Question

Given the IPv4 address in CIDR notation 215.200.110.50/25, identify the subnet IDs.

Answer 1

Answer:

c. 215.200.110.00, 215.200.110.64, 215.200.110.128, 215.200.128.19

Explanation:

IP Address: 215.200.110.50

The first process is to convert all octets to binary individually.

215 to binary

i.e. to divide by 2 as follows:

NOTE:

rem means (remainder)

215/2 = 107 rem 1

107/2 = 53 rem 1

53/2 = 26 rem 1

26/2 = 13 rem 0

13/2 = 6 rem1

6/2 = 3 rem 0

3/2 = 1 rem 1

1/2 = 0 rem1

Then count the remainders from down to up:

So, 215 of decimal is 11010111 in binary

Converting 200 to binary, we divide 200 by 2 until the quotient is 0

200/2 = 100 rem 0

100/2 = 50 rem 0

50/2 = 25 rem 0

25/2 = 12 rem is 1

12/2 = 6 rem 0

6/2 = 3 rem 0

3/2 = 1 rem 1

1/2 = 0 rem 1

Then count the remainders from down to up:

So, 200 of the decimal is 11001000 in binary

Converting 110 to binary; we divide 110 by 2 until the quotient is 0

110/2 = 55 rem 0

55/2 = 27 rem 1

27/2 = 13 rem 1

13/2 = 6 rem 1

6/2 = 3 rem 0

3/2 = 1 rem 1

1/2 = 0 rem 1

Then count the remainders from down to up:

So, 110 of the decimal is 1101110 in binary

In binary,110 = 01101110

We divide 50 by 2 until the quotient is 0

50/2 = 25 rem 0

25/2 = 12 rem 1

12/2 = 6 rem 0

6/2 = 3 rem 0

3/2 = 1 rem 1

1/2 = 0 rem 1

Then count the remainders from down to up:

So, 50 of the decimal is 110010 in binary

To determine the network ID, we keep the first 25 bits of 11010111.11001000.01101110.00110010 and locked all remaining bits to 0.

Thus, network ID in binary is 11010111.11001000.01101110.00000000

Therefore;

11010111.11001000.01101110.00000000 as a decimal notation = 215.200.110.50

Thus, the four subnet IDs are:

11010111.11001000.01101110.00000000 ⇒ 215.200.110.0 0

11010111.11001000.01101110.01000000 ⇒ 215.200.110.64

11010111.11001000.01101110.10000000 ⇒ 215.200.110.128

11010111.11001000.01101110.11000000 ⇒ 215.200.128.19