Sunday, December 9, 2012

.::VULMSIT::.eNoxel.com CS604 MID TERM SUBJECTIVE WITH REFERENCE SOLVED BY UMAR SAULAT

CS604- OPERATING SYSTEM

SOLVED SUBJECTIVE FOR MID TERM EXAM

 

 QNo.1 List and define the different metrics by which might evaluate a scheduler (List at least 4). 5 marks

Answer:-

When evaluating a scheduler's performance, utilization and Throughput are traded off for better Response Time. Response time is important for OS's that aim to be user-friendly.

List of Schedulers include:

  1. First-Come, First-Served  (FCFS)
  2. Round-Robin   (RR)
  3. Shortest-Job-First  (SJF)
  4. Priority Scheduling  (PS)

REF :: handouts Page No. 80

 

QNo.2   Write brief the multilevel feedback queue scheduling. 5 marks

Answer:-

Multilevel feedback queue scheduling allows a process to move between queues. The idea is to separate processes with different CPU burst characteristics. If a process uses too much CPU time, it will be moved to a lower-priority queue. This scheme leaves I/O bound and interactive processes in the higher-priority queues. Similarly a process that waits too long in a lower-priority queue may be moved o a higher priority queue. This form of aging prevents starvation.

 

In general, a multi-level feedback queue scheduler is defined by the following parameters:

  • Number of queues
  • Scheduling algorithm for each queue
  • Method used to determine when to upgrade a process to higher priority queue
  • Method used to determine when to demote a process
  • Method used to determine which queue a process enters when it needs service

REF :: handouts Page No. 89

 

Qno.3    Assumption made while formulating a solution to the critical section problem. 2 marks

Answer:-

A solution to the critical section problem must satisfy the following three requirements:

Assumptions made While formulating a solution, we must keep the following assumptions in mind:

  • Assume that each process executes at a nonzero speed
  • No assumption can be made regarding the relative speeds of the N processes.

REF :: handouts Page No. 98


Qno.4    There are many commands. Write the method through which these commands can communicate with each other. 3 marks

Answer:-

There are many commands some of these are given below:-

Command Name              $ pw

Details

You can display the absolute pathname of your working directory with the pwd command

Syntax

/home/students/nadeem/courses/cs604/programs

 

Command Name              cp

Details

You can use the cp command for copying files. You can use the cp file1

file2 command to copy file1 to file2.

Syntax

 cp ~/file1 ~/memos/file2

 

Command Name              mv

Details

You can use the mv command for moving files. You can use the mv file1

file2 command to move file1 to file2.

Syntax

$ mv ~/file1 ~/memos/file2

 

Command Name              rm

Details

You can use the rm command to remove files. You can use the rm file1

command to remove file1

Syntax

$ rm ~/courses/cs604/programs/test.c $               For single file

rm *.o                                                                                   For all files                        

REF :: handouts Page No. 27

 

Qno.5. Write Difference between SJF and Shortest Remaining Time First Scheduling algorithm. 3 marks

Answer: 

Shorted Job First (SJF) Scheduling

Shortest Remaining Time First Scheduling

For long term scheduling in a batch system, we can use as the length the process time limit that a user specifies when he submits the job

Whenever a new job comes in, check the remaining service time on the

current job.

Preemptive SJF scheduling is sometimes called shortest-remaining-time-first scheduling.

For all but the longest jobs, SRT better than SJF

The response ratio is good (Fast)

The response ratio is good (low)

Waiting time is fast

Waiting time is also quite low for most processes

REF :: handouts Page No. 82

Qno6.    Write formula for calculating waiting time in preemptive Shortest Job First Algorithm. 2 marks

Answer:-

Preemptive SJF scheduling is sometimes called shortest-remaining-time-first scheduling.

We illustrate the working of the SJF algorithm by using the following system state.

