How many 8-bit characters can be transmitted per second over a 9600 baud serial communication link using asynchronous mode of transmission with one start bit,eight data bits, two stop bits and one parity bit?
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Solution
A 2km long broadcast LAN has 107bps bandwidth and uses CSMA/CD. The signal travels along the wire at 2 × 108m/s. What is the minimum packet size that can be used on this network?
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Solution
Now, Distance = Speed ×Time
⇒ 2 × 103= 2 × 108× Time
⇒ Time = 10(–5)s or 10 × 10– 6s or 10ms
Minimum Packet size =l × Time
= 107 × 10 × 10–6
= 100 bits = 12 Byte
In serial data transmission, every byte of data is padded with a ‘0’ in the beginning and one or two 1’sat the end of byte because
The Address Resolution Protocol (ARP) is used for
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Solution
Address Resolution Protocol (ARP) is a computer networking protocol used by the Internet Protocol (IP)for determining a network host’s link layer or hardware address (MAC Address) when only its Internet Layer(IP) or Network Layer address is known. The protocol operates as a part of the interface between the OSI network and OSI link layer below the network layer.The term address resolution refers to the process of finding an address of a computer in a network. In this process, a piece of information is sent by a client process executing on the local computer to a server process executing on a remote computer to resolve the address.The information which is received by the server enables it to uniquely identify the network system for which the address was required and hence, to provide the required address. The address resolution procedure is successfully completed when the client receives the required address as the response from the server.Reverse of ARP is called Reverse Address Resolution Protocol (RARP).
Consider the following functional dependencies in a database:
Date_of_Birth→Age Age→Eligibility
Name→Roll_number Roll_number→Name
Course_number→Course_name Course_number→Instructor(Roll_number, Course_number)→Grade
The relation(Roll_number, Name, Date_of_birth, Age) is
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Solution
From the given,
Date of birth → Eligibility
Since, Date_of_Birth→Age;Age→Eligibility
Also,
Course-number→Instructor
Since, Course_number→Course_name;
Course_number→Instructor (from the given)Checking for 2nd normal form,
It is second normal form, as all the non-prime attributes are functionally dependent on the given relation keys.
Checking for 3rd normal form,
Here, since the non-prime attributes are not dependent on the candidate key, so it is not the third normal form.
Consider the set of relations shown below and the SQL query that follows :
Students :(Roll_number, Name, Data_of_birth)
Courses : (Course number, Course_name, Instructor)
Grades : (Roll_number, Course_number, Grade)
Select distinct Name
from Students, Courses, Grades
where Students. Roll_number = Grades Roll_number and Grades.grade
and courses. Instructor = korth
and Courses.course – number = Grades.course – number
Roll_number = Grades.Roll_number and Grades.grade = A
Which of the following sets is computed by the above query?
Consider a relation geq which represents “greater than or equal to”, that is, (x, y) ∈ geq only if y ≥ x.create table geq
(lb integer not null
ub integer not null
primary key lb
foreign key (ub) references geq on delete cascade)
Which of the following is possible if a tuple (x, y) is deleted?
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Solution
When the tuple (x,y) is deleted, consider the tuple (z, w).
Now, in tuple (z, w),
z is the lower bound and hence > X.and w is the upper bound and hence > Y.
→(X, Y) will be easily deleted.
Therefore, Y >X
Which of the following will ensure that the output string never contains a sub s tring of the form 01n0 or 10n1 where n is odd?
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Solution
To ensure this condition that substring of form 01n0 or 10n1,where n is odd S should be initially 1, we will case only 1 semaphore S.
So at W P(s), at X V(s) whereas at Y P(s), at Z V(s)Hence (c) is correct option.
Consider line number 3 of the following C program:
int min () { /* Line 1 */
intI, N; /* Line 2 */
fro (I =0, I <N, I ++); /* Line 3 */
Identify the compiler’s response about this line while creating the object-module.
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Solution
There are no lexical errors for C because all the wrong spelled keywords would be considered as identifiers until the syntax is checked.So the compiler would give syntax errors.
Which of the following grammar rules violate the requirement of an operator grammar? P, Q, Rare non-terminals, and r, s, t are terminals.
1.P → Q R2 .P → Q S R
3.P → ε 4.P → Q t R r
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Solution
1.P → QR is not possible since two NT should include one operator as Terminal.
2.Correct
3.Again incorrect. (4) Correct.
Hence (b) is correct option.