Question on reducing load on MailScanner machine

[Christopher Fisk]

>  The previous thread about this didn't go very far because
>  I suspect
>  no-one is brave enough to actually try this.  Most of us
>  just either
>  optimise our installations to prevent the queue build-up
>  in the first
>  place or just add another box - it's the far less
>  dangerous and the most
>  travelled path.  That's also the reason it's not covered
>  in the book.

I can add another mx, but would prefer to have a single entry point for email on the network.  I think I would upgrade to a single more powerful machine rather than adding additional boxes for incoming messages.  We're running a relatively slow single processor machine at the moment and have a backup server that has been powered off sitting under it.

>  E-mail isn't instant messaging; a queue of 300 would
>  impose nothing more
>  than a few minutes delay at most which is perfectly
>  acceptable to most
>  people here.

I did a typo, the queue is reaching 3000+.  Sorry about that.  Your point still remains, but you would be suprised at how many of our customers send an email to someone while on the phone and having even a few minute wait annoys them.  I'd rather throw hardware at the problem to get the queue down to 0.


>  You're welcome to try NFS mounting your 'hold' directory
>  and running
>  another box on it at the same time; but you get to keep
>  all the pieces
>  if it breaks and to answer the phone to your users when
>  it goes wrong
>  and they get duplicate messages delivered to them or if
>  their important
>  mail get nuked.

There is actually a good chance I will test this out.  If I do I will inform of the results.


>  So my recommendation would be to avoid this; but if you
>  are going to try
>  it - do it on virtual machines and test it thoroughly
>  (and document it
>  for others too if it works!).

How do the various child processes of MailScanner know when another child process is scanning a message in the queue?  Does the parent process keep track?  It seems like this would logically work.

Looks like I will have to test it =)


Christopher Fisk

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