Thursday, 17 April 2014

Important Descriptive Question Part 1

Q1-Discuss in detail Table management Techniques?
Ans-An Assembler uses the following tables:
OPTAB: Operation Code Table Contains mnemonic operation code and its machine
language equivalent.
SYMTAB: Symbol Table maintains symbolic label, operand and their corresponding
LITTAB is a table of literals used in the program
For efficiency reasons SYMTAB must remain in main memory throughout passes I
and II of the assembler. LITTAB is not accessed as frequently as SYMTAB, however
it may be accessed sufficiently frequently to justify its presence in the memory. If
memory is at a premium, only a part of LITTAB can be kept in memory. OPTAB
should be in memory during pass I

Q2-What is parsing? Write down the drawback of top down parsing of backtracking?
Ans-Parsing is the process of analyzing a text, made of a sequence of tokens, to
determine its grammatical structure with respect to a given formal grammar. Parsing
is also known as syntactic analysis and parser is used for analyzing a text. The task
of the parser is essentially to determine if and how the input can be derived from the
start symbol of the grammar. The input is a valid input with respect to a given formal
grammar if it can be derived from the start symbol of the grammar.
Following are drawbacks of top down parsing of backtracking:
(i) Semantic actions cannot be performed while making a prediction. The actions
must be delayed until the prediction is known to be a part of a successful parse.
(ii) Precise error reporting is not possible. A mismatch merely triggers
backtracking. A source string is known to be erroneous only after all predictions
have failed.

Q3-Write down different system calls for performing different kinds of tasks.
Ans-A system call is a request made by any program to the operating system for
performing tasks -- picked from a predefined set -- which the said program does not
have required permissions to execute in its own flow of execution. System calls
provide the interface between a process and the operating system. Most operations
interacting with the system require permissions not available to a user level process,
e.g. I/O performed with a device present on the system or any form of communication
with other processes requires the use of system calls.
The main types of system calls are as follows:
Process Control: These types of system calls are used to control the processes.
Some examples are end, abort, load, execute, create process, terminate process etc.
File Management: These types of system calls are used to manage files. Some
examples are Create file, delete file, open, close, read, write etc.
Device Management: These types of system calls are used to manage devices.
Some examples are Request device, release device, read, write, get device attributes

Q4-Differentiate between pre-emptive and non-pre-emptive scheduling?
Ans-In a pre-emptive scheduling approach, CPU can be taken away from a process if
there is a need while in a non-pre-emptive approach if once a process has been
given the CPU, the CPU cannot be taken away from that process, unless the
process completes or leaves the CPU for performing an Input Output.
Pre-emptive scheduling is more useful in high priority process which requires
immediate response, for example in real time system. While in nonpreemptive
systems, jobs are made to wait by longer jobs, but treatment of all processes is

Q5- What is a semaphore? Explain busy waiting semaphores.
Ans- A semaphore is a protected variable or abstract data type which constitutes the
classic method for restricting access to shared resources such as shared memory in a
parallel programming environment.
Weak, Busy-wait Semaphores:
? The simplest way to implement semaphores.
? Useful when critical sections last for a short time, or we have lots of CPUs.
? S initialized to positive value (to allow someone in at the beginning).
? S is an integer variable that, apart from initialization, can only be accessed through 2
atomic and mutually exclusive operations:
while (s.value != 0);
All happens atomically i.e. wrap pre and post protocols.

Q6- What are the four necessary conditions of deadlock prevention?
Ans-Four necessary conditions for deadlock prevention:
1. Removing the mutual exclusion condition means that no process may have
exclusive access to a resource. This proves impossible for resources that cannot be
spooled, and even with spooled resources deadlock could still occur. Algorithms that
avoid mutual exclusion are called non-blocking synchronization algorithms.
2. The "hold and wait" conditions may be removed by requiring processes to request
all the resources they will need before starting up. Another way is to require processes
to release all their resources before requesting all the resources they will need.
3. A "no preemption" (lockout) condition may also be difficult or impossible to avoid
as a process has to be able to have a resource for a certain amount of time, or the
processing outcome may be inconsistent or thrashing may occur. However, inability
to enforce preemption may interfere with a priority algorithm. Algorithms that allow
preemption include lock-free and wait-free algorithms and optimistic concurrency
4. The circular wait condition: Algorithms that avoid circular waits include "disable
interrupts during critical sections", and "use a hierarchy to determine a partial
ordering of resources" and Dijkstra's solution.

Q7- Draw an expression tree for the string.
f + (x+y) *((a+b)/(c-d))
Indicate the register requirement for each node and list out the evaluation order for
the expression tree.
Ans- An expression tree for the string “f + (x+y) *((a+b)/(c-d))” is given below:
Maximun register requirement is 2.

Q8-fork-join are primitives in a higher level programming language for implementing interacting processes. 

The syntax is as follows:
join ;

variable. A statement fork label1 causes creation of a new process that starts
executing at the statement with the label label1. This process is concurrent with the
process which executed the statement fork label1.A join statement synchronizes the
birth of a process with the termination of one or more processes.
Fork-Join provide a functionally complete facility for control synchronization.

Q9- List and explain the three events concerning resource allocation. Define the
(i) Deadlock.
(ii) Resource request and allocation graph (RRAG)
(iii)Wait for graph (WFG)
Ans-Deadlock: Each process in a set of processes is waiting for an event that only
a process in the set can cause.
(ii) Deadlocks can be described by a directed bipartite graph called a Resource-
Request-Allocation graph (RRAG).A graph G = (V,E) is called bipartite if V
can be decomposed into two disjoint sets V1 and V2 such that every edge in E
joins a vertex in V1 to a vertex in V2.Let V1 be a set of processes and V2 be a set
of resources. Since the graph is directed we will consider:
? an edge (Rj,Pi) (an assignment edge) to mean that resource Rj has been
allocated to process Pi
? an edge (Pi,Rj) (called a request edge) to mean that process Pi has requested
resource Rj
1. Use a resource allocation graph to derive a wait-for graph.
2. Wait-for graph obtained by making an edge from p1 to p2 iff p1 is waiting for a
resource that is allocated to p2.
3. Deadlock exists iff a cycle exists in resulting wait-for graph.

Q10- What is a race condition? Explain how does a critical section avoid this condition.
What are the properties which a data item should possess to implement a critical
Ans-Race condition: The situation where several processes access – and manipulate
shared data concurrently. The final value of the shared data depends upon which
process finishes last. To prevent race conditions, concurrent processes must be
Data consistency requires that only one processes should update the value of a data
item at any time. This is ensured through the notion of a critical section. A critical
section for data item d is a section of code, which cannot be executed concurrently
with itself or with other critical sections for d. Consider a system of n processes (P0,
P1,…, P n-1), each process has a segment of code called a critical section, in which
the proceses may be changing common variables, updating a table, writing a file, and
so on. The important feature of the system is that, when one process is executing in its
critical section, no other process is to be allowed to execute in its critical section.
Thus the execution of critical sections by the processes is mutually exclusive in time.
Solution to the Critical Section Problem must satisfy the following three
1. Mutual Exclusion. If process Pi is executing in its critical section, then no
other processes can be executing in their critical sections.
2. Progress. If no process is executing in its critical section and there exist some
processes that wish to enter their critical section, then the selection of the processes
that will enter the critical section next cannot be postponed indefinitely.
3. Bounded Waiting. A bound must exist 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.
—Assume that each process executes at a nonzero speed
—No assumption concerning relative speed of the n processes.
a solution to the Dining philosopher problem so that no races arise.


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