Question

Assume that a large number of consecutive IP addresses are available starting at 198.16.0.0 and suppose that two organizations, A, B, C, and D, request 4000, 2000, 4000, and 8000 addresses, respectively, in that order. For each of these, give the first possible IP address within the assigned subnet, the last possible IP address within the assigned subnet, and the subnet representation in slash notation (w.x.y.z/s).

Answer 1

Answer & Explanation:

An IP version 4 address is of the form w.x.y.z/s

where s = subnet mask

w = first 8 bit field, x = 2nd 8 bit field, y = 3rd 8 bit field, and z = 4th 8 bit field

each field has 256 decimal equivalent. that is

binary                                        denary or decimal

11111111      =        2⁸      =             256

w.x.y.z represents

in binary

11111111.11111111.11111111.11111111

in denary

255.255.255.255

note that 255 = 2⁸ - 1 = no of valid hosts/addresses

there are classes of addresses, that is

class A = w.0.0.0 example 10.0.0.0

class B = w.x.0.0 example 172.16.0.0

class C = w.x.y.0 example 198.16.8.1

where w, x, y, z could take numbers from 1 to 255

Now in the question

we were given the ip address : 198.16.0.0 (class B)

address of quantity 4000, 2000, 8000 is possible with a subnet mask of type

255.255.0.0 (denary) or

11111111.11111111.00000000.00000000(binary) where /s =  /16 That is no of 1s

In a VLSM (Variable Length Subnet Mask)

Step 1

we convert the number of host/addresses for company A to binary

4000 = 111110100000 = 12 bit

step 2 (subnet mask)

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

subnet mask: 255.255.16.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1        1       1       1      1     1     1     1

128  64     32    16    8    4     2    1

step 3

in the ip network address: 198.16.0.0/19 (subnet representation) we increment this using 16

that is 16 is added to the 3rd field as follows

That means the ist Valid Ip address starts from

          Ist valid Ip add: 198.16.0.1 - 198.16.15.255(last valid IP address)

Company B starts+16: 198.16.16.0 - 198.16.31.255

                  +16: 198.16.32.0- 198.16.47.255 etc

we repeat the steps for other companies as follows

Company B

Step 1

we convert the number of host/addresses for company B to binary

2000 = 11111010000 = 11 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 11 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11111000.000000                /21

now we have added 5 1s in the third field to reserve 11 0s

subnet mask: 255.255.8.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1        1       1       1       1    1     1     1

128  64     32    16    8   4     2    1

Step 3

Starting from after the last valid Ip address for company A

in the ip network address: 198.16.16.0/21 (subnet representation) we increment this using 8

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.16.1 - 198.16.23.255(last valid IP address)

Company C starts +16: 198.16.24.0- 198.16.31.255

                            +16: 198.16.32.0- 198.16.112.255 etc

Company C

Step 1

we convert the number of host/addresses for company C to binary

4000 = 111110100000 = 12 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 12 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11110000.000000                /20

now we have added 4 1s in the 3rd field to reserve 12 0s

subnet mask: 255.255.16.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1        1       1       1       1     1     1     1

128  64     32    16    8    4     2    1

Step 3

Starting from after the last valid ip address for company B

in the ip network address: 198.16.24.0/20 (subnet representation) we increment this using 16

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.24.1 - 198.16.39.255(last valid IP address)

Company C starts +16: 198.16.40.0- 198.16.55.255

                             +16: 198.16.56.0- 198.16.71.255 etc

Company D

Step 1

we convert the number of host/addresses for company D to binary

8000 = 1111101000000 = 13 bit

Step 2

vary the fixed subnet mask to reserve zeros (0s) for the 13 bit above

fixed subnet mask: 11111111.11111111.00000000.00000000            /16

variable subnet mask: 11111111.11111111.11100000.000000                /19

now we have added 3 1s in the 3rd field to reserve 13 0s

subnet mask: 255.255.32.0 (where the 1s in each field represent a denary number as follows)

1st 1 = 128, 2nd 1 = 64 as follows

1        1       1      1       1     1     1     1

128  64     32    16    8    4     2    1

Step 3

Starting from after the last valid ip address for company C

in the ip network address: 198.16.40.0/20 (subnet representation) we increment this using 32

That means the ist Valid Ip address starts from

           Ist valid Ip add: 198.16.40.1 - 198.16.71.255(last valid IP address)

Company C starts +16: 198.16.72.0- 198.16.103.255

                             +16: 198.16.104.0- 198.16.136.255 etc