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Question 79

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)?

    Correct Answer: A

    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.

Discussion
ojgarcia0608Option: A

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

Alandt

You are correct, all the others have it wrong.

madgeezerOption: 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)

AlandtOption: 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.

sid0308Option: C

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

gilofernandesOption: C

150*2+2=302

Ak_2020Option: C

C is correct as total number of app is 150

kanishka_Option: A

A is correct

sanni27Option: D

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