AC Flag: This flag is set if thee is a carry to fifth bit from the fourth bits if two numbers.

In 8085 microprocessor the address bus is of 16 bits and data bus is 8 bits.

Processing unit→ ALU
Arithmetic logic unit(ALU) is a fundamental building block of many types of computing circuits, including the central processing unit (CPU) of computers, FPUs, and graphics processing units (GPUs).
Instruction unit →Timing and control
Timing and Control unit is responsible for hardwired control unit and Microprogrammed control unit
Storage and Interface unit → General purpose Register
The general purpose register can store a data (or) a memory location address.

Index mode:
The address of the operand is obtained by adding to the contents of the general register (called index register) a constant value. The number of the index register and the constant value are included in the instruction code.
→ Index Mode is used to access an array whose elements are in successive memory locations (or) contiguous memory locations.

FALSE: In memory - mapped I/O, the CPU can manipulate I/O data residing in interface registers that are used to manipulate memory words.
TRUE: The isolated I/O method isolates memory and I/O addresses so that memory address range is not affected by interface address assignment.
FALSE: In asynchronous serial transfer of data the two units share a different clock.
FALSE: In synchronous serial transmission of data the two units have common clocks.

Given data, -- Micro-ops field F₁ = 7 distinct micro operations
-- Micro-ops field F₂ = 7 distinct micro operations
-- Micro-ops field F₃ = 7 distinct micro operations
-- Condition field(CD) = 4 status bits
-- Branch Field(BR) = 4 options
-- Address space =128 memory locations
-- Size of micro operation=?
Given microinstruction format

If a schedule is conflict serializable then no cycle in precedence graph should be present.
But BFS and DFS are also possible for cyclic graphs.
And topological sort is not possible for cyclic graph.
Moreover option (D) is also wrong because in a transaction with more indices might come before lower one.

1NF : A relation is in 1NF if it does not contain multi-value attributes and composite attributes.
2NF: A relation is in 2NF is there is no partial dependency exist in the relation. i.e.
primary_key → non-key attribute
prime_attribute → non-key attribute.
3NF : A relation is in 3NF if there is no transitive dependency or we can say that a relation is in 3NF if either LHS of a functional dependency is a super key or the RHS of the functional dependency is a prime key attribute.
4NF : A relation is in 4NF if there is no multivalued dependency exist in the relation.
Hence the correct answer is option (C)
NOTE: A prime key attribute is a attribute which is the part of primary key attributes. For example: If ABC is the the primary key then A,B,C, AB, AC, BC are the prime key attributes.

The attribute D is not part of any FD's. So, D can be a candidate key or it may be part of the candidate key.
Now D+ = {D}.
Hence we have to add A,B,C,E,F,G,H to D and check which of them are Candidate keys of size 2.
AD+ = {ABCDEFGH}
BD+ = {ABCDEFGH}
ED+ = {ABCDEFGH}
FD+= {ABCDEFGH}
But CD+, GD+ and HD+ does not give all the attributes hence CD, GD and HD are not candidate keys.
Here Candidate keys are AD, BD, ED and FD.
A → BC, B → CFH and F → EG etc are partial dependencies.
So given relation is in 1NF, but not in 2NF.

Option(A): Cartesian product with a empty table will result in zero tuple because we can't have any ordered pair with a empty table. Hence option(A) is incorrect option.
Option(B):

(A, B) (B, C) - common attribute is (B) and due to B→C, B is a key for (B, C) and hence ABC can be losslessly decomposed into (A, B)and (B, C).
(A, B, C) (B, D) - common attributes is B and B→D is a FD (via B→C, C→D), and hence, B is a key for (B, D). So, decomposition of (A, B, C, D) into (A, B, C) (B, D) is lossless.
Thus the given decomposition is lossless.
The given decomposition is also dependency preserving as the dependencies A→B is present in (A, B), B→C is present in (B, C), D→B is present in (B, D) and C→D is indirectly present via C→B in (B, C) and B→D in (B, D).

