Cpu scheduling calculate with line, graphy, bar

CPU scheduling determines which process gets the CPU and for how long, aiming to optimize system performance. Here's how to calculate and visualize scheduling with an example. Let's use the following processes: | Process | Arrival Time (AT) | Burst Time (BT) | |:--------|:------------------|:----------------| | P1 | 0 | 5 | | P2 | 1 | 3 | | P3 | 2 | 2 | | P4 | 3 | 4 | We will calculate: Completion Time (CT): Time when a process finishes execution. Turnaround Time (TAT): $CT - AT$. Waiting Time (WT): $TAT - BT$. 1. First-Come, First-Served (FCFS) Processes are executed in the order they arrive. Gantt Chart: ` | P1 | P2 | P3 | P4 | 0 5 8 10 14 ` Calculations: P1: Starts at 0, Finishes at 5 $CT = 5$ $TAT = 5 - 0 = 5$ $WT = 5 - 5 = 0$ P2: Starts at 5, Finishes at $5 + 3 = 8$ $CT = 8$ $TAT = 8 - 1 = 7$ $WT = 7 - 3 = 4$ P3: Starts at 8, Finishes at $8 + 2 = 10$ $CT = 10$ $TAT = 10 - 2 = 8$ $WT = 8 - 2 = 6$ P4: Starts at 10, Finishes at $10 + 4 = 14$ $CT = 14$ $TAT = 14 - 3 = 11$ $WT = 11 - 4 = 7$ Average Metrics: $$ \text{Average Waiting Time} = \frac{0 + 4 + 6 + 7}{4} = \frac{17}{4} = \boxed{4.25} $$ $$ \text{Average Turnaround Time} = \frac{5 + 7 + 8 + 11}{4} = \frac{31}{4} = \boxed{7.75} $$ 2. Shortest Job First (SJF) - Non-Preemptive The process with the shortest burst time among the available processes is executed next. Gantt Chart: ` | P1 | P3 | P2 | P4 | 0 5 7 10 14 ` Calculations: P1: (Only process at AT=0) Starts at 0, Finishes at 5 $CT = 5$ $TAT = 5 - 0 = 5$ $WT = 5 - 5 = 0$ P3: (At time 5, P2, P3, P4 are available. P3 has BT=2, which is shortest) Starts at 5, Finishes at $5 + 2 = 7$ $CT = 7$ $TAT = 7 - 2 = 5$ $WT = 5 - 2 = 3$ P2: (At time 7, P2, P4 are available. P2 has BT=3, which is shortest) Starts at 7, Finishes at $7 + 3 = 10$ $CT = 10$ $TAT = 10 - 1 = 9$ $WT = 9 - 3 = 6$ P4: (At time 10, only P4 is available) Starts at 10