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
poster:Klavot
thread:766717
URL: http://www.dr-bob.org/babble/admin/20070605/msgs/766768.html