In this programming problem and the
next you'll code up the greedy algorithms from lecture for minimizing the
weighted sum of completion times..This file describes a set of jobs with
positive and integral weights and lengths. It has the format
[number_of_jobs]
[job_1_weight] [job_1_length]
[job_2_weight] [job_2_length]
...
For example, the third line of the file is "74 59", indicating that the second job has weight 74 and length 59. You should NOT assume that edge weights or lengths are distinct.
[job_1_weight] [job_1_length]
[job_2_weight] [job_2_length]
...
For example, the third line of the file is "74 59", indicating that the second job has weight 74 and length 59. You should NOT assume that edge weights or lengths are distinct.
Your task in this problem is to run
the greedy algorithm that schedules jobs in decreasing order of the difference
(weight - length). This algorithm is not always
optimal. IMPORTANT: if two jobs have equal difference (weight - length), you
should schedule the job with higher weight first. Beware: if you break ties in
a different way, you are likely to get the wrong answer. You should report the
sum of weighted completion times of the resulting schedule --- a positive
integer --- in the box below.
Sol:
we will compute weighted sum with the following formula :
And our goal is to minimize that function:
Algo is straight forward, we make a job with length, weight and difference (weight - length). then we keep all the jobs in a list and sorted them in non- decreasing order. If two jobs have same difference we break the tie by choosing a job with higher weight. This tie breaking is achieved by implementing Comparable in Job class.
public int compareTo(Object o) {
// TODO Auto-generated method stub
Job jobOther=(Job)o;
int delta= jobOther.getDifference().intValue()-this.difference.intValue() ;
if(delta==0){
delta=this.weight.intValue()-jobOther.getWeight().intValue();
return delta;
}
return delta;
}
Source code of above problem is available in the following link along with input file that contains 10,000 jobs with different weights and lengths.
yaar where is the answer
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