Operating Systems Exam for Cork Institute of Technology, Semester 1, 2009/10, Exams of Operating Systems

The exam paper for the operating systems module (soft 7006) of the bachelor of science in software development & computer networking – stage 2 and bachelor of science (honours) in software development – stage 3 programs at the cork institute of technology. The exam consists of five questions covering various topics such as process interruption, disk access speed, process scheduling, memory management, and deadlock.

Typology: Exams

2012/2013

Uploaded on 03/25/2013

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CORK INSTITUTE OF TECHNOLOGY
INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ
Semester 1 Examinations 2009/10
Module Title: Operating Systems
Module Code: SOFT 7006
School: Science
Programme Title: Bachelor of Science in Software Development & Computer
Networking Stage 2
Bachelor of Science (Honours) in Software Development – Stage 3
Programme Code: KDNET_8_Y2
KSDEV_8_Y3
KITSM_6_Y2
External Examiner(s): Mr Ken Carroll
Internal Examiner(s): Mr. Gerard McSweeney
Instructions: Answer FOUR questions. All questions carry equal marks.
Duration: 2 Hours
Sitting: Winter 2009
Requirements for this examination:
Note to Candidates: Please check the Programme Title and the Module Title to ensure that you have
received the correct examination paper.
If in doubt please contact an Invigilator.
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CORK INSTITUTE OF TECHNOLOGY

INSTITIÚID TEICNEOLAÍOCHTA CHORCAÍ

Semester 1 Examinations 2009/

Module Title: Operating Systems

Module Code: SOFT 7006

School: Science

Programme Title: Bachelor of Science in Software Development & Computer Networking – Stage 2Bachelor of Science (Honours) in Software Development – Stage 3

Programme Code: KDNET_8_Y KSDEV_8_Y KITSM_6_Y

External Examiner(s): Internal Examiner(s): Mr Ken CarrollMr. Gerard McSweeney

Instructions: Answer FOUR questions. All questions carry equal marks.

Duration: 2 Hours Sitting: Winter 2009

Requirements for this examination:

Note to Candidates: received the correct examination paper. Please check the Programme Title and the Module Title to ensure that you have If in doubt please contact an Invigilator.

(a) Write down an example of when a process might be interrupted. ( 2 Marks ) (b) What is an interrupt handler? ( 2 Marks ) (c) What is the difference between sequenced and nested interrupts? ( 2 Marks ) (d) Apart from normal completion, give five reasons why a process might be terminated. ( 5 Marks ) (e) With reference to processing, draw and label the seven-state transition table. ( 8 Marks ) (f) What is meant by a suspended state? ( 2 Marks ) (g) What is a Linux swap space? ( 2 Marks ) (h) What are the two forms of swap space? ( 2 Marks )

(a) With reference to process scheduling, distinguish between: Long term scheduling Medium term scheduling Short term scheduling ( 6 Marks )

(b) Explain the difference between a pre-emptive and a non pre emptive scheduling algorithm. ( 2 Marks ) (c) What is the purpose of the dispatcher? ( 2 Marks ) (d) Consider the following set of processes: Arrival Service P1 0 3 P2 2 4 P3 4 3 P4 6 2 Draw a time line to show the execution of each of these processes for the shortestprocess next scheduling algorithm. ( 4 Marks )

(e) Calculate the Turnaround Time and the Normalised Turnaround Time. ( 4 Marks ) (f) Explain what is meant by concurrency? ( 2 Marks ) (g) What are the requirements for Mutual Exclusion? ( 5 Marks )

(a) In memory management, explain what is meant by the logical address and thephysical address. ( 4 Marks )

(b) Paging is a form of memory address translation, name another. ( 1 Mark ) (c) With reference to paging, explain the terms frame and page. ( 2 Marks ) (d) With reference to paging, what is the offset? ( 1 Mark ) (e) If a logical address is 0001110101101001 and the page table is as follows: 0 1011 1 0010 2^1001 3 0111 4 1110 5 0100 Calculate the physical address. ( 4 Marks ) (f) How does disk caching work? ( 2 Marks ) (g) Name two disk caching algorithms. ( 2 Marks ) (h) What is the name of the standard Linux file system? ( 1 Mark ) (i) With the aid of a diagram, describe the UNIX block addressing scheme. ( 5 Marks ) (j) If a block can hold 256 addresses and each block is 1K, calculate the maximum filesize. ( 3 Mark )