ENG201 Final Term MCQ,s

ENG201

ENG101 Final Term MCQ,s

ENG101

Windows _shotcut_keys

Widows shotcut keys

Assignment_of_CS201

#include<iostream.h>
#include<conio.h>

main()

{
char i;
int a, b;

do
{
system("cls");
cout<<"******Challan Calculator for Traffic Violations******";
cout<<endl;
cout<<endl;
cout<<"Enter '1' for Breaking Signal:";
cout<<endl;
cout<<"Enter '2' for Over Speed:";
cout<<endl;
cout<<"Enter '3' for Wearing not Seat Belt:";
cout<<endl;
cout<<endl;
cout<<"Enter the type of traffic violation 1, 2 or 3:";
cin>>a;
cout<<endl;
switch(a)
{
case 1:
cout<<"Enter the numbers of days passed till challan date:";
cin>>b;
cout<<endl;

if(b<=10)cout<<"The total fine is Rs: "<<500;
else if (b<=30)cout<<"The total fine is Rs: "<<500*2;
else if (b>30)cout<<"The total fine is Rs: "<<(500*2)+(500/2);
break;
case 2:
cout<<"Enter the numbers of days passed till challan date:";
cin>>b;
cout<<endl;

if(b<=10)cout<<"The total fine is Rs: "<<300;
else if (b<=30)cout<<"The total fine is Rs: "<<300*2;
else if (b>30)cout<<"The total fine is Rs: "<<(300*2)+(300/2);

break;
case 3:
cout<<"Enter the numbers of days passed till challan date:";
cin>>b;
cout<<endl;

if(b<=10)cout<<"The total fine is Rs: "<<200;
else if (b<=30)cout<<"The total fine is Rs: "<<200*2;
else if (b>30)cout<<"The total fine is Rs: "<<(200*2)+(200/2);
break;
default:
cout<<endl;
cout<<endl;
cout<<"Please Enter a Valid Type of Traffic Violation:";
cout<<endl;
cout<<endl;
getch();
cout<<endl;
cout<<endl;
system("cls");
main();
break;

}

cout<<endl;
cout<<endl;
cout<<"Do you want another challan calculation? (y/n):";
cin>>i;
}
while(i != 'N' && i != 'n' );

}

Solution_CS601_3

Assignment No. 3 
Semester Spring 2012
Data Communication - CS601

Q. 1. Suppose you have been asked to give a proposal of a Local Area Network for a company which requires easy reconfiguring, efficient in speed and economical in cost network. Considering these three factors, which guided media you will prefer for this type of network? Give three good reasons to support your answer.

(2 Marks for suggestion/name of media and 3 marks for three reasons) [2+3 = 5]

Solution:

              Unshielded Twisted Pair (UTP) Cable.

Q. 2. Analog signals are considered more prone to Noise as compared to digital signals. Suggest a proper method or technique with its complete steps for a telecommunication company which wants to convert its analog signals in digital signal to avoid from noise.

Note: just name the technique and its steps, detail is not required.  [5 Marks]

Solution:

             Pulse Amplitude Modulation

             Pulse Code Modulation

Q. 3. In Data Communication, situations exist where we need to change the Digital Signals into Analog Signals.  Select the technique(s) which is/are used by modems for this purpose. Choose all that apply. [10 Marks]

1.      B8ZS

2.      Quantization Amplitude Modulation

3.      Amplitude Shift Keying

4.      Frequency Shift Keying

5.      Frequency Modulation

6.      Frequency Reducing Key

7.      Frequency Division Multiplexing

8.      Phase Shift Keying

9.      Phase Modulation

10.  Quadrature Amplitude Modulation

11.  Manchester Encoding Scheme

12.  4 PSK

Solution:

             Amplitude Shift Keying.

             Frequency Shift Keying.

             Phase shift Keying.

            Quardrature Amplitude Modulation.

