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1

$begingroup$

I'm forming an algorithm to assign priority weights to certain jobs that need to be done. I need to integrate a factor of Must be done by time. Meaning, this job HAS to be run by a specific time.

I'm calculating the minutes until that time, e.g. 120 minutes from now, but how do I turn the larger the number into a lesser weight.

Example

Job 1: 300 minutes from now

Job 2: 12 minutes from now

Job 3: 25 minutes from now

I'm trying to just take the minutes from now values, and multiply them by an arbitrary multiplier. The issue is obvious. The 300 minutes would weight the most and the 25 minutes would weigh the least. I need the opposite. What's a good solution to this issue? I'm sure there's a pretty basic equation for this, so excuse my non-math-knowing skills.

FYI - the values that I'm assigning are arbitrary right now. I'll just tweak them and run scenarios until I feel the bias' are correct.

linear-algebra algorithms scoring-algorithm

share|cite|improve this question

asked Nov 7 '18 at 15:57

JamesJames

1061

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could divide by the number of minutes instead of multiplying.
  $endgroup$
  – gandalf61
  Nov 7 '18 at 17:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just tried to take the upper bound by grabbing Max minutes, then subtracted the minutes from now from each job from the upper bound. It seems to be giving me what I'm looking for, though I feel like dividing may be the best bet. Any input?
  $endgroup$
  – James
  Nov 7 '18 at 18:02
  

  
  
  
  

add a comment | 

1

$begingroup$

I'm forming an algorithm to assign priority weights to certain jobs that need to be done. I need to integrate a factor of Must be done by time. Meaning, this job HAS to be run by a specific time.

I'm calculating the minutes until that time, e.g. 120 minutes from now, but how do I turn the larger the number into a lesser weight.

Example

Job 1: 300 minutes from now

Job 2: 12 minutes from now

Job 3: 25 minutes from now

I'm trying to just take the minutes from now values, and multiply them by an arbitrary multiplier. The issue is obvious. The 300 minutes would weight the most and the 25 minutes would weigh the least. I need the opposite. What's a good solution to this issue? I'm sure there's a pretty basic equation for this, so excuse my non-math-knowing skills.

FYI - the values that I'm assigning are arbitrary right now. I'll just tweak them and run scenarios until I feel the bias' are correct.

linear-algebra algorithms scoring-algorithm

share|cite|improve this question

asked Nov 7 '18 at 15:57

JamesJames

1061

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You could divide by the number of minutes instead of multiplying.
  $endgroup$
  – gandalf61
  Nov 7 '18 at 17:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I just tried to take the upper bound by grabbing Max minutes, then subtracted the minutes from now from each job from the upper bound. It seems to be giving me what I'm looking for, though I feel like dividing may be the best bet. Any input?
  $endgroup$
  – James
  Nov 7 '18 at 18:02
  

  
  
  
  

add a comment | 

$begingroup$

I'm forming an algorithm to assign priority weights to certain jobs that need to be done. I need to integrate a factor of Must be done by time. Meaning, this job HAS to be run by a specific time.

I'm calculating the minutes until that time, e.g. 120 minutes from now, but how do I turn the larger the number into a lesser weight.

Example

Job 1: 300 minutes from now

Job 2: 12 minutes from now

Job 3: 25 minutes from now

I'm trying to just take the minutes from now values, and multiply them by an arbitrary multiplier. The issue is obvious. The 300 minutes would weight the most and the 25 minutes would weigh the least. I need the opposite. What's a good solution to this issue? I'm sure there's a pretty basic equation for this, so excuse my non-math-knowing skills.

FYI - the values that I'm assigning are arbitrary right now. I'll just tweak them and run scenarios until I feel the bias' are correct.

linear-algebra algorithms scoring-algorithm

share|cite|improve this question

asked Nov 7 '18 at 15:57

JamesJames

1061

$endgroup$

share|cite|improve this question

asked Nov 7 '18 at 15:57

JamesJames

1061

share|cite|improve this question

asked Nov 7 '18 at 15:57

JamesJames

1061

asked Nov 7 '18 at 15:57

JamesJames

1061

asked Nov 7 '18 at 15:57

JamesJames

1061

1 Answer
1

active

oldest

votes

0

$begingroup$

I think you intended inverse proportionality.

Given a set of jobs indexed with $j in {1, 2, ldots, n}$, and given time limits $t_j in mathbb{R}$ for each job $j$, the weights $w_j$ can be defined in the following way:
$$
w_j = displaystyle frac{1}{t_j}.
$$

Example

Job $1$: $300$ minutes from now

Job $2$: $12$ minutes from now

Job $3$: $25$ minutes from now

In this case the weights will be:
$$
begin{align}
w_1 &= frac{1}{300} approx 0.00333\
w_2 &= frac{1}{12} approx 0.08333\
w_3 &= frac{1}{25} = 0.4\
end{align}
$$

share|cite|improve this answer

answered Mar 14 at 21:00

simonetsimonet

370111

$endgroup$

add a comment | 

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0

$begingroup$

I think you intended inverse proportionality.

Given a set of jobs indexed with $j in {1, 2, ldots, n}$, and given time limits $t_j in mathbb{R}$ for each job $j$, the weights $w_j$ can be defined in the following way:
$$
w_j = displaystyle frac{1}{t_j}.
$$

Example

Job $1$: $300$ minutes from now

Job $2$: $12$ minutes from now

Job $3$: $25$ minutes from now

In this case the weights will be:
$$
begin{align}
w_1 &= frac{1}{300} approx 0.00333\
w_2 &= frac{1}{12} approx 0.08333\
w_3 &= frac{1}{25} = 0.4\
end{align}
$$

share|cite|improve this answer

answered Mar 14 at 21:00

simonetsimonet

370111

$endgroup$

add a comment | 

0

$begingroup$

I think you intended inverse proportionality.

Given a set of jobs indexed with $j in {1, 2, ldots, n}$, and given time limits $t_j in mathbb{R}$ for each job $j$, the weights $w_j$ can be defined in the following way:
$$
w_j = displaystyle frac{1}{t_j}.
$$

Example

Job $1$: $300$ minutes from now

Job $2$: $12$ minutes from now

Job $3$: $25$ minutes from now

In this case the weights will be:
$$
begin{align}
w_1 &= frac{1}{300} approx 0.00333\
w_2 &= frac{1}{12} approx 0.08333\
w_3 &= frac{1}{25} = 0.4\
end{align}
$$

share|cite|improve this answer

answered Mar 14 at 21:00

simonetsimonet

370111

$endgroup$

add a comment | 

$begingroup$

I think you intended inverse proportionality.

Given a set of jobs indexed with $j in {1, 2, ldots, n}$, and given time limits $t_j in mathbb{R}$ for each job $j$, the weights $w_j$ can be defined in the following way:
$$
w_j = displaystyle frac{1}{t_j}.
$$

Example

Job $1$: $300$ minutes from now

Job $2$: $12$ minutes from now

Job $3$: $25$ minutes from now

In this case the weights will be:
$$
begin{align}
w_1 &= frac{1}{300} approx 0.00333\
w_2 &= frac{1}{12} approx 0.08333\
w_3 &= frac{1}{25} = 0.4\
end{align}
$$

share|cite|improve this answer

answered Mar 14 at 21:00

simonetsimonet

370111

$endgroup$

share|cite|improve this answer

answered Mar 14 at 21:00

simonetsimonet

370111

answered Mar 14 at 21:00

simonetsimonet

370111

answered Mar 14 at 21:00

simonetsimonet

370111