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.
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