Question

3) a) You have been allocated a network address 192.168.100.0. You need to create atleast 12 subnets. Find the flowing (4 points) 1) Total number of subnets 2) New subnetmask 3) Total number of valid hosts per subnet (6 points) 1) Write the first five subnets range with prefix 2) Write the last subnet range with prefix (3 points) Write the following from first range of the subnet 1. Network address 2. Broadcast address 3. First valid host address 4. Last valid host address

Answer 1

z. Allocated network = 192.168.100.0 This given address belongs to Class C. class C addresses have 24 network and & host bits. Here we need to create 12 subnets. It requires extra 4 bits ( 24 =16). 8 bits are not sufficient because 3 = 8 212. so, we borrow extra 4 bits from the host bits. Now, we have 28 network and a host bits- ®. Total No. of subned = 2 = 16 subnets New submet Mark - 255.255.255, 240 Total No, of valid hats per subnet = 24-2= 14 hosts ho subnet Range © 199.168.100.00000000 > 192.168.100.4/28 to 192.168, 100.14/28 © 199.168.100.00010000=192.168;100.16/28 to 192.168.100.30/28 192.168.100.00100000 →192.168.100.33/28 to 192.168.100.46/28 In 192.168.100.00110000 =198.168.100.49/28 to 192.168.100.62/28 G 192.168.100.0d0D 0000 > 192.168.100.65/28 to 192.168.106,78/28 2. last subnet Range 192.168.100.11110000@ > 192.168.100.94118 to 192.168.100.15428 6. For the first subnet: ① Network Address : 192.168. 100.0/28 & Broadcast Addrew! 192, 168, 100.15/28 6 first valid host: 192.168.100.1/28 I 6 Last valid host: 192.168.100.14/28 82 30 Public Address: 172. 6. 10.0 20.10.10.1, 11. 10.10.5,200,10.10. 1 152. 10.10.10 25.10.10.10, 218.10.15 Bivate Addves! 149.16.10.0, 192.168.10.10, 10.10.10.10 172. 23. 15.16