 

Process       Arrival Time   Burst Time

P1           0.0          7

P2           2.0          4

P3           4.0          1

P4           5.0          4

REF :: handouts Page No. 83

 

Qno7.    Define race condition and how prevent this condition. 2 marks

Answer:-

A situation like this, where several processes access and manipulate the same data
concurrently and the outcome of the manipulation depends on the particular order in
which the access takes place, is called a race condition

REF :: handouts Page No. 97

 

Qno.8    What is Convoy Effect?
Answer:-

When FCFS scheduling algorithm is used, the convoy effect occurs when short

Processes wait behind a long process to use the CPU and enter the ready queue in a

Convoy after completing their I/O.

REF :: handouts Page No. 82


Qno.9    What are the common data structures in Bakery Algorithm?
Answer:-

The bakery algorithm is due to Leslie Lamport and is based on a scheduling algorithm

commonly used in bakeries, ice-cream stores.

REF :: handouts Page No. 103

 

Qno.10 How a pipe can be created?
Answer:-

The pipe() system call   creates a   pipe and returns two file descriptors, one for reading and second for writing,

REF :: handouts Page No. 103

 

Qno.11 Define Progress and Bounded Waiting.
Answer:-

Process:-

If no process is executing in its critical section and some processes wish to enter their critical sections, then only those processes that are not executing in their remainder section can participate in the decision on which will enter its critical section next, and this selection cannot be postponed indefinitely.

 

 

Bounded Waiting:-

There exists a bound on the number of times that other processes are allowed to enter their critical sections after a process has made a request to enter its critical section and before that request is granted.

REF :: handouts Page No. 99

 

Qno.12 What is Starvation and how it occures
Answer:-

Starvation:-

A process that is ready to run but lacking the CPU can be considered blocked-waiting for the CPU.

How it occur:-  

If a deadlock occurs, it can be resolved if one car backs up (preempt resources
and rollback). Several cars may have to be backed up if a deadlock occurs. Starvation is
possible.

REF :: handouts Page No. 129

 


Qno.13 What are the advantages of Round Robin Scheduling Algorithm?
Answer:-

  • If time slice is too short then you will have freq switching btw processes
  • if time slice is too long then you will have less switching btw processes and high priority may have to wait for low priority tasks leading to starvation
  • The average waiting time under the RR policy however is often quite long.
  • The CPU scheduler goes around the ready queue, allocating the CPU to each process for a time interval of up to 1 time quantum
  • The process may have a CPU burst of less than 1 time quantum, in which case the process itself will release the CPU voluntarily.
  • It is the most simple scheduling algorithm
  • It is easy to implement in software
  • If the processes are of varied length then it becomes slow.

REF :: handouts Page No. 86


Qno.14 Analyze the following algorithm to solve the critical section problem and explain whether it satisfies the Mutual Exclusion Characteristic

Flag[i] = True;
Turn = j;
do{
while(Flag[j] = True && turn==j);
critical section
Flag[i] = False;
remainder section
} While(1)


Answer:-

 

To enter its critical section, Pi sets flag[i] to true, and sets 'turn' to j, asserting that if the other process wishes to enter its critical section, it may do so. If both try to enter at the same time, they will attempt to set 'turn' to i and j. However, only one of these assignments will last, the other will occur but be overwritten instantly. Hence, the eventual value of 'turn' will decide which process gets to enter its critical section.

 

To prove mutual exclusion, note that Pi enters its critical section only if either

flag[j]=false or turn=i. Also, if both processes were executing in their critical sections at
the same time, then flag[0]= = flag[1]= = true. These two observations suggest that P0 and
P1 could not have found both conditions in the while statement true at the same time,
since the value of 'turn' can either be 0 or 1. Hence only one process say P0 must have
successfully exited the while statement.  Hence mutual exclusion is preserved.

