Defining routing topology with RGCs

Hello all.

Exchange Server 2003.
Based on the image below, what would be the next-hop server for a server in Routing Group X who is sending an e-mail to a server in Routing Group Z?

My considerations:
the RGC X-Z has an higher cost then the X-Y, but the total cost for reaching the Routing Group Z, is the same at the last;
so, what will be the next hop Routing Group in this case?
Has the hop count some value?

Found these particulars nowhere.

If someone can answer is a god.

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Comments

meadIT

Logically, I would think it would take the one hop route because there would be less of a chance of failure (1 chance versus 2 chances), but I really have no idea. Now where's royal?

hollow666

I know in exch 2007 it would take the lowest hop count route if all the costs tie not exactly sure about 2003 let me dig up some of my whitepapers.

For 2007:

Examine all possible routing paths across routing group connectors and select the routing path that has the least total routing group cost (this ignores the AD site link costs).
If more than one routing path has the same cost, examine all possible routing paths across IP site links to reach the first routing group connector, and select the routing path that has the lowest total IP site link cost. In our example, if the Dallas routing group connector has the same cost as the London routing group connector, since the IP site link for the London-based Exchange 2007 server is less for a London Exchange 2007 server, the London routing group connector would be used.
If more than one routing path has the same routing group cost, and it has the same IP site link cost, select the routing path that includes the least number of hops.
If more than one routing path has the same routing group cost, the same IP site link cost, and the same number of hops, select the routing path where the name of the last AD site before the destination site has the lowest alphanumeric value.

hollow666

This specifially mentions SMTP connectors not RGC but I'm assuming the behaviour would be the same here it is:

If you use two connectors with the same cost, Exchange servers randomly select which bridgehead server and connector to use.

Of course once it chooses one it will always use that connector until it becomes unavailable then it will fail over to the next

source technet
Exchange Server 2003 Message Routing

For me the answer is murky because I haven't really read anything that proves one way or the other. I'd be instrested to see what anyone else has. You would assume it would try to pick the lowest hop route.

In an organization with multiple routing groups, various routes might lead to the same destination. Typically, the most efficient (that is, the shortest or cheapest) route is used for message transfer, and additional routes stand by, in case the best route is temporarily unavailable. For example, in the topology shown in the following figure, multiple transfer routes exist between all routing groups."

source technet
Exchange Server 2003 Message Routing

rjbarlow

Great contribution hollow.

Claymoore

Here is a reference to Exchange 2000 using a simplified form of Dijkstra's algorithm to build the Link State routing table:

Think of Status node connector information as roughly equivalent to a snapshot of the routing group's Link State Table, which the SMTP Routing Engine maintains in memory and which typically isn't visible to an administrator. (You can use the Winroute utility, in the \support\utils\i386 directory of the Exchange 2000 CD-ROM, to gain a more comprehensive view of the Link State Table.) Exchange 2000 uses a mechanism called Link State Routing as the basis for routing decisions for messages. This device replaces the more static view of available routes that the Exchange Server 5.5 Gateway Address Routing Table (GWART) implements. Link State Routing uses a simplified form of Dijkstra's algorithm to ensure that messages follow the optimal path to their destination. The SMTP Routing Engine bases the decisions it takes to find that path on the data in the Link State Table, which Exchange 2000 updates dynamically as the underlying network changes or as you add new connections. For example, if a network link becomes inoperative and prevents an SMTP connector from sending messages to a specific SMTP domain, the routing group that discovers the failure sends link state messages reporting the failure to all the other routing groups in the Exchange Server organization. Then, the routing master in each group generates new routing data.
Managing Exchange 2000

Dijkstra'a algorithm (or some variant) is used by most Link-State routing protocols like OSPF. Here is the key point to remember - For any given destination node, the best next hop for that destination is the node which is the first step from the root node, down the branch in the shortest-path tree which leads toward the desired destination node.
Link-state routing protocol - Wikipedia, the free encyclopedia

I say the direct route will be used, but if you want to lab it out to be sure, you could prove it with the Winroute utility mentioned in the Technet article.

rjbarlow

Thank You much.

doom969

This is some really interesting info.
Its got me curious, i think i will lab it this week.
Maybe if we're lucky, royal will pitch some useful info too.

thanks to all who contributed !

rjbarlow

Let us know if you proceed.