IP classes and subnetting.
- 1. ASSIGNMENT #1 SUBMITTED TO: MISS SARA SARWAR SUBMITTED BY: RABIA ZAFAR 17581556-045 BS IT(5TH SEMESTER) SECTION A
- 2. Question 1: Write the default Masks for the Class A Class B and Class C IP addresses. Answer: Class A default subnet mask (255.0.0.0). Class B default subnet mask (255.255.0.0). Class C default subnet mask (255.255.255.0). Question 2: How we can distinguish Class A, Class B, Class C, Class D and Class E IP addresses from each other. Write the range of first octet in decimal and Binary for all the 5 IP address classes. Answer: IP addresses are 32-bit binary numbers that are separated by a dot. The number is generally represented as 4 octets of numbers from the ranges 0-255. Range: Classes Range Class A 0-126 Class B 128-191 Class C 192-223 Class D 224-239 Class E 240-255 Class A: Decimal Binary 1 00000001 Class B: Decimal Binary 128 10000000 Class C: Decimal Binary 192 11000000 Class D:
- 3. Decimal Binary 224 11100000 Class E: Decimal Binary 240 11110000 Question 3: Write the default subnet Masks for the following IP addresses: a. 179.65.225.4 b. 222.35.20.18 c. 111.7.80.0? Answer: a. 179.65.225.4, this IP belongs to class B. The default subnet mask of class B is 255.255.0.0. b. 222.35.20.18, this IP belongs to class D. the default subnet mask of class D is 255.255.255.0. c. 111.7.80.0, this IP belongs to class A. The default subnet mask of class A is 255.0.0.0. Question 4: Write down the three available ranges for assigning Private IP addresses recommended by IANA(Internet Assigned Number Authority). Answer: The Internet Assigned Numbers Authority (IANA) has assigned several address ranges to be used by private networks. Address ranges to be use by private networks are: Class A: 10.0.0.0 to 10.255.255.255 Class B: 172.16.0.0 to 172.31.255.255 Class C: 192.168.0.0 to 192.168.255.255 Question 5: A broadcast address is the one that addresses to all the hosts in any network. State that to create a broadcast address, all the bits of network ID portion or all the bits of host ID portion are set to 1? Write down the broadcast addresses of the networks to which the
- 4. following IP addresses belong, write network addresses and ranges of their valid IP address too. (No subnetting) a. 129.65.225.4 b. 211.35.20.18 c. 180.47.115.6 Answer: (a) 192.65.225.4 Class B 129.65.0.0 Network ID 129.65.255.255 Broadcast ID (b) 211.35.20.18 Class C 211.35.20.0 Network ID 211.35.20.255 Broadcast ID (c) 180.47.115.6 Class B 180.47.0.0 Network ID 180.47.255.255 Broadcast ID Question 6: Subnet Mask or Custom mask tells us that how many bits are used for Subnet ID portion and how many for host ID portion. Identify how many bits are used for subnetting in the following IP address using its subnet mask: a. IP Address: 135.65.225.4 Subnet Mask: 255.255.240.0 b. IP Address: 210.35.20.18 Subnet Mask: 255.255.255.248 c. IP Address: 190.47.115.6 Subnet Mask: 255.255.254.0 Answer: (a) IP address: 135.65.255.4 from class B Subnet mask: 255.255.240.0
- 5. 11111111.11111111.11110000.00000000 Bits for subnet ID portion=20 Bits for host ID portion=12 (b) IP address: 210.35.20.18 from class C Subnet mask: 255.255.255.248 11111111.11111111.11111111.11111000 Bits for subnet ID portion=29 Bits for host ID portion=3 (c) IP address: 190.47.115.6 from class B Subnet mask: 255.255.254.0 11111111.11111111.11111110.00000000 Bits for subnet ID portion=23 Bits for host ID portion=9 Question 7: Extract the Network Addresses of the given IP addresses in question number 8, using the subnet masks given with them. (Remember that ANDing the IP address with the Mask extracts the network address from the given IP address). Answer: (a) IP address: 135.65.225.4
