Approximation Algorithms: Job Scheduling

1. Approximation Algorithms: Job Scheduling Lecture 16: March 15

2. Job Scheduling: General Setting • There are n jobs, each job has: • a processing time p(i) (the time to finish this job) • a release time r(i) (can only process this job after time r(i)) • a deadline d(i) (have to finish this job before time d(i)) • a weight w(i) (e.g. the profit of finishing job i) There are m machines available. • Task: to scheduling the jobs to optimize the following criterion. • To minimize the completion time of all jobs (the makespan) • To finish as many jobs as possible meeting their deadlines • To minimize the weighted average completion time, etc… Practically infinitely many possibilities...

3. Job Scheduling: One Machine • There are n jobs, each job has: • a processing time p(j) (the time to finish this job) There are 1 machine available. Task: to scheduling the jobs to -To minimize the completion time of all jobs (the makespan) Trivial: just find an arbitrary ordering of the jobs.

4. Job Scheduling: One Machine • There are n jobs, each job has: • a processing time p(j) (the time to finish this job) There are 1 machine available. Task: to scheduling the jobs to -To minimize the average completion time of all jobs Optimal solution: Find an ordering which schedules long jobs first. Proof: swapping

5. Job Scheduling: Multiple Machines • There are n jobs, each job has: • a processing time p(j) (the time to finish this job) There are m machines available. Task: to scheduling the jobs -To minimize the completion time of all jobs (the makespan) NP-hard even for 2 machines: Partition problem Polynomial time approximation scheme by dynamic programming. Homework: Find a (very simple) 2-approximation algorithm.

6. Job Scheduling: Unrelated Multiple Machines • There are n jobs, each job has: • a processing time p(i,j) (the time to finish this job j on machine i) There are m machine available. Task: to scheduling the jobs -To minimize the completion time of all jobs (the makespan) NP-hard to approximate within 1.5 times of the optimal solution. We’ll design a 2-approximation algorithm for this problem together.

7. Why Unrelated? For example, different processors have different specialties. Computational jobs, display images, etc…

8. Why Unrelated? 2 “processors” Jobs: housework

9. Job Scheduling: Unrelated Multiple Machines • There are n jobs, each job has: • a processing time p(i,j) (the time to finish this job j on machine i) There are m machine available. Task: to scheduling the jobs -To minimize the completion time of all jobs (the makespan) Approach: Linear Programming. How to formulate this problem into linear program?

10. Linear Programming Relaxation whether job j is scheduled in machine i Each job is scheduled in one machine. for each job j Each machine can finish its jobs by time T for each machine i Relaxation for each job j, machine i

11. How good is the relaxation? Example for each job j One job of processing time K for each machine for each machine i Optimal solution = K. Optimal fraction solution = K/m. for each job j, machine i The LP lower bound could be as bad as you want.

12. How good is the relaxation? Example for each job j One job of processing time K for each machine for each machine i Optimal solution = K. Optimal fraction solution = K/m. for each job j, machine i Problem of the linear program relaxation: an optimal solution T could be even smaller than the processing time of a job!

13. How to tackle this problem? Problem of the linear program relaxation: an optimal solution T could be even smaller than the processing time of a job! Ideally, we could write the following constraint: but this is not a linear constraint… Idea? To enforce this constraint by hand!

14. Preprocessing Fix T. Consider the decision problem instead of an optimization problem Call the resulting linear program LP(T). Note that different T have different linear programs. Fancy name: parametric pruning.

15. Decision Problems Fix T Let S(T) be the set of jobs with p(i,j) < T. for each job j for each machine i for each job j, machine i Use binary search to find the minimum T* such that this LP is feasible. For example, if all variables with job j are deleted, then we know that this problem is not feasible.

16. Lower Bound We will use T* as the lower bound on the value of an optimal solution, clearly T* <= OPT, since LP(OPT) is feasible. Observation: Suppose there are m machines and at most m jobs. If we can assign all jobs to the m machines so that each machine is assigned at most 1 job, then the completion time (makespan) is at most T* <= OPT.

