Following questions have been asked in GATE CS 2007 exam.
1) Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively:
(A) 256 Mbyte, 19 bits
(B) 256 Mbyte, 28 bits
(C) 512 Mbyte, 20 bits
(D) 64 Gbyte, 28 bits
Capacity of the disk = 16 surfaces X 128 tracks X 256 sectors X 512 bytes = 256 Mbytes.
To calculate number of bits required to access a sector, we need to know total number of sectors. Total number of sectors = 16 surfaces X 128 tracks X 256 sectors = 2^19
So the number of bits required to access a sector is 19.
2) Group 1 contains some CPU scheduling algorithms and Group 2 contains some applications. Match entries in Group 1 to entries in Group 2.
Group I Group II (P) Gang Scheduling (1) Guaranteed Scheduling (Q) Rate Monotonic Scheduling (2) Real-time Scheduling (R) Fair Share Scheduling (3) Thread Scheduling
(A) P – 3 Q – 2 R – 1
(B) P – 1 Q – 2 R – 3
(C) P – 2 Q – 3 R – 1
(D) P – 1 Q – 3 R – 2
Gang scheduling for parallel systems that schedules related threads or processes to run simultaneously on different processors.
Rate monotonic scheduling is used in real-time operating systems with a static-priority scheduling class. The static priorities are assigned on the basis of the cycle duration of the job: the shorter the cycle duration is, the higher is the job’s priority.
Fair Share Scheduling is a scheduling strategy in which the CPU usage is equally distributed among system users or groups, as opposed to equal distribution among processes. It is also known as Guaranteed scheduling.
3) An operating system uses Shortest Remaining Time first (SRT) process scheduling algorithm. Consider the arrival times and execution times for the following processes:
Process Execution time Arrival time P1 20 0 P2 25 15 P3 10 30 P4 15 45
What is the total waiting time for process P2?
At time 0, P1 is the only process, P1 runs for 15 time units.
At time 15, P2 arrives, but P1 has the shortest remaining time. So P1 continues for 5 more time units.
At time 20, P2 is the only process. So it runs for 10 time units
At time 30, P3 is the shortest remaining time process. So it runs for 10 time units
At time 40, P2 runs as it is the only process. P2 runs for 5 time units.
At time 45, P3 arrives, but P2 has the shortest remaining time. So P2 continues for 10 more time units.
P2 completes its ececution at time 55
Total waiting time for P2 = Complition time - (Arrival time + Execution time) = 55 - (15 + 25) = 15
Please see GATE Corner for all previous year paper/solutions/explanations, syllabus, important dates, notes, etc.
Please write comments if you find any of the answers/explanations incorrect, or you want to share more information about the topics discussed above
- Operating Systems | Set 13
- Operating System | Process Synchronization | Introduction
- Operating Systems | Set 1
- Operating Systems | Set 11
- Operating System | Process Management | CPU Scheduling
- Computer Organization | Instruction Formats (Zero, One, Two and Three Address Instruction)
- Digital Logic | Implicants in K-Map
- Computer Network | Open shortest path first (OSPF) protocol States
- Compiler Design | Syntax Directed Definition
- Computer Network | Open shortest path first (OSPF) protocol fundamentals