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Dr. Bob is Robert Hsiung, MD,
bob@dr-bob.orgShown: posts 1 to 9 of 9. This is the beginning of the thread.
Posted by Klavot on June 29, 2007, at 14:37:01
I propose a block length formula of the following form:
for a first offence,
B = S,
while for a repeat offence,
B = S(exp(-P/D) + k),
where exp(x) refers to the exponential function, B is the block length to be determined, S is the severity of the current incivility, P is the time passed since the previous block ended, D is the duration of the previous block, and k >= 1 is a constant to be fixed. All times are measured in weeks, and rounding only happens after the entire term has been calculated.
In nature, growth and decay happen exponentially. It seems to me that the extent to which previous incivility can be used against a poster should diminish exponentially, hence the term exp(-P/D). This means that after a long period of good behaviour, the poster will essentially have a clean slate again. Serious prior incivility will be reflected through larger values for D and will thus still factor adversely.
In criminal justice systems, previous crimes committed can always be used in aggrevation of sentencing. Hence adding the term k. Thus, as exp(-P/D) tends to 0, previous incivility can still be factored. The precise value of k can be determined by running some hypothetical cases and finding a value that works nicely.
I feel that the severity of the *present* incivility should be the main factor in determining block length, hence the multiplication with S. S can be any numerical value that Dr Bob feels fits the severity of the offence, and this allows for a certain amount of subjectivity. Greater values of S correspond with more serious incivility.
Klavot
Posted by Klavot on June 29, 2007, at 14:46:12
In reply to Block length formula, posted by Klavot on June 29, 2007, at 14:37:01
On second thought, the formula should look something more like
B = S(D*exp(-P) + k).
(In the original formula, D would have factored to the benefit of the poster, which it clearly shouldn't.)
Klavot
Posted by Klavot on June 29, 2007, at 14:54:06
In reply to Re: Correction, posted by Klavot on June 29, 2007, at 14:46:12
Actually the formula should look more like this:
B = S(D*exp(-P/r) + k),
where r is another constant that must be fixed based on running some hypothetical cases and finding a value that works nicely. Without diluting the value P, the term exp(-P) vanishes too quickly.
Klavot
Posted by Sigismund on June 29, 2007, at 15:49:32
In reply to Re: Correction, posted by Klavot on June 29, 2007, at 14:54:06
I'd love to know more about this.
Could you arrange to give an online tutorial on it, with a bridging element for those who have forgotten their maths?
Posted by muffled on June 29, 2007, at 16:13:28
In reply to Re: Correction » Klavot, posted by Sigismund on June 29, 2007, at 15:49:32
work for a wide range of stuff, and still keep the blocks relatively short generally speaking?
I'm not good at math. Maybe you could post an example or two?
thanks,
M
Posted by Sigismund on June 29, 2007, at 16:57:35
In reply to Hey Klavot, you think these number things might » Sigismund, posted by muffled on June 29, 2007, at 16:13:28
I check the boards from the bottom up.
Your post (but not the maths, of course) makes sense to me now.
Posted by Klavot on June 29, 2007, at 17:40:40
In reply to Re: Hey Klavot, you think these number things migh, posted by Sigismund on June 29, 2007, at 16:57:35
In general, lower values of r and k will give more lenient blocks.
Suppose we fix r = 10 and k = 1. Thus the formula becomes
B = S(D*exp(-P/10) + 1).
Here are a few examples.
(1) (Serious incivility against a history of poor behaviour.) John Doe commits a serious incivility and Dr Bob assigns S = 3. John's previous block length was 26 weeks and this block expired 12 weeks ago. Then
B = 3(26*exp(-12/10) + 1) = 26 weeks.
(2) (Minor incivility against a history of poor behaviour.) Jane Doe commits a minor incivility and Dr Bob assigns S = 1. Jane's previous block length was 26 weeks and this block expired 12 weeks ago. Then
B = 1(26*exp(-12/10) + 1) = 9 weeks.
