Computer Networks Test 3

IP address: 173.91.43.43./25.

Here, no. of mask bits = n = 25

In classless addressing, no. of address are given by 232-n

No. of addresses = 232-25 = 27 = 128

For first address set 32 – n rightmost bits to zero.

i.e. set 7 rightmost bits to zero.

43 → 0010 1011 

First address → 173.91.43.0

For last address set 32 – n rightmost bits to 1.

i.e. set 7 rightmost bits to 1.

43 → 0010 1011 

last address → 173.91.43.127

Minimum header size of IPv4 = 20 byte

Maximum header size of IPv4 = 60 byte

Header length field in IPv4 header is of 4 bits.

Maximum possible value (1111) = 15

Scaling factor of \(\frac{{60}}{{15}} = 4\) is introduced

Since 50 is not divisible by 4, append in 0’s to increase the length to a multiple of 4

Header size = 50 + 0’s append = 52 Byte

\(\frac{{52}}{4} = 13\)

Represent 13 in header length field = 1101

Concept:

In class B, 16 bits are reserved for Network id and 16 for host id.

Explanation

5 bit more is added due to subnetting

Total Network Id = 16 + 5 = 21

Number of bit in host id = 32 - 21 = 11 bits

The number of available hosts = 2h − 2,

= 211 - 2 = 2048 - 2 = 2046

H×N =11 × 2046 = 22506

Important Point:

Since first address (Network) and last address (broadcast address) cannot be assigned to a Host

The IPv6 multicast traffic is same as that of IPv4 broadcast. The packet intended to multiple hosts is sent on a special IPv6 multicast address. Any host or interface that are member of that multicast address will receive the  multicast information or packet and process it.

IPv4 address: 32 bits

IP address = Network ID + Host ID

Network ID + Host ID =32

PREFIX is the part of the Network ID

Class	PREFIX (bits reserved from 1st byte)	Network ID (bits)	Range of 1st byte	No. of IP address consumed	Number of networks	IP address per network	Host per network	Percentage of IPV4 address consumed​
Class A	0	8	0 -127	231	27	224	224 – 2	50%
Class B	10	16	128-191	230	214	216	216 – 2	25%
Class C	110	24	192 -223	229	221	28	28 – 2	12.5%
Class D	1110	-	224- 239	228	MULTICAST ADDRESS			6.25%
Class E	1111	-	240-255	228	RESERVED FOR FUTURE			6.25%

 

The address between the range 127. 0. 0. 0 to 127.255.255.255 (exclude 127. 0. 0. 0 and 127.255.255.255) can be used as a loopback address that falls in the range of Class A network. Therefore, the loopback (IP) address is a member of class A network. Therefore, statement P is false.

A loopback address is primarily used to validate that the locally connected physical network card is working properly and the TCP/IP stack installed. A data packet sent on a loopback address, never leaves the host system and is sent back to the source application (to itself). Therefore, statement Q is true.

Hence option 2 is correct

Concept:

• Time to live at S = (0001 0011)2 = 19 

• Time to Live is 8 bits field present in IPv4 header.
• TTL is a mechanism that limits the lifespan of a packet in a computer network.
• TTL prevents a data packet from circulating indefinitely which is present in IPv4 header
• Every host (operates at Network layer) that passes the packet must reduce the TTL by at least one unit.
• Routers can view the TTL field and decrease the value of if when a packet is passed through it.

Explanation:

There are 7 routers and 1 receiver ∴

Therefore, the final value of TTL = 19 – 8 = (11)₁₀ = (0000 1011)₂ = (1011)2

Tips and Tricks:
TTL (Time to Live) is present in IPv4 header and the IP protocol is present in the Network layer, So devices that have access at least up to the network layer can decrement the value of TTL. 

