Computer Networks Final Syllabus Review: SUMMER 2016 Syllabus: Chapter 2: Application Layer Chapter 3: Transport Layer Chapter 4: Network Layer Chapter 5: Data Link Layer [Book: Computer Nerworking A Top-Down Approach By Kurose]
Important Topics: Chapter 2: DNS services and protocols WWW Service and HTTP Email service and SMTP/POP protocols Dynamic Host Configuration Protocol (DHCP) Chapter 3: TCP Server Process and port address working module. Web Client-Server Interaction: TCP and HTTP. UDP Server Process and port address working module TCP Congestion Control and dynamic window size UDP – Low Overhead vs. Reliability UDP vs TCP * Chapter 4: Sub-netting * IP Datagram Format / IPv4 packet header Why need Default gateway.* Routing table with Route Summarization* Routing protocols: Distance Vector: RIP, Link State NAT: Public and private addresses Chapter 5: ARP: IP corresponding to MAC * Switch Forwarding Table * VLAN (Theory Based) Short Note: Logical Link Control and Media Access Control Example of the Use of Various Data Link Protocols Real life scenario of passing data through internetwork * Or, A Day in the Life of a Web Page Request
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Overall Few Important Questions: 1. A University has acquired the 150.60.130.0/24 public address from the local ISP to use in its campus network. Each building has a specific number of devices that are required to be publicly accessible as shown in following figure. Distribute the IP's to the required number of hosts in such a way that number of wasted IP will be minimum.
Answer: The allocated IP Address is 150.60.130.0/24 and required hosts for university buildings are: (ascending order) Server Farm (SF) College of Education (CE) College of Business (CB) Administration (Admin)
= = = =
100 hosts 25 hosts 20 hosts 10 hosts
According to the requirements and acquired IP, the Sub-netting Table will be, Subnet Name SF CE CB Admin
Needed Size 100 25
20 10
Allocated Address Size 126 150.60.130.0 (2^7) – 2
Mask /25
Assignable Range 150.60.130.1
30 (2^5 ) – 2
150.60.130.128
/27
150.60.130.126 150.60.130.129
30 (2^5) – 2
150.60.130.160
14 (2^4) – 2
150.60.130.192
Broadcast Address 150.60.130.127
-----
150.60.130.159
-----
/27
150.60.130.158 150.60.130.161
150.60.130.191
-----
/28
150.60.130.190 150.60.130.193
150.60.130.207
-----
150.60.130.206
2. A router with IPv4 address 125.45.23.12 and Ethernet physical address 23:45:AB:4F:67:CD has received a packet for a host destination with IP address 125.11.78.10. Show the entries in the ARP request packet sent by the router and explain each field. Assume there is no sub-netting.
Answer: According to question, we get Router associated IP address : 125.45.23.12 Interface Physical Address (MAC) : 23:45:AB:4F:67:CD Destination Host IP address : 125.11.78.10
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Router IP: 125.45.23.12
Host IP: 125.11.78.10
23:45:AB:4F:67:CD In ARP request, we know that forwarding packet header from router holds the information of 6 byte router interface’s MAC address as source MAC and 4 byte router IP address as source IP. But for destination MAC, it contains default MAC (00:00:00:00:00) and 4 byte destination host IP. As a result, host response to router by its own MAC address associated with own IP address and that’s the main purpose of ARP. So, the entries in the ARP request packet sent by the router is: Ox 0001 Ox 0800 Ox 06 Ox 04 Ox 0001 Ox 23:45:AB:4F:67:CD Ox 7D:2D:17:0C (hexadecimal IP of source) Ox 00:00:00:00:00:00 Ox 7D:0B:4E:0A (hexadecimal IP of destination)
3. Why IPv6 use simplified header format compared to IPv4 header? Explain. Answer: The reason for this was to increase processing performance coz, simple constant size headers can be processed quickly at or very close to wire-speed. Such as The IPv4 header format contains a lot of fields including some unpredictable optional ones leading to fluctuating header sizes. So, IPv6 shows a different approach - the basic header is minimized and with constant/fix size. Only important fields are included in IPv6 headers. Everything else has been separated from header and formed extension headers, which are attached on requirements in IPv6 headers. IPv4 Header Version IHL Type Total Length
IPv6 Header Version Traffic Flow Label
Identification
Payload Length
Class
TTL
Flags
Fragment Offset Protocol Error Checking Source Address Destination Address
Next Header Source Address
Hop Limit
Destination Address
These figures show the similarities and differences between IPv4 header format and IPv6 header format.
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4. What problems will occur if centralized DNS is used? Answer: DNS is a service in the same way HTTP is. It is designed to be decentralized and redundant. The problems of centralized DNS: single point of failure traffic volume distant centralized database maintenance To a large extent, DNS is not fully centralized because of there are many other roots that are fully independent of a single highest root and the main DNS root. 5. Describe the major components of electronic mail. Suppose you have received an email from your friend. Draw the header format of the e-mail message. Analyze each of the header lines in the message.
