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Ch1:
Explain how operating system works as a Resource
manager with a suitable diagram:
Answer:
OS manager all resource:
- process management
Memory management -
- I\O management
- File management
Evolution of operating system:
1) simple batch system:
-batch job together
- job read via punch card
2)Multiprogramming OS:
- when one job needs to wait for I\O the C.P.U can switch to
other job.
Program A :
Program B :
Combined :
3)Time Sharing: 1980 – now
C.P.U time is shared among multiple job.
Multiple user access the system .
Examble:
In time sharing multiple user access system simultaneously.
Wait Wait RUN RUN
Wait RUN Wait RUN
Wait RUN A
A
RUN B
Q:what is system call: write the types of system call:
Answer:
system call is : interface between process and operating system.
Types of system call:
- Process control.
- Device management.
- File management.
- Communication.
Q:What is kernel:
Answer : portion of operating system that is in main
memory (RAM).
It is also called "Nucleus".
Example: kernel: is also called "Nucleus"
Memory types:
- Ram
- Rom
- Cache
"The end ch1 "
RAM Cache CPU
Ch2: " File Management System"
Q1: What is file , write the properties of file.?
Answer:
File is the central element of an application.
Properties of file:
- Long term existence.
- Sharable.
- Structure.
Q2:Write the operations on file: also on directory
- Create.
- Delete.
- Modify.
- Open.
- Read.
- Close.
Q3: Explain structure of file:
Field: basic element of data.
File: collection of similar records.
Record: collection of related fields.
Database: collection of related files.
FILE
file record database field
Example:
Q4: What is directory.? Write the information of directory?
Answer:
Directory: is a file owned by operating system.
Information of directory:
File name - File types - File organization
Q5:What are address information of directory.?
Answer:
- Volume
- Size used
- Size allocated
Q6:What are the access control information of directory.?
Answer:
- Owner: grant / deny access to user
- Access information: include username and password.
- Action’s: reading , writing , sharable
STUDENT
Name Address ID Course
ALI
HMAD
20110012
20104098
Jazan
abha
OS
OS
Records
Q7:Explain file organization and its types.?
Answer:
- File management refers to logical structure of file.
- A way to access the file.
Types of file organization:
- Pile
- Indexed file
- Sequential file
- Indexed sequential file
- Direct file
Pile:
- Data are collection in the order they arrive.
- N0 structure.
- Search is difficult.
Indexed file:
- May contain a partial index
Sequential file:
- Fixed format used for record.
- Record are stored in key sequence.
Index sequential file :
- Maintain the key of sequential of file.
- It is random accessed.
Direct or hashed file:
- Directly access a block at known address.
- Key field required for each record.
Q8:Describe UNIX file management.?
Answer:
*UNIX have 6 types of files:
- Regular.
- Directory.
- Special.
- Named pipes.
- Links.
- Symbolic links.
Q9: What are the features of NTFS(windows file system).?
Answer:
- Recoverable.
- Security.
- Large disk or large file.
- Compression and encryption.
Q10:What is block.? Write the common approaches.?
Answer:
Block are the unit for I\O with secondary storage.
*Approaches:
- Fixed length blocking.
- Variable length spanned blocking.
- Variable length un spanned blocking.
Discuss file allocation method.
Answer:
There are three methods:
Contiguous: single set of block is allocated to a file at the
time of creation.
Chained: allocation on basis of individual block.
Indexed: file allocation table contains a separate one level
index for each file.
"The end ch2 "
Ch31 : " process management".
Process: A process is a program in execution .
A instance of program running in computer.
What are the process element:
- Program code.
- Data.
- Attributes.
Q1:When process is running , What are the elements.?
Answer:
- Identifier
- State
- Priority
- Program counter
- Memory pointer
- I\O status information
- Accounting information
Q2:What is PCB "Process Control Block".?
- PCB is data structure to store the process elements.
- Create and manage by OS.
Diagram " process control block ".>>>
process program CPU
Q3: PCB Used to store the process element.
Two state process model:
- Running
- Not running
Q4:What is dispatcher.?
Dispatcher is a program which is used to switch the process.
Q5:Explain the five state process model.?
Diagram " Five state process model "
Five state:
1) New : creation of process.
