Lookup based partitioned scheduling

A Lookup-table Driven Approach to Partitioned

Scheduling

Bipasa ChattopadhyaySanjoy Baruah

Department of Computer ScienceThe University of North Carolina at Chapel Hill

Partitioned Scheduling

Given - a task system and m processors

Objective - partition the tasks onto the processors

Advantages - Good cache affinity, no migration delay

Implicit Deadline Sporadic Task System

Each task has a - Worst Case Execution Time, Ci Time Period, Ti Deadline, Di = Ti Utilization, ui =

CiTi

Partitioned Scheduling

Implicit Deadline Sporadic Task System Utilization, ui =

CiTi

Each processor is identical and has computing capacity = 1

Scheduling algorithm on each processor is pre-emptive EDF optimal uniprocessor scheduling algorithm if sum of utilizations

of tasks 1

Partitioning tasks is equivalent to bin-packing Bin-packing is NP-hard in the strong sense

Hochbaum and Shmoys designed a polynomial timeapproximation scheme (PTAS)

Partitioning tasks is equivalent to bin-packing Bin-packing is NP-hard in the strong sense

Hochbaum and Shmoys designed a polynomial timeapproximation scheme (PTAS)

If an optimal algorithm can partition a task system ontom processors each of computing capacity 1 then a

PTAS algorithm can partition the task system onto mprocessors each of computing capacity 1 + (0 < < 1)

in polynomial time.

Proposed PTAS partitioning algorithm has large run time

We suggest a lookup-table driven approach

Pre-compute the lookup-table The results of the expensive computation are stored in a

lookup-table The table is consulted for task partitioning

Computing the Lookup-table

The lookup-table is computed only once for a given value ofm and


Choosing

Determining Utilization Values

Single Processor Configurations

Multiprocessor Configurations

Choosing

If an optimal algorithm can partition a task system ontom processors each of computing capacity 1

then a PTAS algorithm can partition the task systemonto m processors each of computing capacity 1 +

in polynomial time (0 < < 1).

Choosing

If an optimal algorithm can partition a task system ontom processors each of computing capacity 1 ( 11+)

then a PTAS algorithm can partition the task systemonto m processors each of computing capacity 1 + (1)

in polynomial time (0 < < 1).

For example -system utilization loss = 10%= 11+ 0.9

= 19 0.1


= 19 0.1

= 0.1


= 19 0.1

= 0.1

The size of the lookup-table is a function on m and If decreases, the size of the lookup-table and the time needed

to compute it increases


Choosing





Let tasks have fixed utilization values

- Fixed utilization values




V () - set of fixed utilization values




V () - set of fixed utilization values

V () = {, (1 + ), (1 + )2, ... (1 + )j} j is a non-negative integer (1 + )j 1

Let = 0.3 (running example)

j util. value (0.3) (1.3)j

0 0.30001 0.39002 0.50703 0.65914 0.8568

5 1.114

V (0.3) = {0.3000, 0.3900, 0.5070, 0.6591, 0.8568}

Let = 0.3 (running example)

j util. value (0.3) (1.3)j

0 0.30001 0.39002 0.50703 0.65914 0.8568

5 1.114

V (0.3) = {0.3000, 0.3900, 0.5070, 0.6591, 0.8568}

As decreases the number of utilization values in V ()increases

If = 0.1, size of V () = 25

As decreases - The solution is closer to optimal - Less inflation, less utilization

loss

As decreases - The solution is closer to optimal - Less inflation, less utilization

loss

The size of the lookup-table and the time needed to computeit increases


Choosing





How many different tasks with utilization in V () can beassigned to a single processor?


How many different tasks with utilization in V () can beassigned to a single processor?

