a) IPv4 and IPv6 are both Internet Protocol versions used to identify devices on a network.

a) IPv4 and IPv6 are both Internet Protocol versions used to identify devices on a network. • IPv4 uses 32-bit addresses, typically represented in dotted-decimal notation (e.g., 192.168.1.1). It supports approximately 4.3 × 10^9 unique addresses. • IPv6 uses 128-bit addresses, represented in hexadecimal notation with colons (e.g., 2001:0db8:85a3:0000:0000:8a2e:0370:7334). It supports a vastly larger number of addresses, approximately 3.4 × 10^38. b) To compute the appropriate subnet mask for each department, we need to find the smallest number of host bits (h) such that 2^h - 2 is greater than or equal to the number of hosts required. The subnet mask will then be 32 - h. The base network is 194.70.25.0/24. Step 1: Calculate subnet mask for Department A (52 hosts). We need 2^h - 2 52 2^h 54. The smallest power of 2 greater than or equal to 54 is 2^6 = 64. So, h=6 host bits. The subnet mask is 32 - 6 = /26. In dotted decimal: 255.255.255.(256 - 2^6) = 255.255.255.(256 - 64) = 255.255.255.192. The subnet mask for Department A is 255.255.255.192. Step 2: Calculate subnet mask for Department B (29 hosts). We need 2^h - 2 29 2^h 31. The smallest power of 2 greater than or equal to 31 is 2^5 = 32. So, h=5 host bits. The subnet mask is 32 - 5 = /27. In dotted decimal: 255.255.255.(256 - 2^5) = 255.255.255.(256 - 32) = 255.255.255.224. The subnet mask for Department B is 255.255.255.224. Step 3: Calculate subnet mask for Department C (28 hosts). We need 2^h - 2 28 2^h 30. The smallest power of 2 greater than or equal to 30 is 2^5 = 32. So, h=5 host bits. The subnet mask is 32 - 5 = /27. In dotted decimal: 255.255.255.(256 - 2^5) = 255.255.255.(256 - 32) = 255.255.255.224. The subnet mask for Department C is 255.255.255.224. Step 4: Calculate subnet mask for Department D (14 hosts). We need 2^h - 2 14 2^h 16. The smallest power of 2 greater than or equal to 16 is 2^4 = 16. So, h=4 host bits. The subnet mask is 32 - 4 = /28. In dotted decimal: 255.255.255.(256 - 2^4) = 255.255.255.(256 - 16) = 255.255.255.240. The subnet mask for Department D is 255.255.255.240. c) If Department D grows to 28 employees (hosts). We need 2^h - 2 28 2^h 30. The smallest power of 2 greater than or equal to 30 is 2^5 = 32. So, h=5 host bits. The subnet mask is 32 - 5 = /27. In dotted decimal: 255.255.255.(256 - 2^5) = 255.255.255.(256 - 32) = 255.255.255.224. The new subnet mask for Department D is 255.255.255.224. Send me the next one 📸