• Document: 2015 Handout: Subnetting Question
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3201 Computer Networks – 2014/2015 Handout: Subnetting Question Subnetting Questions with Answers Question1: Given the following: Network address: 192.168.10.0 Subnet mask: 255.255.255.224 1. How many subnets? Ans: 6 2. How many hosts? Ans: 30 3. What are the valid subnets? Ans: 32, 64, 96, 128, 160, 192 4. Fill in the table below. Question2: Write the subnet, broadcast address and valid host range for the following: 1. 172.16.10.5 255.255.255.128 Ans: Subnet is 172.16.10.0, broadcast is 172.16.10.127 and valid host range is 172.16.10.1 to 126. You need to ask yourself, “Is the subnet bit in the fourth octet on or off?” If the host address has a value of less than 128 in the fourth octet, then the subnet bit must be off. If the value of the fourth octet is higher than 128, then the subnet bit must be on. In this case, the host address is 10.5 and the bit in the fourth octet must be off. The subnet must be 172.16.10.0. 3201 Computer Networks – 2014/2015 Handout: Subnetting Question 2. 172.16.10.33 255.255.255.224 Ans: Subnet is 172.16.10.32, broadcast is 172.16.10.63 and valid host range is 172.16.10.33 to 10.62. 256-224=32. 32+32=64. The subnet is 10.32 and the next subnet is 10.64, so the broadcast address must be 10.63. 3. 192.168.100.17, with 4 bits of subnetting Ans: Subnet is 192.168.100.16, broadcast is 192.168.100.31 and valid host range is 192.168.100.17 to 30. 256-240=16. 16+16=32. The subnet is, then, 100.16 and the broadcast is 100.31 because 32 is the next subnet. 4. 192.168.100.66, with 3 bits of subnetting Ans: Subnet is 192.168.100.64, broadcast is 192.168.100.95 and valid host range is 192.168.100.65 to 94. 256-224=32. 32+32=64, plus 32=96. The subnet is 100.64 and the broadcast is 100.95. Question3: You have been asked to create a subnet that supports 16 hosts. What subnet mask should you use? 1. 255.255.255.252 2. 255.255.255.248 3. 255.255.255.240 4. 255.255.255.224 4 is correct. A will only support 2 hosts; B only 6 and C only 14. Watch out for the minus 2 in the host calculation! Answer C creates 16 hosts on the subnet, but we lose 2 -- one for the NET ID and one for the Broadcast ID. Question4: What valid host range is the IP address 172.29.217.11/22 a part of? 3201 Computer Networks – 2014/2015 Handout: Subnetting Question Network= 172.29.216.0 Range= 172.29.216.1 to 172.29.219.254 Broadcast= 172.29.219.255 Next Network= 172.29.220.0 Question5: What is the Network ID, Broadcast Address, First Usable IP, or Last Usable IP on the subnetwork that the node 192.168.1.15/26 belongs to? subnet mask is 255.255.255.192 Magic Number is 64  Network ID (First IP in the subnet): 192.168.1.0  Broadcast address (last IP in the subnet): 192.168.1.63  First Usable IP (the address after the network ID): 192.168.1.1  Last Usable IP (the address before the broadcast address): 192.168.1.62 networks are 192.168.1.0, 192.168.1.64, 192.168.1.128, and 192.168.1.192. so it belongs to first network Question6: Enter the last valid host on the network that the host 172.30.118.230/23 is a part of: subnet mask is 255.255.254.0. Magic Number is 2  Network ID (First IP in the subnet): 172.30.118.0  Broadcast address (last IP in the subnet): 172.30.119.255  First Usable IP (the address after the network ID): 172.30.118.1  Last Usable IP (the address before the broadcast address): 172.30.119.254  this is the answer Question7: How many subnets and hosts per subnet can you get from the network 192.168.1.0 255.255.255.224?  Subnet Bits = 2^3 = 8  Host Bits = 2^5-2 = 30 The answer is 8 subnets and 30 hosts per subnet. 3201 Computer Networks – 2014/2015 Handout: Subnetting Question Question8: Given an IP address & Subnet Mask, 192.168.1.58 255.255.255.240 Identify the original range of addresses that this IP address belongs to 255.255.255.240 = 11111111.11111111.11111111.11110000 - As before, the last possible network bit is the Magic Number 16 - Use this magic # to find the network ranges until passing the given IP address: 192.168.1.0 192.168.1.16 192.168.1.32 192.168.1.48 192.168.1.64 (passed given IP address 192.168.1.58) - Now, fill in the end ranges to find the answer to the scenario: 192.168.1.0 – 192.168.1.15 192.168.1.16 – 192.168.1.31 192.168.1.32 – 192.168.1.47 192.168.1.48 – 192.168.1.63 (IP address 192.168.1.58 belongs to this range) Question9: XYZ Company would like to subnet its network so that there are five separate subnets. They will need 25 computers in each subnet. Complete each ofthe following: NOTE: If you create more than five subnets, list the extra ones too. 3201 Computer Networks – 2014/2015 Handout: Subnetting Question Question10: The Acme Company would like to subnet its network (195.5.5.0) so that there are 50 separate subnets. They will need only 2 hosts in each subnet. Complete each ofthe following: NOTE: Because ther

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