2) Ready : is in process.
3) Running : is in CPU or memory.
4) Blocked : suspended.
5) 5) Exit
Dispatcher
ENTER
Pause
Not Running Running
EXIT
Admit Release Dispatch New Ready Running Exit
Blocked
Time out
Event
Wait
Event
Occurs
Q6:What are reason for process suspension.?
- Swapping
- Timing
- Other OS reason
- Parent process request
Q7:Process Table.?
Answer:
Current state
Process ID
Location in memory
Q8: What are the process attributes.?
- Process ID.
- Process state information.
- Process control information.
Q9:What are the role of PCB.?
- It defines the state of the OS.
- It requires protection.
Q10:Define UNIX process.?
A process in UNIX is a state of data structure that provide
OS with all the information to manage.
- User level context.
- Register level context.
- System level context.
Ch32: " Thread "
Q1:What is thread:
- A thread is a light weight process.
- It provides a way to improve application performance
through parallelism.
Stack
Data
Code
Q2:Explain user level thread (ULT).?
Answer:
Diagram " U.L.T "
Stack
Data
Code
Thread
Multithread Single thread
Q3:What are the difference between thread and process.?
Thread
Process
1) It is light weight. 1) I t is heavy weight.
2) Thread switching doesn't
need interaction with OS.
2) Process switching needs
interaction with OS.
3) Use less resources. 3) Use more resources.
Q4:What are advantages of threads.?
- Minimize context switching time.
- Provide concurrency within a process.
- Efficient communication.
- More economical to create context switch thread.
*types of thread:
- User level threads (ULT) – user manage thread.
- Kernel level threads (KLT) – OS manage thread.
Q5:Advantage of ULT (user level thread).?
- Doesn't require kernel mode permission.
- Run on any OS.
- Fast to create and manage.
Q6:Explain multithreading model.?
Some OS manage more ULT and KLT.
Example : solaris
1) Many to many.
- User thread is equal or smaller than kernel thread.
User space
Kernel space
thread thread
CPU CPU
CPU
CPU
CPU
CPU
2) Many to one model:
- Map many user level thread to one kernel thread .
- Management is done in user space .
- Only one thread can access the kernel at time.
Example "in exam "
- In many to one mode only one thread can access the
kernel at a time.
3) One to one.
CPU CPU CPU
Kernel space
User space
thread thread
User space
Kernel space
CPU CPU
thread thread
- One ULT and K LT.
CPU
Q4:What are the difference between ULT and KLT.?
ULT
KLT
1) Faster to create and manage
1) Slower to create and manage
2) Implementation is by user
level.
2) OS create KLT.
3) Run on any OS
3) Specific to OS.
4) Multithread application
can not take advantage of
multiprocessing.
Kernel routine can be
multithreaded.
Q5:Draw the diagram of LINUX process thread model.?
Answer:
Diagram " UNIX process model "
"The end ch3
Ch4: " scheduling "
Q1:What is scheduling.?
Scheduling is: the process which allows to utilization the
CPU until it completes and other process to hold or wait.
scheduling
non preempting keep
CPU until it completes
Example :
1) Non preemitive:
- FCFJ
- SJF
2) Preemitive :
- Round robin
- Priority scheduling
Preempting release CPU
even thought it is neither
completed nor blocked
Scheduling Algorithm:
1-FCFS First Come First Serve.
*grant chart:
P1 P2 P3 P4
0 40 23 13 19
Waiting time of P1=0
Waiting time of P2=40
Waiting time of P3=40+23=63
Waiting time of P4=63+13=76
*Average waiting time= 0+40+63+76 / 4
= 179 / 4 = 44.75 m\sec
Process Burst time
P1 40
P2 23
P3 13
P4 19
Q2: Assume that there are three process P1,P2,P3, that
arrive at CPU for processing .Each process has its own burst
time as shown in table draw grant chart and calculate the
average waiting time in FCFS.?
Process Burst time
P1 24
P2 3
P3 3
*Grant chart:
P1 P2 P3
0 24 3 3
Waiting time of process P1=0
Waiting time of process P2=24
Waiting time of process P3=24+3= 27
*Average waiting time= 0+24+27 / 3
= 51/3 = 17 m/sec
- FCFS: In this scheduling CPU is assigned the process in
order " First Come First Serve".
