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How to calculate a 4-week period of the billing year [closed]

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-1

I'm refactoring a method that should return 4-week periods. A period is calculated as the number of 4-weekly blocks in the current billing year. There are thirteen periods per billing year. First billing year began on 24/08/2008.

I have created a formula:

def period
 period = (((self.utc.to_time - Time.new(2008,8,23,0,0,0,0)) / 60 / 60 / 24 / 7 / 52).modulo(1) * 13).ceil
 period == 0 ? 13 : period
end

Over the years, it has become inaccurate, and we're now seeing a shift of a whole period: bookings in Period 1 are showing as Period 2. I tried re-calculating the formula with no success. I feel the formula is too complex anyway, and can probably be achieved more simply.

ruby

share|improve this question

edited Mar 27 at 11:36

sawa

135k3131 gold badges220220 silver badges317317 bronze badges

asked Mar 27 at 10:58

Peter FealeyPeter Fealey

344 bronze badges

closed as unclear what you're asking by sawa, Umair, cegfault, Henry Woody, Adil B Mar 27 at 16:42

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Define 4-week period. What day of week does a week start on? How are the first and the last week of a (billing) year defined?
  
  – sawa
  Mar 27 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This would be a good example to provide some assertions/specs for, along with the additional rules (What about leap years? And the extra day at the end of the year?)
  
  – Jay Dorsey
  Mar 27 at 12:26
  

  
  
  
  

add a comment | 

-1

I'm refactoring a method that should return 4-week periods. A period is calculated as the number of 4-weekly blocks in the current billing year. There are thirteen periods per billing year. First billing year began on 24/08/2008.

I have created a formula:

def period
 period = (((self.utc.to_time - Time.new(2008,8,23,0,0,0,0)) / 60 / 60 / 24 / 7 / 52).modulo(1) * 13).ceil
 period == 0 ? 13 : period
end

Over the years, it has become inaccurate, and we're now seeing a shift of a whole period: bookings in Period 1 are showing as Period 2. I tried re-calculating the formula with no success. I feel the formula is too complex anyway, and can probably be achieved more simply.

ruby

share|improve this question

edited Mar 27 at 11:36

sawa

135k3131 gold badges220220 silver badges317317 bronze badges

asked Mar 27 at 10:58

Peter FealeyPeter Fealey

344 bronze badges

closed as unclear what you're asking by sawa, Umair, cegfault, Henry Woody, Adil B Mar 27 at 16:42

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Define 4-week period. What day of week does a week start on? How are the first and the last week of a (billing) year defined?
  
  – sawa
  Mar 27 at 11:30
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  This would be a good example to provide some assertions/specs for, along with the additional rules (What about leap years? And the extra day at the end of the year?)
  
  – Jay Dorsey
  Mar 27 at 12:26
  

  
  
  
  

add a comment | 

I'm refactoring a method that should return 4-week periods. A period is calculated as the number of 4-weekly blocks in the current billing year. There are thirteen periods per billing year. First billing year began on 24/08/2008.

I have created a formula:

def period
 period = (((self.utc.to_time - Time.new(2008,8,23,0,0,0,0)) / 60 / 60 / 24 / 7 / 52).modulo(1) * 13).ceil
 period == 0 ? 13 : period
end

Over the years, it has become inaccurate, and we're now seeing a shift of a whole period: bookings in Period 1 are showing as Period 2. I tried re-calculating the formula with no success. I feel the formula is too complex anyway, and can probably be achieved more simply.

