[Operating System] {ud923} P3L1: Scheduling

• Fedorova, Alexandra, et. al.
  "Chip Multithreading Systems Need a New Operating System Scheduler".
  Proceedings of the 11th workshop on ACM SIGOPS European workshop (EW 11). ACM, New York, NY, 2004.

Visual Metaphor 

 

FIFO first in first out

FCFS first come first serve

SJF: shortest job first 

Scheduling Overview 

 

 Run To Completion Scheduling

 

std::priority_queue??

 

0.25;

5;

1;

Throughput Formula:

• jobs_completed / time_to_complete_all_job

Avg. Completion Time Formula:

• sum_of_times_to_complete_each_job / jobs_completed

Avg. Wait Time Formula:

• (t1_wait_time + t2_wait_time + t3_wait_time) / jobs_completed

You do not have to include units in your answers. Also, for decimal answers, please round to the hundredths.

• time_to_complete_all_job / jobs_completed

 Preemptive Scheduling: SJF + Preempt

 

Preemptive Scheduling: Priority 

 

 8；10；11

 Priority Inversion

boost T3 to T1 priority temporarily, so that the lock will be released before T2 gets finished. 

Round Robin Scheduling 

 

 Timesharing and Timeslices

 

 no prior knowledge about the completion time for tasks is needed.

 How Long Should a Timeslice Be

 

 CPU Bound Timeslice Length

 

Full Calculations + ERRATA (for avg. wait time)

• Timeslice = 1 second
  • throughput = 2 / (10 + 10 + 19*0.1) = 0.091 tasks/second
  • avg. wait time = (0 + (1+0.1)) / 2 = 0.55 seconds
  • avg. comp. time = 21.35 seconds
• Timeslice = 5 seconds
  • throughput = 2 / (10 + 10 + 3*0.1) = 0.098 tasks/second
  • avg. wait time = (0 + (5+0.1)) / 2 = 3.05 seconds
  • avg. comp. time = 17.75 seconds
• Timeslice = ∞
  • throughput = 2 / (10 + 10) = 0.1 tasks/second
  • avg. wait time = (0 + (10)) / 2 = 5 seconds
  • avg. comp. time = (10 + 20)/2 = 15 seconds

I/O Bound Timeslice Length 

 

Full Calculation + ERRATA

• for Timeslice = 1sec
  • avg. comp. time = (21.9 + 20.8) / 2 = 21.35

• Timeslice = 5 second*
  • throughput = 2 / 24.3 = 0.082 tasks/second
  • avg. wait time = 5.1 / 2 = 2.55 seconds
  • avg. comp. time = (11.2 + 24.3) / 2 = 17.75 seconds

Summarizing Timeslice Length

 

 

For the case when we have a round robin scheduler with a 10 ms timeslice, as we're going through the 10 I/O bound tasks,

every single one of them will run just for 1ms and then we will have to stop because of issuing an I/O request, so we'll have them context switch in that case.

From CPU utilization perspective, having a large timeslice that favors the CPU bound task is better.

Quiz Help

CPU Utilization Formula:

• [cpu_running_time / (cpu_running_time + context_switching_overheads)] * 100
• The cpu_running_time and context_switching_overheads should be calculated over a consistent, recurring interval

Helpful Steps to Solve:

1. Determine a consistent, recurring interval
2. In the interval, each task should be given an opportunity to run
3. During that interval, how much time is spent computing? This is the cpu_running_time
4. During that interval, how much time is spent context switching? This is the context_switching_overheads
5. Calculate!

 Runqueue Data Structure

 Linux O(1) Scheduler

 

cfs: Completely Fair Scheduler

 Linux CFS Scheduler

 

CFS is the default scheduler for non-realtime tasks (=> so there's a realtime scheduler.)

If you need more information about the red-black tree data structure, then see this red-black tree explanation.

 

The first statement is not correct. Linux O(1) scheduler takes constant amount of time to select and schedule a task regardless of the load.

The second statement is sort of correct inthe sense that as long as there were continuously arriving higher priority tasks, it was possible for low priority tasks to keep waiting an unpredictable amount of time.

But this is not the main reason for O(1) scheduler to be replaced.

 Scheduling on Multiprocessors

LLC: last level cache

 

right one is more common on PC

Hyperthreading 

other names of SMT in purple

• Fedorova, Alexandra, et. al.
  "Chip Multithreading Systems Need a New Operating System Scheduler".
  Proceedings of the 11th workshop on ACM SIGOPS European workshop (EW 11). ACM, New York, NY, 2004.

Scheduling for Hyperthreading Platforms 

IPC: instructions per cycle

 CPU Bound or Memory Bound

 

 Scheduling with Hardware Counters

CPI Experiment Results 

Quiz Help

For analyzing the graph, it might be helpful to use the full experiment chart.

Core 0

• (a) 1, 6, 11, 16
• (b) 1, 6, 6, 6
• (c) 1, 1, 6, 6
• (d) 1, 1, 1, 1

Core 1

• (a) 1, 6, 11, 16
• (b) 1, 6, 11, 11
• (c) 1, 1, 6, 6
• (d) 6, 6, 6, 6

Core 2

• (a) 1, 6, 11, 16
• (b) 1, 11, 11, 16
• (c) 11, 11, 11, 11
• (d) 11, 11, 11, 11

Core 3

• (a) 1, 6, 11, 16
• (b) 1, 16, 16, 16
• (c) 16, 16, 16, 16
• (d) 16, 16, 16, 16