- Grant chare : grant chart is a bar chart that shown start
and finish time of each process.
2- SJF Shortest Job First
Process Burst time
P1 40
P2 23
P3 13
P4 19
*Grant chart:
40 23 19 13 0
Waiting time of P1=13+19+23=55
Waiting time of P2=13+19=32
Waiting time of P3=0
Waiting time of P4=13
*Average waiting time= 55+32+0+13 / 4
= 100 / 4 = 25 m/sec
P4 P3 P2 P1
Q3:Assume that three are four process P1,P2,P3,P4,that
arrive at CPU for processing . Each process has its own
burst time as shown in table. Calculate average waiting time
and grant chart in SJF.?
*Grant chart:
P4 P1 P3 P2
0 3 6 7 8
Waiting time of P1=3
Waiting time of P2=3+6+7=16
Waiting time of P3=3+6=9
Waiting time of P4=0
*Average waiting time = 3+16+9+0 / 4
= 28 /4 = 7 m/sec
SJF : In this scheduling , when CPU is available . it is assigned
to smallest process first.
Process Burst time
P1 6
P2 8
P3 7
P4 3
3- Priority scheduling :
Process Burst time Priority
P1 10 3
P2 1 1
P3 2 4
P4 1 5
P5 5 2
*Grant chart:
P2 P5 P1 P3 P4
0 1 5 10 2 1
Waiting time for P1=6
Waiting time for P2=0
Waiting time for P3=16
Waiting time for P4=18
Waiting time for P5=1
*Average waiting time = 6+0+16+18+1 / 5
= 41 / 5 = 8.2 m/sec
Q4:Explain priority scheduling.?
A priority is associated with each process and CPU is
allocated to the priority with highest priority.
4-Round Robin Time sharing
Time quantum= 4
Process Burst time
P1 24
P2 3
P3 3
*Grant chart:
P1 P2 P3 P1
0 4 4+3=7 7+3=10 10+20=30
Waiting time for P1=10-4=6
Waiting time for P2=4
Waiting time for P3=7
*Average waiting time = 6+4+7 / 3
= 17 / 3 = 5.6 m/sec
Calculate average waiting time in Round Robin for below.?
Time quantum= 2
Process Burst time
P1 24
P2 3
P3 3
*Grant chart:
P1 P2 P3 P1 P2 P3 P1
0 2 4 6 8 9 10 30
Waiting time for P1=(6-2)+(10-6)=8
Waiting time for P2=8-4=4
Waiting time for P3=9-6=3
*Average waiting time = 8+4+3 / 3
= 15 / 3 = 5 m/sec
Q5:Calculate average waiting time in Round Robin for
below.?
Time quantum= 2
Process Burst time
P1 8
P2 4
P3 9
P4 5
*Grant chart:
P1 P2 P3 P4 P1 P2 P3 P4 P1 P3 P4 P1 P3
0 2 4 6 8 10 12 14 16 18 20 21 23
Waiting time for P1= (8-2) + (16-8) + (21-16) =19
Waiting time for P2= (10-4) = 6
Waiting time for P3= (12-6) + (18-12) + (23-18) =17
Waiting time for P4= (14-8) + (20-14) = 12
*Average waiting time =19+6+17+12/ 4
= 54/4 = 13.5 m/sec
" The end ch4 "
26
Ch5: " Dead Lock "
Q1:Define Deadlock :
A process request resources , if the resource is not available
that time ,the process enter a wait state .This situation is
called deadlock .
Q2: What is the sequence in which resource may utilized .
Answer:
1) Request
2) Use
3) Release
Q3:What are the condition under which a deadlock in round
robin for below arise.?
- Mutual Exclusion.
- Hold and wait.
- Non preemption.
- Circular wait.
Q4:What are the methods for dealing deadlock problem.?
- Use a protocol that ensure there should be no deadlock
- Allow the system to recover from deadlock.
- Ignore the problem all together and pretend that
deadlock never happen.
Q5: What are the reasons for deadlock . Explain any two .
1) Mutual Exclusion.
Only one process at time can use resources
2) Hold and wait.