Memory tube display components:
1. Connector Pins
2. Electron Gun
3. Base
4. Focusing System
5. Control Grid Voltage
6. X/Y Deflect
7. Phosphor
8. Collector
9. Ground
Note: Liquid crystal is not a component of memory tube display
.

→ A pumping lemma (or) pumping argument states that, for a particular language to be a member of a language class, any sufficiently long string in the language contains a section, or sections, that can be removed, or repeated any number of times, with the resulting string remaining in that language.
→ The proofs of these lemmas typically require counting arguments such as the pigeonhole principle.
→ Hence, the logic of pumping lemma is a good example of the pigeonhole principle.

→ Heap allocation is required for languages that support dynamic data structures. The heap is managed via calls to new, delete, callac, realloc, malloc, free, etc.
→ Stack allocation is required for local variables. Space on the stack is reserved for local variables when they are declared.

Pumping lemma for regular language is generally used for proving a given grammar is not regular.

→ The given language is Context free language(CFL) because we can push a's onto the top of stack as a's come as input and when a "b" come as input don't push it on the top of stack just change the state and after that when c's come come as input pop one "a" from top of stack for each "c".
→ In this way the given language can be accepted by the deterministic pushdown automata and hence the language is deterministic context free language. And when a language is CFL it is also a recursive language.
Hence the correct answer is option (C)

Given data,
-- Number of Pages(P)=100 per second
-- Number of Lines(L)= 24
-- Number of characters(C) =80
-- One character(O) =8 bits
-- Bit rate of channel=?
Step-1: Bit rate of channel= P*L*C*O
= 100*24*80*8 bits per second
= 1536000
Step-2: Given options in Megabits per seconds.
= 1.536 Mbps

Quadrature Amplitude Modulation means changing both amplitude and phase of the carrier.

Given data,
-- Total number of characters=50,000
-- One character=8 bits
-- Total time to send 50,000 characters= 40 sec
-- Data rate=?
Step-1: Data rate= (Total number of characters* one character bits)/ Total time
= (50,000*8)/40
= 4,00,000/40
= 10,000 bits per seconds
Step-2: We calculated bits per second but given options in Kbps.
= 10 Mbps

Data link layer→ Framing
Network layer→ Routing
Transport layer→ Connection control
Presentation layer→ Encryption

It is a class B address, so there 16-bits for NID and 16-bits for HID.
From HID, we took 6-bits for subnetting.
Then total subnets possible = ( 2⁶ ) - 2 = 64
Total hosts possible for each subnet = (2¹⁰) - 2 = 1022

→ A node's path length is the number of links (or branches) required to get back to the root.
→ The root has path length zero and the maximum path length in a tree is called the tree's height.
→ The sum of the path lengths of a tree's internal nodes is called the internal path length and the sum of the path lengths of a tree's external nodes is called the external path length.
External Path Length:

The sum over all external (square) nodes of the lengths of the paths from the root of an extended binary tree to each node. For example, in the tree above, the external path length is 25 (Knuth 1997, pp. 399-400). The internal and external path lengths are related by
E = I + 2n,
where n is the number of internal nodes.

There are n/k subsequences and each can be ordered in k! ways. This makes a (k!)^n/k outputs. We can use the same reasoning:
(k!)^n/k ≤ 2^h
Taking the logarithm of both sides, we get:
h ≥ lg(k!)^n/k
= (n/k)lg(k!)
≥ (n/k)(k/2)lg(k/2)
= (1/2)*(nlogk)-(1/2)*n
= Ω(nlogk)

The above recurrence is in the form of masters theorem.
a=8,b=2,k=0 and p=0
Case-1: a>b^k = 8>2⁰
T(n)=O(n^{logb^a})
= O(n³)

Method-1:

A well-formed document in XML is a document that "adheres to the syntax rules specified by the XML 1.0 specification in that it must satisfy both physical and logical structures".
Well formed documents requirements:
1. Content be defined.
2. Content be delimited with a beginning and end tag
3. Content be properly nested (parents within roots, children within parents)

TRUE: A Java Servlet is a server-side component that runs on the web server and extends the capabilities of a server.
FALSE: A Servlet can not use the user interface classes like AWT or Swing.
→ Servlets can respond to any types of requests, they most commonly implement web containers for hosting web applications on web servers and thus qualify as a server-side servlet web API.

