FinalTerm Examination
Spring 2010
CS604 – Operating Systems
Question No: 1 ( M a r k s: 1 ) http://vuzs.net
You can display the
contents (names of files and directories) of a directory in UNIX/Linux
directory structure with the --------------- command.
► l
► s
► ls
► none of the given options
Question No: 2 ( M a r k s: 1 ) http://vuzs.net
------------- spend more time doing IO than computations
► short CPU bursts
► CPU bound processes
► IO bound processes
► None of the given options
Question No: 3 ( M a r k s: 1 ) http://vuzs.net
------------buffer places no practical limit on the size of the buffer
► Bounded
► Unbounded
► Both Unbounded & bounded
► Bounded
► Bounded
Unbounded
None of the given options
Question No: 4 ( M a r k s: 1 ) http://vuzs.net
With ----------you use condition variables.
► Semaphores
► Read/Write Locks
► Swaps
► Monitor
Question No: 5 ( M a r k s: 1 ) http://vuzs.net
Deadlocks can be described more precisely in terms of a directed graph called a system ----------
► Directed graph
► Critical path
► Resource allocation graph
► Mixed graph
Question No: 6 ( M a r k s: 1 ) http://vuzs.net
Addresses generated relative to part of program, not to start of physical memory are
► Virtual
► Physical
► Relocatable
► Symbolic
Question No: 7 ( M a r k s: 1 ) http://vuzs.net
Object files and libraries are combined by a ------------- program to produce the executable binary
► Compiler
► Linker
► Text editor
► Loader
Question No: 8 ( M a r k s: 1 ) http://vuzs.net
The set of all physical addresses corresponding to the logical addresses is a ----------------- of the process
► Physical address space
► Process address space
► None of the given options
► Logical address space
Question No: 9 ( M a r k s: 1 ) http://vuzs.net
---------------- indicates size of the page table
► translation look-aside buffers
► Page-table length register (PTLR)
► Page-table base register (PTBR)
► Page offset
Question No: 10 ( M a r k s: 1 ) http://vuzs.net
If validation bit is 0, it indicates a/an --------- state of segment.
► protected
► shared
► legal
► illegal
► shared
► legal
► illegal
Question No: 11 ( M a r k s: 1 ) http://vuzs.net
In ______ allocation scheme free frames are equally divided among processes
► Fixed Allocation
► Propotional Allocation
► Priority Allocation
► None of the given options
Question No: 12 ( M a r k s: 1 ) http://vuzs.net
________ is used to store data on secondary storage device, e.g., a source program(in C), an executable program.
► Block Special File
► Link File
► Ordinary File
► Directory
Question No: 13 ( M a r k s: 1 ) http://vuzs.net
The ___________ method requires each file to occupy a set of contiguous blocks on the disk.
► Contiguous Allocation
► Linked Allocation
► Indexed Allocation
► None of the given options
Question No: 14 ( M a r k s: 1 ) http://vuzs.net
The
basic purpose of ________________is to help the users to utilize the
hardware resources for completing different tasks in a simplified manner
► Operating system
► Application software
► All Software
► All of the given
Question No: 15 ( M a r k s: 1 ) http://vuzs.net
User mode can run the Privileged instructions
► True
► False
Question No: 16 ( M a r k s: 1 ) http://vuzs.net
_______________ wastes CPU cycles and hence is a problem in real multiprogramming system.
► Busy waiting
► Spinlock
► Critical section
► Mutex
Question No: 17 ( M a r k s: 1 ) http://vuzs.net
The __________________
requires that no reader will be kept waiting unless a writer has already
obtained permission to use the shared object.
► first readers-writers problem
► second readers-writers problem
► third readers-writers problem
► fourth readers-writers problem
Question No: 18 ( M a r k s: 1 ) http://vuzs.net
The process of holding at least one resource and waiting to acquire
additional resources that are currently being held by other processes is
known as _________________.
► Mutual exclusion
► Hold and wait
► No preemption
► Circular wait
Question No: 19 ( M a r k s: 1 ) http://vuzs.net
If a process continues to
fault, replacing pages, for which it then faults and brings back in
right away. This high paging activity is called ___________.
► paging
► thrashing
► page fault
► CPU utilization
Question No: 20 ( M a r k s: 1 ) http://vuzs.net
Banker’s algorithm is used for ________________
► Deadlock avoidance
► Deadlock detection
► Deadlock prevention
► Deadlock removal
Question No: 21 ( M a r k s: 1 ) http://vuzs.net
A program can not execute unless whole or necessary part of it resides in the main memory.