For More info Contact With me.
03155522233

STA 301 _Assignment 1_Spring_2012

Question # 1
Frequency Distribution:

Class Limit    Frequency ( f )    Class Boundaries    Class-Mark(Midpoint) X      
20 – 24    1    19.5 – 24.5    22      
25 – 29    4    24.5 – 29.5    27      
30 – 34    9    29.5 – 34.5    32      
35 – 39    10    34.5 – 39.5    37      
40 – 44    15    39.5 – 44.5    42      
45 – 49    9    44.5 – 49.5    47      
50 – 54    2    49.5 – 54.5    52      
Total    50           

Question # 2 (i)
     _
Mean ( X )

Classes    Frequency ( f )    Class Mark (Midpoint) X    fX      
0.7312 – 0.7313    10    0.73125    7.3125      
0.7314 – 0.7315    15    0.73145    10.97175      
0.7316 – 0.7317    20    0.73165    14.633      
0.7318 – 0.7319    25    0.73185    18.29625      
0.7320 – 0.7321    30    0.73205    21.9615      
0.7322 – 0.7323    8    0.73225    5.858      
0.7324 – 0.7325    2    0.73245    1.4649      
Total    110        80.4979   

Applying Formula   
    _      ∑fx
    X = -------
            ∑f
        = 80.4979 / 110
        = 0.7317991
Question 2
Part ii     ^
Mode ( X )

     Classes    Frequency ( f )    Class Boundaries ( C.B)      
0.7312 – 0.7313    10    0.73115 – 0.73135      
0.7314 – 0.7315    15    0.73135 – 0.73155      
0.7316 – 0.7317    20    0.73155 – 0.73175      
0.7318 – 0.7319    25    0.73175 – 0.73195      
0.7320 – 0.7321    30    0.73195 – 0.73215      
0.7322 – 0.7323    8    0.73215 – 0.73235      
0.7324 – 0.7325    2    0.73235 – 0.73255      
Total    110       
Applying Formula
       ^        ( fm – fi )
Mode =  X = L + -------------------------- * h
           ( fm – fi ) + ( fm – f2)

                            (30 - 25)
            = 0.73195 +--------------------------- * 0.0002
              (30 – 25) + (30 – 8)
                5
         = 0.73195 + ----------------- *0.0002
                  (5) + (22)
                  5
         = 0.73195 + ----------- * 0.0002
                  27
        = 0.73195 + 0.1851852 * 0.0002
        = 0.73195 + 0.000037
        = 0.731987

Working:
Class Mark is obtained by adding the sum of two limits of the and dividing by 2.
e.g. 0.7312 + 0.7313 /2 = 0.73125
fX is obtained X multiplying with (Frequency) f.
e.g. 10*0.73125 =7.3125
The Mode lies somewhere between 0.73195 and 0.73215.
By applying formula we noted that
fm = 30, fi = 25, f2 = 8

CS601 _ Assignment 1 Spring 2012 Solution

Q: 1
Solution:

STAR would be more appropriate as it is good with multiple computers connected in a big network plus it can also perform troubleshooting and new devices can be easily added in this network at any time.
According to me STAR would be more suitable as it is good with many computers connected in a network plus can also perform troubleshooting and new devices can be easily added in the network.
Q.2
Solution:

The physical describe of devices on a network, each and every LAN has a topology or the mode that the devices on a network are arranged and how they communicate with each other. They want that the workstations are linked to the network through the real cables that transmit data. The physical structure of the network is called the physical topology. The logical topology, is contrast, is the way that the signals act on the network media. On the way that the data passes through the network from one device to the another without regard to the physical interconnection of the devices.

      Because multipoint topologies share an ordinary channel, each device needs a way to identify itself and the device to which it wants to send information. The method used to categorize senders and receivers is called addressing four types of physical topologies are often used in computer networking:

    1.Star Topology
    2.Bus Topology
    3.Ring Topology
    4.Hybrid Topology
Q.3
Solution:
Following devices can be shared on network.

Scanner
CD-ROM drives
Light pen
Printer