This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.
ScanSolve AI Answer
Here's how to calculate the IPv4 subnet details: Given: Host IP Address: 192.168.200.139 Original Subnet Mask: 255.255.255.0 New Subnet Mask: 255.255.255.224 First, let's convert the new subnet mask to binary and determine its CIDR notation. 255.255.255.224 In binary: 11111111.11111111.11111111.11100000 The number of '1's is 8 + 8 + 8 + 3 = 27. So, the new subnet mask is /27. The original subnet mask 255.255.255.0 is /24. Step 1: Calculate the Number of Subnet Bits. The number of subnet bits is the difference between the new network bits and the original network bits. Number of Subnet Bits = New CIDR - Original CIDR = 27 - 24 = 3 • Number of Subnet Bits: 3 Step 2: Calculate the Number of Subnets Created. The number of subnets created is 2^number of subnet bits. Number of Subnets Created = 2^3 = 8 • Number of Subnets Created: 8 Step 3: Calculate the Number of Host Bits per Subnet. An IPv4 address has 32 bits. The number of host bits is 32 - new network bits. Number of Host Bits per Subnet = 32 - 27 = 5 • Number of Host Bits per Subnet: 5 Step 4: Calculate the Number of Hosts per Subnet. The number of usable hosts per subnet is 2^number of host bits - 2 (subtracting the network address and broadcast address). Number of Hosts per Subnet = 2^5 - 2 = 32 - 2 = 30 • Number of Hosts per Subnet: 30 Step 5: Determine the Network Address of this Subnet. To find the network address, perform a bitwise AND operation between the Host IP Address and the New Subnet Mask. Host IP: 192.168.200.139 New Subnet Mask: 255.255.255.224 The first three octets are straightforward: 192.168.200. For the fourth octet: Host IP fourth octet (139): 10001011_2 Subnet Mask fourth octet (224): 11100000_2 Bitwise AND: c 10001011_2 \\ AND 11100000_2 \\ 10000000_2 10000000_2 is 128 in decimal. So, the Network Address is 192.168.200.128. • Network Address of this Subnet: 192.168.200.128 Step 6: Determine the IPv4 Address of First Host on this Subnet. The first host address is the network address plus one. First Host Address = 192.168.200.128 + 1 = 192.168.200.129 • IPv4 Address of First Host on this Subnet: 192.168.200.129 Step 7: Determine the IPv4 Broadcast Address on this Subnet. The block size for the fourth octet is 256 - 224 = 32. The subnets in the fourth octet are multiples of 32: 0, 32, 64, 96, 128, 160, 192, 224. Since the host IP 139 falls into the 128 block, the next network address is 160. The broadcast address for the 192.168.200.128 subnet is one less than the next network address. Broadcast Address = 192.168.200.160 - 1 = 192.168.200.159 • IPv4 Broadcast Address on this Subnet: 192.168.200.159 Step 8: Determine the IPv4 Address of Last Host on this Subnet. The last host address is the broadcast address minus one. Last Host Address = 192.168.200.159 - 1 = 192.168.200.158 • IPv4 Address of Last Host on this Subnet: 192.168.200.158 Here is the completed table: | Find: | Answer: | | :---------------------------------- | :---------------------- | | Number of Subnet Bits | 3 | | Number of Subnets Created | 8 | | Number of Host Bits per Subnet | 5 | | Number of Hosts per Subnet | 30 | | Network Address of this Subnet | 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 |