MCIA - Level 1 Exam - Question 79

Question

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)?

Examice · Accepted Answer

Based on the scenario provided, the sizing calculation for the Anypoint VPC needs to consider 150 Mule applications across three production environments, resulting in 450 Mule applications (150 applications * 3 environments). Each application is configured with two workers, leading to a total of 900 workers (450 applications * 2 workers). With the zero-downtime feature, an additional 50% of the original worker count should be included for redundancy, which adds 450 workers (900 original workers * 0.5). This results in a total of 1350 IP addresses required (900 + 450). The smallest usable private IP address range that can support 1350 IP addresses is 10.0.0.0/21, which provides 2048 IPs. Therefore, the correct answer is 10.0.0.0/21.

ojgarcia0608 · Answer

150 applications 3 environment 2 workers (in each environment) 150*3*2=900 +50% zero downtime 900+450 = 1350 
answer A

madgeezer · Answer

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)

sanni27 · Answer

It should be D. IP require = no of worker/2

kanishka_ · Answer

A is correct

Ak_2020 · Answer

C is correct as total number of app is 150

gilofernandes · Answer

150*2+2=302

sid0308 · Answer

150*2 + 150 = 450, so 512 will suffice

Alandt · Answer

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.

MCIA - Level 1 Exam - Question 79

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