question 4 answer

The software name commview for wifi does not support to my wireless device. After checking the driver list of the devices provided by the commview for wifi, what I found is that commview for wifi could not support my wireless device. My wireless network adaptor is Atheros AR9285 wireless Network Adapter and the commview for wifi does not have any driver for this adaptor.

a. due to this reason I am using the software inSSIDer wireless capture to capture the wireless network and to view the SID of the capture network, Following is the screenshot of the captured network.
As shown in the above screenshot there are various types of wireless network. Among these network MyNepalMyPride is the default wireless network. While comparing SID of all these network found there is no same SID on this capture. That's why the same SID on this screenshot is zero (0).

b. Following screenshot shows the signal strength of the different network.

Here the signal strength of the default network is -40 which is the most strongest comparing to other network within the range of the access point.

c. Following is the screenshot of the capture of the network:

From the above screenshot the total number of the wireless network and the access point found is 16. All these 16 access point and the wireless have the different SID for their own.

question 1 Answer

As we all know that the image file size depends upon the amount of information on the page and the colour content. It also depends on the resolution of the image, high resolution image took the more space to store the information comparing to the image using low resolution. The use of resolution depends upon the user.

a. The total size needed to store the images raw information having 800x600 with 24-bit color depth are shown as below:

File size= horizontal pixel * vertical pixel * bit depth/8*1024

i.e. file size= 800*600*24/8

=1440000 Bytes

For this byte the image should have the resolution of 150dpi.

b. CLUT is known as colour look-up table. The main function of this table is to transform a range of input colour into another range of colour. This could be either the hardware device which is built into an imaging system or the software function which is built into an image processing applications.a. 480768 bytes is the total size needed to store the image by using the clut encoding,

c. The ratio is 3:1.

question 5 Answer

According to the question the

Given

Total Bandwidth = 33 MHz

Channel bandwidth = 25 KHz simplex channels

i.e. duplex channels= 2*25 =50KHz

Total available channels= (33*1000)/50=33000/50=660 channels

a.

1. For N=4

Total number of channels available per cell= 660/4 =165 channels

2. For N= 7

Total number of channels available per cell = 660/7 = 94.285 i.e. 95 channels

3. For N=12

Total number of channels available per cell = 660/12= 55 channels

b.

We can assume that a 1 MHZ spectrum for control channels implies that there are 1000/50 = 20 control channels out of the 660 channels available. To evenly distribute the control and voice channels, simply allocate the same number of channels in each cell wherever possible. Now the 660 channels must be evenly distributed to each cell within the cluster. In practice, only the 640 voice channels would be allocated, since the control channels are allocated separately as 1 per cell.

1. For N = 4, we can have 5 control channels and 160 voice channels per cell. In practice, however, each cell only needs a single control channel (the control channels have a greater reuse distance that the voice channels). Thus, one control channel and 160 voice channels would be assigned to each cell.

2. For N = 7, 4 cells with 3 control channels and 92 voice channels, 2 cells with 3 control channels and 90 voice channels, and 1 cell with 2 control channels and 92 voice channels could be allocated. In practice, however each cell would have one control channel, four cells would have 91 voice channels, and three cells would have 92 voice channels.

3. For N = 12, we can have 8 cells with 2 control channels and 53 voice channels, and four cells with 1 control channel and 54 voice channels each. In an actual system, each cell would have 1 control channel, 8 cells would have 53 voice channels, and 4 cells would have 54 voice channels.

Frequency reuse depends on the following factors:

a. The power of the transmitted signal.

b. The frequencies used.

c. The type of antenna.

d. The height of the antenna

e. Weather

f. The terrain over which the signal is sent.

There are two set of of 7 cells. The set of frequeincies used by cell 1 of one set is reused by the cell 1 of the second set. These cells must maintain a minimum geographical distance, which is referred to as frequency reuse distance. Frequency reuse distance is computed as:

D/R= Ö3N

Where D= the distance between cells using the same frequency. R is the cell radius and N is the reuse pattern. Thus for a 7 cell group with cell radius r = 3 miles, the frequency reuse distance D is 13.74 miles. The equation indicates that an increase in D reduces the chance of co- channel interference from cells using the same frequencies. This also means that the number of channels assigned to each cell becomes smaller resulting in inefficiencies in managing the spectrum and the trunk.

question 4 answer

The software named commview for wifi does not support to my wireless device. After checking the driver list of the devices provided by the commviw for wifi, what I found is that commview for wifi could not support my wireless device. My wireless network adaptor is Atheros AR9285 wireless Network Adapter and the commview for wifi does not have any driver for this adapter.

question 3 answer

SONET is a Synchronous Optical Network, which is standard for connecting fiber-optic transmission systems. This is proposed by Bellcore in the middle of 1980s and is now an ANSI standard.

