correcting my last statement "So basically the problem needs to be optimized on two aspects, minimizing costs and maximizing the total weight of executed jobs. so *Maximize ( 1 + (Total Weight of executed jobs in j-th iter/ Total Cost incurred in j-th iter) )"*
On Sat, Apr 20, 2013 at 3:49 AM, Piyush Grover <piyush4u.iit...@gmail.com>wrote: > I have a practical problem, need an optimal solution for this > > *What is given?* > Given *N* sets, each containing some jobs to be executed, such that no > two sets are subsets of each other and number of jobs in *i-th* set is *ni > << N*. > The jobs can have values between *1...k where k << N*. Priority of each > job is in the order of their value. i.e *priority(1) > priority(2).....> > priority(k) *so as the weights > *w1 > w2 > .....wi...> wk* > > *What are the constraints?* > -> Every job in every set is executed independent of others. > -> No job can be executed independent of its set i.e if a job needs to be > executed, any > set containing the job will be executed > -> The cost of execution of *i-th* set is *ni* > -> The probability of failing each job during execution is equal and > unknown. > > *What is needed?* > The jobs need to be executed but can be done in multiple iterations. > So* *return the number of sets *mj <=M << N* (and set itself) to be added > for the execution in the j*-th iteration. > *Each iteration adds the additional cost to each set to be executed in a > way such that > *cost of execution of i-th set in the j-th iteration = ni + (j-1)*max( > weights of failed jobs in i-th set)* > If a set is executed (with/without failed jobs) that can not be used in > further iterations > > > So basically the problem needs to be optimized on two aspects, minimizing > costs and maximizing the number of jobs to be executed. > > > Regards > Piyush > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To unsubscribe from this group and stop receiving emails from it, send an email to algogeeks+unsubscr...@googlegroups.com. For more options, visit https://groups.google.com/groups/opt_out.