V (0.3) = {0.3000, 0.3900, 0.5070, 0.6591, 0.8568}

Configurations determine the number of tasks of eachutilization

3, 0, 0, 0, 0 - 3 tasks of ui = 0.3 are assigned to a processor 1, 1, 0, 0, 0 - 1 task of ui = 0.3 and 1 task of uj = 0.39 are

assigned to a processor

V (0.3) = {0.3000, 0.3900, 0.5070, 0.6591, 0.8568}

Valid configuration - sum of utilizations of tasks 1 3, 0, 0, 0, 0 (0.3 + 0.3 + 0.3 1) 1, 1, 0, 0, 0 (0.3 + 0.39 1)

V (0.3) = {0.3000, 0.3900, 0.5070, 0.6591, 0.8568}

Valid configuration - sum of utilizations of tasks 1 3, 0, 0, 0, 0 (0.3 + 0.3 + 0.3 1) 1, 1, 0, 0, 0 (0.3 + 0.39 1)

Maximal Configuration - no other configuration can fit moretasks

3, 0, 0, 0, 0 1, 1, 0, 0, 0 is not a maximal configuration

(0.3 + 0.39 + 0.3 = 0.99)

2, 1, 0, 0, 0 is a maximal configuration. We say that2, 1, 0, 0, 0 dominates 1, 1, 0, 0, 0

The lookup-table for = 0.3 Maximal single processor configurations

Config ID 0.3000 0.3900 0.5070 0.6591 0.8568

A 3 0 0 0 0B 2 1 0 0 0C 1 0 1 0 0D 1 0 0 1 0E 0 2 0 0 0F 0 1 1 0 0G 0 0 0 0 1

The lookup-table for = 0.3 Maximal single processor configurations

Config ID 0.3000 0.3900 0.5070 0.6591 0.8568

A 3 0 0 0 0B 2 1 0 0 0C 1 0 1 0 0D 1 0 0 1 0E 0 2 0 0 0F 0 1 1 0 0G 0 0 0 0 1

The lookup-table for = 0.1 has 9604 maximal singleprocessor configurations


Choosing





Maximal single processor configurations are used to determinemaximal m processor configurations



If m = 2, combine any pair of maximal single processorconfigurations

Single-proc. Config ID< A > 3 0 0 0 0< B > 2 1 0 0 0< C > 1 0 1 0 0< D > 1 0 0 1 0< E > 0 2 0 0 0< F > 0 1 1 0 0< G > 0 0 0 0 1





Example - 1, 2, 1, 0, 0 < C , E >





Example - 1, 2, 1, 0, 0 < C , E > 2, 2, 1, 0, 0 < B, F >

V (0.3) = {0.3000, 0.3900, 0.5070, 0.6591, 0.8568}

Valid Configuration - sum of utilization of tasks on eachprocessor 1

1, 2, 1, 0, 0 < C , E > 2, 2, 1, 0, 0 < B, F >

Combining 2 maximal single processor configurations results ina valid m = 2 processor configuration.

V (0.3) = {0.3000, 0.3900, 0.5070, 0.6591, 0.8568}

Valid Configuration - sum of utilization of tasks on eachprocessor 1

1, 2, 1, 0, 0 < C , E > 2, 2, 1, 0, 0 < B, F >

Combining 2 maximal single processor configurations results ina valid m = 2 processor configuration.

Maximal Configuration - A configuration is maximal if noother configuration dominates it

Configuration < B, F > dominates < C , E >

Remove configurations that are not maximal

The lookup-table consists of only maximal m processorconfigurations

The lookup-table consists of only maximal m processorconfigurations

We compute maximal m processor configurations frommaximal k processor configurations, k < m

If m = 4, combine any pair of maximal m = 2 processorconfigurations

0.3000 0.3900 0.5070 0.6591 0.8568 Single-proc. Config ID

3 2 1 2 0 < D, D, E , C >3 4 2 0 0 < F , F , E , A >0 3 3 0 1 < F , F , F , G >4 1 1 1 1 < G , D, C , B >4 0 1 3 0 < D, D, D, C >

Partitioning Tasks in Polynomial Time

Utilization Inflation

Inflate the utilization values of the task to equal the nearestutilization value V ()

Tasks with ui

1+ (large tasks) are inflated by at most

(1 + )

