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1. What are the three addressing schemes used in a typical TCP/IP operation? Explain the functionality and purpose of each briefly.
A. The three addressing schemes used in a typical TCP/IP operation are given below.
1. MAC Addresses
2. IP Addresses
3. Port Addresses
Functionalities:
1. MAC Addresses: Media Access Control (MAC) is an important address used in computer networking to identify the particular machine. MAC addresses are globally unique. MAC addresses are assigned to the network connecting devices by the manufacturer of a network interface controller (NIC). This is Extended Unique Identifier (EUI-48). It contains 48 bits of address space. The format of the MAC address is like 64-27-37-E8-5D-C8 hexadecimal digits. Here the first 24 bits manufactural ID and last 24 bits shows serial number for the unit.
2. IP Addresses: Internet protocol addresses are the numbers assigned to computer network interfaces. In place of IP addresses we are using name of the website like www.ipaddress.com. Computer translate these names into numerical addresses so they can send data to the right location. These IP addresses are assigned by the internet. These are used by devices called routers on the internet to forward message from one device to another through the internet. There are three types of classes in IP address. Those are class A, class B and class C. The format of IP address is like 50.153.215.89
There are two versions of IP addresses in use. Those two are IPv4 and IPv6. IPv4 addresses are written as a string of four numbers like 192.0.0.58. IPv6 addresses are longer strings of numbers like 2001:0dc5::57. Here two colons side by side which means it contains only zeros. IPv4 contains 32 bits or 4 bytes of space and IPv6 contains 128 bits or 6 bytes of space.
3. Port Addresses: It contains 2 bytes of space. The port address specifies the address of the particular application running on your system at any point in a time. The combination of IP address and the port address is a socket. Three types of port addresses. Well known port numbers (0-1023), registered port numbers (1024-49151) and dynamic port numbers (49152-65536).
Purpose:
1. MAC Addresses: The purpose of media access control addresses is to provide a unique hardware or physical address for every point at which a device is connected to the network on a wide area network or other network. MAC addresses or not used for packet transfer. This addresses function at the data link layer that is second layer in the OSI layer.
2. IP Addresses: It is classified as an address primarily because it serves the same purpose as a home address or an office address, it allows each device present on the internet to be located. It acts as a locator for one IP device to find another and interact with it. It is under control of Internet Assigned Numbers Authority (IANA).
3. Port Addresses: The port address specifies the protocol of the socket. Ports are used for computer hardware control. These ports are merely reference numbers used to define a service. For instance, port 23 is used for telnet services, and HTTP uses port 80 for providing web browsing service.
2. The usual response to packet loss is the retransmission of the same packet. Given that the packet loss ratio is ” and that the acknowledgements are always received, what is the expected number of transmissions to get one packet successfully delivered. Be clear, show your intuition and computation.
A. By the given data
Packet loss ratio = ”
Packet loss ratio = (Number of packets lost)/(Total number of packets)
If one packet is transmitting then the number of packet lost is = ”
Success ratio is = 1- ”
If n number of packets transmitting then the number of packet lost is = x ”
Success ratio is = 1/(1-”)
3. An image is 1600×1200 pixels with 3 bytes/pixel. Assume the image is uncompressed. How long does it take to transmit it over a 56-kbps modem channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over 100-Mbps Ethernet? Over Gigabit Ethernet?
A. The image is 1600×1200 pixels with 3 bytes/pixel
That is 1600*1200*3 = 57, 60,000 bytes.
= 4, 60, 80,000 bits.
Transmission speed is 56Kbps. i.e., 56,000 bits/sec.
That is 4, 60, 80,000 / 56,000 sec
The time taking to transmit it over 56Kbps modem channel is = 822.8571 sec
Transmission speed is 1Mbps. i.e., 1,000,000 bits/sec.
The time taking to transmit it over 1Mbps modem channel is = 46.08 sec
Transmission speed is 10Mbps. i.e., 10,000,000 bits/sec.
