FCFS (First Come First Served) in Operating Systems - Time & Space Complexity

Time Complexity: FCFS (First Come First Served)

O(n^2)

Understanding Time Complexity

Analyzing time complexity helps us understand how the FCFS scheduling algorithm performs as the number of processes increases.

We want to know how the work done grows when more processes arrive.

Scenario Under Consideration

Analyze the time complexity of the following FCFS scheduling code snippet.

for each process in process_list:
    wait_time = sum of burst times of all previous processes
    turnaround_time = wait_time + current process burst time
    record wait_time and turnaround_time

This code calculates waiting and turnaround times for each process in the order they arrive.

Identify Repeating Operations

Look at what repeats as the input grows.

• Primary operation: For each process, summing burst times of all previous processes.
• How many times: For each of the n processes, it sums up to n-1 previous burst times.

How Execution Grows With Input

As the number of processes increases, the total work grows faster because each process sums more burst times.

Input Size (n)	Approx. Operations
10	About 55 sums (1+2+...+9)
100	About 4,950 sums
1000	About 499,500 sums

Pattern observation: The total operations grow roughly like the square of the number of processes.

Final Time Complexity

Time Complexity: O(n2)

This means the work needed grows quickly as more processes are added, because each process waits for all before it.

Common Mistake

[X] Wrong: "Calculating waiting times in FCFS is always fast and simple, so it must be O(n)."

[OK] Correct: Because each process's wait time depends on all previous processes, the total work adds up more than just once per process.

Interview Connect

Understanding how FCFS scales helps you explain scheduling choices clearly and shows you can think about how algorithms behave as systems grow.

Self-Check

"What if we stored cumulative burst times as we go instead of summing each time? How would the time complexity change?"

Practice

1. What does FCFS (First Come First Served) scheduling mean in operating systems?

easy

A. Processes with the shortest time are handled first.

B. Processes are handled in the order they arrive.

C. Processes are handled randomly.

D. Processes with the highest priority are handled first.

Solution

1. Step 1: Understand FCFS concept

   FCFS means tasks are processed in the order they come, like a queue.

2. Step 2: Compare options

   Only Processes are handled in the order they arrive. describes this order-based processing correctly.

3. Final Answer:

   Processes are handled in the order they arrive. -> Option B

4. Quick Check:

   FCFS = Order of arrival [OK]

Hint: Remember: FCFS is like waiting in line [OK]

Common Mistakes:

• Confusing FCFS with priority scheduling
• Thinking shortest tasks go first
• Assuming random order

2. Which of the following is the correct way to describe FCFS scheduling?

easy

A. Processes are scheduled in the order they arrive without preemption.

B. Processes are scheduled based on their priority levels.

C. Processes are scheduled by shortest remaining time first.

D. Processes are scheduled randomly to balance load.

Solution

1. Step 1: Identify FCFS scheduling traits

   FCFS schedules tasks in arrival order and does not interrupt running tasks.

2. Step 2: Match options to traits

   Processes are scheduled in the order they arrive without preemption. correctly states no preemption and order of arrival.

3. Final Answer:

   Processes are scheduled in the order they arrive without preemption. -> Option A

4. Quick Check:

   FCFS = Arrival order + no preemption [OK]

Hint: FCFS runs tasks fully in arrival order [OK]

Common Mistakes:

• Mixing FCFS with priority or shortest job scheduling
• Assuming tasks can be interrupted
• Thinking scheduling is random

3. Given three processes arriving at times 0, 2, and 4 with burst times 5, 3, and 1 respectively, what is the completion time of the second process using FCFS?

medium

A. 10

B. 5

C. 8

D. 9

Solution

1. Step 1: Calculate completion of first process

   Process 1 arrives at 0 and runs for 5 units, finishing at time 5.

2. Step 2: Calculate completion of second process

   Process 2 arrives at 2 but waits until process 1 finishes at 5, then runs for 3 units, finishing at 8.

3. Final Answer:

   8 -> Option C

4. Quick Check:

   Process 2 finishes at 5+3=8 [OK]

Hint: Add burst times in arrival order [OK]

Common Mistakes:

• Starting second process at its arrival time instead of after first finishes
• Adding arrival times incorrectly
• Ignoring waiting time

4. A student wrote this FCFS scheduling code but it gives wrong completion times. What is the likely error?

processes = [(0, 4), (1, 3), (2, 1)]  # (arrival, burst)
completion = []
current_time = 0
for arrival, burst in processes:
    if arrival > current_time:
        current_time = arrival
    completion.append(current_time)
    current_time += burst
print(completion)

medium

A. Not updating current_time after appending completion time.

B. Using a list instead of a queue for processes.

C. Not considering arrival time when updating current_time.

D. Appending completion time before updating current_time.

Solution

1. Step 1: Analyze code logic

   Completion time is appended before current_time is updated with burst time, causing wrong values.

2. Step 2: Identify correct order

   We must update current_time by adding burst before appending completion time to reflect actual finish time.

3. Final Answer:

   Appending completion time before updating current_time. -> Option D

4. Quick Check:

   Update time before recording completion [OK]

Hint: Update current time before saving completion [OK]

Common Mistakes:

• Appending completion time too early
• Ignoring arrival time adjustments
• Confusing process order

5. In an FCFS system, three processes arrive at times 0, 1, and 2 with burst times 4, 2, and 6. If the first process takes longer than expected and runs for 8 units instead of 4, how does this affect the waiting time of the third process?

hard

A. The third process waits longer because the first process delays the queue.

B. The third process waiting time remains the same.

C. The third process starts earlier due to the delay.

D. The third process is skipped and runs immediately.

Solution

1. Step 1: Understand FCFS impact of longer burst

   FCFS runs processes fully in arrival order, so a longer first process delays all others.

2. Step 2: Analyze waiting time effect on third process

   The third process must wait until the first and second finish, so longer first process increases its waiting time.

3. Final Answer:

   The third process waits longer because the first process delays the queue. -> Option A

4. Quick Check:

   Longer first task = longer wait for later tasks [OK]

Hint: Long first task delays all later tasks [OK]

Common Mistakes:

• Assuming later tasks start earlier
• Ignoring impact of burst time changes
• Thinking FCFS skips tasks