PROGRAM - MCA
SEMESTER -
FOURTH
SUBJECT CODE & NAME - MCA4010- MICROPROSESSOR
1. Write short notes on:
a) Central Processing Unit
b) Memory Unit
a) Central Processing Unit
The central processing
unit (CPU) is the electronic brain of the computer. CPU consists of Arithmetic
Logic Unit (ALU) and Control Unit (CU).
(a) Arithmetic
Logic Unit (ALU):
The arithmetic logic
unit (ALU) is responsible for arithmetic and logical
operations. Basically an ArithmeticLogic Unit (ALU) is an electronic circuit
used to carry out the arithmetic operations like addition, subtraction,
multiplication and division. It also carries out logical operations like
greater than, less than, equal to etc. It performs the operation on the data
provided by
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Memory Unit:
Memory unit
stores the data,
instructions, intermediate
results and output temporarily, during
the processing of data. This memory is also called the main memory
or primary memory of the computer. The input data
that is to
be processed will
be usually brought
into the main memory
before processing. It
also stores the
instructions required for processing of
data and any
intermediate results. The
output is stored
in memory before being transferred to the output device. CPU can work
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2. Write short notes on:
a) Bus Interface Unit (BIU)
b) Execution Unit (EU)
a) Bus Interface Unit (BIU)
Bus
Interface Unit (BIU): The BIU handles all the data and address on
the buses for the execution unit (EU). It performs all bus operations such as instruction
fetching, reading and
writing operands for
memory and calculating the
addresses of the
memory operands. The
instruction bytes are transferred
to the instruction
queue. Execution unit
use the instruction queue
to execute the
instructions. Both BIU
and EU work asynchronously to
execute instructions by
using Pipelining
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b) Execution Unit (EU)
Execution Unit (EU)
It decodes
the instructions fetched
by BIU, generates
control signals and executes the instruction. The main parts
of Execution unit are:
i) Control system,
ii) Arithmetic and Logic Unit (ALU)
iii) Instruction decoder unit.
iv) Flag Register
v) General purpose registers and
vi) Pointers and index registers
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3. Write short notes on:
a) REP Prefix
b) Table Translation
a) REP Prefix
Because string
operations inherently involve
looping, the 8086
machine language includes a
prefix that considerably
simplifies the use
of string primitives with loops.
This prefix has the machine code
1 1 1 1 10 0 1 Z
where, for the CMPS and
SCAS primitives, the Z bit helps control the loop. By prefixing MOVS, LODS and
STOS, which do not affect the flags, with the REP prefix 11110011, they are
repeated the number of times indicated by the CX register according to the
following steps:
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b) Table Translation
It is
sometimes necessary to translate from
one code to another. A terminal may communicate with the computer using the
EBCDIC (Extended Binary Coded Decimal
Interchange Code) alphanumeric
code even though
the computer's software is designed to work with the ASCII (American
Standard Code for Information
Interchange) code, or
vice versa. Code
conversions involving fewer than 8 bits (which accommodates up to 256
distinct entities) can be performed most easily by storing the desired code in
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4. Describe about
Key-code Data Formats
and FIFO Status Word formats.
Key-code Data Formats:
After a valid Key
closure, the key code is entered as a byte code into the FIFO RAM,
in the following
format, in scanned
keyboard mode. The
Key code format contains
3-bit contents of
the internal row
counter, 3-bit contents of
the column counter
and status of the SHIFT
and CNTL Keys The
data format of
the Key code
in scanned keyboard
mode is shown
in below figure.
In the
sensor matrix mode, the data from the return lines is directly entered into an
appropriate row of sensor RAM that identifies the row of the sensor that changes
its status. The
SHIFT and CNTL
Keys
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5. Write a note on
(a) RS 232 standard
(b) IEEE 488 standard
RS232 Standard
RS232
standard is developed by the Electronic Industry Association (EIA) in 1962 and
was revised and
renamed as RS232C.
RS stands for "recommended standard." This
is a standard
hardware interface used
for implementing
asynchronous serial data
communication ports on
devices such as CRT terminals, printers, modems and keyboards. RS-232 can be plugged straight
into the computer’s
serial port (know
as COM or
Comm port). RS-232 has been
around as a standard for decades as an
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IEEE-488 Standard
IEEE-488
refers to the Institute of Electrical and Electronics Engineers (IEEE) Standard
number 488. IEEE-488 was created as HP-IB
(Hewlett-Packard Interface Bus),
and is commonly
called GPIB (General Purpose
Interface Bus). The
IEEE-488, General Purpose Interface Bus (GPIB), is a general
purpose digital interface system that can be used to transfer data between two
or more devices. The GPIB system is a
parallel communication system,
which can communicate
with several devices through
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6. Write short note on:
a) Parallel Printer Interface
(LPT)
b) Universal Serial Bus (USB)
a) Parallel Printer Interface
(LPT)
The parallel printer
interface (LPT) is located on the rear panel of the PC. The
LPT stands for
line printer. The
Parallel Port Interface
on the PC compatible computer is one of the most
flexible interfaces for connecting the PC to a wide range of devices. The
interface was originally intended purely
for
connection to printers but due to the simple nature of the digital control
lines it has found many other uses.
