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NEW QUESTION: 1
Which of the following best captures how Focal Point supports Portfolio Management across the organization?
A. A workflow tool where you can specify tasks by role
B. Deliver higher efficiencies by automating the software delivery process.
C. Enable collaboration for geographically distributed teams.
D. Enable strategic assessment of enterprise modernization initiatives.
Answer: B
NEW QUESTION: 2
A technician is asked to install three servers: a 1U, a 4U, and a 6U. Which of the following is the MOST correct order to install these servers starting from the top of the rack?
A. 6U, 4U, 1U
B. 1U, 4U, 6U
C. 4U, 6U, 1U
D. 4U, 1U, 6U
Answer: B
NEW QUESTION: 3
Refer to the exhibit.
Which type of VMware backup configuration is shown?
A. Consolidated backup
B. Guest level backup
C. Console backup
D. Image level backup with VADP
Answer: D
NEW QUESTION: 4
Refer to the exhibit. Customer is planning to deploy a clustering over the WAN UCM topology with 2 subscribers at site 1 and 2 subscribers at site 2.
How much bandwidth would be required between site 1 and site 2 to for database replication?
A. 1.544 Mbps
B. 4.632 Mbps
C. 6.176 Mbps
D. 7.772 Mbps
E. 3.088 Mbps
Answer: B
Explanation:
Consider two sites, Site 1 and Site 2, with Unified CM clustered over the WAN across these two sites that are 80 msec round-trip time apart. Site 1 has one publisher, one combined TFTP and music on hold (MoH) server, and two Unified CM subscriber servers. Site 2 has one TFTP/MoH server and two Unified CM subscriber servers. Site 1 has 5000 phones, each having one DN; and Site 2 has 5000 phones, each having one DN. During the busy hour, 2500 phones in Site 1 call
2500 phones in Site 2, each at 3 BHCA. During that same busy hour, 2500 phones in Site 2 also call 2500 phones in Site 1, each at 3 BHCA. In this case:
Total BHCA during the busy hour = 2500*3 + 2500*3 = 15,000
Total bandwidth required between the sites = Total ICCS bandwidth + Total database bandwidth Because total BHCA is 15,000 (greater than 10,000), we can use the formula to calculate:
Total ICCS bandwidth = (15,000/10,000) * (1 + 0.006*80) = 2.22 Mbps
Total database bandwidth = (Number of servers remote to the publisher) * 1.544 = 3 * 1.544 =
4.632 Mbps
http://www.cisco.com/c/en/us/td/docs/voice_ip_comm/cucm/srnd/7x/uc7_0/models.html
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