Network Layer – Subnetting Question or MCQ-1

Q.1 On Ethernet 1, your router’s IP address is 172.16.2.2/23. which of the following host IDs can be true?

(p) 172.16.1.100
(q) 172.16.1.198
(r) 172.16.2.254
(s) 172.16.3.1

(a) Both p and q (b) Both r and s (b) All of the Above (d) None of the Above

Answer :- (b)

Solution Given IP address is 172.16.2.1/23

CIDR value of above IP Address is 23 it means 23 Network bits will be 1 (One) and 9 Host bit will Zero.

It means 23 N/W bits+9 Host bit =32 Bit IP Address

Sub net Mask of the Above IP address is 11111111.11111111.11111110.00000000

That is 255.255.254.0

Total Number of Sub nets will be 2⁷ =128 Total Sub nets

Total Host in each Sub nets will be 2⁹ =512 Total Hosts in Each Sub net.

Total Usable Host in each Sub nets will be 2⁹-2 =512-2=510 Total Usable Hosts in Each Sub net. First Host reserve for N/W ID and last for Broad cast ID

Calculate the Range of Network is

172.16.2.2 in Binary is 10101100.00010000.00000010.00000010

From	To
Left First 23 Bits and Change Zero to last Nine Bits.	Left First 23 Bits and Change One to last Nine Bits.
10101100.00010000.00000010.00000000	10101100.00010000.00000011.11111111
172.16.2.0	172.16.3.255

Range Will be 172.16.2.0 to 172.16.3.255

Now Checking above Options

(p) 172.16.1.100 Not in range
(q) 172.16.1.198 Not in range
(r) 172.16.2.254 With in range
(s) 172.16.3.1 With in range

Correct Answer is (b) Both r and s 172.16.2.254 and 172.16.3.1 with in range.