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IIS Progress Report 2015/10/12. Problem Revisit Given a set of virtual machines, each contains some virtual cores with resource requirements. Decides.

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Presentation on theme: "IIS Progress Report 2015/10/12. Problem Revisit Given a set of virtual machines, each contains some virtual cores with resource requirements. Decides."— Presentation transcript:

1 IIS Progress Report 2015/10/12

2 Problem Revisit Given a set of virtual machines, each contains some virtual cores with resource requirements. Decides the amount of time and execution order of these virtual cores on heterogeneous multi-cores, such that the power consumption is minimized while the resource requirement is satisfied.

3 Model Revisit For every time interval ◦ Virtual core i with resource requirement v i. ◦ Physical core j with frequency f j. ◦ The power consumption of a core with given f j is linear to its load L j. Some constraints: ◦ vCPU can not run on two cores simultaneously. ◦ A core cannot execute two vCPUs simultaneously

4 The Reviewer Said … Not practical. ◦ The 3-phase solution we proposed is too complex. ◦ The computation for generating a scheduling plan is to heavy.  A scheduling plan per second.

5 Improvement Instead of general workload, focus on scenarios that our model/solution is applicable. ◦ Less work load fluctuation. ◦ Instead of generating scheduling plan periodically, compute a new plan on workload changing.

6 Target Scenario Stable workloads. ◦ Computation-intensive ◦ Generate new scheduling plan during (drastic) workload changing.  Reduce overhead ◦ Continuously ◦ Medium amount of workload  Single (ARM) server Virtualized environment ◦ Or not(?)

7 Target Scenario(Cont.) Current target: streaming data processing ◦ Continuous data input ◦ Pre-processing before storing or send to the next stage  Computation-intensive ◦ Example:  Monitoring data, such as smart homes/building.  VOD decoding on multi-user NAS  …etc.

8 Three-phases Solution Still too complex. Since the new scenario does not require responsiveness. ◦ Remove the concept of “time slot”, therefore we don’t need the 2 nd and 3 rd phase. Focus on the 1 st phase.

9 First-phase Revisit Decide the amount of time each virtual core should run on the physical cores. ◦ Meet the requirement of each virtual core. ◦ The frequencies of physical cores are given. ◦ Can be solved using Linear/Integer Programming.

10 First-phase Revisit Decide the amount of time each virtual core should run on the physical cores. ◦ Meet the requirement of each virtual core. ◦ The frequencies of physical cores are given. ◦ Can be solved using Linear/Integer Programming. Generate the allocation along with the core frequencies.

11 The New Problem Generate the amount of time each virtual core should run on the physical cores along with the core frequencies. ◦ Not sure if it is a NP-C problem, still working on this. ◦ Notice that the constraints still hold.  The total time allocates to a virtual core must be less or equal to the length of a time interval.  How to guarantee?

12 Observation Given a vCPU with requirement vi. ◦ The physical cores hosting this vCPU must provide frequency f j where f j ≧ v i.  Ex: v i = 1200M 1200M800M1600M1200M 40% 60% 40% 70%

13 Heuristic #1 Given a set of virtual core with resource requirement v i. Group two vCPUs with the smallest v i into a new vCPU, where is v’ i the sum of the two vCPU. Repeat this process until the number of vCPU (group) is less or equal than the number of cores. Decide the frequencies of cores according to v’ i of each vCPU group.

14 Heuristic #2 B= {}, L = {v i }, i = 0, …, n-1 Compute ∑ v i, ∑ F j ◦ F j is the largest available frequency of an energy-efficient core. while(∑ v i > ∑ F j ) remove the largest v i from L, add to B

15 Heuristic #2(Cont.) while(∑ v i > ∑ f j ) In crease the smallest f j by 1 level. ◦ // v i ∈ L, f j = max(v i ) initially Foreach v i ∈ L ◦ Start from the smallest v i, assign it to the smallest f j with spare resource. ◦ If(v i > s j ) // s j : spare resource on core j assign (v i - s j ) to core j+1, which is the second smallest core with spare resource.

16 Example 900M 1800M

17 Discussion


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