From: Richard Whitehouse Date: Tue, 17 May 2011 13:30:12 +0000 (+0100) Subject: Add captions X-Git-Url: https://git.richardwhiuk.com/?a=commitdiff_plain;h=14de17e5a046bb54273f420fd76104d5baab1389;p=ii-diss.git Add captions --- diff --git a/evaluation.tex b/evaluation.tex index 8c5ff50..322c0cf 100644 --- a/evaluation.tex +++ b/evaluation.tex @@ -10,6 +10,7 @@ The first section of trials was done on the following network. These were done t \begin{figure}[h] \includegraphics[width=8cm]{simulation-1} +\caption{Operational Validation Network} \end{figure} In this set of tests, the first sends a packet, and then a second later, the second host replies. @@ -40,12 +41,14 @@ Here we have a ring network topology with 25 switches in the ring. Switch 12 and \begin{figure} \includegraphics[width=6cm]{simulation-4} +\caption{Unicast Routing Network} \end{figure} The network has been designed to exploit the worst case of the Ethernet protocol. Namely the spanning tree is constructed such that the the nodes are at the bottom of different branches of the tree. As such, the direct link between switches 12 and 13 is disabled, and the packets must traverse the entire ring. \begin{figure} \includegraphics[width=2cm]{simulation-4-tree} +\caption{Unicast Routing Spanning Tree} \end{figure} Due to this, the Ethernet protocol performs very badly here, requiring 26 frames in each direction for the UDP packet and the ARP Response packet - giving a total of 52 unicast UDP packets, and 52 unicast ARP packets, combining to make a total of 104 unicast frames. @@ -60,6 +63,7 @@ Here we have a mesh network topology with 25 switches in the mesh. Switch 12 and \begin{figure} \includegraphics[height=10cm]{simulation-5} +\caption{Broadcast Network} \end{figure} Using broadcast in a mesh network is problematic for MOOSE, as it utilises Reverse Path Forwarding to prevent broadcast storms, whereas Ethernet uses the Rapid Spanning Tree Protocol. s such, each broadcast packet is sent to the next switch, before it is culled in MOOSE, while in Ethernet, it doesn't need to send it any further as it knows the entire topology.