To prove bounded wait and progress requirements, we note that a process Pi can be
prevented the critical section only if it is stuck in the while loop with the condition
flag[j]= =true and turn=j. If Pj is not ready to enter the critical section, then flag[j]=flase
and Pi can enter its critical section. If Pj has set flag[j]=true and is also executing its while
statement then either turn=i or turn=j. If turn=i then Pi enters its critical section, otherwise
Pj. However, whenever a process finishes executing in its critical section, lets assume Pj,
it resets flag[j] to false allowing Pi to enter its critical section. If Pj resets flag[j]=true, then
it must also set 'turn' to i, and since Pi does not change the value of 'turn' while
executing in its while statement, Pi will enter its critical section (progress) after at most
one entry by Pj (bounded waiting).

REF :: handouts Page No. 145

 

Qno.15 which command is used in Linux environment to get process information? (2)

Answer:-

ps gives a snapshot of the current processes. Without options, ps prints information
about processes owned by the user. Some of the commonly used options are -u, -e, and
-l.

  • -e selects all processes
  • -l formats the output in the long format
  • -u displays the information in user-oriented format

REF :: handouts Page No. 63

 

Qno.16 resource sharing is disadvantage of multithreading....explain?(5)

Answer:-

1.  Resource sharing: Whereas resource sharing is one of the major advantages of
                threads, it is also a disadvantage because proper synchronization is needed
                between threads for accessing the shared resources (e.g., data and files).

2.  Difficult programming model: It is difficult to write, debug, and maintain multi-
                threaded programs for an average user. This is particularly true when it comes to
                writing code for synchronized access to shared resources.

REF :: handouts Page No. 70

Qno.17 Synchronization process is needed in resource sharing and data sharing (3)

Answer:-

Often access to shared data and shared resources, if there is no controlled access to shared data, it is often possible to obtain an inconsistent state of this data. Maintaining data consistency requires mechanisms to ensure the orderly execution of cooperating processes, and hence various process synchronization methods are used.

REF :: handouts Page No. 70


Qno.17 how threads overcome these issues (5)

Answer:-

A thread, sometimes called a lightweight process (LWP), is a basic unit of CPU utilization and executes within the address space of the process that creates it. It comprises a thread ID, a program counter, a register set, errno, and a stack. It shares with other threads belonging to the same process its code sections, data section, current working directory, user and group IDs, signal setup and handlers, PCB and other operating system resources, such as open files and system.

REF :: handouts Page No. 68

 

Qno.18 If process in background then we want its move in foreground then what unix/inux command is use to moving. 3 marks

Answer:-

Moving a process into background

You can use the bg command to put the current or a suspended process into the background. Here is the syntax of the command.

bg [%job_id]

If %job_id is omitted the current job is assumed.

REF :: handouts Page No. 65


Qno.19 How the open source help us to test the algorithm?

Answer:-

The Open Source software licensing has made it possible for us to test various algorithms by implementing them in the Linux kernel and measuring their true performance. The major difficulty is the cost of this approach. The expense is incurred in coding the algorithm and modifying the operating system to support it, as well as its required data structures. The other difficulty with any algorithm evaluation is that the environment in which the algorithm will change.
REF :: handouts Page No. 95

 

 

Qno.20                 Using C for compiling means?

Answer:-

To compile a source file titled program.c, type:
$ gcc program.c

You can run the executable program generated by this command by typing./a.out and
hitting the <Enter> key, as shown in the following session.
$ ./a.out
[ ... program output ... ]

You can store the executable program in a specific file by using the –o option. For
example, in the following session, the executable program is stored in the assignment file.
$ gcc program.c -o assignment

 

You can link a library                   
explicitly by using the –l option. In the following session, we are asking the compiler to
link the math library with our object file as it creates the executable file.
$ gcc program.c –o assignment -lm

$ assignment

[ ... program output ... ]

REF :: handouts Page No. 28


Qno.21 what is difference b/w preemptive and non-preemptive (2)

