Test: Introduction, Process, Threads & CPU Scheduling- 1 - Computer Science Engineering (CSE) MCQ

Spool means simultaneous peri-pheral operations on line, a printer is a spooling device.

Concurrent processes are processes that share the CPU and memory. They do overlap in time while execution. At a time CPU entertain only one process but it can switch to other without completing it as a whole.

Real-time system are very fast and quick respondents systems. Response time of such systems is very low, to the tune of 10 ms or 100 ms or even less.

• Such systems are used where real-life scenario are being implemented like missile system, aircraft system extra.
• In process control system, the time quantum in reality is very less (to give a feel of multiprocessing so real time system is needed.

In time-sharing operating system (example in Round-Robin), whenever the time slot given to a process expires, it goes back to READY state and if it requests for same I/O operation, then it goes to BLOCKED state.

MS-DOS (Microsoft Disk Operating System) is a single-user, single-tasking computer operating system that uses a command line interface. In spite of its very small size and relative simplicity, it is one of the most successful operating systems that has been developed to date.

• Multiprogramming environment means processor is executing multiple processes simultaneously by continuously switching between one-another.
• Therefore, multiple processes should reside in memory. However, processor can't executes more than one process at a time.

When process in blocked state waiting for some I/O services, whenever the services is completed it goes in the ready queue of the ready state.

According to FCFS process solve are p1 p2 p3 p4 p5 so  

For p1 waiting time = 0 process time = 10 then 

For p2 waiting time = (process time of p1-arrival time of p2) = 10 - 0 = 10 then 

For p3 waiting time = (pr. time of (p1+p2) - arrival time of p3) = (10 + 5) - 2 = 13 and 

Same for p4 waiting time = 18 - 5 = 13 

Same for p5 waiting time = 38 - 10 = 28 

So total average waiting time = (0 + 10 + 13 + 13 + 28) / 5

= 12.8

Hence option A.