(3) (Serious incivility against a history of good behaviour.) Jill Doe commits a major incivility and Dr Bob assigns S = 3. Jill's previous block length was 2 weeks and this block expired 16 weeks ago. Then
B = 3(2*exp(-16/10) + 1) = 4 weeks.
(4) (Minor incivility against a history of good behaviour.) James Doe commits a minor incivility and Dr Bob assigns S = 1. James's previous block length was 2 weeks and this block expired 16 weeks ago. Then
B = 1(2*exp(-16/10) + 1) = 1 week.
I gather that Dr Bob's current formula is ((duration of previous block) - (time since previous block) / 10) * 3, although I am not entirely sure about this. Sometimes he triples the block, sometimes not. Under this formula, we would get the following block lengths:
(1) (26 - 12/10)*3 = 75 weeks.
(2) (26 - 12/10)*3 = 75 weeks.
(3) (2 - 16/10)*3 = 3 weeks.
(4) (2 - 16/10)*3 = 3 weeks.
Klavot
Posted by Klavot on June 30, 2007, at 2:54:18
In reply to Re: Examples, posted by Klavot on June 29, 2007, at 17:40:40
To recap, the formula would look something like this:
B = S(D*exp(-P/r) + k),
where r,k >= 1 are constants to be fixed in advance. Lower values of r and k will give more lenient blocks. It makes sense that k = 1. The values for r and k might be adjusted anually or bi-anually if the need arises.
This formula works even for a first block, since then D = 0 = P so that D*exp(-P/r) = 0 and we are left with B = S.
In the previous examples I took r = 10, k = 1. Suppose we feel that those blocks are too severe. Here are those same cases but involving r = 7 and k = 1.
(1) (Serious incivility against a history of poor behaviour.) John Doe commits a serious incivility and Dr Bob assigns S = 3. John's previous block length was 26 weeks and this block expired 12 weeks ago. Then
B = 3(26*exp(-12/7) + 1) = 17 weeks.
In the previous case we got 26 weeks.
(2) (Minor incivility against a history of poor behaviour.) Jane Doe commits a minor incivility and Dr Bob assigns S = 1. Jane's previous block length was 26 weeks and this block expired 12 weeks ago. Then
B = 1(26*exp(-12/7) + 1) = 6 weeks.
In the previous case we got 9 weeks.
(3) (Serious incivility against a history of good behaviour.) Jill Doe commits a major incivility and Dr Bob assigns S = 3. Jill's previous block length was 2 weeks and this block expired 16 weeks ago. Then
B = 3(2*exp(-16/7) + 1) = 4 weeks.
In the previous case we got 4 weeks as well.
(4) (Minor incivility against a history of good behaviour.) James Doe commits a minor incivility and Dr Bob assigns S = 1. James's previous block length was 2 weeks and this block expired 16 weeks ago. Then
B = 1(2*exp(-16/7) + 1) = 1 week.
In the previous case we got 1 week as well.
As you can see from the above examples, the effect of the exponential function is to reduce what would otherwise be excessively long blocks, while leaving shorter blocks in tact.
Klavot
Posted by Klavot on June 30, 2007, at 6:30:18
In reply to Re: Examples, posted by Klavot on June 30, 2007, at 2:54:18
Since my last post, I've had some more ideas on how to improve the block length formula. The formula I previously proposed was
B = S(D*exp(-P/r)+k).
Multiplying out the terms gives
B = S*D*exp(-P/r) + S*k.
The first terms implies that the severity of the current incivility is also applied to previous incivility, which clearly isn't right. So I propose a slightly different formula, one which is both simpler and eliminates the need for the constant k:
B = S + D*exp(-P/r).
Again exp(x) refers to the exponential function, r is a constant >= 1 to be determined in advance (lower values of r will give more lenient blocks), B is the block length to be determined, S is the severity of the current incivility (a numerical value >= 1 to be assessed by Dr Bob; greater values of S correspond with more serious acts of incivility), D is the duration of the previous block, and P is the time passed since the previous block expired. All time lengths are measured in weeks and rounding only happens after the entire term has been calculated.
Klavot
This is the end of the thread.
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