Question: begin { tabular } { l } hline multicolumn { 1 } { | l | } { Table 8 - 1

\begin{tabular}{l}
\hline \multicolumn{1}{|l|}{ Table 8-12 Subnet information for Packet Tracer network }\\
\begin{tabular}{|l|l|l|l|}
\hline & \\
\hline Subnet number & Network ID & Range of host addresses & Broadcast address \\
\hline 1 & 192.168.43.0 & & \\
\hline 2 & & & \\
\hline 3 & & & \\
\hline 4 & & & \\
\hline 5 & & & \\
\hline 6 & & & \\
\hline 7 & & & \\
\hline 8 & & & \\
\hline 9 & & & \\
\hline
\end{tabular}
\end{tabular}
Three of these subnets only need two host addresses, because they connect only two routers. Let's take the first subnet here and divide it again into three additional, smaller subnets. Answer the following questions:
15. If you borrow one more bit from the host portion of the IP address in Subnet 1, how many smaller subnets will this create? Is this enough? 16. If you borrow two more bits from the host portion of the IP address in Subnet 1, how many smaller subnets will this create? Is this enough? 17. What's the new subnet mask for these smaller subnets? 18. How many hosts can each of these smaller subnets have? 19. Fill in Table 8-13 with the smaller subnets' information. The first one is filled in for you.
\begin{tabular}{l}
\hline \multicolumn{1}{|l|}{ Table 8-13 Smaller subnets for router-to-router connections }\\
\begin{tabular}{|l|l|l|l|}
\hline Subnet number & Network ID & Range of host addresses & Broadcast address \\
\hline 1A & 192.168.43.0 & \(192.168.43.1-192.168.43.2\) & 192.168.43.3\\
\hline 1B & & & \\
\hline 1C & & & \\
\hline 1D & & & \\
\hline
\end{tabular}
\end{tabular}
\ begin { tabular } { l } \ hline \ multicolumn {

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