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Question 1
(a) Describe the processing that a Router performs on a datagram received at one of its ports
(b) What are the advantages of doing Reassembly at the ultimate destination instead of doing it after the datagram travels across one network
(c) Discuss the need of Back-up RARP servers for a network with a large number of diskless machines.
(d) What is the minimum network MTU required to send an IP datagram that contains only one octet of data?
(e) How many computers would answer to a PING with a network broadcast address?
Question 1 Solution:
1a

Decrements TTL by one

If TTL = 0, drops the datagram

Otherwise works out the Header Checksum

If not OK, drops the datagram

Else uses the Routing Table to either deliver the datagram to its destination if it is direct delivery (by binding physical address through ARP, or, to send it to the next hop. For both the cases, if the physical h/w requires a smaller size of message, the received message may be fragmented into a number of datagram. The checksum is recomputed and the IP datagram is sent out, after encapsulating it in an appropriate frame.

1b
  1. Routers need not bother about whether the datagram is a fragment or not.
  2. The fragments may travel to the destination through different routers
1c

All the diskless machines, at boot-up, have to find out their IP address using RARP. The back-up server(s) is/are required because

  1. A number of machines may boot-up at the same time and there may be congestion at a single serer.
  2. If the single server should go down, all the machines would not be able to communicate if the back-up server is not there.
1d MTU = 21 octets 1 (data byte) + 20 (IP header bytes)
1e All the machines connected to the network, except the Sender
Question 2
(a)

Please refer to fig.1

Draw the routing table for R3. The table should have 3 columns as follows:

netid, netmask, next hop

(b) State and explain the routing Algorithm used by Routers
Question 2 Solution:
2a netid netmask Next hop
178.080.0.0 255.255.0.0 151.100.01
151.100.0.0 255.255.0.0 Direct Delivery
131.108.0.0 255.255.0.0 223.240.129.7
223.240.129.0 255.255.255.0 Direct Delivery
78.0.0.0 255.0.0.0 223.240.129.5
2b

Routing Algorithm

Extract destination IP address, D, from the datagram and computer the network prefix, N;

If N matches any directly connected network address, the deliver datagram to destination D over that network (by resolving D to a physical address encapsulating the datagram and sending the frame.)

else if the table contains a host-specific route for D, send the datagram to next hop specified in the table.

else if the table contains a route for network N, send the datagram to the next hop specified in the table.

else if the table contains a default route, send the datagram to the next hop specified in the table.

else declare a routing error.

Question 3 Answer the following giving reasons for each of your answers:
(a) Should ARP update the cache if an old entry already exists for a given IP address?
(b) Should ARP software modify the cache even when it receives information without specifically requesting it?
(c) Suppose machine C receives an ARP request sent from A looking for target B, and suppose C has the binding from the IP address o B to the physical address of B. Should C answer the request?
(d) Is the IP address space of Ver 4 sufficient to assign a unique network number to every home in North America?
(e) In what special case(s) does a host connected to an Ethernet not need to use ARP or an ARP cache before transmitting an IP datagram?
Question 3 solution:
3a Yes, since the old entry may be outdated.
3b Yes, when ARP request is broadcast, if the Senders address entry exists in the cache, it is updated. It is to see that the old entry is updated.
3c No. Only B is required to respond. Since other machines may have outdated address information about B.
3d No. The total number of networks in ver 4 can be as follows: Class A 127 Class B 16,384 Class C 2,097,152 Total 2,113,663 The population of North America may be 300 million. Considering an average home to be of 5 persons, there would be 60 million homes, which is much larger than about 2 million available net addresses.
3e For a broadcast message. Since for a broadcast, the netid is required and not the physical address of the hosts.

Question 4. Please refer to Fig 2

A 2400 bytes long message is delivered by TCP to the IP layer of a host machine H1. H1 is located on an Ethernet network and it is to send the message to another host H2. Work out the structure of the datagram(s) that will be created by the IP layer. Specify the value of each one of the following fields in (each of) the datagrams(s): VERS, HLEN, TOTAL LENGTH, FLAGS, FRAGMENT OFFSET, HEADER CHECKSUM.

Please specify the values in the decimal number system only.

IP address of H1 and H2 are 78.111.85.135 and 131.108.85.212 respectively.

Given that IDENTIFICATION is to begin from 257 and that no OPTIONS are to be used. Take SERVICE TYPE = 0, TTL = 10 and PROTOCOL = 6. The four field values are given in decimal numbers.

Question 4 Solution:
Fragment I FragmentII
VERS 4 4
HLEN 5 5
Total length 1500 940
Flags 1 0
Fragment Offset 0 185
Header Checksum 3301 11868
Q4 : Calculation of Check Sum
0100 0101 0000 0000 0000 0101 1101 1100
0000 0001 0000 0001 0010 0000 0000 0000
0000 1010 0000 0110 0000 0000 0000 0000
0100 1110 0110 1111 0101 0101 1000 0111
1000 0011 0110 1100 0101 0101 1101 0100
The 16-bit one’s complement sum:
  4 5 0 0  
  0 5 D C  
  0 1 0 1  
  2 0 0 0  
  0 A 0 6  
  4 E 6 F  
  5 5 8 7  
  8 3 6 C  
  5 5 D 4  
1 F 3 1 9  
        1 end-around carry
  F 3 1 A  
Checksum = OCE5
Decimal = 256 * 12 + 16 * 14 + 5
  = 3301
Second Fragment
0100 0101 0000 0000 0000 0011 1010 1100
0000 0001 0000 0001 0000 0000 1011 1001
0000 1010 0000 0110 0000 0000 0000 0000
0100 1110 0110 1111 0101 0101 1000 0111
1000 0011 0110 1100 0101 0101 1101 0100
The 16 bit one’s complement sum:
  4 5 0 0  
  0 3 A C  
  0 1 0 1  
  0 0 B 9  
  0 A 0 6  
  4 E 6 F  
  5 5 8 7  
  8 3 6 C  
  5 5 D 4  
-- -- -- -- --  
1 D 1 A 2  
        1 end-around carry
  D 1 A 3  
Checksum = 2 E 5 C
Decimal = 4096 * 2 + 256 * 14 + 16 * 15 + 12
  = 11,868
Question 5
(i)

Which of the following technologies or devices has NOT been used for accessing Internet?

  1. ADSL
  2. Cable Modem
  3. Dial-up modems
  4. Multiplexers & Demultiplexers
(ii)

Which of the following is NOT a function of IP?

  1. Fragmentation.
  2. Reassembly at destination
  3. Sending acknowledge-receipt of a packet
  4. Placing IP header on a data packet
(iii)

For sending a message from one computer to the other on a LAN, the packets which are created by various LAN technologies are refered to as

  1. data packs
  2. datagrams
  3. IP-grams
  4. Frames
(iv)

Which of the following is/are valid address(es) which may be used to contact remote host(s) on the Internet?

  1. 78.156.25.31
  2. 127.0.0.0
  3. 358.207.192.5
  4. 248.248.-248.248
(v)

A router

  1. is a dedicated computer which is used to connect two or more networks.
  2. is used to ensure datagrams are sent to destination in the proper order.
  3. is used as an interface between a computer and a network
  4. is a computer used to generate error message only.
Question 5 Solution:
5
  1. d
  2. c
  3. d
  4. a
  5. a

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