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Re: Examples

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


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