FALSE: Entities are used to display characters that have special HTML meaning, such as “<” and “>”.
FALSE: Once a web server returns a cookie to a browser, the cookie will be included in all future requests from the browser to the same server.

FALSE: Constants that cannot be changed are declared using the ‘static’ keyword.
TRUE: A class can only inherit one class but can implement multiple interfaces.

→ Statistical software quality assurance in software engineering involves tracing each defect to its underlying cause, isolating the vital few causes, and moving to correct them.
→ Statistical software quality assurance uses pareto principle to identify vital causes (80% of defects can be traced to 20% of causes) and moves to correct the problems that have caused the defects.

FALSE: Black box tests are based on specifications; better at telling whether program meets specification, better at finding errors of omission.
FALSE: White box tests are based on code; better for finding crashes, out of bounds errors, file not closed errors.
TRUE: Alpha testing is conducted at the developer’s site by a team of highly skilled testers for software that is developed as a product to be used by many customers.

Given data,
-- Software expected to operate after repair(MTBF) = 91.25 days
= 91.25 * 24 hours
= 2190 hours
-- Mean software repair time(MTTR) = 5 mins
= 5/60 hrs
= 0.083 hrs
-- Software Availability=?
Step-1: Software Availability =MTBF / (MTBF + MTTR)
= [2190 / (2190 + 0.083)]*100
= (2190 / 2190.083)*100
= 0.9996211481 * 100
= 99.9962%

Given data,
-- lines of code(LOC)= 20,000 (or 20K)
-- Multiplicative value ‘a’=2.4
-- Exponential ‘b’=1.05
-- Basic COCOMO formula= a*(LOC in K)^b
-- Multiplicative value ‘c’=2.5
-- Exponential ‘d’=0.38
-- Development=c*(basic COCOMO result)^d
Step-1: Basic COCOMO formula= a*(LOC in K)^b
= 2.4*(20)^1.05
= 55.756
Step-2: Development= c*(basic COCOMO result)^d
= 2.5(55.756)^0.38
= 11.52 months

Software Configuration Management (SCM) is a use-case supported by standard version control systems:
1. Develop the next version of a piece of software while fixing problems with the current one.
2. Share code with other team members in a controlled way, allowing you to develop code in parallel with others and join with the current state of the codeline.
3. Identify what versions of code went into a particular component.
4. Analyze where change happened in the history of a component's development.

→ The Bounded buffer problem is also known as producer-consumer problem.
→ The problem describes two processes, the producer and the consumer, who share a common, fixed-size buffer used as a queue.
→ The producer's job is to generate data, put it into the buffer, and start again. At the same time, the consumer is consuming the data (i.e., removing it from the buffer), one piece at a time.
→ The problem is to make sure that the producer won't try to add data into the buffer if it's full and that the consumer won't try to remove data from an empty buffer.

Simple reflex agents[edit]
Simple reflex agents act only on the basis of the current percept, ignoring the rest of the percept history. The agent function is based on the condition-action rule: "if condition, then action".
Model-based reflex agents: A model-based agent can handle partially observable environments. Its current state is stored inside the agent maintaining some kind of structure which describes the part of the world which cannot be seen. This knowledge about "how the world works" is called a model of the world, hence the name "model-based agent".
Utility-based agents: Goal-based agents only distinguish between goal states and non-goal states. It is possible to define a measure of how desirable a particular state is. This measure can be obtained through the use of a utility function which maps a state to a measure of the utility of the state. A more general performance measure should allow a comparison of different world states according to exactly how happy they would make the agent. The term utility can be used to describe how "happy" the agent is.
Learning agents: Learning has the advantage that it allows the agents to initially operate in unknown environments and to become more competent than its initial knowledge alone might allow. The most important distinction is between the "learning element", which is responsible for making improvements, and the "performance element", which is responsible for selecting external actions.

