Serverless vs container decision in AWS - Hands-On Comparison

Serverless vs Container Decision

📖 Scenario: You are working as a cloud architect for a small company. They want to deploy a new web application. You need to decide whether to use a serverless service like AWS Lambda or a container service like AWS Fargate. To help make this decision, you will create a simple data structure to compare the two options based on cost, scalability, and maintenance effort.

🎯 Goal: Create a dictionary with the comparison data, add a threshold for cost, filter the options based on the threshold, and finalize the decision by selecting the best option.

📋 What You'll Learn

Create a dictionary named deployment_options with exact keys and values for AWS Lambda and AWS Fargate

Add a variable named max_cost with the exact value 50

Use a dictionary comprehension named affordable_options to select options with cost less than max_cost

Add a variable named best_option that selects the option with the lowest maintenance effort from affordable_options

💡 Why This Matters

🌍 Real World

Cloud architects often need to compare different deployment methods to choose the best fit for cost, scalability, and maintenance.

💼 Career

Understanding how to structure and filter deployment options helps in making informed decisions for cloud infrastructure design.

1

Create deployment options dictionary

Create a dictionary called deployment_options with these exact entries: 'AWS Lambda': {'cost': 30, 'scalability': 9, 'maintenance': 2} and 'AWS Fargate': {'cost': 70, 'scalability': 8, 'maintenance': 5}.

AWS

# Create the deployment_options dictionary with AWS Lambda and AWS Fargate data
# Your code here

Need a hint?

Use a dictionary with keys 'AWS Lambda' and 'AWS Fargate'. Each key maps to another dictionary with keys 'cost', 'scalability', and 'maintenance'.

2

Add maximum cost threshold

Add a variable called max_cost and set it to the integer 50.

AWS

deployment_options = {
    'AWS Lambda': {'cost': 30, 'scalability': 9, 'maintenance': 2},
    'AWS Fargate': {'cost': 70, 'scalability': 8, 'maintenance': 5}
}
# Set the max_cost variable to 50
# Your code here

Need a hint?

Just create a variable named max_cost and assign it the value 50.

3

Filter affordable deployment options

Create a dictionary comprehension called affordable_options that includes only the entries from deployment_options where the 'cost' is less than max_cost.

AWS

deployment_options = {
    'AWS Lambda': {'cost': 30, 'scalability': 9, 'maintenance': 2},
    'AWS Fargate': {'cost': 70, 'scalability': 8, 'maintenance': 5}
}
max_cost = 50
# Create affordable_options dictionary with options costing less than max_cost
# Your code here

Need a hint?

Use a dictionary comprehension with for name, details in deployment_options.items() and filter with if details['cost'] < max_cost.

4

Select best deployment option

Add a variable called best_option that selects the key from affordable_options with the lowest 'maintenance' value.

AWS

deployment_options = {
    'AWS Lambda': {'cost': 30, 'scalability': 9, 'maintenance': 2},
    'AWS Fargate': {'cost': 70, 'scalability': 8, 'maintenance': 5}
}
max_cost = 50

affordable_options = {name: details for name, details in deployment_options.items() if details['cost'] < max_cost}
# Select best_option with lowest maintenance from affordable_options
# Your code here

Need a hint?

Use the min() function with a key argument to find the option with the lowest maintenance.