A process holding at least one resource is waiting to
acquire additional resource.
3) Non preemption.
A process can be released only until its completed.
4) Circular wait.
Q6: Explain resource allocation Graph ,with examples.
Resource allocation graph tracks which resources is held by
which process and which process is waiting for a resources it
is tool for detecting deadlock .
Process Resource
Examples:
R1
R1
2)
1) P1 is requesting resource R1
R1 P1
P1 is Using resource R1
P1 is requesting resource R2
R2 3)
P1 P2 P2 is Using resource R2
P1 is Using resource R1
P1
Q7: Assuming the operating system detects the system is
deadlocked .What can the OS do to recover from deadlock ?
1) Killing the deadlock process.
2) Rolling Back
3) Preemption
" The end ch5 "
Ch6: " Memory Management "
Q1 : What are the difference between logical and physical address?
Logical Physical
It is the address generated by CPU It is the address seen by Memory
Memory
Volatile example RAM
(Main Memory)
Non Volatile example ROM
(Permanent Memory)
Contiguous allocation Not Contiguous allocation
Q2 :What are the differences between Contiguous and
Not Contiguous allocation ?
Answer :
Contiguous Not Contiguous
A program occupies
contiguous block by
memory
A program occupies
several block and not
necessary to adjacent
block
Faster Slower
More waste in memory Less waste in memory
Q3 : Suppose that we have free segment of size 6,17,25,14
and 19 . Place a program with size 13 kb in free segment
using First Fit , Best Fit , Worst Fit .
Answer :
13 KB
P1
P2
P3
P1
P2
P3
P4
6 KB
17 KB
25 KB
14 KB
19 KB
First Fit
Worst Fit
Best Fit
Q4: Explain page placement Strategies :
Answer :
1) First Fit :
place in first hole that is big enough . (it is faster)
2) Best Fit :
Allocate the smallest hole that is big enough .
3) Worst Fit :
Allocate the largest hole .
In exam :
In ___________ strategy . Wastage of memory is more .
Q5 : What are the difference between segmentation and
paging ?
Answer :
Segmentation Paging
1- Variable size block in memory 1- Fixed size block
2- Address generate by CPU is
divided into segment number .
2- Address generated by CPU is
divided into page number .
3- Physical address = segment+base 3- Physical address =
page size*frame number+off set
Worst Fit
page replacement algorithm
1) FIFO
2) LRU
3) OPTIMAL
FIFO example :
Q6: Consider the following page reference using three
frames.Find the page fault using FIFO algorithm .
8,2,0,1,2,3,5,0,1,3
Answer:
8 8 8 1 1 1 1 0 0 0
2 2 2 2 3 3 3 1 1
0 0 0 0 5 5 5 3
Number of page fault = 9
1 2 3 4 X 5 6 7 8 9
LRU example 1 :
Q7: Consider the following page reference using four
frames.Find the page fault using LRU algorithm .
5,2,5,1,4,5,2,0,4,2,3,1,2,1,0,0,2,4,5,1
5 5 5 5 5 5 5 5 5 5 3 3 3 3 3 3 3 4 4 4
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2
1 1 1 1 0 0 0 0 1 1 1 1 1 1 1 5 5
4 4 4 4 4 4 4 4 4 4 0 0 0 0 0 1
Number of page fault = 11
LRU example 2 :
Q8: Consider the following page reference using three
frames.Find the page fault using LRU algorithm .
8,8,1,0,5,1,2,3,5,1,4,3,2
8 8 8 8 5 5 5 3 3 3 4 4 4
1 1 1 1 1 1 5 5 5 3 3
0 0 0 2 2 2 1 1 1 2
Number of page fault = 11
1 2 X 3 4 X X 5 X X 6 7 X X 8 X X 9 10 11
1 X 2 3 4 X 5 6 7 8 9 10 11
OPTIMAL example :
Q9: Consider the following page reference using three
frames.Find the page fault using OPTIMAL algorithm .
8,8,1,0,5,1,2,3,5,1,4,3,2
8 8 8 8 5 5 5 4 2
1 1 1 1 1 1 1
0 0 2 3 3 3
Number of page fault = 8
1 X 2 3 4 5 6 7 8