Answer: The major components of electronic mail: ⇰ Mail User Agents (MUA) ⇰ Mail servers ⇰ Simple mail transfer protocol: SMTP ⊛ Mail User Agent: It’s actually as known as “mail reader”. Composing, editing, reading mail messages are done by it. e.g., Outlook, Thunderbird, iPhone mail client etc. It managed to outgoing, incoming messages stored on server. ⊛ Mail Servers: Mailbox contains incoming messages for user and message queue of outgoing (to be sent) mail messages. SMTP protocol between mail servers to send email messages where “client” is a sending mail server and “server” is a receiving mail server. ⊛ Electronic Mail SMTP: It uses TCP to reliably transfer email message from client to server and well known port 25 maintains three phases of transfer ⇰ handshaking (greeting) ⇰ transfer of messages ⇰ closure
Header Format of the E-mail Message: In an e-mail, the body (content text) is always preceded by header lines that identify particular routing information of the message, including the sender, recipient, date and subject. Some headers are mandatory, such as the FROM, TO and DATE headers. Others are optional, but very commonly used and viewed headers, and their values: From: sender's name and email address (IP address here also, but hidden) To: recipient's name and email address Date: sent date/time of the email Subject: whatever text the sender entered in the Subject heading before sending.
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Besides this most common identifications, email headers also provide information on the route an email takes as it is transferred from one computer to another by mail transfer agents (MTA). Each time an email is sent or forwarded by the MTA, it is stamped with a date, time and recipient. This is why some emails, if they have had several destinations, may have several RECEIVED headers. In this case the stamp is an email header.
6. What are the parameter to join VLAN as a member'? How does a VLAN provide extra security for a network?
Answer: We need to set the following parameters when we create a VLAN in the management domain: ⇰ VLAN number ⇰ VLAN name ⇰ VLAN type ⇰ VLAN state (active or suspended) ⇰ Maximum transmission unit (MTU) for the VLAN ⇰ Security Association Identifier (SAID) Extra Security: VLANs provide an extra measure of security. People belonging to the same group can send broadcast message with the guaranteed assurance that users in other groups will not receive these messages. Besides, under no circumstances should remote or local access be password-free. For example, configure secure shell (SSH) or Telnet ports for password-only access. Further, access should conform to the roles performed by each person with management responsibilities.
7. Compare and contrast SMTP with HTTP.
Answer: There are three main differences between HTTP and SMTP: a) HTTP is mainly a pull protocol that means someone loads information on a web server and users use HTTP to pull the information from the server. On the other hand, SMTP is primarily a push protocol that means, the sending mail server pushes the file to receiving mail server. b) SMTP requires each message, including the body of each message, to be in seven-bit ASCII format. HTTP does not have this restriction. c) HTTP encapsulates each object of message in its own response message while SMTP places all of the message’s objects into one message.
8. Write the process of TCP connection establishment and termination procedure with appropriate diagram. Answer: TCP Connection Establishment and Termination: When two hosts communicate using TCP, a connection is established before data can be exchanged. After the communication is completed, the sessions are closed and the connection is terminated. The three steps in TCP connection establishment are:
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⇰ The initiating client sends a segment containing an initial sequence value, which serves as a request to the server to begin a communications session. ⇰ The server responds with a segment containing an acknowledgement value equal to the received sequence value plus 1, plus its own synchronizing sequence value. The value is one greater than the sequence number because the ACK is always the next expected Byte or Octet. This acknowledgement value enables the client to tie the response back to the original segment that it sent to the server. ⇰ Initiating client responds with an acknowledgement value equal to the sequence value it received plus one. This completes the process of establishing the connection. Within the TCP segment header, there are six 1-bit fields that contain control information used to manage the TCP processes. Those fields are: URG - Urgent pointer field significant ACK - Acknowledgement field significant PSH - Push function RST - Reset the connection SYN - Synchronize sequence numbers FIN - No more data from sender TCP Connection Establishment:
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TCP connection termination:
These diagrams are representing the TCP connection establishment and termination procedure.
9. Consider the following figure. Host A cannot ping Host B. What could be the cause of this problem? Explain with proper reason.
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Answer: The serial interfaces of the routers are not on the same subnet. Explanation: If we look on the serial interface networks, we get different subnet. Now calculate the range of the networks on serial link: For Router A, serial S0/0, the network: 192.168.1.62/27 Increment: 32, then Network address: 192.168.1.32 Broadcast address: 192.168.1.63 Now, for the network 192.168.1.65/27: Increment: 32 Network address: 192.168.1.64 Broadcast address: 192.168.1.95 So, these two IP addresses do not belong to the same network and they cannot see each other. That’s why route connection does not establish and host A cannot ping host B.
10. Consider the following network. With the indicated link cost apply distance vector 6 routing to compute the value of each node. Now the initialize node 1 to 6 and then update node 4.
Answer: Initialization of tables in distance vector routing: To
Cost Next
1
0
2
3
3
2
4
5
5
∞ ∞
6
1’s Table
-
To
Cost Next
1
3
2
0
3
∞
4
1
5
4
6
∞
-
2’s Table
To
Cost Next
1
2
2
∞
3
0
4
2
5
∞
6
1 3’s Table
-
To
Cost Next
1
5
2
1
3
2
4
0
5
3
6
∞
-
4’s Table
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To
Cost Next
1
∞
2
4
3
∞
4
3
5
0
6
4
To
Cost Next
2
∞ ∞
3
1
4
∞
5
2
6
0
1
-
5’s Table
-
6’s Table
In distance vector routing, each node shares its routing table with its immediate neighbors periodically and when there is a change. Now, updating the node 4 To
Cost
To
1
2
1
4
2
∞
2
∞
3
0
3
2
4
2
4
4
5
∞
5
∞
6
1
6
3
Received from 3’s Table
To
Cost Next
3 3 3 3 3 3
4’s modified Table
Compare
To
Cost Next
1
4
2
1
3
2
4
0
5
3
6
3
3 3
Cost Next
1
5
2
1
3
2
4
0
5
3
6
∞
-
4’s old Table
4’s update Table
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