The time taking to transmit it over 10Mbps modem channel is = 4.608 sec
Transmission speed is 100Mbps. i.e., 100,000,000 bits/sec.
The time taking to transmit it over 100Mbps modem channel is = 0.4608 sec
Transmission speed is 1Gbps. i.e., 1000,000,000 bits/sec.
The time taking to transmit it over 1Gbps modem channel is = 0.04608 sec
4. When an error is detected during a data exchange, what are the options available for recovery? Explain each and elaborate which one you would use under what circumstances.
A. There are two options available for recovery. Those are as fallows,
1. Retransmission
2. Forward error correction
1. Retransmission: Send the information again to the receiver, after it make such a request. The receiver requests that the information be retransmitted whenever it cannot decode the packet, or the result of decoding has been an error. In a variety of circumstances the sender automatically retransmits the data using the retained copy. Things that need to retransmit,
If the receiver knows that expected data has not arrived, and so notifies the sender.
If the sender found, often through some out of bound means, that need retransmission.
If the receiver knows that the data has arrived but is was damages. In that case there is a need of retransmission.
If no such acknowledgement is imminent within a reasonable time.
2. Forward error correction: It is a digital signal processing technique used to enhance data reliability. Once you done receiving packet in error you seems to try to locate the error bit by means of redundant information transmission in conjunction with the data. Forward error correction (FEC) correct those errors on the fly. Its goanna be a beautiful system to correct from the errors on the fly.
The FEC can detect and correct the errors that are exists in the data without retransmitting the data stream. There are two types of forward error correction codes.
Block codes
Convolution codes
If the capability of forward error correction increase, the number of errors that can be corrected also increases. FEC requires encoded data. This technique is used for years to enable high quality data communication channels. It allows for near perfect data transmission accuracy even when faced with a noisy transmission channel.
5. Explain DAS, NAS, and SAN.
A.
DAS: It stands for Direct Attached Storage. It is the digital storage directly attached to the device accessing it, as opposed to storage accessed over a computer network. The perfect example for the direct attached storage is hard disk drives. That is only connected to one device. It is not accessible to other device. The direct attached storage hard disk and may also pen drives. Individual disk drives in a server are also called as direct attached storage.
DAS performs better than the network storage. The main alternatives to direct attached storage are network attached storage (NAS) and the storage area network (SAN).
NAS: It stands for Network Attached Storage. It is a type of dedicated file storage that will provides local area network (LAN) with file based shared storage through a standard cable connection. We can also say it as a file level computer data storage server connected to a computer network providing data access to a heterogeneous groups of clients.
These devices managed with a browser utility program. Each network attached storage on the local area network as an independent network and has its own IP address. It has ability to provide multiple clients on the network with access to the same files. Now a days if we need large storage capacity NAS can simply use with larger disks together to provide both vertical and horizontal scalability.
There are three categories based on the number of drives, drive support and capacity
High end or enterprise NAS
Midmarket NAS
Low end or desktop NAS
SAN: It stands for storage area network. It organizes storage resources on an independent, high performance network, It also a high speed network that interconnects and presents shared pools of storage devices to multiple servers.
The storage area network is typically assembled using three principle components.
Cabling
Host bus adapters (HBAs)
Switches
These storage area network is complex and difficult to manage. We can’t say that storage area networks only purpose is communication between computer and storage. SAN may not uses fibre channel or Ethernet or any other specific interconnect technology.
6. What are the methods to implement WANs? Explain each category together with use cases for each.
A. There are three methods to implement Wide Area Networks. Those three are
Public Switched Telephone Network
Circuit Switching
Packet Switching
Public Switched Telephone Network (PSTN): This is operated by national, regional, or local telephone operators. It contains telephone lines and fibre optic cables. PSTN are dedicated lines. Each line is 64kbps. It contains features such as call waiting, caller ID and so on are usually available at an extra cost. It also referred to as the plain old telephone service.
