1. **Problem Statement:** We are given the IP address 220.230.20.193/26 and need to subnet it to create 2 usable subnets with the maximum number of addresses. 2. **Understanding the original subnet:** The original subnet mask is /26, which means 26 bits are for the network and 6 bits for hosts. 3. **Calculate the number of subnets and hosts:** - Original subnet mask: /26 - Number of host bits: 32 - 26 = 6 - Number of hosts per subnet: $2^6 - 2 = 62$ usable hosts 4. **Subnetting to create 2 subnets:** - To create 2 subnets, we need to borrow 1 bit from the host portion. - New subnet mask: $26 + 1 = 27$ - New subnet mask in dotted decimal: 255.255.255.224 5. **Number of hosts per new subnet:** - Host bits now: $32 - 27 = 5$ - Usable hosts per subnet: $2^5 - 2 = 30$ 6. **Calculate subnet addresses:** - Block size for /27: $2^{32-27} = 32$ - Subnets start at multiples of 32 in the last octet. 7. **Find the first subnet:** - Network address of original: 220.230.20.192 (since 193 falls in 192-223 range) - First subnet: 220.230.20.192/27 - Usable IP range: 220.230.20.193 to 220.230.20.222 - Broadcast address: 220.230.20.223 8. **Find the second subnet:** - Second subnet network address: 220.230.20.224/27 - Usable IP range: 220.230.20.225 to 220.230.20.254 - Broadcast address: 220.230.20.255 9. **Answering the questions:** I. Subnet mask for the first usable subnet: 255.255.255.224 (/27) II. Network address of the first subnet: 220.230.20.192 III. Broadcast address of the second subnet: 220.230.20.255 IV. First usable IP address of the second subnet: 220.230.20.225 **Final answers:** - Subnet mask: 255.255.255.224 - First subnet network address: 220.230.20.192 - Second subnet broadcast address: 220.230.20.255 - First usable IP of second subnet: 220.230.20.225