ruby

share|improve this question

edited Mar 27 at 11:36

sawa

135k3131 gold badges220220 silver badges317317 bronze badges

asked Mar 27 at 10:58

Peter FealeyPeter Fealey

344 bronze badges

def period
 period = (((self.utc.to_time - Time.new(2008,8,23,0,0,0,0)) / 60 / 60 / 24 / 7 / 52).modulo(1) * 13).ceil
 period == 0 ? 13 : period
end

share|improve this question

edited Mar 27 at 11:36

sawa

135k3131 gold badges220220 silver badges317317 bronze badges

asked Mar 27 at 10:58

Peter FealeyPeter Fealey

344 bronze badges

share|improve this question

edited Mar 27 at 11:36

sawa

135k3131 gold badges220220 silver badges317317 bronze badges

asked Mar 27 at 10:58

Peter FealeyPeter Fealey

344 bronze badges

edited Mar 27 at 11:36

sawa

135k3131 gold badges220220 silver badges317317 bronze badges

edited Mar 27 at 11:36

sawa

135k3131 gold badges220220 silver badges317317 bronze badges

asked Mar 27 at 10:58

Peter FealeyPeter Fealey

344 bronze badges

asked Mar 27 at 10:58

Peter FealeyPeter Fealey

344 bronze badges

1 Answer
1

active

oldest

votes

0

I would just use Date instead of Time and the calculation would be much easier.

DAYS_PER_PERIOD = 28

def period
 days = Date.today - Date.new(2008, 8, 23)
 periods = days / DAYS_PER_PERIOD
 periods.to_i # would return a `Rational` instead of an `Integer` otherwise
end

share|improve this answer

answered Mar 27 at 11:32

spickermannspickermann

65.4k77 gold badges6363 silver badges8686 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is great, thanks. What the client threw in was that at some point in the last 11 years they changed the way they calculated periods and added an extra week somewhere. With a change to the start date (2008, 8, 23) to one after the extra week was thrown in, and the code above, we're good!
  
  – Peter Fealey
  Mar 27 at 15:25
  
  

  
  
  
  

add a comment | 

0

I would just use Date instead of Time and the calculation would be much easier.

DAYS_PER_PERIOD = 28

def period
 days = Date.today - Date.new(2008, 8, 23)
 periods = days / DAYS_PER_PERIOD
 periods.to_i # would return a `Rational` instead of an `Integer` otherwise
end

share|improve this answer

answered Mar 27 at 11:32

spickermannspickermann

65.4k77 gold badges6363 silver badges8686 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is great, thanks. What the client threw in was that at some point in the last 11 years they changed the way they calculated periods and added an extra week somewhere. With a change to the start date (2008, 8, 23) to one after the extra week was thrown in, and the code above, we're good!
  
  – Peter Fealey
  Mar 27 at 15:25
  
  

  
  
  
  

add a comment | 

0

I would just use Date instead of Time and the calculation would be much easier.

DAYS_PER_PERIOD = 28

def period
 days = Date.today - Date.new(2008, 8, 23)
 periods = days / DAYS_PER_PERIOD
 periods.to_i # would return a `Rational` instead of an `Integer` otherwise
end

share|improve this answer

answered Mar 27 at 11:32

spickermannspickermann

65.4k77 gold badges6363 silver badges8686 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is great, thanks. What the client threw in was that at some point in the last 11 years they changed the way they calculated periods and added an extra week somewhere. With a change to the start date (2008, 8, 23) to one after the extra week was thrown in, and the code above, we're good!
  
  – Peter Fealey
  Mar 27 at 15:25
  
  

  
  
  
  

add a comment | 

I would just use Date instead of Time and the calculation would be much easier.

DAYS_PER_PERIOD = 28

def period
 days = Date.today - Date.new(2008, 8, 23)
 periods = days / DAYS_PER_PERIOD
 periods.to_i # would return a `Rational` instead of an `Integer` otherwise
end

share|improve this answer

answered Mar 27 at 11:32

spickermannspickermann

65.4k77 gold badges6363 silver badges8686 bronze badges

DAYS_PER_PERIOD = 28

def period
 days = Date.today - Date.new(2008, 8, 23)
 periods = days / DAYS_PER_PERIOD
 periods.to_i # would return a `Rational` instead of an `Integer` otherwise
end

share|improve this answer

answered Mar 27 at 11:32

spickermannspickermann

65.4k77 gold badges6363 silver badges8686 bronze badges

answered Mar 27 at 11:32

spickermannspickermann

65.4k77 gold badges6363 silver badges8686 bronze badges

answered Mar 27 at 11:32

spickermannspickermann

65.4k77 gold badges6363 silver badges8686 bronze badges