delhi belly Essay

Submitted By Kirpal0001
Words: 391
Pages: 2

Student’s Name: Kirpal Singh Grewal Assignment 2: 300782749

NO WORK SHOWN, NO MARKS
[1 mark for each answer box]
Note: Bits coloured in red are related to the subnets. Bits after the subnet ID are the host bits.

Problem
Number of needed usable subnets: 62
Network Address: 107.0.0.0

Answer
Show Your Work Here for Marks

Address class
A

107 in the first octet is between the range of 1-127 of class A network ID's

Default subnet mask
255.0.0.0
1111 1111.000 0000. 0000 0000.0000 0000

Custom subnet mask
255.252.0.0
1111 1111.1111 1100. 0000 0000. 0000 0000

Total number of subnets
64
26=64

Number of usable subnets
62
26= 64-2 =62

Total number of host addresses/subnet
262144
Since there are 18 bits left for the host,
218=262144

Number of usable host addresses/subnet
262142
218= 262144-2=262142

Number of bits borrowed
6
26 = 64

Problem with Solution

Number of needed usable subnets: 62
Number of needed usable hosts: 1022
Network Address: 190.168.0.0

Answer
Show Your Work Here for Marks

Address class

B

190 in the first octet is between the range of 128 and 191 of class B network ID's

Default subnet mask
255.255.0.0

1111 1111.1111 1111.0000 0000.0000 0000

Custom subnet mask
255.255.252.0

1111 1111.1111 1111.1111 1100.0000 0000

Total number of subnets
64

26 = 64

Number of usable subnets
62
26 = 64 - 2 = 62

Total number of host addresses/subnet
1k
Since there are 10 bits left for the hosts,
210 = 1k

Number of usable host addresses/subnet
1022

1k - 2 = 1022

Number of bits borrowed
6

26 yields a total of 64 subnets. Taking away the first and last unusable subnets yields a total of 64-2 = 62 required usable subnets

What is the 2nd usable subnet range, including the invalid host addresses
(state the starting and ending addresses)?
190.168.8.0 to
190.168.11.255
Usable subnet 3rd Octet 4th Octet
1st 0000 0100 0000 0000
2nd 0000 1000 0000 0000
3rd 0000 1100 0000 0000
4th 0001 0000 0000 0000

Thus, the 1st…