考试代码: HP2-K27
考试名称: Supporting and Servicing HP P6000 EVA Solutions
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HP2-K27 考试是 IBM 公司的 Supporting and Servicing HP P6000 EVA Solutions 认证考试官方代号,红宝书的 HP2-K27 权威全真题库是 IBM 认证厂商的授权产品,绝对保证第一次参加 HP2-K27 考试的考生即可顺利通过,否则将全额退款!保证您的利益不受到任何的损失。
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QUESTION NO: 1 What failover mode does the EVA currently use? A. Active B. Passive C. Active-Passive D. Transparent Answer: C Reference:http://h20195.www2.hp.com/v2/GetPDF.aspx/4AA2-8848ENW.pdf(page 10, figure 4) QUESTION NO: 2 In an EVA, what is purpose of virtual disk leveling? A. Dynamically distribute data blocks over many physical spindles to eliminate performance bottlenecks. B. Equalize disk capacity. C. Allocate space per disk group to recover from physical disk group member failure. D. Build virtual disks from physical disks. Answer: C Reference:http://www2.openvms.org/kparris/eva_intro.ppt(slide 11) QUESTION NO: 3 What is the approximate maximum virtual disk size for a disk group with 10 36GB drives and a disk protection level of Single? A. 134 GB B. 202 GB C. 270 GB D. 338 GB Answer: C Reference:http://www.google.com.pk/url?sa=t&rct=j&q=maximum%20virtual%20disk%20size%20f or%20a%20disk%20group%20with%2010%2036gb%20drives%20and%20a%20disk%20protectio n%20level%20of%20single&source=web&cd=1&ved=0CBoQFjAA&url=http%3A%2F%2Fwww2.o penvms.org%2Fkparris%2Feva_intro.ppt&ei=nhTkTrrLMcKZOtKRgLUE&usg=AFQjCNEtQ63RCt RHinCREzeztgQhWFniDQ(slide 14) To find out the maximum virtual disk size, you have to multiply 10 by 36. That makes 360 GB. Now you have to calculate the spare allocation. Normally space allocation is calculated by dividing 36 GB by 4. The result will be 90 GB. Now minus 90 GB from 360 and you will get 270 GB as the maximum virtual disk size for a disk group. QUESTION NO: 4 What is an initialized pair of HSV controllers with a minimum of six or eight physical drives? A. Storage system B. Pstore C. Redundant storage set D. Rstore Answer: C Reference:http://www.google.com.pk/url?sa=t&rct=j&q=maximum%20virtual%20disk%20size%20f or%20a%20disk%20group%20with%2010%2036gb%20drives%20and%20a%20disk%20protectio n%20level%20of%20single&source=web&cd=1&ved=0CBoQFjAA&url=http%3A%2F%2Fwww2.o penvms.org%2Fkparris%2Feva_intro.ppt&ei=nhTkTrrLMcKZOtKRgLUE&usg=AFQjCNEtQ63RCt RHinCREzeztgQhWFniDQ(slide 19) QUESTION NO: 5 What is the maximum number of disk groups supported for any EVA? A. 8 B. 16 C. 32 D. 64 Answer: B QUESTION NO: 6 What is the cache size of each HSV360 controller in a P6500 EVA? A. 1 GB B. 2 GB C. 4 GB D. 8 GB Answer: D Reference:http://h20195.www2.hp.com/v2/GetPDF.aspx/4AA0-6011ENW.pdf(Page 4, Table 2) QUESTION NO: 7 On the web-based OCP, what level of user functionality allows you to set the storage system's networking, security, and system time characteristics?A. Global B. User C. Administrative D. Service Answer: C Reference:http://www.amrila.com/wiki/lib/exe/fetch.php/hp/storageworks/c02076538.pdf(page 7) QUESTION NO: 8 How many SAS lanes are available on a P6500 EVA storage system?A. 6 B. 8 C. 12 D. 16 Answer: D Reference:http://h20195.www2.hp.com/v2/GetPDF.aspx/4AA0-6011ENW.pdf(page 6, topic internal SAS domains) QUESTION NO: 9 Which type of management does the Management Module allow? A. array-based management B. server-based management C. array-based and server-based management D. remote server-based management Answer: C Reference:http://h18000.www1.hp.com/products/quickspecs/13904_div/13904_div.pdf(page 4, 5th bulleted point) QUESTION NO: 10 Which statements are true regarding the M6612 and M6625 drive enclosure I/O modules? (Select two.)A. A and B I/O modules are physically different. B. A and B I/O modules are identified by slot pin. C. Each module contains an Enclosure Management Processor (EMP) and a20-port cut through switch (CTS). D. The modules communicate through the EAB. Answer: C
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