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Exam MCIA - Level 1 topic 1 question 79 discussion

Actual exam question from Mulesoft's MCIA - Level 1
Question #: 79
Topic #: 1
[All MCIA - Level 1 Questions]

An organization is sizing an Anypoint VPC to extend their internal network to CloudHub.
For this sizing calculation, the organization assumes 150 Mule applications will be deployed among three (3) production environments and will use CloudHub's default zero-downtime feature. Each Mule application is expected to be configured with two (2) CloudHub workers. This is expected to result in several Mule application deployments per hour.
What is the minimum number of IP addresses that should be configured for this VPC that results in the smallest usable private IP address range to support the deployment and zero downtime of these 150 Mule applications (not accounting for any future Mule applications)?

  • A. 10.0.0.0/21 (2048 IPs)
  • B. 10.0.0.0/22 (1024 IPs)
  • C. 10.0.0.0/23 (512 IPs)
  • D. 10.0.0.0/24 (256 IPs)
Show Suggested Answer Hide Answer
Suggested Answer: C 🗳️

Comments

Chosen Answer:
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ojgarcia0608
Highly Voted 2 years, 8 months ago
150 applications 3 environment 2 workers (in each environment) 150*3*2=900 +50% zero downtime 900+450 = 1350 answer A
upvoted 6 times
Alandt
2 weeks, 3 days ago
You are correct, all the others have it wrong.
upvoted 1 times
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Alandt
Most Recent 2 weeks, 3 days ago
Selected Answer: A
You are all wrong. According to the official practice exam the correct answer is A. 2048. You can even ask ChatGPT and it will give you the same answer.
upvoted 1 times
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sid0308
7 months, 1 week ago
Selected Answer: C
150*2 + 150 = 450, so 512 will suffice
upvoted 1 times
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gilofernandes
8 months ago
Selected Answer: C
150*2+2=302
upvoted 1 times
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AK2020
1 year, 9 months ago
Selected Answer: C
C is correct as total number of app is 150
upvoted 1 times
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kanishka_
1 year, 10 months ago
Selected Answer: A
A is correct
upvoted 1 times
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madgeezer
1 year, 11 months ago
Selected Answer: C
Please note that 150 Mule applications will be deployed among three (3) production environments. This would suggest 50 on each production environment. Configured with two (2) CloudHub workers CloudHub's default zero-downtime feature. 150*2=300 50% for zero-downtime = 150 Total 300+150=450 C. 10.0.0.0/23 (512 IPs)
upvoted 2 times
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sanni27
2 years, 5 months ago
It should be D. IP require = no of worker/2
upvoted 1 times
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Community vote distribution
A (35%)
C (25%)
B (20%)
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