IST_220_ASSIGN_IPv4_Subnet_P1_2017ChamousSuber

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Greenville Technical College *

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220

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Electrical Engineering

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Apr 3, 2024

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IST 220 Data Communications Name _Chamous Suber____ IPv4 Address Subnetting - #1 Subnetting is all about modifying the subnet mask in order to create smaller networks (subnets) from one large network. Given a network address and a subnet mask:. 1. Determine the number of subnets needed. (In this case, it was given to you. Other times you will look at a layout.) 2. Determine how many host bits you need to borrow from the original subnet mask 3. Determine the new subnet mask, and then determine the increment. 4. Determine the subnet address for each subnet.. 5. Determine the broadcast address for each subnet. 6. Determine the range of usable host IP Addresses for each subnet. a. Determine the IPv4 Address of the First Host on this Subnet. b. Determine the IPv4 Address of the Last Host on this Subnet. 7. Determine number of hosts per subnet Problem A: You need to create 8 subnets. Network Address: 192.168.200.0 Original Subnet Mask -> CIDR -> binary 255.255.255.0 -> /24 -> 11111111 11111111 11111111 00000000 New Subnet Mask: 255.255.255.224 ->/27 -> 11111111 11111111 11111111 111 00000 Find: Number of Subnet Bits borrowed from the original SM to create the new SM 3 Number of Subnets Created Subnets = 2 n where n is the number of bits borrowed. 2 3 = 8 subnets Number of Host Bits left in the new subnet mask 5 # of Hosts per Subnet = 2 n -2 Where n is the number of zeros (host bits) in the new subnet mask. The minus 2 is because the network ID and the Broadcast IP address each reserve an IP address 2 5 -2 = 32-2 = 30 usable host IP addresses +32 increment – the value of the lowest bit borrowed 128 64 32 16 8 4 2 1

IST 220 Data Communications Name _Chamous Suber____ IPv4 Address Subnetting - #1 Network Address of the 1st Subnet This is always the network address you were given to start with 192.168.200.0 with a subnet mask of /27 IPv4 Address of First Host on this Subnet (one more than the network ID) 192.168.200.1 IPv4 Address of Last Host on this Subnet (one less than the Broadcast address) 192.168.200.30 IPv4 Broadcast Address on this Subnet (It is always 1 less than the next subnet ID) 192.168.200.31 Fill-in the remaining tables with the subnet information: Network Address of 2nd Subnet Find the network address of the 2 nd subnet by adding the increment to the previous subnet in the octet that you found it The increment is the values of the lowest bit that you borrowed in to create the new subnet mask.) 192.168.200.0 + 32 increment 192.168.200.32 with a subnet mask of /27 IPv4 Address of First Host on this Subnet 192.168.200.33 IPv4 Address of Last Host on this Subnet 192.168.200.62 IPv4 Broadcast Address on this Subnet (Hint: Find the next subnet.) 192.168.200.63 Network Address of 3rd Subnet Find the network address of the 3rd subnet by adding the increment to the previous subnet in the octet that you found it The increment is the values of the lowest bit that you borrowed in to create the new subnet mask.) 192.168.200.64 /27 IPv4 Address of First Host on this Subnet 192.168.200.65 IPv4 Address of Last Host on this Subnet 192.168.200.94 IPv4 Broadcast Address on this Subnet 192.168.200.95

IST 220 Data Communications Name _Chamous Suber____ IPv4 Address Subnetting - #1 Network Address of 4th Subnet Find the network address of the 4th subnet by adding the increment to the previous subnet in the octet that you found it The increment is the values of the lowest bit that you borrowed in to create the new subnet mask.) 192.168.200.96 IPv4 Address of First Host on this Subnet 192.168.200.97 IPv4 Address of Last Host on this Subnet 192.168.200.126 IPv4 Broadcast Address on this Subnet 192.168.200.127 Network Address of 5th Subnet Find the network address of the 5th subnet by adding the increment to the previous subnet in the octet that you found it. The increment is the values of the lowest bit that you borrowed in to create the new subnet mask. 192.168.200.128 IPv4 Address of First Host on this Subnet 192.168.200.129 IPv4 Address of Last Host on this Subnet 192.168.200.158 IPv4 Broadcast Address on this Subnet 192.168.200.159 Network Address of 6th Subnet Find the network address of the 6th subnet by adding the increment to the previous subnet in the octet that you found it The increment is the values of the lowest bit that you borrowed in to create the new subnet mask.) 192.168.200.160 IPv4 Address of First Host on this Subnet 192.168.200.161 IPv4 Address of Last Host on this Subnet 192.168.200.190 IPv4 Broadcast Address on this Subnet 192.168.200.191

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IST 220 Data Communications Name _Chamous Suber____ IPv4 Address Subnetting - #1 Network Address of 7th Subnet Find the network address of the 7th subnet by adding the increment to the previous subnet in the octet that you found it. The increment is the values of the lowest bit that you borrowed in to create the new subnet mask.) 192.168.200.192 IPv4 Address of First Host on this Subnet 192.168.200.193 IPv4 Address of Last Host on this Subnet 192.168.200.222 IPv4 Broadcast Address on this Subnet 192.168.200.223 Network Address of 8th Subnet Find the network address of the 8th subnet by adding the increment to the previous subnet in the octet that you found it. The increment is the values of the lowest bit that you borrowed in to create the new subnet mask.) 192.168.200.224 IPv4 Address of First Host on this Subnet 192.168.200.225 IPv4 Address of Last Host on this Subnet 192.168.200.254 IPv4 Broadcast Address on this Subnet 192.168.200.255 There are no more subnets after the 8th subnet. So to find the last Broadcast Address, write the last octet of the network address in binary. 192.168.200.224 -> 224 = 111 00000 The host bits are always all zeros in the network address. Now change all the host bits (the zeros) to ones. 192.168.200.??? -> 111 11111 The host bits are always all ones in the broadcast address. Hence the last Broadcast Address for this example = 192.168.200.255

IST 220 Data Communications Name _Chamous Suber____ IPv4 Address Subnetting - #1 Problem B: You have been given the 192.168.10.0/24 network address to subnet, with the following topology. Determine the number of networks needed and then design an appropriate addressing scheme. Step 1: Determine the number of subnets in Network Topology A. 1) How many subnets are there? 2 2) How many bits should you borrow to create the required number of subnets? 2 3) How many usable host addresses per subnet are in this addressing scheme? 62 Note: Usable host addresses = 2 (n) - 2 where n is the number of host bits left in the new subnet mask Router1 (R1) to Router2(R2) is to be counted as a subnet. The G0/1 interface on Router1(R1) is the default gateway to another subnet

IST 220 Data Communications Name _Chamous Suber____ IPv4 Address Subnetting - #1 4) What is the new subnet mask in dotted decimal format? 255.255.255.192 Dotted decimal Binary Original subnet mask = 255.255.255.0 11111111 11111111 11111111 00000000 New subnet mask = 255.255.255.??? 11111111 11111111 11111111 ???????? 5) What is the new CIDR notation for each new subnet? (original network = /24) 192.168.10.0/26 6) What is the increment between subnets? 2 7) How many subnets are available for future use? 1 Step 2: Record the subnet information. Fill in the following table with the subnet information: Subnet Number Subnet Address First Usable Host Address Last Usable Host Address Broadcast Address 1 st subnet Also called “Subnet 0” 192.168.10.0/26 192.168.10.1 192.168.10.62 192.168.10.63 2 nd subnet Also called “Subnet 1” 192.168.10.64/26 192.168.10.65 192.168.10.26 192.168.10.127

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