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Spring 2018 ASSIGNMENT
PROGRAM MCA
SEMESTER FOURTH
SUBJECT CODE & NAME MCA411-
MICROPROSESSOR
CREDIT 4 BK ID B1778 MAX. MARKS 60
SET
1
Q1.
Write short notes on:
a)
Central Processing Unit
b)
Memory Unit 5+5
Answer:
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).
(I)
Arithmetic
2 Write
short notes on:
a) Bus
Interface Unit (BIU)
b)
Execution Unit (EU) 5+5 10
Answer:
A)
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
3 Write
short notes on:
a) REP
Prefix
b)
Table Translation 5+5 10
Answer:
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
SET
2
1
Describe about Key-code Data Formats and FIFO Status Word formats. 5+5 10
Answer:
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
2 Write
a note on
(a) RS
232 standard
(b) IEEE 488 standard 5+5 10
Answer:
A) 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,
3 Write
short note on:
a)
Parallel Printer Interface (LPT)
b)
Universal Serial Bus (USB) 5+5 10
Answer:
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.
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Spring
2018 PROGRAM- MCA(REVISED FALL 2012)
SEMESTER- FOURTH
SUBJECT CODE & NAME- MCA412- PROBABILITY AND STATISTICS
SET 1
Q1. 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,
Q3. The data shows the distribution of weight of students of
1st standard of a school. Find the quartiles.
|
Class Interval
|
13 - 18
|
18 - 20
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20 - 21
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21 – 22
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22 - 23
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23 - 25
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25 – 30
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||||||
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Frequency
|
22
|
27
|
51
|
42
|
32
|
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16
|
10
|
|||||
|
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Solution:
|
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||||||
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Class interval
|
|
Frequency
|
|
Cumulative frequency
|
|
|||||||
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13
|
– 18
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|
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22
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22
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SET
2
Q1. Fit a trend line to the following data by the freehand
method:
|
Year
|
Production
of wheat
|
Year
|
Production
of wheat
|
|
|
(in
tonnes)
|
|
(in
tonnes)
|
|
|
|
|
|
|
1995
|
20
|
2000
|
25
|
|
|
|
|
|
|
1996
|
22
|
2001
|
23
|
|
|
|
|
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|
1997
|
24
|
2002
|
26
|
|
|
|
|
|
|
1998
|
21
|
2003
|
25
|
|
|
|
|
|
|
1999
|
23
|
2004
|
24
|
|
|
|
|
|
Solution:
Graphical
representation of trend line relating to the production of wheat by the
method of free hand curve is shown in figure
Q2. Let X be a random variable and its probability mass
function is 
find the m.g.f. of X and hence its mean and
variance.
find the m.g.f. of X and hence its mean and
variance.
Solution: By definition,
=
=
Q3. The diastolic blood pressures of men
are distributed as shown in table. Find the standard deviation and variance.
|
Pressure(men)
|
78-80
|
80-82
|
82-84
|
84-86
|
86-88
|
88-90
|
|
|
|
|
|
|
|
|
|
No. of Men
|
3
|
15
|
26
|
23
|
9
|
4
|
|
|
|
|
|
|
|
|
Solution:
The table represents
the frequency distribution of data required for calculating the standard
deviation.
|
Class
|
Mid
|
Frequency
|
x-83
|
d = (x-
|
d²
|
fd
|
fd2
|
|
Interval
|
value
X
|
‘f’
|
|
83)/2
|
|
|
|
|
|
|
|
|
|
|
|
|
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Spring 2018 PROGRAM-
MCA
SEMESTER-4
SUBJECT
CODE & NAME-mca413 - OBJECT ORIENTED PROGRAMMING
SET
1
Q1. Describe the following:
a) Multi-threading.
b) Significance of Java Bytecode
Answer.
Multi-threading
Multithreading
is the
ability of an application to perform multiple tasks at the same time. We can
create multithreading programs using Java. The core of Java is also
multithreaded.
Significance of
Java
Q2. Differentiate Break and
Continue statements in Java with example program.(5+5)
Answer.
Break and Continue statements in
Java
Break statements
The
break statement is used inside the switch to terminate a
statement sequence. When a break statement is encountered,
Q3. Differentiate between
packages and Interfaces.(5+5)
Answer.
Packages and Interfaces
Packages
Java
provides a mechanism for partitioning the class name space into more manageable
chunks. This mechanism is the package. The package is both a naming and a
visibility control mechanism. You can define classes inside a package
SET
2
Q1. What are Applets? What are
the restrictions of Applets? Describe about applet class.(2+3+5)
Answer.
Applets
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
Q2. Compare JDBC and ODBC (5*2)
Answer.
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
Q3. Describe about Java Beans and
BeanBox. (5+5)
Answer.
Java Beans and BeanBox
Java
Beans
JavaBeans
is the software component architecture for Java. It allows constructing
applications efficiently by configuring and
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Spring
2018 ASSIGNMENT
PROGRAM - Master of Science in
Information Technology (MSc IT)Revised Fall 2011
SUBJECT CODE & NAME – mca414-
ANALYSIS AND DESIGN OF ALGORITHMS
SET
1
Q1. Write the steps
involved in analyzing the efficiency of non-recursive algorithms.
Answer: Answer:
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
2 Define selection sort and
explain how to implement the selection sort? 3+7=10
Answer:
Definition: 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 list
3 What is mean by
Topological sort? And explain with example. 5+5=10
Answer: 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
sorting.
If the di-graph has a directed cycle present, the problem will not have a
solution.
SET
2
1. Explain good-suffix and
bad-character shift in Boyer-Moore algorithm. 5+5=10
Answer:
Good suffix Shift
This shift helps in shifting a matched
part of the pattern, and is denoted by Q. Good suffix shift Q is applied after
0 < k < m characters are matched.
Q = distance between
2 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. 10
Answer:
Knapsack Problem by Memory
Functions
|
I
|
0
|
1
|
2
|
3
|
|
4
|
5
|
6
|
7
|
8
|
9
|
10
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
0
|
0
|
0
|
0
|
0
|
|
0
|
0
|
0
|
0
|
0
|
0
|
0
|
3 Describe Variable Length
Encoding and Huffman Encoding.
Answer:
Variable Length Encoding
Variable-length
codes can allow sources to be compressed and decompressed with zero error
(lossless data compression) and still be read back symbol by symbol. With the
right coding strategy an independent and identically-distributed source may be
compressed almost arbitrarily
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