176.129.160.0/21

Network ID + Host ID = 32

21 + Host ID = 32

∴ Host ID = 11

1011 0000. 1000 0001. 1010 0000. 0000 0000

NETWORK ID	HOST ID	Classless Representation
1011 0000. 1000 0001. 1010 0	000. 0000 0000	176.129.160.0/21

 

First Divide the IP address in Half (need 1 host ID)

FIRST HALF

NETWORK ID	HOST ID	Classless Representation
1011 0000. 1000 0001. 1010 00	00. 0000 0000	176.129.160.0/22

 

SECOND HALF

NETWORK ID	HOST ID	Classless Representation
1011 0000. 1000 0001. 1010 01	00. 0000 0000	176.129.164.0/22

 

Either FIRST HALF is divided or second half

Assume FIRST HALF division

1^st Quarter

NETWORK ID	HOST ID	Classless Representation
1011 0000. 1000 0001. 1010 000	0. 0000 0000	176.129.160.0/23

 

2^nd Quarter

NETWORK ID	HOST ID	Classless Representation
1011 0000. 1000 0001. 1010 001	0. 0000 0000	176.129.162.0/23

 

SECOND HALF division

1^st Quarter

NETWORK ID	HOST ID	Classless Representation
1011 0000. 1000 0001. 1010 010	0. 0000 0000	176.129.164.0/23

 

2^nd Quarter

NETWORK ID	HOST ID	Classless Representation
1011 0000. 1000 0001. 1010 011	0. 0000 0000	176.129.166.0/23

 

If first half is allocated to TESTBOOK then 1^st quarter of second half or 2^nd quarter of second half is allocated to SCHOLARS

If second half is allocated to TESTBOOK then 1^st quarter of first half or 2^nd quarter of first half is allocated to SCHOLARS

Sender:

Total length = Data length + header length

1160 = Data length + 40

∴ Data length = 1120 byte

Router (MTU):

Total length = Data length + header length

200 = Data length + 40

∴ Data length = 160 byte

Number of fragments = \(\frac{{1120}}{{160}} = 7\) 

Offset uses scaling factor of \(\frac{{{2^{16}}}}{{{2^{13}}}} = 8\)

→ offset value of 1st segmen = 1

→ offset value of 2nd segment = \(1 + \frac{{160}}{8} = 21\)

→ offset value of 3rd segment = \(21 + \frac{{160}}{8} = 21+ 20 = 41\)

→ offset value of 4th segment = \(41 + \frac{{160}}{8} = 41 + 20 = 61\)

→ offset value of 5th segment = \(61 + \frac{{160}}{8} = 61 + 20 = 81\)

Shortcuts:

The number of fragments is 5, since every fragment has 20 more offset than the previous one.

Fragment offset of last fragment = initial offset + (n - 1) × 20

= 1 + 4 × 20 = 81

Subnet mask: 255.255.255.248 = 1111 1111. 1111 1111. 1111 1111. 1111 1000

A’ IP address: 200.175.168.1 = 1100 1000. 1010 1111. 1010 1000. 0000 0001

Subnet Mask & A’ IP address = 1100 1000. 1010 1111. 1010 1000. 000 0000

∴ Subnet Id = 200.175.168.0

B’ IP address: 200.175.168.5 = 1100 1000. 1010 1111. 1010 1000. 0000 0101

Subnet Mask & A’ IP address = 1100 1000. 1010 1111. 1010 1000. 000 0000

∴ Subnet Id = 200.175.168.0

C’ IP address: 200.175.168.5 = 1100 1000. 1010 1111. 1010 1000. 0000 0111

Subnet Mask & A’ IP address = 1100 1000. 1010 1111. 1010 1000. 000 0000

∴ Subnet Id = 200.175.168.0

∴ A, B and C belongs to same subnet 200.175.168.0

Tips and Tricks:

& → bit wise AND

Since Network Mask is 255.255.255.248, not have to check first 3-byte because subnet id will same as first 3 byte of IP address. Only 4^th byte need to bit wise AND with given IP address

ICMP error message should NOT be sent after receiving when:

• After another ICMP error message
• After an IGMP message of any kind.
• After any fragment other than the first fragment from a fragmented IP datagram

ICMP error message should be sent after receiving when:

• If DF( do not fragment) bit is set and a router needs to fragment the datagram to send it further then the ICMP error message must be sent to the sender.
• If TTL is exceeded then also it should report to the sender by sending the ICMP error message.