Its simplicity relies on the fact
that the data to and from the port forms an 8 bit binary on/off pattern. Unlike
serial ports which rely
on a chip
to do the
data transmission, parallel
data
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b) Universal Serial Bus (USB)
The two main
problems associated with peripherals
connected to computer systems today are
"plug and play"
and speed of
data transfer. USB (Universal Serial Bus) is designed to
overcome these problems. Each USB port provides a single connector for any
device that previously used parallel, serial,
keyboard, and mouse
or game ports.
USB provides a
serial bus standard for
connecting peripherals devices
to PC with
simplified addition and
removal. USB can connect
peripherals such as mice, keyboards,
game pads and joysticks,
scanners, digital cameras,
printers, external storage,
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SEMESTER FOURTH
SUBJECT CODE & NAME - MCA4020- PROBABILITY AND
STATISTICS
1. Three machines A, B and C
produce respectively 60%, 30% and 10% of the total number of items of a
factory. The percentage of defective output of these machines are respectively
2%, 3% and 4%. An item is selected at random and is found
to be defective.
Find the probability
that the item
was produced by machine C.
Solution: Let P(A), P(B) and
P(C) denote the probabilities of choosing an item produced
by A, B,
C respectively. Also,
let P(X/A), P(X/B),
P(X/C) denote the probabilities of choosing a defective item from the
outputs of A, B, C respectively. Then, from what is given, we have
We are
required to find
P(C/X), the probability
that the item
selected is
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2. Find the constant k so that
Is a joint probability density function. Are X and Y independent?
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3. The data shows
the distribution of weight of
students of 1st standard of a
school. Find the quartiles.
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4. Fit a trend line to the following data by the freehand method:
Solution: Graphical representation of
trend line relating to the production of
wheat by the method of free hand curve is shown in figure
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5. Let X be a random variable and its probability mass function is 
Find the m.g.f. of X and hence it’s mean and
variance.
Find the m.g.f. of X and hence it’s mean and
variance.
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6. The diastolic blood pressures of men are distributed as shown in
table. Find the standard deviation and variance.
Solution: The table
represents the frequency distribution
of data required for calculating the standard
deviation.
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SEMESTER 4
SUBJECT CODE & NAME - MCA 4030 - PROGRAMMING IN JAVA
1. Describe the following:
a) Multi-threading.
b) Significance of Java Bytecode
Multithreading is the ability of
a program or an operating system process to manage its use by more than one
user at a time and to even manage multiple requests by the same user without
having to have multiple copies of the programming running in the computer.
Central processing units have hardware support to efficiently execute multiple
threads. These are distinguished from multiprocessing systems (such as
multi-core systems) in that the threads have to share the resources of a single
core: the computing units, the CPU caches and the translation lookaside buffer
(TLB). Where multiprocessing
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Java Magic: Byte Code
A programming
language has to be portable
and also provide
enough ground for security.
The key factor
that allows Java
to solve both
the security and the portability problems is that the output of a Java
compiler is not executable code. Rather, it is bytecode. Bytecode
is a highly optimized set of instructions
designed to be
executed by the
Java run-time system, which
is called the
Java
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2. Differentiate Break and
Continue statements in Java with example program.
Break Statement
By using break, you can force immediate termination of
a loop, bypassing the conditional expression and any remaining code in the body
of the loop. When a break
statement is encountered
inside a loop,
the loop is terminated and program control resumes at
the next statement following the loop. Here is a simple example:
// Using break to exit a loop
class BreakLoop {
public static void main(String args[ ]) {
for(int i=0; i<100; i++) {
if(i == 10) break; // terminate loop if i is 10
System.out.println("i: " + i);
}
System.out.println("Loop complete.");
}
}
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3. Differentiate between packages
and Interfaces.
Packages
To create a package is
quite easy: simply include a package command as the first statement in a Java
source file. Any classes declared within that file will belong
to the specified
package. The package
statement defines a name space in which classes are stored. If
you omit the package statement, the class names are put into the default
package, which has no name. (This is
why you haven't
had to worry
about packages before
now.) While the default package is fine for short,
sample programs, it is inadequate for
real applications. Most of the time, you will define a package for your code.
This is the general form of the package statement: package pkg;
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4. What are
Applets? What are the
restrictions of Applets?
Describe about applet class.
An applet is
a Java program
that can be
embedded in a
web page. Java applications are
run by using
a Java interpreter.
Applets are run on any browser
that supports Java.
Applets can also
be tested using
the appletviewer tool included in
the Java Development Kit. In order to run an applet it must be included in a
web page, using HTML tags. When a user browses
a web server
and it runs
applets on the
user’s system.