It is continuous connection between two phones. That begins with a dial tone and ends when the phone is dangle up. The public switched telephone network is based on the principles of circuit switching. At present PSTN is also providing data communications.
Circuit Switching: It is connection oriented. It is a line switched network. Originally developed for the analogue-based telephone system. There are three basic elements in circuit switching. End ‘stations, transmission media and switching. Four dedicated services.
Data rate (bits per second)
Delay
Jitter
Loss rate
Three phases of circuit switching communication system are circuit establishment, data transfer and circuit disconnect. Each node must have available internal switching capacity to handle the requested connection. The switching nodes must have the intelligence to make proper allocation and to establish a route through the network. Example of the circuit switching is public telephone network (PTN). The basic functionalities of circuit switching is signalling, control, switching, and interfacing.
Packet Switching: The main goal of the packet switching is to allow maximal sharing. Two methods in packet switching are,
Virtual circuits: for business. Route once switch many
Datagrams: Route many switch many. Example is internet
In packet switching data are transmitted in short packets. The control information should at least contain destination address and source address. Packets can take any practical route and may arrive out of order and also may go missing. There is a call set up before the exchange of data in virtual switching. Virtual packet switching is not a dedicated path. It ia clear request to drop circuit.
7. Assume that an end-to-end communications system is made up of three networks with date rate in Mbps ”1, ”2, and ”3, respectively. Ignoring the other delay components, what is the total end-to-end delay to transmit a packet size of ”?
A. Data rates of three networks are
”1Mbps, ”2Mbps and ”3Mbps respectively.
The total end-to-end delay to transmit a packet size of ”
= ”/”1+”/”2+”/”3
8. A data channel at 5Gbps is shared by two users. Assume that each user is transmitting at 2.5Gbps data rate continuously. Each user only transmits only 30% of the time.
(a) If circuit switching is used, how many users can be supported?
A. Two users can be supported. Because each required one second of the bandwidth.
(b) If packet switching is used, would there be queuing delay to accommodate these two users? If a third identical user is added will there be a queuing delay?
A. Each user requires 2.5Gbps when transmitting, if two users sending data a maximum of 5Gbps will be required.
The available bandwidth of the share link is 5Gbps. That why there is no queuing delay to accommodate these two users.
If the third identical user added, the bandwidth required will be 7.5Gbpas which is more than the available bandwidth of the shared link. In this case, there will be queuing delay.
(c) What is the probability that a user is sending data?
A. The probability that a given user is transmitting is
30/100 = 0.3
(d) Now assume three users under packet switching. Compute the probability that at any given time all three are transmitting at the same time. What is the fraction of time when the queue is growing?
A. Probability that all three users are sending data at a time is
= (0.3) x (0.3) x (0.3)
= 0.027
The queue grows when all the three users are sending data, the fraction of time during which the queue grows is 0.027
9. Complete the Wireshark lab from Professor Kurose and Ross’s textbook as attached in the following pages.
A.
1. Three protocols that appear in the protocol column in the unfiltered packet listing window are given below,
TCP (Transport Control Protocol)
TLSv1 (Transport Layer Security)
LLMNR (Link-Local Multicast Name Resolution)
2. If we look at the frame section of the GET request we see that the time the packet arrived is 00:51:03.390399000
The same section for the HTTP OK shows an arrival time of 00:51:03.465180000
The difference of these 2 times gives
00:51:03.390399000 – 00:51:03.465180000 = 00:00:00.74781000
3. In the IP section of the GET request, the source and destination are shown.
Source: 192.168.1.233
Destination: 128.119.245.12
The source is the local machine’s address and the destination is the web server’s public
My (local machine’s) address = 192.168.1.233
IP address of www-net.cs.umass.edu is 128.119.245.12
4. HTTP GET message
HTTP OK message
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