Spring 2010 CS604 Operating Systems Midterm Paper
Time: 60 min M a r k s: 40
Question No: 1 ( M a r k s: 1 )
The hardware mechanism that enables a device to notify CPU is called an -------------
► Interrupt
► Signal
► Trap
► Process
Question No: 2 ( M a r k s: 1 )
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: 3 ( M a r k s: 1 )
The -------------- system call suspends the calling process.
► fork
► wait
► exec
► exit
Question No: 4 ( M a r k s: 1 )
In ---------addressing, the recipient is not required to name the sender.
► Symmetric
► Asymmetric
► Both symmetric and asymmetric
► None of the given options
Question No: 5 ( M a r k s: 1 )
-------- command gives a snapshot of the current processes.
► ps
Question No: 3 ( M a r k s: 1 )
The -------------- system call suspends the calling process.
► fork
► wait
► exec
► exit
Question No: 4 ( M a r k s: 1 )
In ---------addressing, the recipient is not required to name the sender.
► Symmetric
► Asymmetric
► Both symmetric and asymmetric
► None of the given options
Question No: 5 ( M a r k s: 1 )
-------- command gives a snapshot of the current processes.
► ps
► top
► who
► ls
Question No: 6 ( M a r k s: 1 )
--------------command to resume the execution of a suspended job in the foreground
► fg
► who
► ls
Question No: 6 ( M a r k s: 1 )
--------------command to resume the execution of a suspended job in the foreground
► fg
► bg
► jobs
► kill
Question No: 7 ( M a r k s: 1 )
You can use the ------------- command to display the status of suspended and background processes
► fg
► bg
► jobs
► kill
Question No: 8 ( M a r k s: 1 )
You can terminate a foreground process by pressing --------------
► <Ctrl-A>
► <Ctrl-C>
► <Ctrl-Z>
► None of the given options
Question No: 9 ( M a r k s: 1 )
A time sharing system is
► Multi tasking
► Interactive
► All of these
► Multi user
Question No: 10 ( M a r k s: 1 )
The main characteristic of a Real time system is
► Efficiency
► Large Virtual Memory
► Large secondary storage device
► Usability
Question No: 11 ( M a r k s: 1 )
Shared libraries and kernel modules are stored in _____________ directory
► /bin
► /dev
► /boot
► /lib
Question No: 12 ( M a r k s: 1 )
__________ scheduler selects the process from the job pool and put them in main memory.
► Long term
► Short term
► Medium term
► Swapper
Question No: 13 ( M a r k s: 1 )
In indirect inter process communication, a sender ___ mention the name of the recipient.
► do
► do not
Question No: 14 ( M a r k s: 1 )
A ______________ is an integer variable that, apart from initialization is accessible only through two standard atomic operations: wait and signal.
► Semaphore
► Monitor
► Critical region
► Critical section
Question No: 15 ( M a r k s: 1 )
A semaphore that cause Busy-Waiting is termed as ___________.
► Spinlock
► Monitor
► Critical region
► Critical section
Question No: 16 ( M a r k s: 1 )
The execution of critical sections must NOT be mutually exclusive
► True
► False
Question No: 17 ( M a r k s: 1 )
The performance of Round Robin algorithm does NOT depends heavily on the size of the time quantum.
► True
► False
Question No: 18 ( M a r k s: 1 )
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: 19 ( M a r k s: 1 )
The critical section problem can be solved by the following except
► Software based solution
► Firmware based solution
► Operating system based solution
► Hardware based solution
Question No: 20 ( M a r k s: 1 )
___________ is also called Swapper.
► Swap space
► Medium term scheduler
► Short term scheduler
► Long term scheduler
Question No: 21 ( M a r k s: 2 )
Write the formula/ procedure for calculating the waiting time in preemptive Shortest Job First scheduling.
Question No: 22 ( M a r k s: 2 )
What are the common data structures in Bakery Algorithm?
Question No: 23 ( M a r k s: 3 )
If a process exits and there are still threads of that process running, will they continue to run?
Question No: 24 ( M a r k s: 3 )
What are the important characteristics of TestAndSet? What will be its advantage.
Question No: 25 ( M a r k s: 5 )
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.
Question No: 26 ( M a r k s: 5 )
Analyze the given algorithm proposed to solve the critical section problem. Identify the shortcomings of this algorithm.
do{
while(turn!=j);
critical section
turn=j;
remainder section
} while(1)
► jobs
► kill
Question No: 7 ( M a r k s: 1 )
You can use the ------------- command to display the status of suspended and background processes
► fg
► bg
► jobs
► kill
Question No: 8 ( M a r k s: 1 )
You can terminate a foreground process by pressing --------------
► <Ctrl-A>
► <Ctrl-C>
► <Ctrl-Z>
► None of the given options
Question No: 9 ( M a r k s: 1 )
A time sharing system is
► Multi tasking
► Interactive
► All of these
► Multi user
Question No: 10 ( M a r k s: 1 )
The main characteristic of a Real time system is
► Efficiency
► Large Virtual Memory
► Large secondary storage device
► Usability
Question No: 11 ( M a r k s: 1 )
Shared libraries and kernel modules are stored in _____________ directory
► /bin
► /dev
► /boot
► /lib
Question No: 12 ( M a r k s: 1 )
__________ scheduler selects the process from the job pool and put them in main memory.
► Long term
► Short term
► Medium term
► Swapper
Question No: 13 ( M a r k s: 1 )
In indirect inter process communication, a sender ___ mention the name of the recipient.
► do
► do not
Question No: 14 ( M a r k s: 1 )
A ______________ is an integer variable that, apart from initialization is accessible only through two standard atomic operations: wait and signal.
► Semaphore
► Monitor
► Critical region
► Critical section
Question No: 15 ( M a r k s: 1 )
A semaphore that cause Busy-Waiting is termed as ___________.
► Spinlock
► Monitor
► Critical region
► Critical section
Question No: 16 ( M a r k s: 1 )
The execution of critical sections must NOT be mutually exclusive
► True
► False
Question No: 17 ( M a r k s: 1 )
The performance of Round Robin algorithm does NOT depends heavily on the size of the time quantum.
► True
► False
Question No: 18 ( M a r k s: 1 )
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: 19 ( M a r k s: 1 )
The critical section problem can be solved by the following except
► Software based solution
► Firmware based solution
► Operating system based solution
► Hardware based solution
Question No: 20 ( M a r k s: 1 )
___________ is also called Swapper.
► Swap space
► Medium term scheduler
► Short term scheduler
► Long term scheduler
Question No: 21 ( M a r k s: 2 )
Write the formula/ procedure for calculating the waiting time in preemptive Shortest Job First scheduling.
Question No: 22 ( M a r k s: 2 )
What are the common data structures in Bakery Algorithm?
Question No: 23 ( M a r k s: 3 )
If a process exits and there are still threads of that process running, will they continue to run?
Question No: 24 ( M a r k s: 3 )
What are the important characteristics of TestAndSet? What will be its advantage.
Question No: 25 ( M a r k s: 5 )
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.
Question No: 26 ( M a r k s: 5 )
Analyze the given algorithm proposed to solve the critical section problem. Identify the shortcomings of this algorithm.
do{
while(turn!=j);
critical section
turn=j;
remainder section
} while(1)