Question

Subnetting Questions (5pts) - Complete the following problems. You must show ALL WORK. You can write up, take pix from a phone, and embed images in your submission. 1) What valid host range is the IP address 192.168.123.148/29 a part of? 2) Which subnetwork does host 192.168.94.170 255.255.255.192 belong to? 3) How many subnets and hosts per subnet can you get from the network 172.22.0.0/23? 4) What is the broadcast address of the network 192.168.18.64/27? 5) How many subnets and hosts per subnet can you get from the network 10.0.0.0 255.255.240.0? 6) (2pts XC) What is the broadcast address of the network 172.17.80.0 255.255.254.0? IPv6 Expansion / Contraction (2pts) - Complete the following problems. You must show ALL WORK. You can write up, take pix from a phone, and embed images in your submission. Each one is 0.5pts. 1) Contract the following - 2000:00b8:65a3:0000:0000:8a2e:0370:7334. Explain how you arrived at your answer 2) Contract the following - 2001.cdba:0000:0000:0000:0000:3257:9652. Explain how you arrived at your answer 3) Expand the following - FE80::202:0:FE1E:832. Explain how you arrived at your answer 4) Expand the following - 21cd:53::e8bb:42:30:c394. Explain how you arrived at your answer

Answer 1

Solution

Before Solving

NOTE:

Rnages of classful ip addresses

Class A: 1 – 126

Class B: 128 – 191

Class C: 192 – 223

Class D: 224 – 239

Class E: 240 – 254

Subnetting Questions:
1. Ans

IP Address: 192.168.123.148

Binary ID:  11000000 10101000 01111011 10010100

Binary Subnet Mask: 11111111.11111111.11111111.11111000

Bitwise AND operation b/w Binary ID and Binary Subnet Mask yields

11000000 10101000 01111011 10010000

Therefore IP range is : 192.168.123.144 - 192.168.123.151

And valid host IP address range is: 192.168.123.145 - 192.168.123.150

2. Ans

IP Address: 192.168.94.170

Binary ID:  11000000 10101000 01011110 10101010

Binary Subnet Mask:  11111111.11111111.11111111.11000000

Bitwise AND operation b/w Binary ID and Binary Subnet Mask yields

11000000 10101000 01011110 10000000
Hence Subnet Address is:  192.168.94.128

3. Ans

IP Address:  172.22.0.0

Since the n/w class is B, there are 23-16 = 7 bits for subnets

Therefore the number of subnets = 2^7 = 128

Since there are 9 (32 - 23) bits in the host part, therefore, there are 2^9 - 2 = 510 hosts per subnet

4. Ans

IP Address:  192.168.18.64

Since the last IP address in this address is 192.168.18.95

Hence broadcast address is: 192.168.18.95

5. Ans

IP Address:  10.0.0.0

Subnet Mask: 255.255.240.0

Since the n/w class is B, there are 20-16 = 4 bits for subnets

Therefore the number of subnets = 2^4 = 16

Since there are 12 (32 - 20) bits in the host part, therefore, there are 2^12 - 2 = 4094 hosts per subnet

6. Ans

IP Address: 172.17.80.0

Since the last IP address in this n/w is 172.17.81.255

Hence Broadcast address is:  172.17.81.255

IPV6 Expansion/Contraction

1.   2000:00b8:65a3:0000:0000:8a2e:0370:7334

Compressing

2000:b8:65a3:0:0:8a2e:0370:7334

Compressed Address: 2000:b8:65a3::8a2e:370:7334

2. 2001:cdba:0000:0000:0000:0000:3257:9652

Compressing

2001:cdba:0:0:0:0:3257:9652

Compressed Address: 2001:cdba::3257:9652

3.  FE80::202:0:FE1E:832

Expanding

fe80:0:0:0:202:0:fe1e:832

Expanded Address: fe80:0000:0000:0000:202:0000:fe1e:832

4. 21cd:53::e8bb:4f2:3c:c394

Expanding

21cd:0053:0:0:e8bb:4f2:003c:c394

Expanded Address: 21cd:0053:0000:0000:e8bb:4f2:003c:c394