Given data,
-- Equivalence condition(Both RHS and LHS should be TRUE)= P ⇔ (Q V ~Q)
-- Holding condition(Both RHS and LHS may TRUE/FALSE)= Q ⇔ R

"If X then Y unless Z" ⇒ ¬Z → (X→Y)
⇒ Z ∨ ¬X ∨ Y
⇒ ¬X ∨ Z ∨ Y
Option B: (X ∧ ¬Z) → Y
= ¬(X ∧ ¬Z ) ∨ Y
= ¬X ∨ Z ∨ Y
Hence, option (B) is correct.

→ Classical planning environments that are fully observable, deterministic, finite, static and discrete (in time, action, objects and effects).

TRUE: Every recursive language is recursively enumerable.
TRUE: L = {0ⁿ1ⁿ 0ⁿ │n=1, 2 , 3, ....} is recursively enumerable.
TRUE: Recursive languages are closed under intersection.
→ Recursive closed under union,intersection,complementation,etc.., except Homomorphism and substitution.
FALSE: Recursive languages are not closed under intersection.
→ Recursively enumerable languages are closed under union,intersection,etc.., except set difference and complementation.

→ Context-free grammar are closed under Union,Concatenation and Kleene closure.
→ Context-free grammar are not closed under set difference,Complementation and intersection.

L₁={a^m bⁿ | m≠n }
→ It means m could be greater than or less than ‘n’ but m will not be equal to ‘n’.
→ Since for accepting the language L1 we need a storage so then no. of a’s can be stored in it and when b’s come as input a’s and b’s can be compared.
→ Since finite automata do not have a storage element hence the language is not regular.
→ Now state transition diagram to check whether L1 is CFL or not:

→ Hence L₂ is also a CFL.
L3: First simply read one 'a', then push one 'a' in the stack after reading two a's and then pop all the a's by reading the b's. Since can be done by PDA hence CFL.

→ UNIX provides three types of permissions using octal numbers
1. Read(R or 4)
2. Write(W or 2)
3. Execute(X or 1)
→ UNIX provides three sets of permissions. Each set of permission we are having 3 types of permissions.
1. Permission for owner(RWX or 421)
2. Permission for group(RWX or 421)
3. Permission for others(RWX or 421)
→ We can change/modify permissions using chmod, chown,etc.., commands
→ We can see the permissions for every file using ls -l command

Distributed systems are using fixed routing only.
Fixed routing: A path from A to B is specified in advance; path changes only if a hardware failure disables it.
1. Since the shortest path is usually chosen, communication costs are minimized.
2. Fixed routing cannot adapt to load changes
3. Ensures that messages will be delivered in the order in which they were sent
Virtual routing: A path from A to B is fixed for the duration of one session.
1. Different sessions involving messages from A to B may have different paths
2. Partial remedy to adapting to load changes
3. Ensures that messages will be delivered in the order in which they were sent
Dynamic routing: The path used to send a message form site A to site B is chosen only when a message is sent
1. Usually a site sends a message to another site on the link least used at that particular time
2. Adapts to load changes by avoiding routing messages on heavily used path
3. Messages may arrive out of order; This problem can be remedied by appending a sequence number to each message
4. Most complex to set up

In UNIX Wait() system call is equivalent to windows system call is WaitForSingleObject() .
In UNIX Fork() system call is equivalent to windows system call is Create-process().
In UNIX Create() system call is equivalent to windows system call is CreateFile().
In UNIX Close() system call is equivalent to windows system call is CloseHandle()