CPU and I/O BURST Cycle in Operating System | OS Tutorials

In Operating System process execution consist of a cycle of CPU execution and I/O wait. Processes alternate between these two states. Process execution begins with a CPU burst, which is followed by an I/O burst, which in turn followed by another CPU burst, then another I/O burst and so on. Final CPU burst ends with a system request to terminate as shown in Figure 1.

The duration of CPU burst vary greatly from process to process and then from computer to computer. An I/O bound programs typically has many short CPU bursts. A CPU bound program might have a few long CPU bursts.

Example 1 – Consder the following set of process with the arrival time and CPU burst time given in milliseconds:

We need to determine Average Waiting time and Average turn around time with the preemptive SJF scheduling.

Solution – As per the given information the resluting preemptive SJF scheduling is depicted in the following Gantt chart:

As shown in above table process P₁ is started at time 0, since it is the only process in the queue. Process P₂ arrives at time 3. The remaining time for process P₁ (24.3 = 21 miliseconds) is larger than the time required by process P₂ (7 miliseconds), so process P₁ is preempted and process P₂ is scheduled.

Similarly, process P₃ arrives at time 5. But the remaining time for process P₂ (7-2 = 5 miliseconds) is smaller than the time required by process P₃(6 miliseconds), so P₂ will continue its execution. After P₂, the process P₃ is scheduled and then P₄ and then P₁ will continue its execution.

Average waiting time = $\frac{(26-3) + (3-3) + (10-5) + (16-10)}{4} = \frac{23+0+5+6}{4} = \frac{34}{4}$

AWT = 8.5 miliseconds

Average turn around time = Average Waiting Time + Average Execution Time = 8.5 + (47/4) = 20.25 miliseconds

Example 2 – Consider the following set of process with the arrival time and CPU burst time given in miliseconds.

What is Average waiting time for these. Process with the preemptie shorts job first scheduling?

Solution – Shortest-Job-First-Scheduling: A different approach to CPU scheduling is the Shortest – Job First Scheduling. This algorithm associates with each process the length of the process’s next CPU burst. When the CPU is available, it is assigned to the process that has the smallest next CPU burst. In the next CPU burst of two processes are the same, FCFS scheduling is used to break the tie. Note that more appropriate terms for this scheduling method would be the shortest next CPU – burst algorithm because scheduling depends on the length of the next CPU burst of a process, rather than its total length.

From above question SJFS is:

Average waiting time = $\frac{7+2+0+3}{4} = \frac{12}{4} = 3 \space milliseconds$