Re: [algogeeks] Re: Scheduling

2011-06-11 Thread Algoose chase
Will this work ?

Order by choosing the last element of the permutation first.

initially Calculate T = Total time of all tasks
and calculate for all i, (T-Ti)*Ci and choose the task with min among them
as last.
To find the next last element , re-calculate T = T-Ti(i being the element
chosen in prev step) and proceed with the same steps as mentioned above.


On Tue, Jun 7, 2011 at 6:43 PM, ross jagadish1...@gmail.com wrote:

 @Aakash Johari:
 Sorting works fine if all jobs can be completed in a day..
 I have a modification to this question -
 suppose the time to do a job is not one day and is given as Ti for job
 i, then how should one solve it?


 On Jun 7, 11:58 am, Aakash Johari aakashj@gmail.com wrote:
  yes, it's correct. simply sort according to their costs (in decreasing
  order)
 
  On Mon, Jun 6, 2011 at 11:52 PM, sunny agrawal sunny816.i...@gmail.com
 wrote:
 
 
 
 
 
 
 
 
 
   Sort in decreasing order of Ci ?
 
   On Tue, Jun 7, 2011 at 8:22 AM, aanchal goyal 
 goyal.aanch...@gmail.comwrote:
 
   Given n jobs, each ith job has a cost Ci associated with it. The
 waiting
   time for a job i is defined as Ci*Delay, where Delay is the number of
 days
   it takes from the first day to complete a job. Assume each job can be
   completed in 1 day. So, a job started at day 1 will have delay=1, the
 one
   started at day 2 will have delay=2, etc. Order the jobs in such a way
 that
   waiting time is minimum.
 
   --
   Regards,*
   Aanchal Goyal*.
 
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   Indian Institute Of Technology,Roorkee
 
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Re: [algogeeks] Re: Scheduling

2011-06-11 Thread Arpit Sood
can you give the problem link ?

On Thu, Jun 9, 2011 at 12:38 PM, Algoose chase harishp...@gmail.com wrote:

 Will this work ?

 Order by choosing the last element of the permutation first.

 initially Calculate T = Total time of all tasks
 and calculate for all i, (T-Ti)*Ci and choose the task with min among them
 as last.
 To find the next last element , re-calculate T = T-Ti(i being the element
 chosen in prev step) and proceed with the same steps as mentioned above.


 On Tue, Jun 7, 2011 at 6:43 PM, ross jagadish1...@gmail.com wrote:

 @Aakash Johari:
 Sorting works fine if all jobs can be completed in a day..
 I have a modification to this question -
 suppose the time to do a job is not one day and is given as Ti for job
 i, then how should one solve it?


 On Jun 7, 11:58 am, Aakash Johari aakashj@gmail.com wrote:
  yes, it's correct. simply sort according to their costs (in decreasing
  order)
 
  On Mon, Jun 6, 2011 at 11:52 PM, sunny agrawal sunny816.i...@gmail.com
 wrote:
 
 
 
 
 
 
 
 
 
   Sort in decreasing order of Ci ?
 
   On Tue, Jun 7, 2011 at 8:22 AM, aanchal goyal 
 goyal.aanch...@gmail.comwrote:
 
   Given n jobs, each ith job has a cost Ci associated with it. The
 waiting
   time for a job i is defined as Ci*Delay, where Delay is the number of
 days
   it takes from the first day to complete a job. Assume each job can be
   completed in 1 day. So, a job started at day 1 will have delay=1, the
 one
   started at day 2 will have delay=2, etc. Order the jobs in such a way
 that
   waiting time is minimum.
 
   --
   Regards,*
   Aanchal Goyal*.
 
--
   You received this message because you are subscribed to the Google
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   --
   Sunny Aggrawal
   B-Tech IV year,CSI
   Indian Institute Of Technology,Roorkee
 
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  --
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  (IIIT Allahabad)

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-- 
Regards,
Arpit Sood

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Re: [algogeeks] Re: Scheduling

2011-06-11 Thread Akshata Sharma
ya aanchal, kaun si problem hai?

On Sun, Jun 12, 2011 at 1:41 AM, Arpit Sood soodfi...@gmail.com wrote:

 can you give the problem link ?


 On Thu, Jun 9, 2011 at 12:38 PM, Algoose chase harishp...@gmail.comwrote:

 Will this work ?

 Order by choosing the last element of the permutation first.

 initially Calculate T = Total time of all tasks
 and calculate for all i, (T-Ti)*Ci and choose the task with min among them
 as last.
 To find the next last element , re-calculate T = T-Ti(i being the element
 chosen in prev step) and proceed with the same steps as mentioned above.


 On Tue, Jun 7, 2011 at 6:43 PM, ross jagadish1...@gmail.com wrote:

 @Aakash Johari:
 Sorting works fine if all jobs can be completed in a day..
 I have a modification to this question -
 suppose the time to do a job is not one day and is given as Ti for job
 i, then how should one solve it?


 On Jun 7, 11:58 am, Aakash Johari aakashj@gmail.com wrote:
  yes, it's correct. simply sort according to their costs (in decreasing
  order)
 
  On Mon, Jun 6, 2011 at 11:52 PM, sunny agrawal 
 sunny816.i...@gmail.comwrote:
 
 
 
 
 
 
 
 
 
   Sort in decreasing order of Ci ?
 
