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Exam MCIA - Level 1 All Questions

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Exam MCIA - Level 1 topic 1 question 55 discussion

Actual exam question from Mulesoft's MCIA - Level 1
Question #: 55
Topic #: 1
[All MCIA - Level 1 Questions]

Refer to the exhibit. An organization is sizing an Anypoint VPC for the non-production deployments of those Mule applications that connect to the organization's on-premises systems. This applies to approximately 60 Mule applications. Each application is deployed to two CloudHub workers. The organization currently has three non-production environments (DEV, SIT and UAT) that share this VPC. The AWS region of the VPC has two AZs.
The organization has a very mature DevOps approach which automatically progresses each application through all non-production environments before automatically deploying to production. This process results in several Mule application deployments per hour, using CloudHub's normal zero-downtime deployment feature.
What is a CIDR block for this VPC that results in the smallest usable private IP address range?

  • A. 10.0.0.0/26 (64 IPs)
  • B. 10.0.0.0/25 (128 IPs)
  • C. 10.0.0.0/24 (256 IPs)
  • D. 10.0.0.0/22 (1024 IPs)
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Suggested Answer: D 🗳️

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gilofernandes
1 year ago
Selected Answer: D
60 mulapplications * 2 workers * 3 env * 2 AZ + 2 for network and broadcast= 722 Answer is D
upvoted 1 times
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lzrvs
1 year, 11 months ago
Selected Answer: D
The AZ information is misleading. It is only used for the HA feature and zero downtime deployment, but does not imply necessarily a larger number of IPs to account for. My count is: 60 apps * 2 workers/app * 3 env * 2 for Zero Downtime Deployment = 720 ips < 1024 ips
upvoted 2 times
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madgeezer
2 years, 3 months ago
Selected Answer: D
60 Mule applications each application is deployed to 2 CloudHub workers - 60*2=120 3 non-production environments (DEV, SIT and UAT) - 120*3=360 The AWS region of the VPC has 2AZs - 360*2 = 720 zero-downtime deployment feature requires 50% additional - 360+720=1080
upvoted 1 times
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