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 $4.3 \times 10^9$ 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 $3.4 \times 10^{38}$. 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 $2^h - 2 \ge 52 \implies 2^h \ge 54$. 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 $\boxed{255.255.255.192}$. Step 2: Calculate subnet mask for Department B (29 hosts). We need $2^h - 2 \ge 29 \implies 2^h \ge 31$. 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 $\boxed{255.255.255.224}$. Step 3: Calculate subnet mask for Department C (28 hosts). We need $2^h - 2 \ge 28 \implies 2^h \ge 30$. 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 $\boxed{255.255.255.224}$. Step 4: Calculate subnet mask for Department D (14 hosts). We need $2^h - 2 \ge 14 \implies 2^h \ge 16$. 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 $\boxed{255.255.255.240}$. c) If Department D grows to 28 employees (hosts). We need $2^h - 2 \ge 28 \implies 2^h \ge 30$. 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 $\boxed{255.255.255.224}$. Send me the next one 📸