Inflation factor = (1+)j+1

ui4 1 1 1 1 < G , D, C , B >4 0 1 3 0 < D, D, D, C >

Output = < F , F , F , G >

Given task system - 15 ,15 ,

13 ,

720 ,

925 ,

25 ,

12 ,

12 ,

34

Inflate large tasks-15 ,

15 , 0.3900, 0.3900, 0.3900, 0.5070, 0.5070, 0.5070, 0.8568

0, 3, 3, 0, 1(Input) = Lookup-table = < F , F , F , G >(Output)


13 ,

720 ,

925 ,

25 ,

12 ,

12 ,

34


15 , 0.3900, 0.3900, 0.3900, 0.5070, 0.5070, 0.5070, 0.8568


Phase 1 - Large tasks are assigned as specified byconfiguration < F , F , F , G >

0, 1, 1, 0, 0 < F > ; 0, 0, 0, 0, 1< G >

Single-proc. Config ID Task Assignment Remaining Capacity

F 1/3, 2/5 0.2667F 7/20, 1/2 0.15F 9/25, 1/2 0.14G 3/4 0.25


13 ,

720 ,

925 ,

25 ,

12 ,

12 ,

34


15 , 0.3900, 0.3900, 0.3900, 0.5070, 0.5070, 0.5070, 0.8568



0, 1, 1, 0, 0 < F > ; 0, 0, 0, 0, 1< G >


F 1/3, 2/5 0.2667F 7/20, 1/2 0.15F 9/25, 1/2 0.14G 3/4 0.25

Phase 2 - Small tasks are accommodated in the remainingcapacity


13 ,

720 ,

925 ,

25 ,

12 ,

12 ,

34


15 , 0.3900, 0.3900, 0.3900, 0.5070, 0.5070, 0.5070, 0.8568



0, 1, 1, 0, 0 < F > ; 0, 0, 0, 0, 1< G >


F 1/3, 2/5, 1/5 0.0667F 7/20, 1/2 0.15F 9/25, 1/2 0.14G 3/4, 1/5 0.05

Phase 2 - Small tasks are accommodated in the remainingcapacity

Performance Guarantee

If the partitioning algorithm fails to partition the tasks

in , then no algorithm can partition on a platform of

m processors each of computing capacity 11+





Phase 1 fails - tasks with utilization 1+ are inflated byatmost (1 + )






if inflated tasks can not be partitioned onto m processorseach of utilization 1 then unmodified tasks can not bepartitioned onto m processors each of utilization 11+







Phase 2 fails - tasks with utilization < 1+ can not beaccommodated







Phase 2 fails - tasks with utilization < 1+ can not beaccommodated

sum of utilizations of tasks assigned to each processormust be > (1 1+)= total utilization of > m (1 1+)

> m( 11+)

Context and Summary

Build the lookup-table during platform synthesis

Context and Summary

Build the lookup-table during platform synthesis

Ship the platform with the lookup-table

Context and Summary

Store the lookup-table on a central server

Future Work

What happens if the exact configuration does not exist in thetable?

Configuration 0, 3, 3, 0, 1 was easy to find, what ifconfiguration was 0, 2, 3, 0, 0

Need good heuristic for lookup function.

Future Work

What happens if the exact configuration does not exist in thetable?

Configuration 0, 3, 3, 0, 1 was easy to find, what ifconfiguration was 0, 2, 3, 0, 0

Need good heuristic for lookup function.

Partition the tasks onto processors and memory.

Thank You

Does better than First Fit - 0.25, 0.25 + , 0.50, 0.50 + (utilization 1.5 + 2 )

For m = 2 processors first fit can not partition this tasksystem. Our partitioning algorithm can.

= 0.1, = 0.01, 0.25, 0.26, 0.50, 0.51 0.259, 0.285, 0.505, 0.556

Size of lookup-table for = 0.15 is - m = 1, 14KB ( 252) m = 2, 240KB ( 8523) m = 3, 3MB (122599)

Size of lookup-table for = 0.1 is - m = 1, 250KB (9604) m = 2, tens of MB() m = 3, hundreds of MB()

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Lookup based partitioned scheduling