Subject: Re: Suggested resolution - #826: Section 4 - Removal of the IAOC Chair

From: [email-masked] (Scott Bradner)
Date: Fri, 28 Jan 2005 08:19:03 -0500 (EST)

Harald suggests
   The Chair serves at the pleasure of the IAOC, and may be removed from
   that position at any time by a vote of five of the IAOC voting members.

I don't think its a good idea to use absolute numbers - its better
to use fractions '4/5ths of the voting members' for example - in case
you have a situation where some IAOC members have dropped off for some 
reason - using absolute numbers can get into a situation where the
action can not be taken even though all existing members of the
IAOC want to do so

Scott

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From: Harald Tveit Alvestrand <[email-masked]>
Date: Fri, 28 Jan 2005 14:59:36 +0100

> action can not be taken even though all existing members of the

But that slides us straight back into the situation where we must make 
rules for whether or not people who are on holiday are counted or not, what 
constitutes a quorum, and so on. How do you say that 2/3 of a meeting that 
had only 4 of the IAOC members present is not acceptable?

It was the fraction "2/3" that Russ objected to in the first place, 
pointing out that this means 6 out of 8 if everyone's present - which he 
thought was too much of a required majority.

In the case where the IAOC is short 3 members (required for the situation 
you describe), I think we can live with a chair not being removed until the 
selecting bodies have named replacements for members who are no longer 
willing or able to serve (if that is the situation you are worried about).

                         Harald

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From: [email-masked] (Kai Henningsen)
Date: 29 Jan 2005 11:47:00 +0200

> had only 4 of the IAOC members present is not acceptable?

The phrase "all existing members" sort of suggests that we're talking  
about formal membership here, so people on holiday would definitely count  
- as in, "you can't remove the chair just because most of his supporters  
are on vacation". And given that, there's also no question of quorum I can  
see - the question isn't "2/3 of a meeting" but "2/3 of the members" -  
including absent members.

That is, you need enough votes to make it even if everyone who could vote  
was actually there. That way, it doesn't matter how many people actually  
did show up.

> thought was too much of a required majority.

Which just points to a lower fraction, not to absolute numbers.

> willing or able to serve (if that is the situation you are worried about).

That surely depends on the reason for removing that chair.

MfG Kai

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From: Harald Tveit Alvestrand <[email-masked]>
Date: Sun, 30 Jan 2005 21:42:32 +0100

> Which just points to a lower fraction, not to absolute numbers.

I understand that you and Scott both think the same thing - although I 
don't understand why, I'll ask the question in a different way - is using 
the term "5/8 of the voting members" an acceptable phrase?
As long as we have 8 voting members, that translates to 5 members.

For those who want to see what they are discussing under different 
scenarios, here is the number of people needed to remove the chair with 
varying fractions and varying number of members of the IAOC:

Number   Out of 8  Out of 7 Out of 6 Out of 5
3/4        6          6        5        4
2/3        6          5        4        4
5/8        5          5        4        4

                Harald

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From: Jeffrey Hutzelman <[email-masked]>
Date: Sun, 30 Jan 2005 19:52:35 -0500

On Sunday, January 30, 2005 21:42:32 +0100 Harald Tveit Alvestrand 
<[email-masked]> wrote:

>> e I
> the term "5/8 of the voting members" an acceptable phrase? As long as we
> have 8 voting members, that translates to 5 members.
>

Suggestion: Instead of "2/3 of the voting members", why not just say "2/3 
of the voting members, excluding the chair".  So if the IAOC is at full 
strength, it takes 5 people to remove the chair.