Applets have certain restrictions put on them :
·
They can not read or write files on the user’s
system.
·
They can not load or run any programs stored on
the user’s system.
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5. Compare JDBC and ODBC .
Java Data Base Connectivity (JDBC)
The JDBC
API (Java Data
Base Connectivity Application
Program Interface) can access any
kind of tabular data, especially data stored in a Relational Database.
It works on top of
ODBC (Open Data
Base Connectivity) which was the
driver for database connectivity since
age old days but since
ODBC was implemented
in C so
people from the VB
background had some
problems in understanding
the implementation intricacies. Since
JDBC works on top of ODBC we have something called as a JDBC-
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6. Describe about Java Beans and
BeanBox.
Java Beans
JavaBeans is
the software component
architecture for Java.
It allows constructing applications
efficiently by configuring
and connecting components called
Beans. The Beans need to be written and
tested with a rich set of mechanisms for interaction between objects, along
with common actions that most objects will need to support such as persistence
and event handling.
The BeanBox
The
BeanBox is a very simple test container. It allows you to try out both the BDK
example beans and
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SEMESTER FOURTH
SUBJECT CODE & NAME - MCA4040- ANALYSIS AND DESIGN
OF ALGORITHM
1. Write the
steps involved in
analyzing the efficiency
of nonrecursive algorithms.
Analyzing efficiency
of non recursive algorithms
The steps involved
in analyzing the
efficiency of non-recursive
algorithms are as follows:
·
Decide
the input size based on the constraint n
·
Identify
the basic operations of algorithm
·
Check
the number of times the basic operation is executed. Find whether the execution
of basic operation is dependent on input size n or not. If the basic operation
is depending on worst case, best case and average case then analysis of
algorithm needs more attention.
Set up summation formula
for the number of times the basic operation is
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2. Define selection sort
and explain how
to implement the selection sort?
Selection sort is one
of the simplest
and performance oriented
sorting techniques that work
well for small
files. It has
time complexity as O(n2)
which is unproductive on large lists.
Let us see an example for selection
sort. In below figure , we use selection sort to sort five names in
alphabetical order.
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3. Define Topological sort. And explain with example.
Topological sort
is done using
a directed acyclic
graph (DAG), which is a
linear ordering of all vertices G= (V, E) is an ordering of all vertices such
that if G contains
an edge (u,
v), then u
appears before v
in the ordering.
A topological sort of a particular graph can be looked upon as a
horizontal line where all directed
edges travel from
left to right.
Thus, topological sort differs
from the usual
sorting
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4. Explain good-suffix and
bad-character shift in
Boyer-Moore algorithm.
The Boyer-Moore algorithm
uses two heuristics:
good-suffix and badcharacter shift.
Bad character shift
We use this when mismatch occurs. We decide the number
of places to shift by using bad character shift.
·
As in
Horspool’s algorithm if
the rightmost character
does not match, then the pattern is shifted to the
right by its length.
·
When the
rightmost character of the pattern
matches with that
of the text, then each character is compared from right to
left. If at some point a mismatch occurs
after a certain
number of ‘k’
matches with text’s character ‘T’, then bad character
shift is denoted by P. P = max { text (T) – k,1}
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5. Solve the Knapsack problem using memory functions.
Item 1 2 3 4
Weight 2
6 4 8
Value (in Rs.) 12 16 30 40
Knapsack capacity is
given as W=12. Analyze the Knapsack problem using memory functions with the
help of the values given above.
Answer:
If we apply the
recurrence formulas to this set of data, then we will get the following table
|
0
|
1
|
2
|
3
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
|
1
|
0
|
12
|
12
|
12
|
12
|
12
|
12
|
12
|
12
|
12
|
|
2
|
0
|
12
|
12
|
12
|
12
|
16
|
16
|
16
|
16
|
16
|
|
3
|
0
|
12
|
12
|
30
|
30
|
30
|
30
|
30
|
30
|
54
|
|
4
|
0
|
16
|
16
|
30
|
30
|
30
|
30
|
40
|
40
|
64
|
w1= 2, v1=12
w2= 6, v2= 16
w3= 4, v3= 30
w4= 8, v4= 40
We can compute the maximum value of
V[4,10] as Rs.64. We can use the table to track
the optimal subset. Since V[4,10] V[3,10], item 4 is included in an
optimal solution along
with an optimal
subset for filling
10-3=7 remaining units of
the Knapsack
V[3,10], item 4 is included in an
optimal solution along
with an optimal
subset for filling
10-3=7 remaining units of
the Knapsack
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6. Describe Variable Length
Encoding and Huffman Encoding.
Huffman codes
are digital data compression codes
which are the outcome of the
brilliant piece of
work by Prof.
David A. Huffman
(1925-1999). Huffman codes give good compression ratios. Even today, after
50 years, Huffman codes have not only survived but are unbeatable in many cases. Huffman compression is a
compression technique where there is no loss of information when the data is
compressed i.e. after
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