer of domain names. An example of an IQN is iqn.200802.com.example:optional_string. The optional_string provides a serial number, an asset number, or any other device identifiers. An iSCSI Qualified Name enables storage administrators to assign meaningful names to iSCSI devices, and

  therefore, manage those devices more easily.

  -Extended Unique Identifier (EUI): An EUI is a globally unique identifier based on the IEEE EUI-64 naming standard. An EUI is composed of the eui prefix followed by a 16-character exadecimal name, such as eui.0300732A32598D26.

  In either format, the allowed special characters are dots, dashes, and blank spaces. EMC E10-001 Student Resource Guide. Module 6: IP SAN and FCoE

• Question 44:

  What does the area ID of the FC address identify?

  A. Group of ports within a switch
  B. An individual port within a fabric
  C. Location of the name server within the fabric
  D. Unique number provided to each switch in the fabric
  A. Group of ports within a switch

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  FC Addressing in Switched Fabric

  An FC address is dynamically assigned when a node port logs on to the fabric. The FC address has a distinct format, as shown in the slide.

  The first field of the FC address contains the domain ID of the switch. A Domain ID is a unique number provided to each switch in the fabric.

  Although this is an 8-bit field, there are only 239 available addresses for domain ID because some addresses are deemed special and reserved for fabric management services. For example, FFFFFC is reserved for the name server, and

  FFFFFE is reserved for the fabric login service. The area ID is used to identify a group of switch ports used for connecting nodes. An example of a group of ports with common area ID is a port card on the switch. The last field, the port ID,

  identifies the port within the group. Therefore, the maximum possible number of node ports in a switched fabric is calculated as:

  239 domains X 256 areas X 256 ports = 15,663,104

  

  EMC E10-001 Student Resource Guide. Module 5: Fibre Channel Storage Area Network (FC SAN)

• Question 45:

  What is characteristic of a virtual tape library?

  A. Uses disk drives as backup media
  B. Requires changes on the backup software
  C. A virtual tape cannot span multiple LUNs
  D. Uses a special mechanism to address the 'shoe-shining' effect
  A. Uses disk drives as backup media

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  Virtual Tape Library

  

  A virtual tape library (VTL) has the same components as that of a physical tape library, except that the majority of the components are presented as virtual resources. For the backup software, there is no difference between a physical tape

  library and a virtual tape library.

  Virtual tape libraries use disks as backup media.

  Emulation software has a database with a list of virtual tapes, and each virtual tape is assigned space on a LUN. A virtual tape can span multiple LUNs if required. File system awareness is not required while backing up because the virtual

  tape solution typically uses raw devices.

  EMC E10-001 Student Resource Guide. Module 10: Backup and Archive

• Question 46:

  Which benefit does the measured service characteristic provide to a cloud service provider?

  A. Enables control and optimization of resource use
  B. Enables cloud services to communicate with each other
  C. Enables availability of specific resources depending on policy
  D. Enables the reduction and/or elimination of upfront IT expenditures
  A. Enables control and optimization of resource use

• Question 47:

  Which fabric service is responsible for managing and distributing Registered State Change Notifications (RSCNs) to attached node ports?

  A. Fabric controller
  B. Name server
  C. Fabric login server
  D. Management server
  A. Fabric controller

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  Fabric Services

  

  EMC E10-001 Student Resource Guide

  Module 5: Fibre Channel Storage Area Network (FC SAN)

• Question 48:

  Click on the calculator icon in the upper left corner.

  An application requires 3.5 TB of capacity. The application generates 2300 IOPS to disks during peak workloads. The vendor indicates that a 146 GB, 15K rpm drive is capable of performing a maximum of 150 IOPS.

  How many disks are required to meet both capacity and performance requirements?

  A. 15
  B. 22
  C. 24
  D. 60
  C. 24

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  Storage Design based on Application Requirements and Disk Drive Performance

  

  Calculation Dc = 3.5TB / 146GB = 3500 / 146 = 23.97 = 24 Drives Dp = 2300 iops / 150 iops = 15.33 = 15 Drives Disk required = max(Dc,Dp) = max(24,15) = 24

  EMC E10-001 Student Resource Guide. Module 2: Data Center Environment

• Question 49:

  What describes a landing zone in a disk drive?