17. Fractional Solutions Fractional solutions could be very fractional. What have we learnt in the last class? Look at a vertex (basic) solution. Very fractional solutions may only be the average of many solutions.

18. Vertex Solutions for each job j for each machine i for each job j, machine i What can we say about a vertex solution of this LP? Basic solution: unique solution of n linearly independent tight inequalities, where n is the number of variables.

19. Vertex Solutions for each job j for each machine i for each job j, machine i A tight inequality of the last type corresponds to a variable of zero value. There are at most n+m inequalities of the first two types, and hence there are at most n+m nonzero variables.

20. Vertex Solutions Say a job is integral if it is assigned entirely to one machine; otherwise a job is fractional. Each fractional job is assigned to at least two machines. Let p be the number of integral jobs, and q be the number of fractional jobs. There are at most n+m nonzero variables. • p + q = n • p + 2q <= n + m • p >= n – m • q <= m There are at most m fractional jobs.

21. Integral Jobs How to handle integral jobs? Just follow the optimal fractional solution. And so we can schedule all the integral jobs in time at most T* <= OPT, as this schedule (on integral jobs) is just a subset of the fractional solution.

22. Fractional Jobs Remember this observation we made before. Observation: Suppose there are m machines and at most m jobs. If we can assign all jobs to the m machines so that each machine is assigned at most 1 job, then the completion time (makespan) is at most T* <= OPT. There are at most m fractional jobs. If we could find such a “matching”, then we use this matching to schedule all the fractional jobs in time at most T* <= OPT.

23. Approximation Algorithm Goal: to design a 2-approximation algorithm for this problem • Do preprocessing (parametric pruning) and find a smallest T* so that LP(T*) is feasible. • Find a vertex (basic) solution, say x, to LP(T*). • Assign all integral jobs to machines as in x. • Match the fractional jobs to the machines so that each machine is assigned at most one job. Proof (assuming a matching exists): Schedule all integral jobs in time T*, Schedule all fractional jobs in time T*, Schedule all jobs in time 2T* <= 2OPT.

24. Bipartite Matching Task: Match the fractional jobs to the machines so that each machine is assigned at most one job. Create a vertex for each job j, and create a vertex for each machine i, add an edge between machine i and job j if 0 < x(i,j) < 1. Now, the problem is to find a matching so that every job is matched.

25. Bipartite Matching job machine Assume the graph is connected. There are at most n+m nonzero variables. n + m vertices, n + m edges, at most one cycle.

26. Bipartite Matching Leaves must be machines, since each fractional job is adjacent to two machines. Match a leaf machine with its adjacent job, then remove these vertices and repeat. n + m vertices, n + m edges, at most one cycle.

27. Bipartite Matching Match a leaf machine with its adjacent job, then remove these vertices and repeat. Eventually a cycle is left, and we can find a perfect matching. n + m vertices, n + m edges, at most one cycle.

28. Bipartite Matching If the graph is not connected, we apply the same argument to each connected component. Prove: (1) each component has at most n’+m’ edges. (2) each component has a matching.

29. Bad Examples m machines m2 – m + 1 jobs: 1 job of processing time m on all machines remaining jobs have processing time 1 on all machines Optimal solution: the large job on one machine, m small jobs on the remaining m-1 machines, makespan = m LP vertex solution: 1/m of the first job and m-1 other jobs to each machine. Our rounding procedure will produce a schedule of makespan 2m-1.

30. Bad Examples Optimal solution: the large job on one machine, m small jobs on the remaining m-1 machines, makespan = m LP vertex solution: 1/m of the first job and m-1 other jobs to each machine. Our rounding procedure will produce a schedule of makespan 2m-1. Note that there is no integrality gap in this example. Just that the rounding procedure produces the error. Open question: Can we obtain a 3/2 approximation algorithm for this problem?

31. Quick Summary • Look at a vertex solution (not many fractional jobs). • Schedule all the integral jobs as in the fractional solution. • Schedule all the fractional jobs by a matching. • Each set can be scheduled in time at most OPT. • So we have a 2-approximation algorithm. • Get a feeling on job scheduling problems. • Linear programming is (again) a powerful tool.