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Consider the network below, which shows four $10 \hspace{0.1cm} Mbps$ LANs connected by two bridges, labelled $B1$ and $B2$. Assume all users (labelled  $1$ through $7$) are very chatty and equally chatty.

  1. What is the effective data rate seen by user $4$ ?
  2. What is the effective data rate seen by user $5$ ?
  3. What is the effective data rate seen by user $6$ ?
  4. What is the effective data rate seen by user $6$ if the two bridges are replaced with hubs ?
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A bridge is used to connect two networks. It usually has two ports such that a host on one side perceives the entire network on the other side as a single host and hence it reduces the collision domain.

On the other hand, a hub has multiple input and output ports, one each to a host. It forwards all the data it received, hence a hub does not reduce the collision domain and has a lower effective data rate.

Given that the data rate is $10\space Mbps$ for all LANs.

$a.$ The user $4$ see the two Bridges $B1$ and $B2$, and host $3$, and host $4$ itself.

So, the effective data rate is $\frac{10\space Mbps}{4} = 2.5\space Mbps$.

$b.$ The user $5$ can only see the Bridge $B2$, and host $5$ itself.

So, the effective data rate is $\frac{10\space Mbps}{2} = 5\space Mbps$.

$c.$ The user $6$ can see Bridge $B2$, host $7$ and host $6$ itself.

So, the effective data rate is $\frac{10\space Mbps}{3} = 3.33\space Mbps$.

$d.$ If the bridges $B1$ and $B2$ are replaced with hubs, all the hosts could be identified by the user $6$. Hence total number of hosts the user see is $7$.

So, the effective data rate is $\frac{10\space Mbps}{7} = 1.43\space Mbps$.

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