- 6. Subnet mask: 255.255.0.0 Binary of IP address: 10000111.010000011.11100001.00000100 Binary of subnet mask: 11111111.11111111.00000000.00000000 --------------------------------------------------------------------- 10000111.01000001.00000000.00000000 Network address: 135.65.0.0 (b) IP address: 210.35.20.18 Subnet mask: 255.255.255.0 Binary of IP address: 11010010.00100011.00010100.00010010 Binary of subnet mask: 11111111.11111111.11111111.00000000 ------------------------------------------------------------------- 11010010.00100011.00010100.00000000 Network address: 210.34.20.0 (c) IP address: 190.47.115.6 Subnet mask: 255.255.0.0 Binary of IP address: 10111110.00101111.01110011.00000110 Binary of subnet mask: 11111111.11111111.00000000.00000000 --------------------------------------------------------------------- 10111110.00101111.00000000.00000000 Network address: 190.47.0.0 Question 8: Which of the following Subnet masks would allow a class A network to allow subnets to have up to 150 hosts and allow for up to 164 subnets? a. 255.0.0.0
- 7. b. 255.255.255.0 c. 255.255.192.0 d. 255.255.240.0 e.255.255.252.0 Answer: (a) 255.0.0.0 from class A Binary: 11111111.00000000.00000000.00000000 Subnet: 2^s =2^0 =1 Host: 2^h – 2 = 2^24 -2 =16,777,216 -2 =16,777,214 (b) 255.255.255.0 from class C Binary: 11111111.11111111.11111111.00000000 Subnet: 2^s= 2^0 = 1 Host: 2^h-2 =2^8-2=256-2 =254 (c) 255.255.192.0 from class B Binary: 11111111.11111111.11000000.00000000 Subnet: 2^s=2^2=4 Host: 2^s-2=2^14-2=16384-2=16382 (d) 255.255.240.0 from class B Binary: 11111111.11111111.11110000.0000000 Subnet: 2^s =2^4 =16 Host: 2^h-2 =2^12 -2=4096-2=4094 (e)
- 8. 255.255.252.0 from class B Binary: 11111111.11111111.11111100.00000000 Subnet: 2^s = 2^6 =64 Host: 2^h-2 = 2^10-2=1024-2=1022 Question 9: Suppose you have a class C Network 208.94.115.0. Your task is to design a subnet scheme so that we can create 16 Network segments (subnets) within this Network. Each subnet should support 10-14 hosts. 2n >= 14 24 >= 14 16 > 14 So, n = 4. a. How many bits would you use for the subnet ID? 28 bits are used for subnet ID. b. How many bits would you use for the Host ID? 4 bits are used for Host ID c. How many maximum possible subnets will be there? 24 subnets are possible because we have 8 bits for Host ID. So, 16 maximum subnets are possible. d. How many maximum possible hosts will be there in each subnet? 24 subnets are possible because we have 8 bits for Host ID. So, 16 maximum hosts are possible per subnet. e. Write down the Subnet Mask of your scheme. 208.94.115.11110000 255.255.255.240 f. Write down All the valid IP addresses Valid IP’s 208.94.115.0/28 208.94.115.16/28 208.94.115.32/28 208.94.115..48/28 208.94.115.64/28 208.94.115.80/28 208.94.115.96/28 208.94.115.112/28 208.94.115.128/28
- 9. 208.94.115.160/28 208.94.115.176/28 208.94.115.192/28 208.94.115.208/28 208.94.115.224/28 208.94.115.240/28 g. The broadcast address of the First subnet of your scheme. 208.94.115.15 /28 Question 10: Suppose you have a class C Network 220.94.115.0. Your task is to design a subnet scheme so that we can create 28 Network segments (subnets) within this Network. Each subnet should support hosts as given below. a. 2 Network Segment support 30 Hosts b. 4 Network Segment support 14 Hosts c. 8 Network Segment support 6 Hosts d. 14 Network Segment support 2 Hosts Steps 1- IP 220.94.115.0 2- Convert to binary 220.94.115.00000000 3- Find n 2n >= max host value 2n >= 30 25 >= 30 N=5 4- Give bits to network 220.94.115.11100000 5- IP of this network 220.94.115.224 6- Subnet Mask of IP 255.255.255.224/27 7- Subnets 220.94.115.0/27 220.94.115.32/027 220.94.115.64/027
- 10. 220.94.115.96/027 220.94.115.128/027 220.94.115.160/027 220.94.115.192/027 220.94.115.224/027 These are the network segments that can be created. Each subnet will support up to 32 hosts. To make the 28 segments we can use Class A or B for more segments.