Sonet defines interface standards at every physical layer of the OSI seven-layer model. The standard defines a hierarchy of interface rates which allows data stream at different rates to be multiplexed. SONET establishes Optical Carrier levels from 51.8 Mbps to 9.95 GBPS. Prior rate standard used by different countries specified which are not compatible for multiplexing. With the help of SONET communication carriers throughout the world can interconnect their existing digital carrier and fiber optic systems.

SONET was originally designed for the public telephone network. Fiber optic cabling already prevailed for long distance voice traffic transmissions, but the existing networks proved unnecessarily expensive to build and difficult to extend for long haul data and video traffic.

SONET is the technology which transmits data at the speed between 155 megabits per seconds and 2.5 gigabits per seconds. To build these high-bandwidth data stream, it multiplexes the channel together having the bandwidth as low as 64 kilobits per seconds into data frames send at fixed intervals. While comparing the Ethernet cabling which spans distance up to 100 meters i.e. 328 feet, SONET fiber technology can run much further. The most interesting characteristics of the SONET is that it supports the ring topology. The figure below is the example of the ring topology the SONET supports.

From the above figure we can find out that the one piece of fiber the working ring handles all the data traffic while the other piece of fiber the protection ring remains standby. If the working ring fails, SONET includes the capability to automatically detect the failure and transfer control to the protection ring in a very short time period some time even in fraction of a second. Due to this reason SONET can be described as a self-healing network technology. As SONET can carry very karge amount of traffic, it would seem on the surface to be an ideal technology for future voice and data broadband networks. SONET competes with several other viable technologies including ATM and Gigabit Ethernet for this role. Due to all these features SONET can be used to extend to reach of fiber channel data center protocol for international data traffic.

question 2 answer

Electronic mail is a facility which allows users at workstations and terminals to create and exchange messages. As we all know that the simplest form of e-mail is the single –system facility. This facility allows to all the users of all those shared computer system to exchange the messages to each other. This type of facility can be used in a mainframe environment in which multiple users share the same system. Every user should register on the system and should have a unique identifier. One mailbox is associated with each user. Any incoming mail is simply store as a file under that user’s mailbox directory.
Now according to the question that the people like to send same message to three different users instead of sending it individually. While doing this thing we can find some difference such as time saving, can prioritize easily while sending.
While sending the same message to three different users following procedure is to be used:
1. User directory: This is mostly used by the system. It may also be accessible to users to able to look up address on the directory of the user.
2. Multiple addressing: By using this process any user can send the same message to different users at same time. This will helps to copy the message and sent to multiple addresses. By doing so same message will be sent to the entire different user at time. By doing so we can save some time for creating the same messages for multiple times. While sending the message in this method we can create the distribution list and also by the central administrative function.
3. Message priority: While sending the same message to the multiple addresses, user can create the priority as well. After prioritizing the address the message will be delivered rapidly to the high priority address created by the user. This function will help to notify the recipient of the arrival of the high priority messages.
4. Status information: A user may request the notification of delivery by the recipient. Even a user could the query present status of a message. For example: qued for transmission, transmitted but receipt confirmation not yet received.
5. Interface to other facilities: These will include other electronic systems, such as telex, and physical distribution facilities, such as courier and the public mail service.
The process will be same for the single address while sending the same message to the different address. While sending the message to the different address a user has to create the same message different time according to the priority of the address. To perform all these process the SMTP have to follow the following standards:
• Post Office Protocol (POP3): POP3 allows an email client to download an e-mail from an email server. POP3 UAs connect via TCP/IP to the server. The UA enters the username and password. Once the user name and password was been authorized UA can issue POP3 commands to retrieve and delete the mail.
• Internet Mail Access Protocol (IMAP): As with POP#, IMAP also enables an email client to access mail and email server. IMAP also uses TCP/IP, with server TCP. IMAP is bit complex than POP3. It provides stronger authentication than POP3 and provides other functions not supported by POP3.
• Simple Mail Transfer Protocol (SMTP): This type of protocol mainly used to transfer of mail from UA to and MTA and from MTA to another.
• Multipurpose Internet Mail Extensions (MIME): Supplements SMTP allows the encapsulation of multimedia messages inside of standard SMTP message.