April 17, 2019

Network Security - Mid Sem Solution

Mid-Sem Paper Solutions:


Question:
What are the different types of active security attack ?

Answer:
Security Attack: Any action that compromises the security of information owned by an organization. There are 2 types of attack as shown below.

Active Attack:
An Active attack attempts to alter system resources or effect their operations. The attacker changes the data or harms the system.

Passive Attacks:
The attacker’s goal is to just obtain the information. The attack does not harm the system.

Types of active attacks include:
  • Denial of service (DoS) - It prevents normal use of communication facilities. Slowing down or totally interrupt the service of the system. E.g. multiple requests to bring an exam result server down.
  • Session replay - Subsequent retransmission of a captured message to produce an unauthorized effect. E.g. Bill payment fake reminders.
  • Masquerade - Masquerade attack takes place when one entity pretends to be different entity. E.g. Hoax bank sites. 
  • Message modification - It means that some portion of a message is altered or that message is delayed or reordered to produce an unauthorized effect.
  • Repudiation - Sender denies that it sent the message or the receiver denies that it received the message.
  • Trojans - A Trojan horse, or Trojan, is a type of malicious code or software that looks legitimate but can take control of your computer. A Trojan is designed to damage, disrupt, steal, or in general inflict some other harmful action on your data or network.
Types of passive attack include:
  • Snooping - Data is intercepted by an unauthorized person. E.g. Tapping
  • Traffic Analysis - May be the data is masked, so no information can be extracted but some patterns like - sender, receiver, message length, time of the message etc. can be extracted to make intelligent guesses.

April 15, 2019

Office 365 Admin Center - Digital Skill (TrendNXT L1)

Marks: 10
Set: 1
Question:62



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MCQS:




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Software Architecture - Mid Sem Solution

Mid-Sem Paper Solutions:



Question:
Q.1 Choose the most appropriate architectural pattern (one) for the 5 descriptions below. Give reasons for choosing the pattern (in 1-2 line maximum)

1. Wants to split a system into a number of computationally independent execution structures (groups of software and hardware) such as database, business logic, web interface and client, connected by some communication media. The structure is chosen to provide a specific server environment optimized for operational requirements and resource usage.

2. A set of heterogeneous specialized modules which dynamically change their strategies as a response to unpredictable events, and the system has to deal with uncertain knowledge.

3. Wants a system that can be divided into reusable, loosely coupled components that can be flexibly combined and arranged to transform between various data formats.

4. Wants a distributed system with a structure that enables that service users do not need to know the nature or location of service providers.  

5. A program relationship in which one program requests a service or resource from another Program.

Nominees:
(a) Layered 
(b) Broker 
(c) Model-view-controller
(d) Pipe-and-Filter 
(e) Client-Server 
(f) Blackboard 
(g) Service-Oriented 
(h) Publish-Subscribe 
(i)  Map-Reduce

For Practice:
  • Wants a system that quickly can analyze enormous volumes of data by sorting the data and then analyzing the grouped data. -> Map-Reduce
  • Wants to set up a set of equal distributed computational entities that are connected via a common protocol to share their services and provide high availability and scalability. -> Peer-to-peer
Answer:

1. Multi-tier
2. Blackboard
3. Pipe-and-Filter 
4. Broker
5. Client-Server

Question:
Q.2. Write brief notes on followings:

2.1  Discuss whether this statement is true or false: “Architecture is a transferable and reusable model”.
Answer:
Yes, it is true. Architecture can be created as transferable, reusable model that forms the heart of a product line.
Below is some key-point which does prove the importance of this.
Architecture can provide the basis for evolutionary prototyping.
Architecture is the key artifact that allows the architect and project manager to reason about cost and schedule.
By restricting design alternatives, architecture channels the creativity of developers, reducing design and system complexity.
Architecture defines a set of constraints on subsequent implementation.

2.2 What parameters (inputs) are typically used in models for analyzing performance
Answer:
Parameters (inputs) for analytic modeling of performance (candidates):
Arrival rate of events
queuing discipline
scheduling algorithm
service time for events
network topology 
network bandwidth
routing algorithm 

2.3 What is the difference between reference architecture and an architectural pattern? 
Answer:

An important difference between a pattern and reference architecture is that a pattern must be a recurrent solution that was already used on existing implementations. On the other hand, reference architecture can propose new, innovative solutions in its structure, whose usage was not yet proven in existing software. As a consequence, reference architecture is more suitable to propose a structure for new domains, which are not still well established.

Patterns and reference architectures document solutions in different levels of granularity. While a pattern focus on a single problem and in a single recurrent solution, reference architecture usually considers the target domain as a whole.