Answer:-

Preemptive

non-preemptive

in preemptive scheduling we preempt the currently executing process, in non preemptive scheduling we allow the current process to finish its CPU burst time 

in non preemptive scheduling the process at running state can not be forced to leave the CPU until it completes On a preemptive kernel, a process running in kernel mode can be replaced by another process while

in preemptive scheduling the process is forcibly sent to waiting state when a process with higher priority comes to CPU, 

When scheduling takes place only
under
:-

  • When a process switches from the running state to the waiting state (for example,
    an I/O request is being completed)
  • When a process switches from the running state to the ready state (for example
    when an interrupt occurs)
  • When a process switches from the waiting state to the ready state (for example,
    completion of I/O)
  • When a process terminates

 

REF :: handouts Page No. 80

 

Qno.22 Virtual machine V scheduling algorithm different steps

Answer:-

UNIX System V scheduling algorithm is essentially a multilevel feedback priority queues algorithm with round robin within each queue, the quantum being equal to1 second. The priorities are divided into two groups/bands:

  Kernel Group

  User Group

Priorities in the Kernel Group are assigned in a manner to minimize bottlenecks, i.e, processes waiting in a lower-level routine get higher priorities than those waiting at relatively higher-level routines. We discuss this issue in detail in the lecture with an example. In decreasing order of priority, the kernel bands are:

  Swapper

  Block I/O device control processes
  File manipulation

  Character I/O device control processes   User processes

REF :: handouts Page No. 91

 

Qno.23 If a process exits and there are still threads of that process running, will they continue to run?
Answer:-
i
f the thread in the process is running and receives a signal(say Ctrl-C) and the default action of the signal is to terminate a process, does the running thread terminates or the parent process will also terminate. That happens to the threads if the running process terminates because of some signal

REF :: http://stackoverflow.com/questions/3754018/threads-some-questions


Qno.24 What are the important characteristics of TestAndSet? What will be its advantage.
The important characteristic is that this instruction is executed atomically. Thus if two TestAndSet instructions are executed simultaneously, they will be executed sequentially in some arbitrary order.

Advantages :-

  • Simple to implement and understand
  • One memory location for arbitrarily many CPUs

REF :: handouts Page No. 106

 

Qno.25 Considering the Resource sharing feature of thread, what do you think is 'resource sharing' an advantage of a thread or disadvantage of a thread. Explain yours answer briefly.

Answer:-
The Advantages and Disadvantages of Threads

Four main advantages of threads are:

1.  Responsiveness:  Multithreading an interactive application may allow a program
                to continue running even if part of it is blocked or is performing a lengthy
                operation, thereby increasing responsiveness to the user.

2.  Resource sharing: By default, threads share the memory and the resources of the
                process to which they belong. Code sharing allows an application to have several
                different threads of activity all within the same address space.

3.  Economy: Allocating memory and resources for process creation is costly.
                Alternatively, because threads share resources of the process to which they
                belong, it is more economical to create and context switch threads.

4.  Utilization of multiprocessor architectures: The benefits of multithreading of

multithreading can be greatly increased in a multiprocessor environment, where each thread may be running in parallel on a different processor. A single threaded process can only run on one CPU no matter how many are available.

Multithreading on multi-CPU machines increases concurrency.

Some of the main disadvantages of threads are:

1.  Resource sharing: Whereas resource sharing is one of the major advantages of
                threads, it is also a disadvantage because proper synchronization is needed
                between threads for accessing the shared resources (e.g., data and files).

2.  Difficult programming model: It is difficult to write, debug, and maintain multi-
                threaded programs for an average user. This is particularly true when it comes to
                writing code for synchronized access to shared resources.

REF :: handouts Page No. 70

 


--
Zindagi mein 2 Logo ka buhat khayal rahkoooo
Ist woh jiss ney tumhari jeet ke Liye buhat kuch hara hoo
(Father)
2nd woh jiss ko tum ney har dukh me pukaara hoo (Mother)
Regards,
Umair Saulat Mc100403250

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