Question 1: Total Points (10)
Consider performance of FCFS algorithm for three computer-bound processes. If process P1 takes 25 seconds, P2 takes 4 seconds and P3 takes 5 seconds and processes arrive in the given order P1, P2, P3. You need to calculate the following:
a) Turnaround Time per process.
b) Average Turnaround time of processes

Ideas Solution Example

Scheduling Algorithms
The following subsections will explain several common scheduling strategies, looking at only a single CPU burst each for a small number of processes. Obviously real systems have to deal with a lot more simultaneous processes executing their CPU-I/O burst cycles.
6.3.1 First-Come First-Serve Scheduling, FCFS

FCFS is very simple - Just a FIFO queue, like customers waiting in line at the bank or the post office or at a copying machine.
Unfortunately, however, FCFS can yield some very long average wait times, particularly if the first process to get there takes a long time. For example, consider the following three processes:

Process Burst Time P1 24 P2 3 P3 3

In the first Gantt chart below, process P1 arrives first. The average waiting time for the three processes is ( 0 + 24 + 27 ) / 3 = 17.0 ms.
In the second Gantt chart below, the same three processes have an average wait time of ( 0 + 3 + 6 ) / 3 = 3.0 ms. The total run time for the three bursts is the same, but in the second case two of the three finish much quicker, and the other process is only delayed by a short amount.

FCFS can also block the system in a busy dynamic system in another way, known as the convoy effect. When one CPU intensive process blocks the CPU, a number of I/O intensive processes can get backed up behind it, leaving the I/O devices idle. When the CPU hog finally relinquishes the CPU, then the I/O processes pass through the CPU quickly, leaving the CPU idle while everyone queues up for I/O, and then the cycle repeats itself when the CPU intensive process gets back to the ready queue.