  A. Area on which the read/write head rests
  B. Area where the read/write head lands to access data
  C. Area where the data is buffered before writing to platters
  D. Area where sector-specific information is stored on the disk
  A. Area on which the read/write head rests

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  Disk Drive Components

  

  The key components of a hard disk drive are platter, spindle, read-write head, actuator arm assembly, and controller board. I/O operations in a HDD is performed by rapidly moving the arm across the rotating flat platters coated with magnetic particles. Data is transferred between the disk controller and magnetic platters through the read-write (R/W) head which is attached to the arm. Data can be recorded and erased on magnetic platters any number of times.

  Platter: A typical HDD consists of one or more flat circular disks called platters. The data is recorded on these platters in binary codes (0s and 1s). The set of rotating platters is sealed in a case, called Head Disk Assembly (HDA). A platter is a rigid, round disk coated with magnetic material on both surfaces (top and bottom). The data is encoded by polarizing the magnetic area, or domains, of the disk surface. Data can be written to or read from both surfaces of the platter. The number of platters and the storage capacity of each platter determine the total capacity of the drive.

  Spindle: A spindle connects all the platters and is connected to a motor. The motor of the spindle rotates with a constant speed. The disk platter spins at a speed of several thousands of revolutions per minute (rpm). Common spindle speeds are 5,400 rpm, 7,200 rpm, 10,000 rpm, and 15,000 rpm. The speed of the platter is increasing with improvements in technology; although, the extent to which it can be improved is limited. Read/Write Head: Read/Write (R/W) heads, read and write data from or to platters. Drives have two R/W heads per platter, one for each surface of the platter. The R/W head changes the agnetic polarization on the surface of the platter when writing data. While reading data, the head detects the magnetic polarization on the surface of the platter. During reads and writes, the R/W head senses the magnetic polarization and never touches the surface of the platter. When the spindle is rotating, there is a microscopic air gap maintained between the R/W heads and the platters, known as the head flying height. This air gap is removed when the spindle stops rotating and the R/W head rests on a special area on the platter near the spindle. This area is called the landing zone . The landing zone is coated with a lubricant to reduce friction between the head and the platter. The logic on the disk drive ensures that heads are moved to the landing zone before they touch the surface. If the drive malfunctions and the R/W head accidentally touches the surface of the platter outside the landing zone, a head crash occurs. In a head crash, the magnetic coating on the platter is scratched and may cause damage to the R/W head. A head crash generally results in data loss.

  Actuator Arm Assembly: R/W heads are mounted on the actuator arm assembly, which positions the R/W head at the location on the platter where the data needs to be written or read. The R/W heads for all platters on a drive are attached to one actuator arm assembly and move across the platters simultaneously.

  Drive Controller Board: The controller is a printed circuit board, mounted at the bottom of a disk drive. It consists of a microprocessor, internal memory, circuitry, and firmware. The firmware controls the power to the spindle motor and the speed of the motor. It also manages the communication between the drive and the host. In addition, it controls the R/W operations by moving the actuator arm and switching between different R/W heads, and performs the optimization of data access. EMC E10-001 Student Resource Guide. Module 2: Data Center Environment

• Question 50:

  An application requires 1.46 TB of capacity. The application generates 7200 IOPS to disks during peak workloads. The vendor indicates that a 146 GB, 15K rpm drive is capable of performing a maximum of 180 IOPS at 70% utilization. How many disks are required to meet both the application's capacity and performance requirements?

  A. 10
  B. 40
  C. 50
  D. 72
  B. 40

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  Storage Design based on Application Requirements and Disk Drive Performance

  

  Calculation Dc = 1.46TB / 146GB = 1460 / 146 = 10 = 10 Drives Dp = 7200 iops / 180 iops = 40 = 40 Drives Disk required = max(Dc,Dp) = max(10,40) = 40

  EMC E10-001 Student Resource Guide. Module 2: Data Center Environment