► True
► False
Question No: 22 ( M a r k s: 1 ) http://vuzs.net
The size of pages and frames are same in logical memory and physical memory respectively.
► True
► False
Question No: 23 ( M a r k s: 1 ) http://vuzs.net
Which command, Display permissions and some other attributes for prog1.c in your current directory?
► ls –l prog1.c
► ls –d prog1.c
► ls file prog1.c
► ls –l prog1.c /Directory
Question No: 24 ( M a r k s: 1 ) http://vuzs.net
In the C-Scan and C-Look
algorithms, when the disk head reverses its direction, it moves all the
way to the other end, without serving any requests, and then reverses
again and starts serving requests.
► True
► False
Question No: 25 ( M a r k s: 1 ) http://vuzs.net
In paged segmentation, we divide every segment in a process into __________ pages.
► Fixed size
► Variable size
Question No: 26 ( M a r k s: 1 ) http://vuzs.net
Intel 80386 used paged segmentation with _________ level paging.
► One
► Two
► Three
► Four
Question No: 27 ( M a r k s: 1 ) http://vuzs.net
The logical address of Intel 80386 is _________
► 36 bits
► 48 bits
► 64 bits
► 128 bits
Question No: 28 ( M a r k s: 1 ) http://vuzs.net
Following is NOT true about Virtual memory.
► Virtual memory help in executing bigger programs even greater in size that of main memory.
► Virtual
memory makes the processes to stuck when the collective size of all the
processes becomes greater than the size of main memory.
► Virtual memory also allows files and memory to be shared by several different processes through page sharing.
► Virtual memory makes the task of programming easier because the programmer need not worry about the amount of physical memory,
Question No: 29 ( M a r k s: 1 ) http://vuzs.net
The
Swap instruction which is the hardware solution to synchronization
problem does not satisfy the ________ condition, hence not considered to
be a good solution.
► Progress
► Bounded waiting
► Mutual exclusion
► None of the given
Question No: 30 ( M a r k s: 1 ) http://vuzs.net
The following requirement for solving critical section problem is known as ______________.
“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.”
► Progress
► Bounded Waiting
► Mutual Exclusion
► Critical Region
Question No: 31 ( M a r k s: 2 )
When a process is rolled out
of memory, it loses its ability to use the CPU (at least for a while).
Describe another situation where a process loses its ability to use the CPU, but where the process does not get rolled out.
Where the Infinite loop starts it also be the cause to suspend the CPU from working sitution to idle stat.
Question No: 32 ( M a r k s: 2 )
How can you achieve memory protection in paging?
memory protection
in paging is achieved by associating protection bits with each page.
these bits are associated with each page table entry and specify
protection on the corresponding page.
Question No: 33 ( M a r k s: 2 )
What is the basic function of pipe system call?
the pipe system call creates a pipe and returns two file descriptors, one for reading and second for writing.
Question No: 34 ( M a r k s: 2 )
Provide names of common file structures.
FAT32
NTFS
Question No: 35 ( M a r k s: 3 )
What is the structure of Two-Level Page Table?
Question No: 36 ( M a r k s: 3 )
Ignoring CPU overhead,
what are the three primary components of disk access time (the time from
the request of a sector to the time that it is available to the CPU).
Given a transfer of a single sector, rank them according to their
relative average cost.
Question No: 37 ( M a r k s: 3 )
Calculate the Average Waiting Time for four processes. Assume that Processes come in the sequence of P1,P2,P4, P3 at time=0 and scheduling algorithm applied is FCFS.
Processes CPU Burst
P1 4
P2 1
P3 3
P4 2
Question No: 38 ( M a r k s: 3 )
What do you think that why Main Memory is kept as a volatile memory and why not a permanent storage device?
In my view the
Main memory is volatile because it has to maintain current and pending
jobs to do till job done it kept data then transfers to the permanemt
storage device but the permanent stroage has to only store the data so
thats why data cant be on stack always all data cant be in working.
Question No: 39 ( M a r k s: 5 )
Briefly explain Thrashing as it pertains to main memory management.
A process is
thrashing if it is spending more time paging than executing. thrashing
result in severe performance problems: low cpu utilization, high disk
utilization, low utilization of other i/o devices.
Question No: 40 ( M a r k s: 5 )
Summarize the tradeoffs among simple arrays, trees, and hash tables as implementations of a page table.
