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 |
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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
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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 ones 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 ones 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?
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(ii) | Which of the following is NOT a function of IP?
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(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
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(iv) | Which of the following is/are valid address(es) which may be used to contact remote host(s) on the Internet?
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(v) | A router
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Question 5 Solution: | |
5 |
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