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2021 May C9020-461 Study Guide Questions:
Q21. A customer wants to estimate the Total Cost of Ownership (TCO) for a storage project.
Which software tool should be used?
A. Knowledge Center
B. Systems Consolidation Evaluation Tool
C. ISAT Tool
D. System Storage Interoperation Center Tool
Answer: B
Q22. A customer requests that a sales specialist demonstrate the performance improvement that will be achieved on the Windows environment with a proposed IBM Storwize V7000 with Easy Tier.
Which type of Windows data should the sales specialist gather from the customer?
A. Performance Monitor
B. PT Reports
C. iostat
D. esxtop reports
Answer: C
Q23. Which feature of the IBM Storwize family is recommend when additional space for disk drives is unavailable?
A. Virtualization
B. Storage tiering
C. Real-time Compression
D. Deduplication
Answer: A
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Q24. A customer has conducted an analysis of its Vmware environment to determine the capacity savings available by converting from thick to thin provisioning. The results of the analysis are in the exhibit provided.
What is the minimum capacity needed for the solution assuming compression is NOT part of the proposed solution?
A. 51540 GB
B. 17774 GB
C. 33796 GB
D. 27060 GB
Answer: B
Q25. A client’s executive officer has questions about the financial impact of a new information technology project.
Which tool will help analyze cost and benefits for the project?
A. ROInow!
B. Disk Magic
C. TCO!now
D. Capacity Magic
Answer: B
Reference: http://www-304.ibm.com/partnerworld/gsd/showimage.do?id=25691
Q26. A customer provides the information shown in the exhibit for the company’s primary database environment. Solution requirements include meeting of exceeding the current I/O rate of the existing environment.
System name Name Max IOPS Avg IOPS Min IOPS Max BW (MBps) Avg BW (MBps) Min BW (MBps) Max Latency (ms) Avg Latency (ms) pldenxivg3 denoubivs_arch1 1101 20 0 274 5 0 3 2 pldenxivg3 denoubivs_arch2 3 0 0 0 0 0 0 0 pldenxivg3 denoubivs_bin1 322 1 0 1 0 0 32 0 pldenxivg3 denoubivs_bin2 184 0 0 1 0 0 21 0 pldenxivg3 denoubivs_bkup1 2 0 0 0 0 0 0 0 pldenxivg3 denoubivs_bkup2 25 0 0 2 0 0 0 0 pldenxivg3 denoubivs_bkup3 1 0 0 0 0 0 0 0 pldenxivg3 denoubivs_data1 997 39 0 128 3 0 28 2 pldenxivg3 denoubivs_data2 997 38 0 133 3 0 31 2 pldenxivg3 denoubivs_data3 1005 43 0 127 3 0 23 1 pldenxivg3 denoubivs_data4 2217 38 0 175 3 0 19 2 pldenxivg3 denoubivs_data5 1194 37 0 109 2 0 32 1 pldenxivg3 denoubivs_data6 1118 30 0 70 2 0 23 1 pldenxivg3 denoubivs_data7 1111 29 0 70 2 0 21 1 pldenxivg3 denoubivs_data8 2443 24 0 156 1 0 10 0 pldenxivg3 denoubivs_data9 1710 8 0 108 0 0 12 1 pldenxivg3 denoubivs_log1 2107 49 0 46 1 0 2 0 pldenxivg3 denoubivs_log2 2174 44 0 43 1 0 2 0 pldenxivg3 denoubivs_NJP_bin 8 0 0 0 0 0 0 0 pldenxivg3 denoubivs_NJP_data 1452 9 0 93 0 0 0 0 pldenxivg3 denoubivs_NJP_log 65 14 0 3 0 0 0 0 pldenxivg3 denoubivs_os1 16 0 0 0 0 0 0 0 pldenxivg3 denoubivs_os2 0 0 0 0 0 0 0 0 pldenxivg3 denoubivs_pg1 12 0 0 1 0 0 0 0 pldenxivg3 denoubivs_pg2 0 0 0 0 0 0 0 0 pldenxivg3 denoubivs_arch1 501 5 0 122 1 0 3 1 pldenxivg3 denoubivs_arch2 0 0 0 0 0 0 0 0 pldenxivg3 denoubivs_back1 1 0 0 0 0 0 0 0 pldenxivg3 denoubivs_back2 0 0 0 0 0 0 0 0 pldenxivg3 denoubivs_back3 2 0 0 0 0 0 0 0 pldenxivg3 denoubivs_bin1 209 2 0 6 0 0 16 0 pldenxivg3 denoubivs_bin2 283 0 0 17 0 0 18 1 pldenxivg3 denoubivs_bin3 176 1 0 8 0 0 18 1 pldenxivg3 denoubivs_data1 3128 116 0 201 5 0 17 0 pldenxivg3 denoubivs_data2 1117 65 0 71 2 0 16 0 pldenxivg3 denoubivs_data3 1129 69 0 71 2 0 16 0 pldenxivg3 denoubivs_data4 1136 63 0 72 2 0 16 0 pldenxivg3 denoubivs_data5 1147 65 0 72 2 0 16 0 pldenxivg3 denoubivs_data6 3470 60 0 223 2 0 36 0 pldenxivg3 denoubivs_data7 2389 51 0 152 2 0 20 0 pldenxivg3 denoubivs_log1 204 12 0 13 0 0 2 0 pldenxivg3 denoubivs_log2 279 11 0 12 0 0 3 0 pldenxivg3 denoubivs_os1 21 0 0 0 0 0 0 0 pldenxivg3 denoubivs_os2 0 0 0 0 0 0 0 0 pldenxivg3 denoubivs_pg1 2 0 0 0 0 0 0 0
Total= 35458 943 0 2580 44
Average= 788 21 0 57 1 0 10 0
What should the minimum I/O rate be for the new solution?
A. 35458 I/Os per second
B. 943 I/Os per second
C. 21 I/Os per second
D. 788 I/Os per second
Answer: C
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Q27. A customer has a FlashSystem 840 installed with 12 TB of capacity (8 x 2 TB flash modules, RAID 5). The storage solution was purchased for a project that has now been completed and the customers plan is to install additional flash modules and deploy this device to a new project and has asked for you recommendations.
Which advice should be given to this customer?
A. TheFlashSystem 840 is factory sealed and cannot be upgraded.
B. TheFlashSystem 840 can support mixed capacity flash modules and the upgrade is data destructive.
C. TheFlashSystem 840 supports only same capacity flash modules and the upgrade is data destructive.
D. TheFlashSystem 840 can support mixed capacity flash modules and additional new capacity is automatically rebalanced and available on the system.
Answer: D
Q28. A customer is looking for a cost-effective, all-flash solution.
Which benefit would the FlashSystem 840 offer this customer?
A. Increased data availability
B. Improved storage efficiency
C. Enhanced capacity utilization
D. Extended wear life
Answer: C
Q29. Which tool can be used to obtain an interactive image of Storwize system while using a mobile device?
A. IBM 3D Product Catalog
B. IBM Storage Solutions Handbook
C. IBM Remote
D. Quick Reference for IBM System Storage
Answer: C
Q30. A customer has been experiencing sporadic performance issues on its IBM Storwize
V7000 system. Analysis has shown that during peak workloads it is CPU constrained.
What can be done to alleviate the CPU contention?
A. Add IBM Spectrum Control
B. Add a second I/O group
C. Implement Real-time Compression
D. Perform a Capacity Magic study
Answer: B