   On Tue, Jun 7, 2011 at 8:22 AM, aanchal goyal 
 goyal.aanch...@gmail.comwrote:
 
   Given n jobs, each ith job has a cost Ci associated with it. The
 waiting
   time for a job i is defined as Ci*Delay, where Delay is the number
 of days
   it takes from the first day to complete a job. Assume each job can
 be
   completed in 1 day. So, a job started at day 1 will have delay=1,
 the one
   started at day 2 will have delay=2, etc. Order the jobs in such a
 way that
   waiting time is minimum.
 
   --
   Regards,*
   Aanchal Goyal*.
 
--
   You received this message because you are subscribed to the Google
 Groups
   Algorithm Geeks group.
   To post to this group, send email to algogeeks@googlegroups.com.
   To unsubscribe from this group, send email to
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   For more options, visit this group at
  http://groups.google.com/group/algogeeks?hl=en.
 
   --
   Sunny Aggrawal
   B-Tech IV year,CSI
   Indian Institute Of Technology,Roorkee
 
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  --
  -Aakash Johari
  (IIIT Allahabad)

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 --
 Regards,
 Arpit Sood

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[algogeeks] Re: Scheduling dp problem - MSFT interview

2011-06-08 Thread pschaus
sort  the jobs increasingly according to ratio: duration/penalty.
It is easy to prove it's optimal for two jobs.
Then you generalize the proof for n jobs easily because you can swap
two consecutive jobs i,j that have duration(i)/penalty(i)duration(j)/
penalty(j) to decrease the total penalty of the schedule (same
reasoning as for two jobs + a constant for the ones before and after
that remain unchanged).

On May 30, 7:00 pm, ross jagadish1...@gmail.com wrote:
 You are given a sequence of jobs. The no. of days which each job takes
 to complete is also provided.
 You are also given the penalty amount to be paid per day each day a
 job left done. Give an optimal ordering
 among jobs to minimize penalty. There are no concurrent jobs.

 eg:
 Jobs:                                  J1 J2 J3
 no. of days to complete:       1  10  4
 Penalty incurred each day    1000 30 40
 the job is pending

 output:
 Schedule is J1 J3 J2
 hence, J1 goes for 1st day. J3 for subsequent 4 days. J2 for the next
 10 days.

 Penalty incurred = (delay for job i ) * (penalty per day of job i) =
                         = (0)(1000) + (1)(40) + 5(30) = 190

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Re: [algogeeks] Re: Scheduling dp problem - MSFT interview

2011-06-08 Thread Aakash Johari
Yes, pshaus is absolutely right. I have just confirmed it by exchange
arguments. Just schedule the one first having (duration/panelty) ratio
lesser.

On Wed, Jun 8, 2011 at 12:01 AM, pschaus psch...@gmail.com wrote:

 sort  the jobs increasingly according to ratio: duration/penalty.
 It is easy to prove it's optimal for two jobs.
 Then you generalize the proof for n jobs easily because you can swap
 two consecutive jobs i,j that have duration(i)/penalty(i)duration(j)/
 penalty(j) to decrease the total penalty of the schedule (same
 reasoning as for two jobs + a constant for the ones before and after
 that remain unchanged).

 On May 30, 7:00 pm, ross jagadish1...@gmail.com wrote:
  You are given a sequence of jobs. The no. of days which each job takes
  to complete is also provided.
  You are also given the penalty amount to be paid per day each day a
  job left done. Give an optimal ordering
  among jobs to minimize penalty. There are no concurrent jobs.
 
  eg:
  Jobs:  J1 J2 J3
  no. of days to complete:   1  10  4
  Penalty incurred each day1000 30 40
  the job is pending
 
  output:
  Schedule is J1 J3 J2
  hence, J1 goes for 1st day. J3 for subsequent 4 days. J2 for the next
  10 days.
 
  Penalty incurred = (delay for job i ) * (penalty per day of job i) =
  = (0)(1000) + (1)(40) + 5(30) = 190

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-- 
-Aakash Johari
(IIIT Allahabad)

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[algogeeks] Re: Scheduling

2011-06-07 Thread ross
@Aakash Johari:
Sorting works fine if all jobs can be completed in a day..
I have a modification to this question -
suppose the time to do a job is not one day and is given as Ti for job
i, then how should one solve it?


On Jun 7, 11:58 am, Aakash Johari aakashj@gmail.com wrote:
 yes, it's correct. simply sort according to their costs (in decreasing
 order)

 On Mon, Jun 6, 2011 at 11:52 PM, sunny agrawal sunny816.i...@gmail.comwrote:









  Sort in decreasing order of Ci ?

  On Tue, Jun 7, 2011 at 8:22 AM, aanchal goyal 
  goyal.aanch...@gmail.comwrote:

  Given n jobs, each ith job has a cost Ci associated with it. The waiting
  time for a job i is defined as Ci*Delay, where Delay is the number of days
  it takes from the first day to complete a job. Assume each job can be
  completed in 1 day. So, a job started at day 1 will have delay=1, the one
  started at day 2 will have delay=2, etc. Order the jobs in such a way that
  waiting time is minimum.

  --
  Regards,*
  Aanchal Goyal*.

   --
  You received this message because you are subscribed to the Google Groups
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  For more options, visit this group at
 http://groups.google.com/group/algogeeks?hl=en.

  --
  Sunny Aggrawal
  B-Tech IV year,CSI
  Indian Institute Of Technology,Roorkee

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 --
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 (IIIT Allahabad)

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