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[CQ-Contest] For max total ponts go M/2 or two M/S?

To: cq-contest reflector <cq-contest@contesting.com>
Subject: [CQ-Contest] For max total ponts go M/2 or two M/S?
From: Jim Smith <jimsmith@shaw.ca>
Date: Fri, 11 Jan 2008 17:59:44 -0800
List-post: <mailto:cq-contest@contesting.com>
If I want to maximize the total points made by a station with 2 op 
positions, there is a choice between M/2 or M/M or two M/s under 
separate calls.

We're going here for max total points generated by the two op positions 
for our own obscure reasons.  I do understand that one cannot combine 
the scores from 2 M/S entries for any known (other than Poisson d'Avril) 
awards.

So, let's compare 2 M/S stations to a M/2 just for fun.  This assumes 
that both stations work on all bands, that the rigs and antennas are 
identical, the ops all have comparable skills, band conditions are as we 
find them these days at the bottom of the cycle and that there are no 
band change rules.  If there are, well you just go M/M.

For a start, 2 M/S is clearly more fun as the ops can get in more op 
time.  But, remember that we're trying to maximize score here.

If the M/2 can make 3,000 Qs then the same number of ops split between 2 
M/S stations should easily be able to make more than 3,000 between them 
because they have different calls and are, therefore, fresh meat when 
they take over a band, particularly if they alternate on the good bands 
instead of one stn working out 20m while the other one twiddles its 
thumbs on 15 or 40. Let's say the 2 M/S's make a total of 3500 Qs (1750 
each).

I think a similar argument holds for mults.  If the M/2 can make 200 
mults then the same number of ops on 2 M/S stations should easily be 
able to make more than 200 between them.  Let's say they make a total of 
300 (150 each).

Assuming that points = Qs times Mults

M/2 makes 3000 Qs times 200 mults = 600,000 points
Each M/S makes 1750 Qs times 150  = 262,500 points
So 2 M/S make                       525,000 points.

Well, the difference isn't as large as I expected.

If the M/S's each make 2,000 Qs then it's dead even.

Now, if the M/S's each make 2,000 Qs and 200 mults then they make 
400,000 times 2 = 800,000 points.  But, to get the 200 mults they have 
to be really efficient at swapping bands and knocking off the mults.

If the M/S's can operate on the same band at the same time and not cause 
any problems at all for each other then maybe they could each get pretty 
well the same number of Qs and mults as the M/2 as they could both be on 
the high rate bands at the same time.  This gives a score of 1,200,000.

Hmm... I see many filters and miles of coax stubs in the future.

Anyway, these are pretty idealistic assumptions.

Which way do you think will maximize the total points?

73, Jim VE7FO
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