Question:
Q.3. Write any five availability tactics for fault detection. Explain each one briefly. 

Answer
Below are the 5 availability tactics for fault detection:
Ping
– Client (or fault-detector) pings the server and gets response back
– To avoid less communication bandwidth- use hierarchy of fault-detectors,
the lowest one shares the same h/w as the server
Heartbeat
– Server periodically sends a signal
– Listeners listen for such heartbeat. Failure of heartbeat means that the server is dead
– Signal can have data (ATM sending the last txn)
Exception Detection
– Adding an Exception handler means error masking
Voting (TMR)
– Three identical copies of a module are connected to a voting system which compares outputs from all the three components. If there is an inconsistency in their outputs when subjected to the same input, the voting system reports error/inconsistency.
– Majority voting, or preferred component wins
Timer and Time-stamping
– If the running process does not reset the timer periodically, the timer triggers off and announces failure
- Time-stamping: assigns a timestamp (can be a count, based on the local clock) with a message in a decentralized message passing system. Used to detect inconsistency.


March 18, 2019

BDD - Digital Skill (TrendNXT L1)

Marks: 20
Set: 2


Note: Kindly prepare yourself to score good marks in BDD. This is for TrenDNXT level 1 dumps.

MCQS:




Code MCQS:





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February 15, 2019

Data Mining - Comprehensive Paper Solution



Note: This is just a reference paper which you can go through,  we are facing some issue with the website. If you have any more important question/answer, let us know. 
Share it on our Email - 1trickyworld1@gmail.com


Question:
For the following vectors x and y, calculate the cosine similarity and euclidean distance measures:
x =(4,4,4,4), y=(2,2,2,2)

Solution:

Cosine
x ● y = 4*2 + 4*2 + 4*2 + 4*2 = 32
||x|| = sqrt(4*4 + 4*4 + 4*4 + 4*4) = sqrt (64)   = 8
||y|| = sqrt(2*2 + 2*2 + 2*2 + 2*2) = sqrt (16) = 4
cos(x,y) = (x ● y) /  (||x||*||y||) = (32)/ (8*4) 
cos(x,y) = 1

Euclidean
d(x, y) = sqrt((4-2)^2 + (4-2)^2 + (4-2)^2 + (4-2)^2) 
Euclidean distance = 4

Question:
Consider the one-dimensional data set shown on the below table

X  
0.6  
3.2  
4.5  
4.6  
4.9  
5.2  
5.6  
5.8  
7.1  
9.5  
Y
-
-
+
+
+
-
-
+
-
-

Classify the data point x=5.0 according to its 3- and 9- nearest neighbors (Using majority Vote)

Answer:
We need to first find the difference of each data set with respect to x=5.0, Refer the below table for the same.

x
X
Difference (x & X)
Y
5.0
0.6
4.4
5.0
3.2
1.8
5.0
4.5
0.5
+
5.0
4.6
0.4
+
5.0
4.9
0.1
+
5.0
5.2
0.2
5.0
5.6
0.6
5.0
5.8
0.8
+
5.0
7.1
2.1
5.0
9.5
4.5

As asked,
Using 3- nearest neighbors method, 3 Closest points to the point x=5.0 will be the one who has least difference among them - > 4.9, 5.2, 4.6
Classes ->   +
Using Majority Vote, 3-nearest neighbor: +

Using 9- nearest neighbors method, 9 Closest points to the point x=5.0 will be the one who has least difference among them - > 4.9, 5.2, 4.6, 4.5, 5.6, 5.8, 3.2, 7.1, 0.6
Classes -> +  + +  +   
Using Majority Vote, 9-nearest neighbor: 

Question:
Suppose a group of 12 sales price records has been sorted as follows:
5; 10; 11; 13; 15; 35; 50; 55; 72; 90; 204; 215:
Partition them into three bins by each of the following methods.
(a) equal-frequency partitioning
(b) equal-width partitioning
(c) clustering

Answer:
(a) equal-frequency (equidepth) partitioning:
Partition the data into equidepth bins of depth 4: [given as n=4]
Bin 1: 5, 10, 11, 13
Bin 2: 15, 35, 50, 55
Bin 3: 72, 90, 204, 215

(b) equal-width partitioning:
Partitioning the data into 3 equi-width bins will require the width to be (215−5)/3 = 70.
We get interval like- (1,70),(71,140),(141,210),(211,280)
Bin 1: 5, 10, 11, 13, 15, 35, 50, 55
Bin 2:72, 90
Bin 3: 204
Bin 4: 215

(c) clustering:
Using K-means clustering to partition the data into three bins we get
Bin 1: 5, 10, 11, 13, 15, 35
Bin 2: 50, 55, 72, 90
Bin 3: 204, 215